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160 PARTI ® FUNDAMENTALS (4) = 0 O(a) = 0 + x a) = Ư May = 0 (@) (6) (©) (@)

Figure 4.5 Boundary conditions: (a) fixed end; () simply supported end; (c) free end; (d) guided or sliding support

The deflection v of a beam can be found by solving any one of the foregoing equations by successive integrations The choice of equation depends on the ease with which an expres~

sion of load, shear, or moment can be formulated and individual preference The approach

to solving the deflection problem beginning with Eq (4.16c) or (4.16b) is known as the multiple-integration method When Eq, (4.16a) is used, because two integrations are required to obtain the 0, this is called the double-integration method The constants of the integration are evaluated using the specified conditions on the ends of the beam, that is, the boundary conditions Frequently encountered conditions that may apply at the ends (x = a) of a beam are shown in Figure 4.5 We see from the figure that the force (static) vari- ables M, V, and the geometric (kinematic) variables v, 6 are 0 for common situations

If the beam has a cross-sectional width b that is large compared to the depth h (i.c., b > h), the beam is stiffer, and the deflection, is less than that determined by Eqs (4.16) for narrow beams The large cross-sectional width prevents the lateral expansion and con- traction of the material, and the deflection is thereby reduced, as shown in Section 4.9 An improved value for the deflection uv of wide beams is obtained by multiplying the result given by the equation for a narrow beam by (1 — v), where v is Poisson’s ratio

EXAMPLE 4.5 Finding Beam Deflections by the Double-Integration Method

A simply supported beam is subjected to a concentrated load at a distance a from the left end as

shown in Figure 4.6 Develop

(a) The expressions for the elastic curve

(b):: The deflection at point C for the case in which a = b = L/2

Figure 4.6 Exariple 4.5

CHAPTER 4 @ DEFLECTION AND IMPACT 161 Solution: The reactions are noted in the figure

@)

(b)

The moments for the segments AC and CB of the beam are expressed as

m= Oe O<x <a)

Pb

Mạ = ox ~ P(e =a) (@<x<L)

Double integrations of these equations give the results

For segment AC For segment CB

EH(= 2 Eng = x — pea) En = Fee be Ett = Fox = 2Œ =4)? +ó

Elyse barte Ely = Te 2a) tox ten From the boundary and the continuity conditions, we easily obtain

tị (4) = 02 (4): €L = đ

Đị (2) = U2(4): C2 = C4

1¡(0) =0: 0=“, v(L) = 0: ¬

The elastic curves for the left- and right-hand segments are therefore

=, PPX an ye a

vp SE be — x*) (O<x <a) aan

POX an yy ca Pa ,

ves epee EO Be) ea @sxsl)

Then, through the use of Eq (4.15) the slopes for the two parts of the beam can readily be found

Force P acts at the middle of the beam span (a = b = L/2) and hence Eqs (4.17) result in

PL

Umax = Uc = ~ ABET {4.18}

Comments: The minus sign means that the deflection is downward In this case, the elastic curve is symmetric about the center of the beam

Trang 2

162 PARTE @ FUNDAMENTALS

4.5 BEAM DEFLECTIONS BY SUPERPOSITION

The elastic deflections (and slopes) of beams subjected to simple loads have been solved and are readily available (see Tables A.9 and A.10) In practice, for combined load config- urations,-the method of superposition may be applied to simplify the analysis and design The method is valid whenever displacements are linearly proportional to the applied loads ‘This is the case if Hooke’s law holds for the material and deflections are small

To demonstrate the method, consider the beam of Figure 4.7a, replaced by the beams depicted in Figures 4.7b and 4.7c At point C, the beam undergoes deflections (v)p and (v)y, due to P and M, respectively Hence, the deflection uc due to combined loading is ve = (uc)p + (vc) From the cases 1 and 2 of Table A.9, we have

_ 5PLỀ ML2

°c = §pT T §E1 ga

Similarly, the deflection and the angle of rotation at any point of the beam can be found by

the foregoing procedure

The method of superposition can be effectively applied to obtain deflections or reactions for statically indeterminate beams In these problems, the redundant reactions are considered unknown loads and the corresponding supports are removed or modified accordingly, Next, superposition is employed: The load diagrams are drawn and expressions are written for the deflections produced by the individual loads (both known and unknown); the redundant reactions are computed by satisfying the geometric boundary conditions Fol- lowing this, all other reactions can be found from equations of static equilibrium

The steps described in the preceding paragraph can be made clearer though the illus- tration of a beam statically indeterminate to the first degree (Figure 4.8a) Reaction Rg is selected as redundant and treated as unknown load by eliminating the support at B Decomposition of the loads is shown in Figures 4.8b and 4.8c Deflections due to w and the redundant Rg are (see cases 5 and 8 of Table A.9):

(up) so 5w1 ; (op) Velen Ral?

7”2AET pie 6EI

From geometry of the original beam,

SwL4 Ral? 0s = =ZTÐT * oar ~° or Rg = 5 L ee aw (4.20) 4.20) PB P 4 4 4 4 ) =f + 4 ) a A £ je Biz, a A c B AA C đu 2 a} (a) (6) ©)

Figure 4.7 Deflections of a cantilevered beam

one

CHAPTER 4 = & «= DEFLECTION AND Impact 163

Ww °

A = ct af - le

®) ()

Figure 4.8 Deflections of a two-span continuous bear,

The remaining reactions are Ra = R, = 3wL/8, as determined by applying the equa-

tions of equilibrium Having the reactions available, deflection can be obtained using the

method discussed in the preceding section

Determination: of Interaction Force between Two Beams Placed EXAMPLE 4.6 One on Top of the Other

Two simply supported beams are situated one‘on top of the other as shown in Figure 4.9 The top

beam:is: subjected to.a' uniformly: distributed load of intensity w Find

(a), The interaction force R'at midspan C acting upward on the beam AB and acting downward

on Đeam ĐE,

: (b)- The maxinum miortenf añd deflection in beam AB

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164 PARTIE ® FUNDAMENTALS

Assumption: The beams are supported such a way that they are to deflect by the same amount at the junction’ C:

Solution: Both beams are statically indeterminate to the first degree We select R as redundant and

treat it'as an unknown load Considering the two beams in turn and using the data in Table A.9, the deflections at the center are as follows For beam AB, due to load w and owing to R,

_ 58041: RGa)? DU” 384B KT “§Đ1,

The total downward deflection is therefore

- Swat Ra?

ve MEL, 6Ely

For beam DE, due to R; the downward deflection is

RB

1= PT

(a) Equating the two expressions for deflections, vg = vp, and solving for the interaction force, we have _ 1.25wa ` R (4.21) where bY? I onn() 5

Comment: If beam DE is rigid (i.e., 1, > o), thena = | and R = 1.25wa, which is equal to the central reaction for the beam resting on three simple supports

{b) Maximum bending moment and deflection in beam AB occurring at the center are,

respectively, ¬ 1, ky 1.25 Mẹ = 3 4ˆ — bu = 2a 1 a {4,22a) 5w(2a)* i Swat 1 = j )= _— 4.22b Yc = S84ET, a) = 348, OTS 4.226)

Comments: We see from the preceding results that, as beam DE is made stiffer by either reduc- ing its span 2b or increasing its moment of inertia J), the value of a decreases and hence the value of R increases This decreases the deflection and also reduces the bending moment in the loaded beam AB

CHAPTER 4 @ DEFLECTION AND Impacr 165

Case Study 4-1

A schematic representation of the frame of a winch crane is shown in Figure 1.5 Determine the deflection under the

load using the method of superposition

Given: ‘The geometry and loading of the critical portion of the frame are known from Case Study 3-1

Data:

E = 200 GPa, = 2.01(10~°) m*

Assumptions: The loading is static Deflection due to

transverse shear is neglected

Solution: See Figure 3.31 and Table A.9

When the load P and the weight w of the cantilever

depicted in the figure act alone, displacement at D (from

WINCH CRANE FRAME DEFLECTION ANALYSIS

cases | and 3 of Table A.9) are PL? /3EJ and wlth /SEI, respectively It follows that, the deflection vp at the free

end owing to the combined loading is 3E1 SEI

Substituting the given numerical values into the pre-

ceding expression, we have

en 1 3000(1.5)? + 130.5

2720010301 3 8

= —8.6 mm

Here minus sign means a downward displacement

Comment: Since vp </A/2, the magnitude of the

deflection obtained is well within the acceptable range (see Section 3.7)

Case Study 4-2 | Bout CUTTER DEFLECTION ANALYSIS

Members 2 and 3 of the bolt cutter shown in Figure 3.32 are critically stressed Determine the deflections employ- ing the superposition method

Given: The dimensions (in inches) and loading are known from Case Study 3-2 The parts are made of AISI

1080 HR steel having E = 30 x 10° psi

Assumptions: The loading is static The member 2

can be approximated as a simple beam with an overhang Solution: See Figures 3.32 (repeated here), 4.10 and Table A.9

Member 3 The elongation of this tensile link

(Figure 3.32a) is obtained from Eq (4.1) So, due to

{a) @)

Figure 3.32 (repeated) Some free-body diagrams of bolt cutter shown in Figure 1.6:

(a) link 3; (b) jaw 2 (continued)

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166 PARTI ® » FUNDAMENTALS

Case Study (CONCLUDED)

Đ - ` The angle @, at the support A (from case 7 of Table A.9) is

vol 8; A B 6, = Mb

+ 38T

Q 122

E—z—Ì—————»————

Figure 4.10 Deflection of simple beam with an overhang

symmetry in the assembly, the displacement of each end

point A is gw Lf PL) _ Fabs 4® 2\AE) 2AE 128(1.25) = =569(0759 in, 2($) (4) G0 x 109) 6.900") in

Member 2 This jaw is loaded as shown in Fig- ure 3.32b The deflection of point D is made up of two parts: a displacement v, owing to bending of part DA act-

ing as a cantilever beam and a displacement v, caused by

the rotation of the beam axis at A (Figure 4.10) The deflection v; at D (by case | of Table A.9) is

where M = Qa The displacement v2 of point D, due to

only the rotation at A, is equal to @4@, or _ Qba?

SEI

The total deflection of point D, v, + v2, is then

- 8

Up = 3p7 at Tn the foregoing, we have

1

l= Tạ

1/31/39) ân ga 4

-3(3) (3) = 0.824(107) in

Substitution of the given data results in

96(12)(1 +3)

= ———————————=Ố Fin

GO x 105) 0.824 x Tos) PTS LOT in

Up

Comment: Only very small deflections are allowed in

members 2 and 3 to guarantee the proper cutting stroke,

and the values found are acceptable

| 4.6 BEAM DEFLECTION BY THE MOMENT-AREA METHOD

In this section, we consider a semigraphical technique called the moment-area method for determining deflections of beams The approach uses the relationship between the derivatives of the deflection v and the properties of the area of the bending moment diagram Usually it

gives more rapid solution than integration methods when the deflection and slope at only one point of the beam are required The moment-area method is particularly effective in the

analysis of beams of variable cross sections with uniform or concentrated loading (3-5]

MOMENT-AREA THEOREMS

Two theorems form the basis of the moment-area approach These principles are developed by considering a segment AB of the deflection curve of a beam under an arbitrary loading The sketches of the W/E/ diagram and greatly exaggerated deflection curve are shown in

Cee

CHAPTER 4 @ DEFLECTION AND ÍMPACT

M — B El 1/8 °C A @ oT AIX 1Í { + G4 II 43 ị = {| 1g db i Noe ».z fag net x9 Ps LAL Tangents atA and B Figure 4.11 Moment-area method: (a) M/E! diagram; (b) elastic curve

Figure 4.11, Here M is the bending moment in the beam and £/ represents the flexural

rigidity The changes in the angle dO of the tangents at the ends of an element of length dx and the bending moment are connected through Eqs (4.14) and (4.15):

M

đổ = —~ đx EI @)

The difference in slope between any two points A and & for the beam (Figure 4.11) can be expressed as follows:

a8 fide a

Qua — Ôn == | a area of 1 diagram between A-and B} (4.23)

oe da El Elo |

This is called the first moment-area theorem: The change in angle 024 between the tangents to the elastic curve at two points A and B equals the area of the M/EJ diagram between those points Note that the angle 6g4 and the area of the M/EJ diagrams have the same sign That means a positive (negative) area corresponds to a counterclockwise (clockwise) rotation of the tangent to the elastic curve as we proceed in the x direction Hence, 0g,

shown in Figure 4.11b is positive

Inasmuch as the deflection of a beam are taken to be small, we see from Figure 4.11b that the vertical distance dt due to the effect of curvature of an element of length dx equals x d@, where dO is defined by Eq (a) Therefore, vertical distance AA’, the tangential devi- ation tag of point A from the tangent at B is

Be Máy

_ “ae [aca ot diagram between Aand 2] T1 (4.24)

tag =

in which x is the horizontal distance to the centroid C of the area from A This is the second moment-area theorem: The tangential deviation t4g of point A with respect to the tangent at B equals the moment with respect to A of the area of the M/EJ diagram between A and B

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168 PART fo @.’ FUNDAMENTALS Likewise, we have tye (4.25)

[me of S diagram between A and B] ”

The quantity X2 represents the horizontal distance from point B to the centroid C of the area (Figure 4.11a) Note that tan 4 fga generally, Also observe from Eqs (4.24) and (4.25) that the signs of t,g and tg4 depend on the sign of the bending moments In many beams, it is obvious whether the beam deflects upward or downward and whether the slope is clockwise or counterclockwise, When this is the case, it is not necessary to follow the sign conventions described for the moment-area method: We calculate the absolute values and find the directions by inspection

APPLICATION OF THE MOMENT-AREA METHOD

Determination of beam deflections by moment-area theorems is fairly routine They are equally applicable for rigid frames In continuous beams, the two sides of a joint are 180° to one another, whereas in rigid frames the sides of a joint often are at 90° to one an- other Our discussion is limited to beam problems A correctly constructed M/EI diagram and a sketch of the elastic curve are always necessary Table A.3 may be used to obtain the

areas and centroidal distances of common shapes The slopes of points on the beam with

respect to one another can be found from Eq (4.23) and the deflection, using Eq (4.24) or (4.25), The moment-area procedure is readily used for beams in which the direction of the tangent to the elastic curve at one or more points is known (e.g., cantilevered beams) For computational simplicity, often M/EJ diagrams are drawn and the formulations made in terms of the quantity E/; that is, numerical values of EJ may be substituted in the final step of the solution

For a statically determinate beam with various loads or an indeterminate beam, the displacements determined by the moment-area method are usually best found by super- position This requires a series of diagrams indicating the moment due to each load or reaction drawn on a separate sketch In this manner, calculations can be simplified, because the areas of the separate M/EJ diagrams may be simple geometric forms When treating statically indeterminate problems, each additional compatibility condition is expressed by a moment-area equation to supplement the equations of statics

= EXAMPLE 4.7 Shaft: Deflection by the Moment-Aiea Method

A-simple shaft carries its own weight of intensity w, as depicted in Figure 4.12a Determine the slopes

at the ends and center deflection:

Assumption: Bearings act as simple supports

Solution: : Inasmuch as the flexural rigidity EJ is constant, the M/E diagram has the same para-

bolic shape as the bending-moment diagram (Figure 4.12b), where the area properties are taken from

Table'A:3:.Thé elastic curve is depicted in Figure 4.12c, with the tangent drawn at A

CHAPTER4 @ DEFLECTION AND IMPACT ¥ W + thị li li tk - AL Cc eae x Lp HhJ—/2— {ay 278 `: [oy La cC x 51/16 31/16 (6) ” ce B A oe mủ€ ì Tangent’ ci Tangent

atA ae

©) Figure 4.12 Example 4.7

The shaft and loading aré symmetric about the cénter C; hence, the tangent to the elastic curve at C is horizontal: @¢ = 0 Therefore, 6¢4 = 0-64 or 04 = ~Oc, and

Ava LY (wi? _ wis

P= 3\2/\8EI)~ 24871

By the first moment-area theorem, 6¢4 = Ay: wl}

ee 4.26

oa =~ SRR 8 (4.26)

The minus sign means that the end A of the beam rotates clockwise, as stiown in the figure

Through the use of the second moment-area theorem, Eq (4.25), =A 4E wl

fea = 1\ 56) = T2RET

in which ¢¢, = CC’ and @4,L/2 = C’C” (Figure 4.12c) The maximum deflection, vax = CC”, is 2 wh? (Lh wit _ _ Swl4

Umax 28T (2 128EI 384ET

(4.27)

The minus sign indicates that the deflection is downward

Comment: Alternatively, the moment of area A, about point A, Eq (4.24), readily gives the nu-

merical value of Umax -

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FUNDAMENTALS 170 PARTIL.®

Displa ents of a Stepped Cantilevered Beam by the Moment-Area Method

nonprismatic cantilevered beam with two different moment of inertia caries a concentrated load P

it its free end (Figure 4.13a) Find the slope at B and deflection at C

P EI Seed Be eA c ep p_ (} SpiJagr.~PL/ABI —PL/2EI (6) () Figure 4.13 Example 4.8

Solution: The M/EJ diagram is divided conveniently into its component parts, as shown in Figure 4.13b:

PL? PL? PL?

Pape 8EI Age >, 16B! Ỷ 38m SEI

The elastic curve is in Figure 4.13c Inasmuch as 04 = 0 and vq = 0, we have Oc = Oc4, 0g = Opa, ĐC = te4, and vg = tag

Applying the first moment-area theorem,

5PL?

Op = Art Ag + Ag = ~T6ET (4.28a)

The minus sign means that the rotations are clockwise From the second moment-area theorem,

i L\ 5PB

= A; ([— Agfa fe

" (G) + 2(5) 96E7

The minus sign shows that the deflection is downward

(4.28b)

CHAPTER 4 e DEFLECTION AND Impact 171

Reactions of a Propped up Cantilever by the Moment-Area Method

‘A propped cantilevered beam is loaded by a concentrated force P acting at the position shown in Eieure 4144 Determine the reactional forces and moments at the ends of the beam 8

Figure 4.14 Example 4.9

Solution: The reactions indicated in Figure 4.14a shaws that the beam is statically indeterminate to the first degree We select Rg as a redundant (or unknown) load and remove support B (Fig-

uré 4.14b) The corresponding M/EJ diagram is in Figure 4.14c, with the component areas Pa

2E1

Ải=——

ET

One displacement compatibility condition is required to find the redundant load Observe that the stope at the fixed end and the deflection at the supported end are 0; the tangent to the elastic curve at A passes through B, or fg, = 0 Therefore, by the second moment-area theorem,

Ral? (2L\_ Pa? -3)= 2ET (=) 287V 3 5 A = Solving, Pa? Rg=è-(3L—a 8 = 5736 ) (4.29)

Comments: The remaining reactions are obtained from equations of statics Then, the slope and deflection are found as needed by employing the usual moment-area procedure

EXAMPLE 4.9

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172 PARTI @ FUNDAMENTALS 4.7 IMPACT LOADING

A moving body striking a structure delivers a suddenly applied dynamic force that is called an impact or shock load Details concerning the material behavior under dynamic loading are presented in Section 2.9 and Chapter 8 Although the impact load causes elastic mem- bers to vibrate until equilibrium is reestablished, our concern here is with only the influence

of impact or shock force on the maximum stress and deformation within the member

Note that the design of engineering structures subject to suddenly applied loads is complicated by a number of factors, and theoretical considerations generally serve only qualitatively to guide the design [6-8] In Sections 4.8 and 4.9, typical impact problems are analyzed using the energy method of the mechanics of materials theory together with the following common assumptions:

1 The displacement is proportional to the loads

2 The material behaves elastically, and a static stress-strain diagram is also valid under impact

3 The inertia of the member resisting impact may be neglected

No energy is dissipated because of local deformation at the point of impact or at the

supports

Obviously, the energy approach leads to an approximate value for impact loading It presupposes that the stresses throughout the impacted member reach peak values at the same time In a more exact method, the stress at any position is treated as a function of time, and waves of stress are found to sweep through the elastic material at a propagation rate This wave method gives higher stresses than the energy method However, the former

is more complicated than the latter and not discussed in this text The reader is directed to

references for further information [9-12]

4.8 LONGITUDINAL AND BENDING IMPACT

Here, we determine the stress and deflection caused by linear or longitudinal and bending impact loads In machinery, the longitudinal impact may take place in linkages, hammer-type power tools, coupling-connected cars, hoisting rope, and helical springs Examples of bending

impact are found in shafts and structural members, such as beams, plates, shells, and vessels

FREELY FALLING WEIGHT

Consider the free-standing spring of Figure 4.15a, on which is dropped a body of mass m from a height 4 Inasmuch as the velocity is 0 initially and 0 again at the instant of maximum deflection of the spring (8nax), the change in kinetic energy of the system is 0 Therefore, the work done by gravity on the body as it falls is equal to the resisting work done by the spring:

WÁ + 8imas) = ts max (4.30)

in which & is the spring constant

CHAPTER4 ® DEFLECTION AND IMPACT

Figure 4.15 (a) Freely falling body (b) Horizontal moving body

The deflection corresponding to a static force equal to the weight of the body is simply

W/k This is called the static deflection, 5 The general expression of maximum dynamic deflection is, using Eq (4.30),

Smax = DI + (8s)? + 2h {4.31 a}

This may be written in the form

4 Srna

(: + lt XS) da = KS (4.31b)

The impact factor K, is defined by

Kal+ +e (432)

& W5

Multiplying the K by W gives an equivalent static, or maximum dynamic load:

Prnax = KW (4.33)

To compute the maximum stress and deflection resulting from impact loading, P may be used in the formulas for static loading

Two extreme situations are clearly of particular interest When A >> dmax, the work term, Wax, in Eq (4.30) may be neglected, reducing the expression to

Smax = VA 28sh (4.34)

On the other hand, when A = Q, the load is suddenly applied, and Eq (4.30) reduces to

Snax = 25s (4.35)

HorIZONTALLY MOVING WEIGHT

An analysis similar to the preceding one may be used to develop expressions for the case of a mass (m = W/g) in horizontal motion with a velocity v, stopped by an elastic body, In this case, kinetic energy 2, = mu? /2 replaces W(h + Smax), the work done by W, in

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174 PARTI:: @ FUNDAMENTALS

Eq (4,30) By so doing, the maximum dynamic deflection and load are „2 2E:

Smax = Sse IF =,/ 7 (4.36a)

vg

Prax = My) y= = VDE ck (4.36b)

st

The quantity dy is the static deflection caused by a horizontal force W Note that m is mea- sured in kg in SI or Ib-s’/in in U.S units, Likewise expressed are v (in m/s or in./s), the gravitational acceleration g(in m/s? or in./s”), and E; (in N-m or in.- ib)

When the body hits the end of a prismatic bar of length L and axial rigidity AE (Fig- ure 4.15b), we have k = AE/L and hence 6, = mgL/AE Equations (4.36) are therefore

ime

Snax = AE (4.37a)

mraz

Prnax = L (4.37b)

The corresponding maximum dynamic compressive stress, taken to be uniform through the

bar, is :

J0 4.38,

Gmáx S1 me Ay AL ườn ( ) The foregoing shows that the stress can be reduced by increasing the volume AL or decreasing the kinetic energy E, and the modulus of elasticity E of the member We note that the stress concentration in the middle of a notched bar would reduce its capacity and

tend to promote brittle fracture This point has been treated in Section 2.9

Impact Loading ona Rod

EXAMPLE 4.10

‘The prismatic rod depicted in Figure 4.16 has length L, diameter d, and modulus of elasticity E A rubber compression washer of stiffness k and thickness ¢ is installed at the end of the rod

(a) Calculate the maximum stress in the rod caused by a sliding collar of weight W that drops froma height A onto the washer

(b) : Redo part'a, with the washer removed ~

Given: “Le Sf hk = 3 ft, đ=‡in te din a

E = 30 x 10% psi, k == 25 Ib/in., W = 8 Ib

CHAPTER4 © DEFLECTION AND IMPACT

175

Solution: The cross-sectional area of the rod A = z(l /23?/4 = 1/16 in? pa RR

(a) For the rod with the washer, the static deflection is pe 5 = Whe a W W[ | Collar

: AE i y

8(16)(5 x 12) 8 3

= + — = 0.081 x 107° + 0.32 in: Rod

zQ@0 x 109 25 h 4

The maximum dynamic stress, from Eqs (433) and (4.32); | 4

mis | / WK + + Om A Rubber washer 2 12 : = - : i+ _ ] = 653 psi Figure 4.16 x ` Example 4.10

(by “In the absence of the washer, this equation results in

8x 16 23 x 12 + 0.25) `

= “Tem I=386ks

Ome | + oR 103 »

Comments: The differencẻ in stress for the preceding two solutions is large This suggests the

need for flexible’ systems for withstanding impact loads Interestingly, bolts subjected to dynamic

Toads, stich as those used to attach the ends to the tube in pneumatic cylinders, are often designed with long’ grips (see Section 15.9) to take advantage of the more favorable stress conditions

EXAMPLE 4.11

Impact Loading on a Beam

Aweight W is dropped a height A, striking at midspan a simply supported steel beam of length L The

beam is of rectangular cross section of width b and depth d (Figure 4.17) Calculate the maximum

deflection and maxiraum stress for these two cases:

(a) The beam is rigidly supported at each end, (b) The beam is supported at each end by springs

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176 PART! ® FUNDAMENTALS

Given: W = 100N2A= 150 mm, L =2m, b= 30mm, andd = 60 mm

: Assumptions: Modulus of elasticity 2 = 200 GPa and spring rate k = 200 kN/m

Solution: | We have Moax = WL/4 at point C and 7 = bd? /12 The maximum deflection, due to a

static: load, is (ftom case Sof Table A.9):

W72 100(2)3(12)

Spe eS 0.154

98 G8ET ~ 48000 x 10.0300 04m

The maximuni static stress equals

Cy Mrax C = 100(2) 0.03) (12) = 2,778 MPa

1 ` -4(0.03)(0.06)2

(a): The impact factor, using Eq (4.32),

2(0.15) K=t F01540 45.15 Theréfore, Sinax’ = 45.15(0.154) = 6.95 mm Ønax = 45.15(2.778) = 125 MPa

(b) The static deflection of the beam due to its own bending and the deformation of the

springs is 50

Sq = 91 + 208” 9.404 mm The impact factor is then

2(0.15) K=l+jt+ 0:404(10-3) 28.27 Hence, Bmax == 28.27(0.404) = 11.42 mm Ginax == 28.27(2.778) == 78.53 MPa

Comments: Comparing the results, we observe that dynamic loading considerably increases de- flection and stress in a beam Also noted is a reduction in stress with increased flexibility, owing to the spring added to the supports However, the values calculated are probably somewhat higher than

the actual quantities, because of our simplifying assumptions 3 and 4

CHAPTER 4 ° DEFLECTION AND IMPACT 17?

4.9 TORSIONAL IMPACT

In machinery, torsional impact occurs in rotating shafts of punches and shears, in geared drives, at clutches, brakes, torsional suspension bars, and numerous other components Here, we discuss the stress and deflection in members subjected to impact torsion The problem is analyzed by the approximate energy method used in the preceding section

Advantage will be taken of the analogy between linear and torsional systems to readily

write the final relationships

Consider a circular prismatic shaft of flexural rigidity GJ, length L, fixed at one end and subjected to a suddenly applied torque T at the other end (Figure 4.18) The shaft stiff- ness, from Eq (4.10), is k = GJ/L, where J = zZ4/32 and d is the diameter The maxi- mum dynamic angie of twist (in rad), from Eq (4.36a), is

`

max = a : (4.39a)

in which E, is the kinetic energy Similarly, the maximum dynamic torque, referring to

Eg (4.36b), is

: 2E:G1

Tag = \J 7 : (4.39b)

The maximum dynamic shear stress, Tax = 16T max /xd?, is therefore

: EG

Truy 2 4} AL (4.40)

Here A represents the cross-sectional area of the shaft

Recall from Section 1.11 that, for a rotating wheel of constant thickness, the kinetic energy is expressed in the form

(4.41)

Trang 10

178 PARTI ® FUNDAMENTALS with 1 2 T= ~mb 2 (4.42a) m = 1b*to (4.42b)

In the foregoing, we have

7 = mass moment of inertia (N-s?-m or Ib- ?-in.) @ = angular velocity (rad/s)

m= mass (kg or lb - s2ñn.)

b = radius t = thickness

p = mass density (kg/m? or Ib- s*/in.*)

As before, W and g are the weight and acceleration of gravity, respectively A detailed treat- ment of stress and displacement in disk flywheels is given in Section 16.5

Note that in the case of wheel of variable thickness, the mass moment of inertia may conveniently be obtained from the expression

I= mr? (4.43)

The quantity r is called the radius of gyration for the mass It is hypothetical distance from the wheel center at which the entire mass could be concentrated and still have the same moment of inertia as the original mass

Impact Loading ‘on‘a Shaft

EXAMPLE 4.12

A shaft of diameter d and length L has a flywheel (radius of gyration r, weight W, modulus of rigid-

ity G, yield strength in shear S,,) at oné end and runs at a speed of n If the shaft is instantly stopped

at the other end, determine

(a).: The maximum shaft angle of twist {b) The maximuin shear stress

Given: 4= 3 in, E = 2.5 ft, W = 1201b, = 10 in., n = 150 rpm

Assumption: The shaft is made of ASTM-A24? steel So, by Table B.1, G = 11.5 x 10° psi and

Sys = 30 ksi

Solution:: The area properties of the shaft are

#3)? s2 z@% A= q =T, 069 in.*, J = 3 = 7 7.952 in 3

CHAPTER 4 e DEFLECTION AND IMPACT

The atigular velocity equals 2z Qn

o=n (5) == 150 (5) = Sx rad/s

(a): The kinetic energy of the flywheel must be absorbed by the shaft So, substituting

Bq (4.43) into Eq (4.41), we have

Wor? 28 1205x302 2686) Ey = (a) = 3835 in-Ib From Eq (4.39a);

`, REL: fF 23835)2.5 x 12) 7!”

Pons = EF = (115 x 1059)(7.952)

== 0.05 rad = 2.87°

(b) Through the use of Eq (4.40),

<a [kể _ ;[ G835)015 x 109 11

Fax = AT” “| 06905 x 12) == 28.84 ksi

Comment: The stress is within the elastic range, 28.84 < 30, and hence assumption 2 of Sec- tion 4.7 is satisfied

179

*4.10 BENDING OF THIN PLATES

A plate is an jnitiaily flat structural member with thickness small compared with remaining dimensions it is usual to divide the plate thickness f into equal halves by a plane parallel to the faces This plane is called the midsurface of the plate The plate thickness is mea-

sured in a direction normal to the midsurface at each point under consideration Plates of technical importance are usually defined as thin when the ratio of the thickness to the smaller span length is less than 1/20 Here, we discuss briefly the bending of thin plates For a detailed treatment of the subject, see [13~15]

Basic ASSUMPTIONS

Consider a plate before deformation, depicted in Figure 4.19a, where the xy plane coin-

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180 PARTI @ FUNDAMENTALS + m t2 PA LF te } aes {4 x ff w ì i SAS TW ‘dw % `, ax ; od = ~z ow z fu eo {a) (6)

Figure 4.19 Plate in bending: (a) before deformation; (b) rotation ofa plane section due to deflection

the coordinate system shown, the basic assumptions of the small deflection theory of bend- ing for isotropic, homogenous, thin plates may be summarized as follows:

1 The deflection w of the midsurface is small in comparison with the thickness f of the plate; hence, the slope of the deflected surface is much less than unity

2 Straight lines (such as mn) initially normal to the midsurface remain straight and nor-

mal to that plane after bending:

3 No midsurface straining occurs due to bending This is equivalent to stating that Strains 1⁄s¿, Yee, and €, are negligible

4 The component of stress normal to the midsurface, o,, may be neglected

These presuppositions are analogous to those associated with the simple bending theory of beams

STRAIN-DISPLACEMENT RELATIONS

On the basis of the foregoing assumptions, the strain displacement relations of Eq (3.61) reduce to

eo, = ot ea tt = You ta)

xạ J9” ay ấy

and it can be shown that

aw » aw (e)

u=x—Z—, =—z—

ax : ay

Combining Eqs (a) and (b), we have ‘

: 32w a aw 8?

& = Tay?” y sẽ ay? Te eae ay (4.44)

CHAPTER4 © DEFLECTION AND IMPACT

Because, in small deflection theory, the square for a slope may be considered negligible, the partial derivatives of these relations represent the curvatures of the plate The curva-

tures at the midsurface in planes parailel to the zx (Figure 4.19b), yz, and xy planes are,

respectively,

Law 1 wy 1 aw

ry 8x2? ry ay?’ ny Oxy (4-45)

The quantity r represents the radius of curvature Clearly, the curvatures are the rates at which the slopes vary over the plate

Examining the preceding relationships, we are left to conclude that a circle of curva-

ture can be constructed similarly to Mohr’s circle of strain The curvatures hence trans-

form in the same manner as strains It can be verified by using Mohr’s circle that

(1/7) + (1/ry) = VẦw, The sum of the curvatures in perpendicular directions, called the

average curvature, is invariant with respect to rotation of the coordinate axis This asser- tion is valid at any location on the midsurface

PLATE STRESS, CURVATURE, AND MOMENT RELATIONS

For a thin plate, substituting Eqs (4.44) into Hooke’s law, we obtain

——— Ez {aw 82w x=—r Ay + hy oa we y=~TT (8y † hy aw 82 (4.46) Ez 8?w ty = “Teo axdy

where tyy = ty We see from these relations that the stresses are 0 at the midsurface and vary linearly over the thickness of the plate

The stresses distributed over the side surfaces of the plate, while producing no net force, result in bending and twisting moments These moment resultants per unit length (in SI units N- m/m, or simply N) are designated M,, My, and M,, With reference to Figure 4.20a,

1/2 1/2 / oy dy dz = ay f 20, dz = M, dy 2 —/2 or 4/2 My = 20x dz (4.47) tf

Here, t is the thickness of the plate Substituting into this expression the stress o, given by Eqs (4.46), we obtain M, in terms of curvatures Expressions involving M, and

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182 PARTI-: ® FUNDAMENTALS {a)

Figure 4.20 (a) Plate segment in pure bending {b) Positive stresses on an element in the bottom half of a plate

Myy = My, are derived in a like manner In so doing, we have

(4.48)

The quantity D represents the flexural rigidity:

Ep

D=— 120 — v2 449

Interestingly, if a plate element of unit width were free to expand sidewise under the given loading, the flexural rigidity would be Et*/12 The remainder of the plate does not allow this action, however Because of this, a plate shows greater stiffness than a narrow beam by

a factor 1/(1 — v) or about 10% for v = 0.3

According to the sign convention, a positive moment is one that results in positive stresses in the positive (bottom) half of the plate (see Section 1.13), as depicted in Fig- ure 4.20b The maximum stresses occurring on the surface of the plate are obtained by sub- stituting z = 1/2 into Eqs (4.46), together with the use of Eqs (4.48), as

6M, 6M, Sàn 6M,

Onan qe Oy man 2 SExy max Ste 2 4.50)

Since there is a direct correspondence between the moments and stresses, the equation for transforming the stresses should be identical with that used for the moments Mohr’s circle

therefore may be applied to moments as well as to stresses

CHAPTER 4 ° DEFLECTION AND IMPACT 183

4.11 DEFLECTION OF PLATES BY INTEGRATION

Consider an element dx dy of the plate subject to a uniformly distributed load per unit area p

(Figure 4.21) In addition to the moments M,, My, and M,y previously discussed, we now find vertical shear forces Q, and Q, (force per unit length) acting on the sides of the element With the change of location, for example, by dx, one of the moment components, say, M,, acting on the negative x face, varies in value relative to the positive x face as M, + (0M,,/0x) dx Treating all components similarly, the state of stress resultants shown

in the figure is obtained :

Variations in the moment and force resultants are governed by the conditions of equi- librium Application of the equations of statics to Figure 4.21 leads to a single differential equation in terms of the moments This, when combined with Eqs (4.48), results in [13]

Hw be

Oxt (4.514)

or, in concise form,

WỶw= (4.51b)

The preceding expression, first derived by Lagrange in 1811, is the governing differential

equation for deflection of thin plates

BOUNDARY CONDITIONS

Two common boundary conditions that apply along the edge at x = a of the rectangular plate with edges parallel to the x and y axes (Figure 4.22) may be expressed as follows

In the clamped edge (Figure 4.22a), both the slope and the deflection must vanish: aw — 3x = 0, (x =a) (4.52) eo } yy Section a tw we % Plan view (a) {b)

Figure 4.21 Positive moments and shear forces per Figure 4.22 Two common boundary unit length and distributed loads per unit area on a conditions: (a) fixed edge; (b) simply

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184 PARTI @ FUNDAMENTALS

In the simply supported edge (Figure 4.22b), both the deflection and bending moment are 0 The latter implies that, at edge x = a, dw/dy = 0 and d°w/y? = 0 Therefore,

: 8?w

w= Ô, mm =0, (x =a) (4.53)

Other typical conditions at the boundaries may be expressed similarly

To determine the deflection w, we must integrate Eq (4.51) with the constants of inte- gration dependent on the appropriate boundary conditions Having the deflection available, the stress (as well as strain and curvature) components are obtained using the formulas derived in the preceding section This is illustrated in the solution of the following problem

EXAMPLE 4.13 | Determination of Deflection and Stress in'a Plate

A long, narrow plate of width b and thickness ¢, a so-called plate strip, is simply supported at edges y= O-and y= b, as depicted in Figure 4.23 The plate carries the loading of the form

op Q)= pe sin (=) (a) TE 'Ƒigure:4.23.' Example 4.13

The quantity: p, répresents the load intensity along the x’ axis Let v = 1/3 Determine (a=: The equation of the deflection surface and maximum stresses

(b): The maximum, deflection: and stresses for p, = 15 kPa, b = 0.5 m, ¡ = 12.5 mm, and

G.=210 GPa

Solution: Due'to symmetry in the loading and end restraints about the x axis, the plate deforms

into: a cylindtical’ surface with its generating line parallel to the x’ axis Since, for this situation, 8w/8x = Ú and 82w/2xôy =s 0, Eqs (4.48) reduce to

Pw Pw

Mỹ= T19: My == gã (4.54)

Also Bg (4:51) becomes

aw pee `

Se ae D (4.55) 4.55

This equation is the same as the wide beam equation, and we conclude that the solution proceeds as inthe case of a beam:

CHAPTER 4 8 DEFLECTION AND IMPACT

(a): Introducing Eq (a) into Eq (4.55), integrating, and satisfying the boundary conditions

Pig

w= 0, Se = 0, (y = 0, y= b)

we obtain the deflection b pe (ay

Woes 6) 5 sin(72) (4.56)

‘The largest deflection of the plate occurs at y = b/2 Therefore,

B\t

Winax = (=) = (4.57) The moments are now readily determined carrying Eq (4.56) into (4.54) Then, the maxi- mum stresses, occurring at y = b/2, are found applying Eqs (4.50) as

2

b\? ` /0\?

Øy max =2 0.6 Po 7}" Os trax = VOy.max = 0.2 Po () > Try = 0 (4.58)

(}- Substituting the given data into Eq (4.49), we have

_ 210(10°)(0.0125)?

we 12(b= 8)

Similarly, Eqs (4.57) and (4.58) lead to

0.5\* 15(403) Wrhax (=) 38.452 ss 0.25 mm == 38,452 N-m 2 Øy ma = 0.6(15 x 108) () = 14.4 MPa 1

Ởy max = 344) =: 4.8 MPa

Comment: The result, Wma, /t = 0.02, shows that the deflection surface is extremely flat, as is

often the case for small deflections

REFERENCES

1 Ugural, A C., and S K Fenster Advanced Strength and Applied Elasticity, 4th ed Upper Saddle River, NJ: Prentice Hall, 2003

2, Young, W C., and R C Budynas Roark’s Formulas for Stress and Strain, 7th ed New York:

McGraw-Hill, 2001

3 Ugural, A C Mechanics of Materials New York: McGraw-Hill, 1991

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186 PARTI- ® - FUNDAMENTALS

5 Wang, C K., and C G Salmon Introductory Structural Analysis Upper Saddle River, NJ: Prentice Hall, 1984

6 Timoshenko, S P., and J N Goodier Theory of Elasticity, 3rd ed New York: McGraw~

Hill, 1970

7 Fliigge, W., ed Handbook of Engineering Mechanics New York: McGraw-Hill, 1962, Sec- tion 27.3

Harris, C M Shock and Vibration Handbook New York: McGraw-Hill, 1988

9 Burr, A H., and J B Cheatham, Mechanical Analysis and Design, 2nd ed Upper Saddle River, NJ: Prentice Hall, 1995

10 Goldsmith, W Impact London: Edward Arnold Ltd., 1960 11, Koisky, H Stress Waves in Solids New York: Dover, 1965

12 Varley, E., ed “The Propagation of Shock Waves in Solids.” AMD-17 Symposium New York:

ASME, 1976

13 Ugural, A C Stresses in Plates and Shells, 2nd ed New York: McGraw-Hill, 1999

14 Szilard, R Theory and Analysis of Plates Upper Saddle River, NJ: Prentice Hall, 1974 15 Timoshenko, S P., and S Woinowsky-Krieger Theory of Plates and Shells, Ind ed New York:

McGraw-Hill, 1959

œ

PROBLEMS

Sections 4.1 through 4.6

4.1 A high-strength steel rod of length L, used in control mechanism, must carry a tensile load of P without exceeding its yield strength S, with a factor of safety n nor stretching more than 4 (a) What is the required diameter of the rod?

(b) Calculate the spring rate for the rod

Given: P = 10 KN, E = 200 GPa, S$, = 250 MPa, L = 6m, 6 == S mm Design Decision: The rod will be made of ASTM-A242 steel Take » = 1.2

4.2 Before loading there is a gap A between the wall and the right end of the copper rod of diameter d (Figure P4.2), Calculate the reactions at A and B, after the rod is subjected to an axial load of P

Given: A = 0.0)4in., d = 4 in, P = 8 kips, E = 17 x 10° psi

E——8in.—— [10 fo, ke A pat,

“A B

Figure P4.2

4.3 Atroom temperature (20°C), a gap A exits between the wall and the right end of the bars shown in Figure P4.3 Determine

(a) The compressive axial force in the bars after temperature reaches 140°C

(b) The corresponding change in length of the aluminum bar

4.5

4.6

47

CHAPTER4 ® DEFLECTION AND Impact 187 300 mm oh 250 mm a kA Figure P4.3 Given: Aq = 1000 mm?, Eq =70GPa, a = 23 x 10°6/°C, A= tmm A, = 500 mm?, E, = 210 GPa, oy = 12 x 107S/°C, Redo Problem 4.3 for the case in which A = 0

Two steel shafts are connected by gears and subjected to a torque 7, as shown in Figure P4.5

Calculate

(a) The angle of rotation in degrees at D (b) The maximum shear stress in shaft AB

Given: G = 79 GPa, T = 500N-m, d; = 45 mm, @ = 35 mm

Figure P4.5

Determine the diameter d, of shaft AB shown in Figure P4.5, for the case in which the maximum shear stress in each shaft is limited to 150 MPa

Design Decisions: d; = 65 mm The factor of safety against shear is n = 1.2

A disk is attached to a 40-mm diameter, 0.5-m long steel shaft (G = 79 GPa) as depicted in

Figure P4.7

40 mm

Trang 15

188 PARTI ® FUNDAMENTALS 48 49 4.10 401 412

Design Requirement: To achieve the desired natural frequency of torsional vibrations, the

stiffness of the system is specified such that the disk will rotate ].5° under a torque of 1 KN-m How deep (A) must a 22-mmi diameter hole be drilled to satisfy this requirement?

A solid round shaft with fixed ends is under a distributed torque of intensity T(x) = 7), Ib- in,/in., as shown in Figure P4.8 Determine the reactions at the walls

Figure P4.8

Resolve Problem 4.8, for the case in which T(x) = (x/L)T)

A simply supported beam AB carries a triangularly distributed load of maximum intensity wy (Figure P4.10)

(a) Employ the fourth-order differential equation of the deflection to derive the expression for the elastic curve

(6) Determine the maximum deflection vax

Figure P4.10

Asimply supported beam is loaded with a concentrated moment M,, as shown in Figure P4.i1 Derive the equation of the elastic curve for the segment AC of the beam

Figure P4.11

Asimple beam of wide~flange cross section carries a uniformly distributed load of intensity w (Figure P4.12) Determine the span length L

Given: A = 12.5 in., B = 30 x 10° psi

Design Requirements: Oyox = 10 ksi, Umax = Ỹ in

4.13

414

4.15

CHAPTER 4 ° DEFLECTION AND IMPACT 189

THHHTHH Figure P4.12

The overhanging beam ABC supports a concentrated load P at the free end, as shown in

Figure P4.13 For the segment BC of the beam,

(a) Derive the equation of the elastic curve (b) What is the maximum deflection Đmax?

(ec) Calculate the value of the Umax for the following data:

I1=5.12x 10 mm, E=200GPa, - P=25kN L=2m, a=0.5m Figure P4.13

Two cantilever beams AB and CD are supported and loaded as shown in Figure P4.14 What is the interaction force R transmitted through the roller that fits snugly between the two

beams at point C? Use the method of superposition and the deflection formulas of the beams

from Table A.9

P B A es = E\l, * Es Cl Ey Dd L 3 2 | Figure P4.14

Figure P4.15 shows a compound beam with a hinge at point B It is composed of two parts: a

beam BC simply supported at C and a cantilevered beam AB fixed at A Apply the superposi- tion method using Table A.9 to determine the deflection vg at the hinge

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190 PARTI- ® ' FUNDAMENTALS

4.16 A propped cantilevered beam carries a uniform load of intensity w (Figure P4.16) Determine

the reactions at the supports, using the second-order differential equation of the beam

deflection 1 w CES Ry _—_ —————EL Re Figure P4,.16

4.17 A fixed-ended beam AB is under a symmetric triangular load of maximum intensity wo, as shown in Figure P4.17 Determine all reactions, the equation of the elastic curve, and the max- imum deflection

Requirement: Use the second-order differential equation of the deflection

Figure P4.17

448 A fixed-ended beam supports a concentrated load P at its midspan (Figure P4,18), Determine all reactions and the equation of the elastic curve

Figure P4,18

4.19 Redo Problem 4.16, using the method of superposition together with Table A.9

4.20 A cantilever beam is subjected to a partial loading of intensity w, as shown in Figure P4.20 Use the area moments to determine

(a) The slope at the free end

(b) The deflection at the free end

4.21

4.22

CHAPTER4 ® DEFLECTION AND IMPACT 191

Ww

5 Đ ,

og E2 a2~~ Figure P4.20

A simple beam with two different moments of inertia is under a center load P, as shown in

Figure P4.21 Apply the area moments to find

(a) The slope at point B (b) The maximum deflection

P 4EI BI A ee B AER he a E—1/2———L/2——] Figure P4.21

and 4.23 A simple beam with an overhang and a continuous beam are supported and loaded,

as shown in Figures P4.22 and P4.23, respectively Use the area moments to determine the sup-

port reactions q = € A : “ sas? ks B, Hom gr J pe 1/2 | Figure P4.23 Figure P4.22 Sections 4.7 through 4.11

4.24 The uniform rod AB is made of steel Collar D moves along the rod and has a speed of

v == 3.5 m/s as it strikes a small plate attached to end A of the rod (Figure P4.24) Determine the largest allowable weight of the collar

Given: S, = 250 MPa, E = 210 GPa

Design Requirement: A factor of safety of = 3 is used against failure by yielding

20 mm

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192 PARTL ~~ @:).° FUNDAMENTALS

4.25 The 20-kg block D is dropped from a height h onto the steel beam AB (Figure P4.25) Determine

(a) The maximum deflection of the beam {b) The maximum stress in the beam

Given: h = 0.5m, E = 210 GPa

Figure P4.25

4,26 Collar of weight W, depicted in Figure P4.26, is dropped from a height A onto a flange at the

end B of the round rod Determine the W

Given: h = 3.5 ft,d = lin, L = 15 ft, E = 30 x 10° psi Requirement: The maximum stress in the rod is limited to 35 ksi

he Collar

Flange Figure P4,26

4,27 The collar of weight W falls a distance 4 when it comes into contact with end B of the round

steel rod (Figure P4.26) Determine diameter d of the rod

Given: W = 201b,h =4 ft, L = 5 ft, E = 30 x 10° psi

Design Requirement: The maximum stress in the rod is not to exceed [8 ksi

4.28 The collar of weight W falls onto a flange at the bottom of a slender rod (Figure P4.26), Calculate the height / through which the weight W should drop to produce a maximum stress in the rod

Given: W = 500N,L=3m,d = 20mm, E = 170 GPa

Design Requirement: Maximum stress in the rod is limited to oy, = 350 MPa

CHAPTER 4 ® DEFLECTION AND IMPACT

4.29 Design the shaft (determine the minimum required length Lmin), described in Example 4.12, so

that yielding does not occur Based on this length and the given impact load, what is the angle

of twist?

4.30 The steel shaft and abrasive wheels A and B at the ends of a belt-drive sheave rotates at

n rpm (Figure P4.30) If the shaft is suddenly stopped at the wheel A because of jamming, determine

(a) The maximum angle of twist of the shaft (b) The maximum shear stress in the shaft

Given: D, =: 125 mm, D, = 150 mm, d=t= 25mm, L=0.3m,

n= 1500 rpm, G = 79 GPa, Sy; = 250 MPa, density of wheels p = 1800 kg/m?

Assumption: Abrasive wheels are considered solid disks

alee ae

Figure P4.30

4.31 Redo Problem 4.30, for the cease in which the shaft runs at a = 1200 rpm and the wheel end Bis suddenly stopped because of jamming

4,32 Arectangular sheet plate of thickness ¢ is bent into circular cylinder of radius r, Determine the diameter D of the cylinder and the maximum moment Myx developed in the plate

Given: t = š in., E = 30 x 10° psi, v = 0.3

Design Assumption: The maximum stress in the plate is limited to 18 ksi 4.33 A long, narrow rectangular plate is under a nonuniform loading

JẾ, p= po sin T”

and clamped at edges y = 0 and y = 6 (Figure P4.33) Determine

(a) An expression for the deflection surface w (b) The maximum bending stress

(c) The values of maximum deflection and stress for the data

b=20in, f=04in, E= 10x 10° psi, v= 0.3, De = 5 psi

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194 PARTI ©: @ © FUNDAMENTALS Figure P4.33

4.34 Figure P4.34 depicts a long, narrow rectangular plate with edge y = 0 simply supported and edge y = b clamped The plate is under a uniform load of intensity p, Determine

(a) An expression for the deflection surface w

(b) The maximum bending stress at y = b

(c) The values of maximum deflection at y = 6/2 and maximum bending stress at y = b, based on the data given in Problem 4.33

Figure P4.34

ENERGY METHODS IN DESIGN|

Outline 5.1 5.2 1.3 5.4 5.5 5.6 5.7 5.8 359 35.10 Introduction o Strain Energy

Components of Strain Energy Strain Energy in Comimon Menbers

TheWork-Energy Merhod _

Castigliano’s Theoteim:

Statically Indeterminate Problems Virtual Work and Potential Enetoy

Use of Trigonometric Series in Eneroy Methods

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196 PARTi @ © FUNDAMENTALS 5.1 INTRODUCTION

As pointed out in Section 1.4, instead of the equilibrium methods, displacements and forces can be ascertained through the use of energy methods The latter are based on the concept of strain energy, which is of fundamental importance in analysis and design Ap- plication of energy techniques is effective in cases involving members of variable cross sections and problems dealing with elastic stability, trusses, and frames In particular, strain energy approaches can greatly ease the chore of obtaining the displacement of members under combined loading

In this chapter, we explore two principal energy methods and illustrate their use with a variety of examples The first deals with the finite deformation experienced by load- carrying components (Sections 5.2 through 5.7) The second, the variational methods, based on a virtual variation in stress or displacement, is discussed in the remaining sections Literature related to the energy approaches is extensive [1-19]

§.2 STRAIN ENERGY

Internal work stored in an elastic body is the internal energy of deformation or the elastic strain energy It is often convenient to use the quantity, called strain energy per unit volume or strain energy density The area under the stress-strain diagram represents the strain energy density, designated U,, of a tensile specimen (Figure 5.1) Therefore

Ũ, = [os de, (5.1a)

The area above the stress-strain curve is termed the complimentary energy density:

UZ= | Ey dex (5.2)

Observe from Figure 5.1b that, for a nonlinearly elastic material, these energy densities have different values

oy ox ———— Proportional _—_ Proportional limit + limit } I } 1 i | ! - oO #y 9 ey (a) (by

Figure 5.1 Work done by uniaxial stress: (a) linearly elastic material; (b) nonlinearly elastic material

CHAPTERS @ KNERGY Meruops iN DESIGN

In the case of a linearly elastic material, from origin up to the proportional limit, sub-

stituting o, /E for ¢,, we have

4

U, Sp Ovey 3 (5.1b)

and the two areas are equal U, = UZ, as shown in Figure 5 1a In SI units, the strain energy

density is measured in joules per cubic meter (J /mŸ}or in pascals; in U.S customary units, it is expressed in inch-pounds per cubic inch (in -Ib/in.*) or psi Similarly, strain energy density for shear stress is given by

(8.3)

When a body is subjected to a general state of stress, the total strain energy density

equals simply the sum of the expressions identical to the preceding equations We have then

1

Uo = 2x8 T ØyẾy + 0/8, + tuy + Ty Vye + Tre¥xe) (5.4)

Substitution of generalized Hooke’s law into this expression gives the following equation,

involving only stresses and elastic constants:

Ll os 2 2 i

Uo = 55 [2 toy + OF ~ WOxoy + ayo, + O,0,) + 2 (rổ, + rổ + sả) (5.5)

When the principal axes are used as coordinate axes, the shear stresses are 0 The preced- ing equation then becomes

1

Ul = splot + oF + ơ? — 2V0(ØiØ; - 0Ø +- 0103) | (5.6) in which ơi, 02, and 03 are the principal stresses,

The elastic strain energy U stored within an elastic body can be obtained by integrat- ing the strain energy density over the volume V Thus

u= 0ay = [lƒ Uy dx dy dz (5.7)

W :

This equation is convenient in evaluating the strain energy for a number of commonly encountered shapes and loading It is important to note that the strain energy is a nonlinear

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198 PARTE: @ 2 FUNDAMENTALS

(quadratic) function of load or deformation The principle of superposition therefore is not valid for the strain energy

5.3 COMPONENTS OF STRAIN ENERGY

The three-dimensional state of stress at a point (Figure 5.2a) may be separated into two parts The state of stress in Figure 5.2b is associated with the volume changes, so-called dilatations In the figure o, represents the mean stress or the octahedral stress Gq, defined

in Section 3.16 On the other hand, the shape changes, or distortions, are caused by the set

of stresses shown in Figure 5.2c

The dilatational strain-energy density can be obtained through the use of Eq (5.6) by letting 0) = 02 = 03 = 0,, In so doing, we have

3q — 2v) 1

on” OB —=2U

(On)? = 6 Firat ay (5.8)

The distortional strain-energy density is readily found by subtracting the foregoing from

Eq (5.6):

(5.9)

' tới Bộ tÚI 0H táo

The quantities G and E are related by Eq (2.10) The octahedral planes where oo and Toct

act are shown in Figure 3.46

Test results indicate that the dilatational strain energy is ineffective in causing failure by yielding of ductile materials The energy of distortion is assumed to be completely

responsible for material failure by inelastic action This is discussed further in Chapter 8

Stresses and strains associated with both components of the strain energy are also very use- ful in describing the plastic deformation [4, 6]

(a)

Figure 5.2 {a) Principal stresses, resolved into (b) dilatational stresses and (c) distortional

stresses

ce

CHAPTER 5 6 Enercy METHODS IN DESIGN 199

Determining the Components of Strain Energy in a Prismatic Bar

A:structutal steel bar having uniform cross-sectional area A carries an axial tensile load P Find the strain energy density and its components

Solution:: The state of stress is uniaxial tension, 0, = = P/A, and the remaining stress compo- nents are 0 (Figure 5.2a) We therefore have the stresses causing volume change c,, = 0/3 and shape change oy + On = 26/3, Oy ~ On =O, — On, = O/3 (Figures 5.2b and 5.2c) The strain energy den- sities for the stresses in Figure 5.2, from Eqs (5.5), (5.8), and (5.9), are

Ỷ oe

U, = 3E

(L—~20)ơø? ` ø?

Yow = Cư = TRE

2 (+ vjo? = $0?

Uod = Tập = TE

Comments: We observe that U, = ¿y + Usa and that 5U,, = Usg That is, to change the shape of a unit volume element subjected to simple tension, five times more energy is absorbed than

to change the volume

EXAMPLE 5.1

5.4 STRAIN ENERGY IN COMMON MEMBERS

Recall from Section 5.2 that the method of superposition is not applicable to strain energy; that is, the effects of several forces or moments on strain energy are not simply additive, as illustrated in Example 5.2 In this section, the following types of loads are considered for the various members of a structure: axial loading, torsion, bending, and shear Note that the

equations derived are restricted to linear material behavior AXIALLY LOADED Bars

The normal stress at any transverse section through a bar subjected to an axial load P is 0, = P/A, where A represents the cross-sectional area and « is the axial axis (Figure 5.3)

——:———+ +—

Figure 5.3 Nonprismatic

bar with varying axial

Trang 21

200

PARTI ® FUNDAMENTALS

Substitution of this and Eq (5.1) into Eq (5.7) and setting dV = Adx, we obtain

1 fh Pedx

Uy == 4 sf AE 546)

For a prismatic bar, subjected to end forces of magnitude P, Eq (5.10) becomes

(5.11)

The quantity E represents the modulus of elasticity and L is the length of the member

52 |

A prismatic bar: suspended: from) one-énd ‘carries, in addition to its own weight, an axial load F

(Figure 5.4) Find the strain energy stored in the member

‘Determination of Strain Energy Stored in a Bar Due to Combined Loading

Solution: The axial joad; in.an element at a distance x from the fixed end, is expressed as

PeyA@—x)+F

- Here is the weight of thé material and A is the cross-sectional area Introducing this equation into

Eg 6.10),

Ly ACh oxy RFR 2A yPL2 ` F2L

" Is 2AE 5 PAD YRL 6E 2E 2AE 542 Comments: | The first and third terms on the right side of this equation are the strain energy of the bar due to its own weight and the strain energy of the bar supporting only axial force F The presence

of the middle term shows that the strain energy produced by the two loads acting simultaneously is not equal'to-the sum of the strain energies associated with the loads acting separately

CIRCULAR TORSION BARS

In the case of pure torsion of a bar, Eq 3.11 for an arbitrary distance r from the centroid of the cross section gives t = Tr/J The strain energy density, Eq (5.3), becomes then

U, = T?r?/2J°G Inserting this into Eq (5.7), the strain energy owing to torsion is L 2

r 2

U, -{ x67 (ƒ: as) dx (5.13) We have dV = dA dx; dA is an element of the cross-sectional area By definition, the term

in parentheses is the polar moment of inertia J of the cross-sectional area Hence,

unl eT? dx

_ GI (5.14)

CHAPTERS © ENERGY Meruops IN DESIGN

For a prismatic bar subjected to end torques T (Figure 3.6), Eq (5.14) appears as

TL :

U2 3G7 (5.15)

in which L is the length of the bar

BEAMS -

Consider a beam in pure bending The flexure formula gives the axial normal stress oy =

My/I Using Eq (5.1), the strain energy density is U, = M?y*/2E/* After carrying this

into Eq (5.7) and noting that M?/2E7? is a function of x alone, we obtain

L Mì 4

Uy = b [ 3n (f> aa) ax (5.16)

Since the integral in parentheses defines the moment of inertia / of the cross-sectional area about the neutral axis, the strain energy due to bending is

e1 LM? dx

v5 F E 7 (5.17)

This, integrating along beam length L, gives the required quantity

For a beam of constant flexural rigidity £7, Eq (5.17) may be written in terms of de- flection by using Eq (4.14) as follows:

El fh (@u\?

une f (oe “a |, (=) dự (5.18)

The transverse shear force V produces shear stress z;„ at every point in the beam The strain

energy density, inserting Eq (3.20) into Eq (5.3), is Up = V?Q?/2G/*b* Integration of

this, over the volume of the beam of cross-sectional area A, results in the strain energy for beams in shear:

_ fo avede

|, 2G (8.19)

In the foregoing, the form factor for shear is

A f Q

a= Pp Fi or dA (5.20)

This represents a dimensionless quantity specific to a given cross-section geometry

Example 5.3 illustrates the determination of the form factor for shear for a rectangular

cross section Other cross sections can be treated similarly Table 5.1 furnishes several cases [4] Subsequent to finding a, the strain energy due to shear is obtained using

Eq (5.19)

Trang 22

202 PARTI © FUNDAMENTALS

Table 5.1 Form factor for shear for various beam cross sections Cross section Form factor o Rectangle 6/5

T section, box section, or channels* A/Ayer

Circle 10/9 Thin-walled circular 2

*A = area of the entire section, Avep = area of the web ht, where h is the

beam depth and tis the web thickness

EXAMPLE 5.3 | Determining the Total Strain Energy Stoted in a Beam

“A

A cantilevered beam with a rectangular cross section supports a concentrated load P as depicted inFigure 5:5: Find: the ‘total: strain energy and compare the values of the bending and shear

contributions: :

P

Eigure.5.5' Examples 5.3 and 5.4

Solution: The 8rst monment of the area, by Eq (3.21), is @ = (b/2)[0/2)2 — y?] Inasmuch as A/P se 144/bh>, Eq: 6.20) sive

¬" Boy “ha _ 6

BB Lian 4 4 vĩ Ue

From the equilibrium requirements, the bending moment M = ~Px and the shear force V = P atx @igure 5:5): Carrying these and a = 6/5 into Eqs (5.17) and (5.19) then integrating, we obtain

b px) PL?

se ao des 5

Ys [ DEI 6EI 6-21)

we £6 V2 ewe BPPL (5.22)

Snn lạ 5 ZAG SAG

Note that [/A-== h?/12 Thé total strain energy stored in the cantilever beam is

P?r3 3E fh\?

Us Up = coe ( = 23

et Us = eae | i0G (7) (5.23)

Through the use of Eqs (5.21) and (5.22), we find the ratio of the shear strain energy to the bending strain energy in the beam‘as follows:

2

Uy BE fhY 2 3 Ay?

a= ie (z) =3¢*(7) cáo

CHAPTER 5 ki ENERGY METHODS IN ĐESIGN

Comments: When, for example, L = 10h and v = 1/3, this quotient is only 1/125; the strain energy owing to the shear‘ is less than | percent For a slender beam, A « L, it is observed that the

energy is due mainly to bending Therefore, it is usual to neglect the shear in evaluating the strain en- ergy jn beams of ordinary proportions: Unless stated otherwise, we adhere to this practice

203

PLATES

Based on the assumptions of Section 4.10, for thin plates ø;, yy;, and 1⁄4; can be disre- garded Hence, introducing Eq (5.4) together with Eq (2.6) into Eq (5.7); we have

Leas i

U= If |zzt +ợi — 20Ø,0y) + | dx dy dz (5.25) In the case of a plate of constant thickness, this equation may be written in terms of

deflection w through the use of Eqs (4.46) and (4.49) In so doing, we obtain

82 aw 82w 83w aw \?

=5/f° (5) + (Gr) +» ?m =s †2v— »(sar) Jaa (5.26) By

The preceding expression, the strain energy for plates in bending, is expressed in the following alternative form:

a? aw Rw ew aw

Le :ÍÍ pm + ay? a) of 2 2 Oy? ay Jew 627)

Integrations over the area A of the plate surface yield the required quantity

Applications of the equations developed in this section are illustrated in the formula-

tion of various energy techniques and in the solution of sample problems

5.5 THE WORK-ENERGY METHOD

The strain energy of a structure subjected to a set of forces and moments may be expressed

in terms of the forces and resulting displacements Suppose that all forces are applied _

ually and the final values of the force and displacement are denoted by Py(k = 1,2, , 1) and 4 The total work w,} 5 SX Pydy, is equal to the strain energy gained by the structure, provided no energy is dissipated That is,

1

U=We=- 5 » Py Sz, Ok (5.28) 5.28

In other words, the work done by the loads acting on the structure manifests as elastic strain

energy

Consider a member or structure subjected to a single concentrated load P Equa- tion (5.28) then becomes

Pb (5.29)

Trang 23

204 PARTI ® FUNDAMENTALS

The quantity 8 is the displacement through which the force P moves In a like manner, it can be shown that

1

U => 2M (5.30) 3

1

U=<T 5 @ (5.31)

Note that M (or 7) and @ (or ¢) are, respectively, the moment (or torque) and the associated slope (or angle of twist) at a point of a structure

The foregoing relationships provide a convenient approach for finding the displace- ment This is known as the method of work-energy In the next section, we present a more general approach that may be used to obtain the displacement at a given structure even when the structure carries combined loading,

EXAMPLE 5.4 Determination of Beam: Deflection by the Work-Energy Method

‘A-cantileveréd beam with'a rectangular cross séction is loaded as shown in Figure 5.5 Find the

deflection’ v, at the free end by: considering the effects of both the internal bending moments and ‘shear force

Solution: |The total strain energy [7 of the beam, given by Eq (5.23), is equated to the work, Woes Pu, /2: Hence

“PL 14 SEB 2 ` (6.32)

"A= 3ET |<" 10G (E ,

Comment: If the effect of shear is disregarded, note that the relative error is identical to that found in the previous example: As already shown, this is less than 1 percent for a beam with ratio L/h = 10

§.6 CASTIGLIANO’S THEOREM

Castigliano’s theorems are in widespread use in the analysis of structural displacements and forces They apply with ease to a variety of statically determinate as well as indeterminate problems Two theorems were proposed in 1879 by A Castigliano (1847-1884) The first theorem relies on a virtual (imaginary) variation in deformation and is discussed in Sec- tion 5.8 The second concerns the finite deformation experienced by a member under load Both theorems are limited to small deformations of structures The first theorem is pertinent to structures that behave nonlinearly as well as linearly We deal mainly with Castigliano’s second theorem, which is restricted to structures composed of linearly elastic materials, Un- less specified otherwise we refer in this text to the second theorem as Castigliano’s theorem Consider a linearly elastic structure subjected to a set of gradually applied external forces P,(k = 1,2, , 7) Strain energy U of the structure is equal to the work done W by

CHAPTERS @ ENERGY METHODS IN DESIGN

the applied forces, as given by Eq (5.28) Let us permit a single load, say, P;, to be increased a small amount dP;, while the other applied forces Py remain unchanged The increase in strain energy is thendU = (0U/0P;) dP;, where 0U/P; represents the rate of change of the strain energy with respect to P; The total energy is

, 9U U's U+ ( iB ) dP;

Alternatively, an expression for U' may be written by reversing the order of loading Suppose that dP; is applied first, followed by the force Py Now the application of dP; causes a small displacement dé; The work, dP; dé;/2, corresponding to this load incre-

ment, can be omitted because it is of the second order The work done during the applica-

tion of the forces P; is unaffected by the presence of dP; But, the latter force dP; performs work in moving an amount 6; Here 4; is the displacement caused by the application of P, The total strain energy due to the work done by this sequence of loads is therefore

ƯỮ sU+dP +5;

Equating the preceding equations, we have the Castigliano’s theorem:

BU

8P

6j (5.33)

The foregoing states that, for a linear structure, the partial derivative of the strain energy with respect to an applied force is equal to component of displacement at the point of application and in the direction of that force

Castigliano’s theorem can similarly be shown to be valid for applied moments M (or torques 7) and the resulting slope @ (or angle of twist ¢) of the structure Therefore,

6; = âU = oe (5.34) 5.34

3U

$= ar (5.35)

In using Castigliano’s theorem, we must express the strain energy in terms of the external forces or moments In the case of a prismatic beam, we have U = f M? dx/2EJ To obtain the deflection v; corresponding to load P;, it is often much simpler to differentiate under the integral sign In so doing, we have

vx

"op gr] ap”

UL (5.36)

Similarly, the slope may be expressed as

oe HS Ma de (5.37)

Trang 24

206 PARTI ® FUNDAMENTALS

Generally, the total strain energy in a straight or curved member subjected to a number of common loads (axial force F, bending moment M, shear force V, and torque T, equals the sum of the strain energies given by Eqs (5.11), (5.17), (5.19), and (5.15) So, applying Eq (5.33), the displacement at any point in the member is obtained in the following con- venient form:

5 3F + [MỀ?⁄&+-L [vệ to [roa PT 3F ETJ “5P “Tao |°l sp OT 1 aM 1 apt (648) 5

i=

Clearly, the last term of this equation applies only to circular bars An expression may be written for the angle of rotation in a like manner If it is necessary to obtain the displace~ ment at a point where no corresponding load acts, the problem is treated as follows We place a fictitious force Q (or couple C) at the point in question in the direction of the de- sired displacement ä (or @) We then apply Castigliano’s theorem and set the fictitious load Q = 0 (orC = 0) to obtain the desired displacements

_ EXAMPLE 55 |

Determining the Deflection of a Simple Beam by Castigliano’s Theorem

A simple beam is'subjected:to a-uniformload of intensity iw, as shown in Figure 5.6a What is the “deflection vc: at-an arbitrary distance @ from the left support?

in seer ï rc ï lui hy i 4 T+.- be x! La Ry @e @) a Roby & - Figure 516°: Example 5.5:

‘Solution: As the deflection is sought, a fictitious force Q is introduced at point C (Figure 5.6b) From the conditions of équilibritum; the reactions are found to be

wh Ob, wh, Ga

Beep hyp Mr bếp

The appropriate moment equations are then

2 Mac =k TT, Mặc = px” a Therefore, Mac be OM nc ax! 00 Ee AQ ek

CHAPTERS © ENERGY METHODS IN DESIGN 207 Be Introducing the foregoing into Eq: (5.38), we have

of whe Ob wx Tf bx

: “=m|[ (e+): 2 St) pe] f 4NE fae

pel pee Oa wx2 (=) ,

di + yee a ae

+f ( Oa), = N(e

setting oe 0Ø leads tỏ

ace tal “ Pye

Integration results in swab

KT rrr

Nok, as: a: check, that; fora = 'b; the preceding equation reduces to 5wL4 /384ET (see case 8 of

Table:A.9):

L@e $b) -3@ +E)

Detenmination of Deflection of a Curved Frame Using Castigliano’s Theorem EXAMPLE 5.6

A load of P is applied to a steel curved frame; as depictéd in Figure 5.7a Develop an expression for

tlie vertical deflection’s; of the free énd by considering the effects of the internal normal and shear forces in:additHön to the bending moment Calculate the value of 5, for the following data:

a= 60 mim; h:<.40 mm, b= l5mm,

Pest 10 KN, E == 210 GPa, G = 80 GPa

Figure.5.7 Exarnple 5:6? (a) steel curved frame;

(6) free-bady diagram:

Solution: “A frée-body diagram of the part of the bar defined by angle @ is depicted in Figure 5.7b,

where the internal forces (F and V) and moment (M4) are positive as shown Referring to the figure, we write

Trang 25

208 PART! @ FUNDAMENTALS `Therefore; SMS : ave oF

3E a= RC cosy, oP = sinổ; ậpP = COS 0

The form factor for shear for the rectangular section is @ = 6/5 (Table 5.1) Substitution of the preceding expressions into Eq (5.38) with dx = R d@ results in

3 R a ad 5, 248 6P [ston 5 | éos2 9 đổ 0 AE Jo z een ` 2 ai 008 0)" dO + SG

Using thé trigonometric identities cos? Ø = (1 + cos 20)/2 and sin? 6 = (1 ~ cos 20)/2, we obtain,

after integration;

“3m PR* 3nPR xPR

=———— —— 5.39

v= 2BET ` SẠO ` 2AE 6.39

The geometric properties of the cross section of the bar are

Á = 0/015(0.03) = 4.5 x 1074 m?, R=0.075m, Em zl00150.08)) = 337.5 x 10”!9 mt

Carrying these values into Eq (5.39) gives

8y = (2.81 + 0.04-+ 0.01) x 1079 = 2.86 mm

Comments: If the effects of the normal and shear forces are neglected, we have 5, = 2.81 mm Then, the error in deflection is approximately 1.7 percent For this curved bar, where R/c = 5, the contribution of V and F to the displacement can therefore be disregarded It is common practice to

negleót thẻ ñrst and the third terms in Eq (5.38) when R/c > 4 (see Section 16.8)

EXAMPLE 5.7

Determining Displacements of a Split Ring by Castigliano’s Theorem

A‘slender, cross-sectional, circular ring of radius R is cut open at @ = 0, as shown in Figure 5.8a The ring is fixed at one end and loaded at the free end by a z-directed force P Find, at the free énd A,

{a): The z-directed displacement (b) The rotation about the y axis

Solution: Since the rotation is sought a fictitious couple C is applied at point A The bending and twisting monients at any section are (Figure 5.8b)

Mẹ = =PRsin8 — Csin0, Ty = P.R(L —~ cos8Ø) — C cos8 (a)

CHAPTERS ® ENERGY METHODS IN DESIGN

Tạ My XS K RÓ ~— cos8) Rsind NY » AY ` Z Aa ` Figure 5.8 - Example 5.7

(a); Substitution of these quantities with C = 0 and dx = R dé into Eq (5.38) gives

2U 1 of <= ap * Ey , (—~PRsim6)(—sin8) dỡ 209 2m

ote GI Jo PR(1—cos@)R(i —~ cos 8) d6 (6.40a)

_ PR? + 3x PR?

— BI G7

(b) Introducing Eqs (a) into (5.37) with dx = để and setting C = 0, we have

ox yo ac ~ EI Le * _pRsi 8)(—sin 0) dØ 0 tt sin

i 2m

+ aif PR —cosé}(—cos 8) đô {5.40b)

G7 Jo

_ zPRỲ, xPR?

— BỊ GI

Case Study 5-1 | WINCH CRANE FRAME DEFLECTION ANALYSIS BY THE ENERGY METHOD

A schematic representation of a winch crane frame is Assumption: The loading is static depicted in Figure 1.5 Find the deflection at the free end

applying the energy method

Given: The dimensions and loading of the frame is transverse shear are considered

known from Case Study 3-1

E=200GPa, G=76.9 GPa, page) and Table 3.1

f= 2.011079 mế

Trang 26

210 PARTE @ FUNDAMENTALS

Case Study (CONCLUDED)

Weight per length w= 130 N/m

bh eb = 50 mm gah = 100 mm P=3kN Figure 3.31 (repeated) Part CD of the crane frame shown in Figure 1.4,

The form factor for shear for the rectangular box sec- tion, from Table 5.1, is

A A

œ= =

Aven fit

Here f is the beam depth and 1 represents the wal} thick- ness The moment and shear force, at an arbitrary section x distance from the free end of the beam, are expressed as follows: 1 M = —Px~ swx!, V=P+wx Therefore, aM 3V “— =—X, ¬_.' aP aP

After substitution of all the preceding equations, Eq (5.38)

becomes 1 ptt am 1 ft ay =— | Moetay— Vind Yo af ap 2p J On oP rope P +] ?] ey de _— —WX Ei Jy \( *”? by Tự , (CP + wx)(1) dx Integrating, 1 (PL? wh} 1 wl wae (oy 1 Ì+ ——~| Pu + “St “án mo Pity 6.41) Substituting the given numerical values into this equation, we obtain Up _ 1 3000(1.5)2 + 130(1.5# — 200 x 103(2.01) 3 8 Le samc + + 100(6)(76.9 x 103) 130(1.5)2 2 = 8.640.107 = 8.7mm

Since P is vertical and directed downward, vp represents

a vertical displacement and is positive downward Comments: if the effect of shear force is omitted,

vp = 8.6 mm; the resultant error in deflection is about 1.2 percent The contribution of shear force to the displace- ment of the frame can therefore be neglected

APPLICATION TO TRUSSES

We now apply Castigliano’s theorem to plane trusses As pointed out in Section 1.9, itis as- sumed that the connection between the members is pinned and the only force in the mem-

ber is an axial force, either tensile or compressive Note that, in practice, members of a

plane truss are usually riveted, welded, or bolted together by means of so-called gusset

plates However, due to the slenderness of the members, the internal forces can often be

computed on the basis of frictionless joints that prevent translation in two directions,

CHAPTERS @ ENERGY METHODS IN DESIGN

corresponding to reactions of two unknown force components The methods of joints and sections are commonly used for the analysis of trusses The method of joints consists of analyzing the truss, joint by joint, to determine the forces in the members by applying the

conditions of equilibrium to the free-body diagram for each joint This approach is most

effective when the forces in all members of a truss are to be determined If, however, the force in only a few members of a truss are desired, the method of sections applied to a por- tion (containing two or more joints) of the truss isolated by imagining that the members in which the forces are to be ascertained are cut

The strain energy U for a truss is equal to the sum of the strain energies of its members In the case of a truss consisting m members of length L,;, axial rigidity Aj £;, and internal axial force F;, the strain energy can be found from Eq (5.11) as

PPL,

U= i (6.42)

2 AE,

The displacement 4; of the joint of application of load P; can be obtained by substituting Eq (5.42) into Castigliano’s theorem, Eq (5.33) Therefore,

(5.43)

The preceding discussion applies to statically determinate and indeterminate linearly elastic trusses

211

Determination of Displacements of a-Truss by Castigliano’s Theorem

A planar truss with pin and roller supports at A and B, respéctively, is subjected to a vertical load P at joint £, as shown in Figure 5.9 Determine

: @) ‘The vertical displacement of point E : (6) The horizontal displacement of point E

Figure 5.9 Example.5:8:

Assumption: All members ate of equal axial rigidity AE:

Trang 27

212 PART! ® FUNDAMENTALS Solution:

(a): The equilibrium conditions for the entire truss (Figure 5.9) results inRa, = 2P,Ray = P,

and Rip = 2P: Applying the method of joints at A, E, C, and D, we obtain Fap = V2P, Fac = Fee = P, Fog = —V2P,

Fac = Fep = 0, Fạp = =2P »

Through the use of Eq (5.43), 1 OF 8 = a5 LPL xp = g2/1)0) +2(V2P)(@/21)2 + 2P)Œ)(—2)] PL z 11.657 —= i AE

The positive sign for 8, indicates that the displacement has the same sign as that assumed for P; it is downward

(b) As the horizontal displacement is sought, a fictitious load Q is applied at point E Fig-

ure 5.9) Now Fac = For = P + Q; all other forces remain the same as given by Eq (b) From Eq (5.43), we have

1 8F 2L

= Loe (P

b= 7g LL PL gg = aE tO

However, Q = 0, and the preceding reduces to

PL

pe 2 5.44)

8h 1E (5.44)

| §.7 STATICALLY INDETERMINATE PROBLEMS

Castigliano’s theorem or the unit load method may be applied as a supplement to the equa- tions of statics in the solutions of support reactions of a statically indeterminate structure Consider, for instance, a structure indeterminate to the first degree In this case, we select one of the reactions as the redundant (or unknown) load, say, Ri, by removing the corre~ sponding support All external forces, including both loads and redundant reactions, must generate displacements compatible with the original supports We first express the strain energy in terms of Ry and the given loads Equation (5.36) may be applied at the removed support and equated to the given displacement:

au = §; = 0 5.45

aR, 1 € }

This expression is solved for the redundant reaction Ry Then, we can find from static equi- librium the other reactions

CHAPTERS @ ENERGY Mxruops IN DESIGN 213

In an analogous manner, the case of statically indeterminate structure with n redundant

reactions, assuming no support movement, can be expressed in the form

:0U DR, 0i, aR = 0 bu : (5.46)

By solving these equations simultaneously, the magnitude of the redundant reactions are obtained The remaining reactions are found from equations of equilibrium Note that Eq (5.46), Castigliano’s second theorem, is also referred to as the principle of least work

in some literature [10]

Analytical techniques are illustrated in solution of the following sample problems

Determining the Reactions of a Propped Cantilevered Beam EXAMPLE 5.9 Using Castigliano’s Theorem

A propped cantilevered beam carries a concentrated load P at its midspan (Figure 5.10a) Find the Support reactions : Po 7 P AE — me P : ; coe up 1/2 là Ra <2 7/2 (ay : @y Figure 5:10 Example 5.9

Solution: |The free-body diagram of Figure 5:10b shows that the problem is statically indetermi- nate to the first degree We select R,:as redundant, hence the expressions for the moments are

: aoe L

May = Rax, Mng = RAx=P ( = 5)

tis: important to! note: that: the remaining unknowns, Ry and My, do not appear in preceding equations: : 4

The deflection v, at A must be 0 Equation (5.36) is thus

: gy £72 & L

n= , ` ố [Ra P( ~ §)]xax| =0 ‘The preceding, after integrating, results in

2

Ry =P ng

Trang 28

214 PARTI ® FUNDAMENTALS EXAMPLE 5.10

: Determination of the Deflection of a Ring by Castigliano’s Theorem

- Ating of radius Ris hinged: and subjected to force P as shown in Figure 5.1 La, Taking into account -only the strain energy due to bending, determine the vertical displacement of point C

(©)

Figure 5.11 : Example 5.10: (a) ring;

(b) quarter segment

Solution: Owing to symmetry, it is necessary to analyze only a quarter segment (Figure 5.118) Thasmuch as A4 and Mp are unknowna, the problem is statically indeterminate The morment at any

sectOn.đ-4 15

i

“Mạ = Mạ — 2PRQ —cos8)

Since the slope is 0 at B, substituting this expression into Eq (5.37) with dx = R d@, we have

repe M 1 PRO 0) | (A) Rdé = 0

=a f [oe - 5 (1 cos | =

from which Ms =-P R[U — (2/m)]/2 The first equation then becomes

My = PR (cose = 2) 2 l #

“Thẻ displacemem of point C; by Eq: G.36) is

4 2: 8Mạ

c= ef Moe Rae

4/22 ER LẦN:

a af Plz [5 (cose 2)| —= R đô Tnfegration of the foregoing results in

PR

bo = 05 (5.47)

‘The: positive sign of 5¢ means that the displacement has the same sense as the force P, as shown

CHAPTERS 9 ENERGY Meruops IN DESIGN 215

Determining the Deflection and Reaction of a Frame Using Castigliano’s Theorem

‘Aframe of constant flexural rigidity E I supports a downward load P, as depicted in Figure 5.12a Find

(ae The deflection at Z,

(b): The horizontal reaction at &, if the point Z is a fixed pin (Figure 5.[2b)

\" B CĐ Cc A HỆ” RDS eR pfs 2 2 ®) Figure 5.12 Example 5.11: Solution:

{a)*: Inasmuch as‘a displacement is sought, a fictitious forcé Q is introduced at point EF (Fig-

ute 5.12a) Because of symmetry about a vertical axis through point C, we need to write expressions for moment associated with segments ED and DC, respectively:

My = Qx, My = Oa

The horizontal displacement at £ is found by substituting these equations into Eq (5.38):

Lif? 2 poe Px

=a f (Qx)x dx + Ty , (ca+ S) aa {a}

Setting O = 0 and integrating, Pab?

8g = SE7 (5.48a)

(B)- The problem is now statically indeterminate to the first degree; there is one unknown reac- tion: after’ satisfying three equations of statics (Figure 5.12b) Since 3g = 0, setting

= +R, we equate the deflection given by Eq (a) to zero In so doing, we have

Lope 4 per Px

—Rx}x dy + —= ~Ra + —— =

af wart ef (-Re+ ) aae 0

The preceding gives

Re 30° PR

5 §a2 + 3ab (6.48)

EXAMPLE 5.11

Trang 29

216 PARTE ® FUNDAMENTALS

5.8 VIRTUAL WORK AND POTENTIAL ENERGY

In connection with virtual work, we use the symbol 6 to denote a virtual infinitesimally small quantity An arbitrary or imaginary incremental displacement is termed a virtual dis- placement A virtual displacement results in no geometric alterations of a member It also must not violate the boundary or support conditions for a structure In the brief develop- ment to follow, p,, py, and p, represent the x, y, and z components of the surface forces per unit area and the body forces are taken to be negligible

THE PRINCIPLE OF VIRTUAL WORK

The virtual work, 6W, done by surface forces on a member under a virtual displacement is given by

ôW = loa + pydu + p,dw)dA (5.49)

The quantity A is the boundary surface area and 8,, 5,, and 5,, represent the x-, y-, and z- directed components of a virtual displacement

In a like manner, the virtual strain energy, 5U, acquired of a member of volume V caused by a virtual straining is expressed as follows:

ôU = Tae + Oyd8y O26, + 1yŠ⁄4y + yyôyy; HF 1y¿Šy„¿) AV (5.50)

V

It can be shown that {1,.4] the total work done during the virtual displacement is 0: SW —8U = 0 Therefore,

ŠW = SU (5.51)

This is called the principle of virtual work

THe PRINCIPLE OF MINIMUM POTENTIAL ENERGY

Since the virtual displacements do not alter the shape of the member and the surface forces are considered constants, Eq (5.51) can be written in the form

és = 6 = W) = 0 (5.52)

In the foregoing, we have

H=U-W (5.53)

The quantity {1 designates the potential energy of the member

Equation (5.52) represents the condition of stationary potential energy of the member An elastic member is in equilibrium if and only if the changes in potential energy SIT are 0 for a virtual displacement It can be verified that, for a stable equilibrium, the potential en- ergy is a minimum For all displacements fulfilling the prescribed boundary conditions and the equilibrium conditions, the potential energy has a minimum value The preceding states the principle of minimum potential energy

CHAPTERS ® ENERGY METHODS IN DESIGN

CASTIGLIANO’S First THEOREM

Consider now a structure subjected to a set of external forces Py(k = 1, 2, ,”) Suppose that the structure experiences a continuous virtual displacement in such a manner that it vanishes at all points of loading except under a single load, say, P; The virtual displace- ment in the direction of this force is denoted by $(;) From Eq (5.51), we have 6U = P;- 8(6;) In the limit, the principle of virtual work results in

au

P=

Đôi (5.54)

This is known as Castigliano’s first theorem: For a linear or nonlinear structure, the partial derivative of the strain energy with respect to an applied virtual displacement is equal to the load acting at the point in the direction of that displacement Similarly, it can be demonstrated that

8U

M.= mã (5.55)

in which 6; is the angular rotation and M; represents the resulting moment

Note that Castigliano’s first theorem is also known as the theorem of virtual work It is the basis for the derivation of the finite element stiffness equations In applying the Castigliano’s first theorem, the strain energy must be expressed in terms of the displacements

217

*3.9 USE OF TRIGONOMETRIC SERIES IN ENERGY METHODS

Certain problems in the structural analysis and design are amenable to solutions by the use of trigonometric series This technique offers a significant advantage because a single ex- pression may apply to the entire length or surface of the member A disadvantage in the trigonometric series is that arbitrary support conditions can make it impossible to write a series that is simple The solution by trigonometric series is applied for variously loaded members in this and the sections to follow [1, 4, 5]

The method is now illustrated for the case of a simple beam loaded as depicted in Figure 5.13 The deflection curve can be represented by the following Fourier sine series:

C _ MIX

vs » Qm Sin

+

(5.56)

ne]

Trang 30

218 PARTIL- ® FUNDAMENTALS

Figure 5.13 Simply supported beam under a force Pat an arbitrary distance ¢ from the left

support

The strain energy of the beam, substituting Eq (5.56) into Eq (5.18), is expressed in the form

2

EL ft (đu? EI ft & mm” max |-

0Í (5) dx = af, » sin Z dx (a)

m

The term in brackets, after expanding, can be expressed as 2

mit

U= ` » ata ) (=) sin —_ sin —

m=| nl

For the orthogonal functions sin(max/L) and sin(nz/L), by direct integration it can be verified that

Ù MTX , nHx 0, (for m #n)

[io sin sin TT dx = là (for m =n) (5.57a)

The strain energy given by Eq (a) is therefore

WEL S

Um ma m 4a (5.58)

mse!

The virtual work done by a force P acting through a virtual displacement at A increases the strain energy of the beam by 6U Applying Eq (5.51),

~—P + dug = 5U tb)

The minus sign means that P and dv, are oppositely directed So, by Eqs (5.58) and (b),

mre

—~P » sin 8a = = oom a,

„ml ars}

The foregoing gives

2PL> 1 3 MITC

Cy = ee SI ———

„ wi EI m+ L

CHAPTERS ® ENERGY METHODSIN DESIGN 219

Carrying this equation into Eq (5.56), we have the equation of the deflection curve 2PR Sl ome | max

“x*ET Đan L 3T m= (5.59)

Using this infinite series, the deflection for any prescribed value of x can be readily

obtained

Determination of the Deflection of a Cantilevered Beam Using the EXAMPLE 5.12 Principle of Virtual Work

A cantilevered beam is subjected to a concentrated load P at its free end, as shown in Figure 5.14

Dérive the equation öÊ the deffecHỏn curve

Assumptions: The origin of the coordinates is placed at the fixed end Deflection is taken in the

form : URES Š 4, ( cos 2) (5.60) SH mm (18/88 2L 1 P Ỹ T—> Me t b Figure 5.14” Example 5.12:

Solution: The boundary: conditions; v(0) = 0 and ‘v'(0) = 0, are satisfied by the preceding

equation Substitution: of this seriés into Eq: (5.18) gives

2 2

1 | < (=) co 4

TS thi | Sợ zs X

kh 2E 2E

Squating the term in brackets and noting the orthogonality relation,

Bo ange nee 0; đồ - im # n)

[ m 2E sa “OL tự = la (for * i = ny (5.576) the strain energy becomes

z*EI.&

= OLS By ah mi 153, iy 6-69

By the principle of virtual work, +P Sua, = dU; we have

= app 2

=P `) (: Cos Bì = a >> _ m°àa?

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