Mechanics of materials 10e global edtion hibbeler 2

If it is subjected to a shear force V, determine the maximum average shear stress in the member using the shear formula.. Determine the maximum shear stress acting in the fiberglass be

pRELImINaRY pROBLEmS

P7–1 In each case, calculate the value of Q and t that are

used in the shear formula for finding the shear stress at A

Also, show how the shear stress acts on a differential volume

element located at point A.

V

A

(b)

0.3 m 0.1 m 0.1 m

0.2 m 0.2 m 0.2 m

A

V

(e)

0.2 m 0.4 m 0.2 m

V

A

0.1 m 0.2 m

0.3 m 0.3 m

(f)

0.1 m

0.1 m 0.5 m

0.1 m

0.1 m

0.3 m 0.2 m

Prob P7–1

F7–1 If the beam is subjected to a shear force of

V = 100 kN, determine the shear stress at point A Represent

the state of stress on a volume element at this point.

200 mm

90 mm

300 mm 20mm

20mm

20 mm V

A

Prob F7–1 F7–2 Determine the shear stress at points A and B if the

beam is subjected to a shear force of V = 600 kN Represent

the state of stress on a volume element of these points.

developed in the beam.

F7–4 If the beam is subjected to a shear force of

V = 20 kN, determine the maximum shear stress in the beam.

F7–5 If the beam is made from four plates and subjected

to a shear force of V = 20 kN, determine the shear stress at point A Represent the state of stress on a volume element

7–1. If the wide-flange beam is subjected to a shear of

V = 20 kN, determine the shear stress on the web at A

Indicate the shear-stress components on a volume element

located at this point.

7–2. If the wide-flange beam is subjected to a shear of

V = 20 kN, determine the maximum shear stress in the

beam.

7–3. If the wide-flange beam is subjected to a shear of

V = 20 kN, determine the shear force resisted by the web of

*7–4. If the beam is subjected to a shear of V = 30 kN,

determine the web’s shear stress at A and B Indicate the

shear-stress components on a volume element located

at these points Set w = 200 mm Show that the neutral axis

is located at y = 0.2433 m from the bottom and

I = 0.5382(10−3 ) m 4

7–5. If the wide-flange beam is subjected to a shear of

V = 30 kN, determine the maximum shear stress in the

Probs 7–4/5

7–6. The wood beam has an allowable shear stress of

tallow = 7 MPa Determine the maximum shear force V that

can be applied to the cross section.

maximum shear stress in the shaft.

*7–8. The shaft is supported by a thrust bearing at A and a journal bearing at B If the shaft is made from a material

having an allowable shear stress of tallow = 75 MPa,

determine the maximum value for P.

7–9. Determine the largest shear force V that the member

can sustain if the allowable shear stress is tallow = 56 MPa.

7–10. If the applied shear force V = 90 kN, determine the

maximum shear stress in the member.

7–13. Determine the shear stress at point B on the web of the cantilevered strut at section a–a.

7–14. Determine the maximum shear stress acting at

section a–a of the cantilevered strut.

7–11. The overhang beam is subjected to the uniform

distributed load having an intensity of w = 50 kN>m

Determine the maximum shear stress in the beam.

*7–12. A member has a cross section in the form of an

equilateral triangle If it is subjected to a shear force V,

determine the maximum average shear stress in the member

using the shear formula Should the shear formula actually

be used to predict this value? Explain.

V

a h

*7–16. Determine the maximum shear stress in the T-beam

at point C Show the result on a volume element at this point.

Probs 7–15/16

7–17. The strut is subjected to a vertical shear of V = 130

kN Plot the intensity of the shear-stress distribution acting

over the cross-sectional area, and compute the resultant shear

force developed in the vertical segment AB.

7–21. Determine the maximum shear stress acting in the fiberglass beam at the section where the internal shear force is maximum.

7–25. Determine the length of the cantilevered beam so that the maximum bending stress in the beam is equivalent

to the maximum shear stress.

A B

section of a rod that has a radius c By what factor is the

maximum shear stress greater than the average shear stress

acting over the cross section?

c

V

y

Prob 7–18 7–19. Determine the maximum shear stress in the strut if

it is subjected to a shear force of V = 20 kN.

*7–20. Determine the maximum shear force V that the

strut can support if the allowable shear stress for the

material is tallow = 40 MPa.

shear-stress components on a volume element located at these

points Set w = 125 mm Show that the neutral axis is located

at y = 0.1747 m from the bottom and INA = 0.2182(10 −3 ) m 4

7–23. If the wide-flange beam is subjected to a shear of

V = 30 kN, determine the maximum shear stress in the

beam Set w = 200 mm.

*7–24. If the wide-flange beam is subjected to a shear of

V = 30 kN, determine the shear force resisted by the web

of the beam Set w = 200 mm.

Prob 7–25

7–29. The composite beam is constructed from wood and reinforced with a steel strap Use the method of Sec 6.6 and calculate the maximum shear stress in the beam when it is

subjected to a shear of V = 50 kN Take Est = 200 GPa,

7–30. The beam has a rectangular cross section and is

subjected to a load P that is just large enough to develop a fully plastic moment Mp = PL at the fixed support If the material is elastic perfectly plastic, then at a distance x 6 L the moment M = Px creates a region of plastic yielding

7–26. If the beam is made from wood having an allowable

shear stress tallow = 3 MPa, determine the maximum

7–27. The beam is slit longitudinally along both sides If it

is subjected to a shear of V = 250 kN, compare the

maximum shear stress in the beam before and after the cuts

were made.

*7–28. The beam is to be cut longitudinally along both

sides as shown If it is made from a material having an

allowable shear stress of tallow = 75 MPa, determine the

maximum allowable shear force V that can be applied

before and after the cut is made.

with an associated elastic core having a height 2y′ This

situation has been described by Eq 6–30 and the moment M

is distributed over the cross section as shown in Fig 6–48e

Prove that the maximum shear stress in the beam is given

by t max = 32(P >A′), where A′ = 2y′b, the cross-sectional

area of the elastic core.

7–31. The beam in Fig 6–48f is subjected to a fully plastic

moment Mp Prove that the longitudinal and transverse

shear stresses in the beam are zero Hint: Consider an element of the beam shown in Fig 7–4d.

*The use of the word “flow” in this terminology will become meaningful as it pertains

to the discussion in Sec 7.4.

7.3 Shear Flow in built-up

MeMberS

Occasionally in engineering practice, members are “built up” from several composite parts in order to achieve a greater resistance to loads An example is shown in Fig 7–13 If the loads cause the members to bend, fasteners such as nails, bolts, welding material, or glue will be needed to keep the component parts from sliding relative to one another, Fig 7–2

In order to design these fasteners or determine their spacing, it is necessary

to know the shear force that they must resist This loading, when measured

as a force per unit length of beam, is referred to as shear flow, q.*

The magnitude of the shear flow is obtained using a procedure similar

to that for finding the shear stress in a beam To illustrate, consider

finding the shear flow along the juncture where the segment in Fig 7–14a

is connected to the flange of the beam Three horizontal forces must act

on this segment, Fig 7–14b Two of these forces, F and F + dF, are the result of the normal stresses caused by the moments M and M + dM, respectively The third force, which for equilibrium equals dF, acts at the juncture Realizing that dF is the result of dM, then, like Eq 7–1, we have

dF = dM I L

A′

y dA′

The integral represents Q, that is, the moment of the segment’s area A′

about the neutral axis Since the segment has a length dx, the shear flow,

or force per unit length along the beam, is q = dF>dx Hence dividing both sides by dx and noting that V = dM >dx, Eq 6–2, we have

Here

q= the shear flow, measured as a force per unit length along the beam

V= the shear force, determined from the method of sections and the equations of equilibrium

I = the moment of inertia of the entire cross-sectional area calculated

about the neutral axis

Q = y′A′, where A′ is the cross-sectional area of the segment that is connected

to the beam at the juncture where the shear flow is calculated, and y′ is the distance from the neutral axis to the centroid of A′

Fig 7–14

Fastener Spacing. When segments of a beam are connected by

fasteners, such as nails or bolts, their spacing s along the beam can be

determined For example, let’s say that a fastener, such as a nail, can

support a maximum shear force of F (N) before it fails, Fig 7–15a If

these nails are used to construct the beam made from two boards, as

shown in Fig 7–15b, then the nails must resist the shear flow q (N>m)

between the boards In other words, the nails are used to “hold” the top

board to the bottom board so that no slipping occurs during bending

(See Fig 7–2a.) As shown in Fig 7–15c, the nail spacing is therefore

determined from

F (N) = q (N>m) s (m)

The examples that follow illustrate application of this equation

Other examples of shaded segments connected to built-up beams by

fasteners are shown in Fig 7–16 The shear flow here must be found at

the thick black line, and is determined by using a value of Q calculated

from A ′ and y′ indicated in each figure This value of q will be resisted by

a single fastener in Fig 7–16a, by two fasteners in Fig 7–16b, and by three

fasteners in Fig 7–16c In other words, the fastener in Fig 7–16a supports

the calculated value of q, and in Figs 7–16b and 7–16c each fastener

supports q >2 and q>3, respectively.

_

¿

A¿

A N

• Shear flow is a measure of the force per unit length along the axis

of a beam This value is found from the shear formula and is used

to determine the shear force developed in fasteners and glue

that holds the various segments of a composite beam together

V V

V V

V V

Fig 7–15

The beam is constructed from three boards glued together as shown in

Fig 7–17a If it is subjected to a shear of V = 850 kN, determine the shear flow at B and B ′ that must be resisted by the glue

SOLUTION

the bottom of the beam, Fig 7–17a Working in units of meters, we have

y = Σy∼A ΣA = 2[0.15 m](0.3 m)(0.01 m)2(0.3 m)(0.01 m)+ [0.305 m](0.250 m)(0.01 m)+ (0.250 m)(0.01 m)

= 0.1956 mThe moment of inertia of the cross section about the neutral axis is thus

I = 2c121 (0.01 m)(0.3 m)3 + (0.01 m)(0.3 m)(0.1956 m - 0.150 m)2d+ c121 (0.250 m)(0.01 m)3+ (0.250 m)(0.01 m)(0.305 m - 0.1956 m)2d = 87.42(10-6) m4

The glue at both B and B ′ in Fig 7–17a “holds” the top board to the

beam Here

Q B = y=

B A=

B = [0.305 m - 0.1956 m](0.250 m)(0.01 m) = 0.2735(10-3) m3

Shear Flow.

q = VQ I B = 850(10

3) N(0.2735(10-3) m3)87.42(10-6) m4 = 2.66 MN>m

Since two seams are used to secure the board, the glue per meter length of beam at each seam must be strong enough to resist one-half

of this shear flow Thus,

have to be recalculated, and the shear flow at C and C′ determined from q = V y′ C A′C >I Finally, this value is divided by one-half to obtain

A box beam is constructed from four boards nailed together as shown

in Fig 7–18a If each nail can support a shear force of 30 N, determine

the maximum spacing s of the nails at B and at C to the nearest 5 mm

so that the beam will support the force of 80 N

SOLUTION

its length, the internal shear required for equilibrium is always

V = 80 N, and so the shear diagram is shown in Fig 7–18b.

area about the neutral axis can be determined by considering a

75@mm * 75@mm square minus a 45@mm * 45@mm square

I = 121 (0.075 m)(0.075 m)3 - 121 (0.045 m)(0.045 m)3 = 2.295(10-6) m4

The shear flow at B is determined using Q B found from the darker

shaded area shown in Fig 7–18c It is this “symmetric” portion of the

beam that is to be “held” onto the rest of the beam by nails on the left

side and by the fibers of the board on the right side

Thus,

Q B = y′A′ = (0.03m)(0.075m)(0.015m) = 33.75(10-6)m3

Likewise, the shear flow at C can be determined using the “symmetric”

shaded area shown in Fig 7–18d We have

Q C = y′A′ = (0.03m)(0.045m)(0.015m) = 20.25(106)m3

Shear Flow.

q B = VQ I B = (80 N)[33.75(10

-6) m3]2.295(10-6) m4 = 1176.47 N>m

q C = VQ I C = (80 N)[20.25(10

-6)m3]2.295(10-6) m4 = 705.88 N>mThese values represent the shear force per unit length of the beam

that must be resisted by the nails at B and the fibers at B′, Fig 7–18c,

and the nails at C and the fibers at C′, Fig 7–18d, respectively Since in

each case the shear flow is resisted at two surfaces and each nail can

resist 30 N, for B the spacing is

Nails having a shear strength of 900 N are used in a beam that can be constructed either as in Case I or as in Case II, Fig 7–19 If the nails are spaced at 250 mm, determine the largest vertical shear that can be supported in each case so that the fasteners will not fail

Since the cross section is the same in both cases, the moment of inertia about the neutral axis is

I = 121 (0.075 m)(0.1 m)3 - 121 (0.05 m)(0.08 m)3 = 4.1167(10-6) m4

flange onto the web For one of these flanges,

Q = y′A′ = (0.045 m)(0.075 m)(0.01 m) = 33.75(10-6) m3

so that

q = VQ I

900 N0.25 m =

V[33.75(10-6) m3]4.1167(10-6) m4

the web Thus

Q = y′A ′ = (0.045 m)(0.025 m)(0.01 m) = 11.25(10-6) m3

q = VQ I

900 N0.25 m =

V[11.25(10-6) m3]4.1167(10-6) m4

V = 1.3173(103) N = 1.32 kN Ans.

F7–6 The two identical boards are bolted together to form

the beam Determine the maximum spacing s of the bolts to

the nearest mm if each bolt has a shear strength of 15 kN The

beam is subjected to a shear force of V = 50 kN.

Prob F7–6

F7–7 Two identical 20-mm-thick plates are bolted to the

top and bottom flange to form the built-up beam If the beam

is subjected to a shear force of V = 300 kN, determine the

maximum spacing s of the bolts to the nearest mm if each

bolt has a shear strength of 30 kN.

F7–8 The boards are bolted together to form the built-up

beam If the beam is subjected to a shear force of V = 20 kN, determine the maximum spacing s of the bolts to the nearest

mm if each bolt has a shear strength of 8 kN.

beam If the beam is subjected to a shear force of V = 75 kN,

determine the allowable maximum spacing of the bolts to the nearest multiples of 5 mm Each bolt has a shear strength

of 30 kN.

100 mm 75mm

Prob F7–9FUNDamENTaL pROBLEmS

*7–32. The double T-beam is fabricated by welding the

three plates together as shown Determine the shear stress in

the weld necessary to support a shear force of V = 80 kN.

7–33. The double T-beam is fabricated by welding the

three plates together as shown If the weld can resist a shear

stress t allow = 90 MPa, determine the maximum shear V

that can be applied to the beam.

together with three rows of nails spaced s = 50 mm apart If

each nail can support a 2.25-kN shear force, determine the

maximum shear force V that can be applied to the beam.The

allowable shear stress for the wood is t allow = 2.1 MPa.

7–35. The beam is constructed from two boards fastened

together with three rows of nails If the allowable shear stress

for the wood is t allow = 1 MPa, determine the maximum shear

force V that can be applied to the beam Also, find the maximum

spacing s of the nails if each nail can resist 3.25 kN in shear.

maximum load P that can be applied to the end of the beam.

thickness of 12 mm If a shear of V = 250 kN is applied to

the cross section, determine the maximum spacing of the bolts Each bolt can resist a shear force of 75 kN.

7–38. The beam is fabricated from two equivalent structural tees and two plates Each plate has a height of 150 mm and a

thickness of 12 mm If the bolts are spaced at s = 200 mm determine the maximum shear force V that can be applied to

the cross section Each bolt can resist a shear force of 75 kN.

7–39. The double-web girder is constructed from two

plywood sheets that are secured to wood members at its top

and bottom If each fastener can support 3 kN in single

shear, determine the required spacing s of the fasteners

needed to support the loading P = 15 kN Assume A is

pinned and B is a roller.

*7–40. The double-web girder is constructed from two

plywood sheets that are secured to wood members at its top

and bottom The allowable bending stress for the wood is

sallow = 56 MPa and the allowable shear stress is

tallow = 21 MPa If the fasteners are spaced s = 150 mm

and each fastener can support 3 kN in single shear,

determine the maximum load P that can be applied to the

beam.

7–42. The simply supported beam is built up from three boards

by nailing them together as shown The wood has an allowable shear stress of tallow = 1.5 MPa, and an allowable bending stress of s allow = 9 MPa The nails are spaced at s = 75 mm,

and each has a shear strength of 1.5 kN Determine the

maximum allowable force P that can be applied to the beam.

7–43. The simply supported beam is built up from three

boards by nailing them together as shown If P = 12 kN, determine the maximum allowable spacing s of the nails to

support that load, if each nail can resist a shear force of 1.5 kN.

P

B

s A

7–41. A beam is constructed from three boards bolted

together as shown Determine the shear force in each bolt if

the bolts are spaced s = 250 mm apart and the shear is

Prob 7–44

7–45. The nails are on both sides of the beam and each can

resist a shear of 2 kN In addition to the distributed loading,

determine the maximum load P that can be applied to the

end of the beam The nails are spaced 100 mm apart and the

allowable shear stress for the wood is tallow = 3 MPa.

nails within region AB of the beam The nails are located on

each side of the beam and are spaced 100 mm apart Each

nail has a diameter of 4 mm Take P = 2 kN

7–47. The beam is made from four boards nailed together

as shown If the nails can each support a shear force of

500 N, determine their required spacing s′ and s to the nearest mm if the beam is subjected to a shear of V = 3.5 kN.

P P

strength of 80 kPa, determine the maximum load P that can

be applied without causing the glue to lose its bond.

7–49. The timber T-beam is subjected to a load consisting

of n concentrated forces, P n If the allowable shear Vnail for each of the nails is known, write a computer program that will specify the nail spacing between each load Show an

application of the program using the values L = 5 m,

7.4 Shear Flow in thin-walled

MeMberS

In this section we will show how to describe the shear-flow distribution

throughout a member’s cross-sectional area As with most structural

members, we will assume that the member has thin walls, that is, the wall

thickness is small compared to its height or width

Before we determine the shear-flow distribution, we will first show

how to establish its direction To begin, consider the beam in

Fig 7–20a, and the free-body diagram of segment B taken from the top

flange, Fig 7–20b The force dF must act on the longitudinal section in

order to balance the normal forces F and F + dF created by the

moments M and M + dM, respectively Because q (and t) are

complementary, transverse components of q must act on the cross

section as shown on the corner element in Fig 7–20b.

Although it is also true that V + dV will create a vertical shear-flow

component on this element, Fig 7–20c, here we will neglect its effects

This is because the flange is thin, and the top and bottom surfaces of

the flange are free of stress To summarize then, only the shear flow

component that acts parallel to the sides of the flange will be

considered

(a)

M V

M  dM

V  dV dx

q¿ is assumed to be zero throughout flange thickness since the top and bottom surfaces of the flange must be stress free

(c)

Fig 7–20

Shear Flow in Flanges The shear-flow distribution along the top

flange of the beam in Fig 7–21a can be found by considering the shear flow q, acting on the dark blue element dx, located an arbitrary distance x from the centerline of the cross section, Fig 7–21b Here

Q = y′A ′ = [d>2](b>2 - x)t, so that

q = VQ I = V [d >2](b>2 - x)t

I = Vt 2I d ab2 - xb (7–5)

By inspection, this distribution varies in a linear manner from q = 0 at

x = b >2 to (qmax)f = Vt db>4I at x = 0 (The limitation of x = 0 is

possible here since the member is assumed to have “thin walls” and so the thickness of the web is neglected.) Due to symmetry, a similar analysis yields the same distribution of shear flow for the other three flange

segments These results are as shown in Fig 7–21d.

The total force developed in each flange segment can be determined

by integration Since the force on the element dx in Fig 7–21b is

dF = q dx, then

F f =

Lq dx = L

b>2 0

Vt d 2I ab2 - x b dx = Vt 16I db2

We can also determine this result by finding the area under the triangle

in Fig 7–21d Hence,

F f = 12 (qmax)fab2 b = Vt 16I db2

All four of these forces are shown in Fig 7–21e, and we can see from

their direction that horizontal force equilibrium on the cross section is maintained

b t

t A

V

N t

t

t

y q

dy d

2

Fig 7–21

Shear Flow in Web A similar analysis can be performed for the

web, Fig 7–21c Here q must act downward, and at element dy

we have Q = Σy′A ′ = [d>2](bt) + [ y + (1>2)(d>2 - y)]t(d>2 - y) =

bt d >2 + (t>2)(d2>4 - y2), so that

q = VQ I = Vt I Jdb2 + 12¢d4 -2 y2≤ R (7–6)

For the web, the shear flow varies in a parabolic manner, from

q = 2(qmax)f = Vt db>2I at y = d>2 to (qmax)w = (Vt d>I)(b>2 + d>8)

Substituting this into the above equation, we see that F w = V, which is to

be expected, Fig 7–21e.

From the foregoing analysis, three important points should be observed

First, q will vary linearly along segments (flanges) that are perpendicular

to the direction of V, and parabolically along segments (web) that are

inclined or parallel to V Second, q will always act parallel to the walls of

the member, since the section of the segment on which q is calculated is always taken perpendicular to the walls And third, the directional sense of

q is such that the shear appears to “flow” through the cross section, inward

at the beam’s top flange, “combining” and then “flowing” downward

through the web, since it must contribute to the downward shear force V,

Fig 7–22a, and then separating and “flowing” outward at the bottom

flange If one is able to “visualize” this “flow” it will provide an easy means

for establishing not only the direction of q, but also the corresponding direction of t Other examples of how q is directed along the segments of thin-walled members are shown in Fig 7–22b In all cases, symmetry

prevails about an axis that is collinear with V, and so q “flows” in a

direction such that it will provide the vertical force V and yet also satisfy

horizontal force equilibrium for the cross section

(a)

V

V

V V

• The shear flow formula q = VQ>I can be used to determine

the distribution of the shear flow throughout a thin-walled

member, provided the shear V acts along an axis of symmetry

or principal centroidal axis of inertia for the cross section

• If a member is made from segments having thin walls, then

only the shear flow parallel to the walls of the member is

• On the cross section, the shear “flows” along the segments so

that it results in the vertical shear force V and yet satisfies

horizontal force equilibrium

ExampLE 7.7

The thin-walled box beam in Fig 7–23a is subjected to a shear of 200 kN

Determine the variation of the shear flow throughout the cross section

SOLUTION

By symmetry, the neutral axis passes through the center of the cross

section For thin-walled members we use centerline dimensions for

calculating the moment of inertia

I = 121 (0.05 m)(0.175 m) 3 +2[(0.125 m)(0.025 m)(0.0875 m) 2 ] = 70.18(10 -6 ) m 4

Only the shear flow at points B, C, and D has to be determined For

point B, the area A′ ≈ 0, Fig 7–23b, since it can be thought of as

being located entirely at point B Alternatively, A′ can also represent

the entire cross-sectional area, in which case Q B = y′A ′ = 0 since

y ′ = 0 Because Q B = 0, then

q B = 0

For point C, the area A ′ is shown dark shaded in Fig 7–23c Here, we

have used the mean dimensions since point C is on the centerline of

each segment We have

The shear flow at D is determined using the three dark-shaded

rectangles shown in Fig 7–23d Again, using centerline dimensions

Q D = Σy′A′ = 2c a0.0875 m2 b(0.025 m)(0.0875 m)d + [0.0875 m](0.125 m)(0.025 m) = 0.4648(10 -3 ) m 3

Because there are two points of attachment,

q D = 12aVQ I Db = 12c[200 (10

3 ) N][0.4648 (10 -3 ) m 3 ] 70.18 (10 -6 ) m 4 d = 662.33(10 3 ) N >m = 662 kN>mUsing these results, and the symmetry of the cross section, the shear-

flow distribution is plotted in Fig 7–23e The distribution is linear

along the horizontal segments (perpendicular to V) and parabolic

along the vertical segments (parallel to V).

(a)

A N

* 7.5 Shear Center For open

thin-walled MeMberS

In the previous section, the internal shear V was applied along a principal

centroidal axis of inertia that also represents an axis of symmetry for the

cross section In this section we will consider the effect of applying the

shear along a principal centroidal axis that is not an axis of symmetry As

before, only open thin-walled members will be analyzed, where the dimensions to the centerline of the walls of the members will be used

A typical example of this case is the channel shown in Fig 7–24a Here

it is cantilevered from a fixed support and subjected to the force P If

this force is applied through the centroid C of the cross section, the channel will not only bend downward, but it will also twist clockwise

(qmax)f

Shear-flow distribution

(b)

d C A

The reason the member twists has to do with the shear-flow distribution

along the channel’s flanges and web, Fig 7–24b When this distribution is

integrated over the flange and web areas, it will give resultant forces of F f

in each flange and a force of V = P in the web, Fig 7–24c If the moments

of these three forces are summed about point A, the unbalanced couple

or torque created by the flange forces is seen to be responsible for

twisting the member The actual twist is clockwise when viewed from the

front of the beam, as shown in Fig 7–24a, because reactive internal

“equilibrium” forces F f cause the twisting In order to prevent this

twisting and therefore cancel the unbalanced moment, it is necessary to

apply P at a point O located an eccentric distance e from the web, as

shown in Fig 7–24d We require ΣM A = F f d = Pe, or

e = F P f d

The point O so located is called the shear center or flexural center

When P is applied at this point, the beam will bend without twisting,

Fig. 7–24e Design handbooks often list the location of the shear center

for a variety of thin-walled beam cross sections that are commonly used

in practice

From this analysis, it should be noted that the shear center will always

lie on an axis of symmetry of a member’s cross-sectional area For

example, if the channel is rotated 90° and P is applied at A, Fig 7–25a, no

twisting will occur since the shear flow in the web and flanges for this

case is symmetrical, and therefore the force resultants in these elements

will create zero moments about A, Fig 7–25b Obviously, if a member has

a cross section with two axes of symmetry, as in the case of a wide-flange

beam, the shear center will coincide with the intersection of these axes

procedure for analysIs

The location of the shear center for an open thin-walled member for

which the internal shear is in the same direction as a principal

centroidal axis for the cross section may be determined by using the following procedure

Shear-Flow Resultants

• By observation, determine the direction of the shear flow through the various segments of the cross section, and sketch the force resultants on each segment of the cross section (For

example, see Fig 7–24c.) Since the shear center is determined by taking the moments of these force resultants about a point, A,

choose this point at a location that eliminates the moments of

as many force resultants as possible

• The magnitudes of the force resultants that create a moment

about A must be calculated For any segment this is done by determining the shear flow q at an arbitrary point on the segment and then integrating q along the segment’s length

Realize that V will create a linear variation of shear flow in segments that are perpendicular to V, and a parabolic variation

of shear flow in segments that are parallel or inclined to V.

Shear Center

• Sum the moments of the shear-flow resultants about point A

and set this moment equal to the moment of V about A Solve

this equation to determine the moment-arm or eccentric

distance e, which locates the line of action of V from A.

• If an axis of symmetry for the cross section exists, the shear

center lies at a point on this axis

Determine the location of the shear center for the thin-walled channel

having the dimensions shown in Fig 7–26a.

SOLUTION

the section causes the shear to flow through the flanges and web as

shown in Fig 7–26b This in turn creates force resultants F f and V in

the flanges and web as shown in Fig 7–26c We will take moments

about point A so that only the force F f on the lower flange has to be

determined

The cross-sectional area can be divided into three component

rectangles—a web and two flanges Since each component is assumed

to be thin, the moment of inertia of the area about the neutral axis is

This same result can also be determined without integration by first

finding (qmax)f , Fig 7–26b, then determining the triangular area

q dx

h

2

Fig 7–26

Determine the location of the shear center for the angle having equal

legs, Fig 7–27a Also, find the internal shear-force resultant in each leg.

When a vertical downward shear V is applied at the section, the shear

flow and shear-flow resultants are directed as shown in Figs 7–27b and 7–27c, respectively Note that the force F in each leg must be

equal, since for equilibrium the sum of their horizontal components must be equal to zero Also, the lines of action of both forces intersect

point O; therefore, this point must be the shear center, since the sum of

the moments of these forces and V about O is zero, Fig 7–27c.

The magnitude of F can be determined by first finding the shear flow

at the arbitrary location s along the top leg, Fig 7–27d Here

Fig 7–27 (cont.)

The moment of inertia of the angle about the neutral axis must be

determined from “first principles,” since the legs are inclined with respect

to the neutral axis For the area element dA = t ds, Fig 7–27e, we have

q = VQ I = V

(tb3>3)J

1

22ab - 2 bs stR = 3V

22b3 s ab - 2 bs The variation of q is parabolic, and it reaches a maximum value when

s = b, as shown in Fig 7–27b The force F is therefore

components of the force F in each leg must equal V and, as stated

previously, the sum of the horizontal components equals zero

7–50. The beam is subjected to a shear force of V = 25 kN

Determine the shear flow at points A and B.

7–51. The beam is constructed from four plates and is

subjected to a shear force of V = 25 kN Determine the

maximum shear flow in the cross section.

7–54. A shear force of V = 18 kN is applied to the box girder Determine the shear flow at points A and B.

7–55. A shear force of V = 18 kN is applied to the box girder Determine the shear flow at point C.

D

Probs 7–50/51

*7–52. The aluminum strut is 10 mm thick and has the

cross section shown If it is subjected to a shear of

V = 150 N, determine the shear flow at points A and B.

7–53. The aluminum strut is 10 mm thick and has the cross

section shown If it is subjected to a shear of V = 150 N,

determine the maximum shear flow in the strut.

7–58. The H-beam is subjected to a shear of V = 80 kN

Determine the shear flow at point A.

7–59. The H-beam is subjected to a shear of V = 80 kN

Sketch the shear-stress distribution acting along one of its

side segments Indicate all peak values.

7–62. The box girder is subjected to a shear of V = 15 kN Determine the shear flow at point B and the maximum shear flow in the girder’s web AB.

7–63 Determine the location e of the shear center, point O,

for the thin-walled member having a slit along its section.

100 mm

100 mm

100 mm

e O

Prob 7–63

*7–64. Determine the location e of the shear center, point O,

for the thin-walled member The member segments have the

*7–60. The built-up beam is formed by welding together

the thin plates of thickness 5 mm Determine the location of

the shear center O.

V = 35 kN. Determine the shear flow at points A and B

and the maximum shear flow in the cross section.

7–65. The angle is subjected to a shear of V = 10 kN

Sketch the distribution of shear flow along the leg AB

Indicate numerical values at all peaks.

section of the thin-walled tube as a function of elevation y

and show that t max = 2V>A, where A = 2prt Hint: Choose

a differential area element dA=Rt du Using dQ = ydA,

formulate Q for a circular section from u to (p - u) and

show that Q = 2R2t cos u, where cos u=2R2 - y2>R.

t

y du ds

R

u

Prob 7–66 7–67. Determine the location e of the shear center, point O,

for the beam having the cross section shown The thickness is t.

*7–68. Determine the location e of the shear center, point O,

for the thin-walled member The member segments have the

same thickness t.

d

d

O d

7–69. A thin plate of thickness t is bent to form the beam

having the cross section shown Determine the location of

the shear center O.

t

O e

r

Prob 7–69

7–70. Determine the location e of the shear center, point O,

for the tube having a slit along its length.

O e

r t

Prob 7–70

CHapTER REVIEW

Transverse shear stress in beams is determined indirectly

by using the flexure formula and the relationship between

moment and shear (V = dM>dx) The result is the shear

formula

t = VQ It

In particular, the value for Q is the moment of the area A′

about the neutral axis, Q = y′A′ This area is the portion

of the cross-sectional area that is “held onto” the beam

above (or below) the thickness t where t is to be

determined.

A

Area  A¿

t N

_

¿

t

If the beam has a rectangular cross section, then the

shear-stress distribution will be parabolic, having a

maximum value at the neutral axis For this special case,

the maximum shear stress can be determined using

Fasteners, such as nails, bolts, glue, or weld, are used to

connect the composite parts of a “built-up” section The

shear force resisted by these fasteners is determined from

the shear flow, q, or force per unit length, that must be

supported by the beam The shear flow is

q = VQ I

A¿

If the beam is made from thin-walled segments, then the

shear-flow distribution along each segment can be

determined This distribution will vary linearly along

horizontal segments, and parabolically along inclined or

Provided the shear-flow distribution in each segment of an

open thin-walled section is known, then using a balance of

moments, the location O of the shear center for the cross

section can be determined When a load is applied to the

member through this point, the member will bend, and

not twist.

P

O e

R7–1 Sketch the intensity of the shear-stress distribution

acting over the beam’s cross-sectional area, and determine

the resultant shear force acting on the segment AB The

shear acting at the section is V = 175 kN Show that

C

V

Prob R7–1

R7–2. The T-beam is subjected to a shear of V = 150 kN

Determine the amount of this force that is supported by the

web B.

200 mm

40 mm

B V= 150 kN

R7–5 Solve Prob R7–4 if the beam is rotated 90° from

the position shown.

CHAPTER 8

The offset hanger supporting this ski gondola is subjected to the combined loadings

of axial force and bending moment.

(© ImageBroker/Alamy)

Cylindrical or spherical pressure vessels are commonly used in industry

to serve as boilers or storage tanks The stresses acting in the wall of these

vessels can be analyzed in a simple manner provided it has a thin wall,

that is, the inner-radius-to-wall-thickness ratio is 10 or more (r >t Ú 10)

Specifically, when r >t = 10 the results of a thin-wall analysis will predict

a stress that is approximately 4% less than the actual maximum stress in

the vessel For larger r >t ratios this error will be even smaller.

In the following analysis, we will assume the gas pressure in the vessel

is the gage pressure, that is, it is the pressure above atmospheric pressure,

since atmospheric pressure is assumed to exist both inside and outside

the vessel’s wall before the vessel is pressurized

CHAPTER OBJECTIVES

■ This chapter begins with an analysis of stress developed in

thin-walled pressure vessels Then we will use the formulas for axial

load, torsion, bending, and shear to determine the stress in a

member subjected to several loadings

Cylindrical pressure vessels, such as this gas tank, have semispherical end caps rather than flat ones in order to reduce the stress in the tank.

Cylindrical Vessels The cylindrical vessel in Fig 8–1a has a wall thickness t, inner radius r, and is subjected to an internal gas pressure p

To find the circumferential or hoop stress, we can section the vessel by

planes a, b, and c A free-body diagram of the back segment along with its contained gas is then shown in Fig 8–1b Here only the loadings in the

x direction are shown They are caused by the uniform hoop stress s1, acting on the vessel’s wall, and the pressure acting on the vertical face of

the gas For equilibrium in the x direction, we require

ΣF x = 0; 2[s1(t dy)] - p(2r dy) = 0

The longitudinal stress can be determined by considering the left portion

of section b, Fig 8–1a As shown on its free-body diagram, Fig 8–1c, s2 acts

uniformly throughout the wall, and p acts on the section of the contained

gas Since the mean radius is approximately equal to the vessel’s inner

radius, equilibrium in the y direction requires

ΣF y = 0; s2(2p rt) - p(p r2) = 0

For these two equations,

s1, s2 =    the normal stress in the hoop and longitudinal directions,

respectively Each is assumed to be constant throughout the

wall of the cylinder, and each subjects the material to tension

p=   the internal gage pressure developed by the contained gas

r=   the inner radius of the cylinder

t=   the thickness of the wall (r >t Ú 10)

2r

t p

By comparison, note that the hoop or circumferential stress is twice

as large as the longitudinal or axial stress Consequently, when fabricating

cylindrical pressure vessels from rolled-formed plates, it is important

that the longitudinal joints be designed to carry twice as much stress as

the circumferential joints

Spherical Vessels We can analyze a spherical pressure vessel

in a similar manner If the vessel in Fig 8–2a is sectioned in half, the

resulting free-body diagram is shown in Fig 8–2b Like the cylinder,

equilibrium in the y direction requires

ΣF y = 0; s2(2p rt) - p(p r 2) = 0

This is the same result as that obtained for the longitudinal stress in the

cylindrical pressure vessel, although this stress will be the same regardless

of the orientation of the hemispheric free-body diagram

Limitations The above analysis indicates that an element of

material taken from either a cylindrical or a spherical pressure vessel is

subjected to biaxial stress, i.e., normal stress existing in only two

directions Actually, however, the pressure also subjects the material to a

value equal to the pressure p at the interior wall and it decreases through

the wall to zero at the exterior surface of the vessel, since the pressure

there is zero For thin-walled vessels, however, we will ignore this stress

component, since our limiting assumption of r >t = 10 results in s2 and

s1 being, respectively, 5 and 10 times higher than the maximum radial

stress, (s3)max = p Finally, note that if the vessel is subjected to an

external pressure, the resulting compressive stresses within the wall may

cause the wall to suddenly collapse inward or buckle rather than causing

the material to fracture

This thin-walled pipe was subjected to an excessive gas pressure that caused it to rupture

in the circumferential or hoop direction The stress in this direction is twice that in the axial direction as noted by Eqs 8–1 and 8–2.

EXAMPLE 8.1

A cylindrical pressure vessel has an inner diameter of 1.2 m and a thickness

of 12 mm Determine the maximum internal pressure it can sustain so that neither its circumferential nor its longitudinal stress component exceeds 140 MPa Under the same conditions, what is the maximum internal pressure that a spherical vessel with a similar inner diameter can sustain?

SOLUTION

circumferential direction From Eq 8–1 we have

s1 = pr t ; 140(106) N>m2 = p0.012 m(0.6 m)

p = 2.80(106) N>m2 = 2.80 MPa Ans.

Note that when this pressure is reached, from Eq 8–2, the stress in the longitudinal direction will be s2 = 12 (140 MPa) = 70 MPa Furthermore,

the maximum stress in the radial direction occurs on the material at the

inner wall of the vessel and is (s3) max = p = 2.80 MPa This value is 50

times smaller than the circumferential stress (140 MPa), and as stated earlier, its effects will be neglected

perpendicular directions on an element of the vessel, Fig 8–2a From

Eq. 8–3, we have

s2 = pr 2t; 140(106)N>m2 = 2(0.012 m)p(0.6 m)

p = 5.60(106) N>m2 = 5.60 MPa Ans.

vessel will carry twice as much internal pressure as a cylindrical vessel

8–5. Air pressure in the cylinder is increased by exerting

forces P = 2 kN on the two pistons, each having a radius of

45 mm If the cylinder has a wall thickness of 2 mm, determine the state of stress in the wall of the cylinder.

8–6. Determine the maximum force P that can be exerted

on each of the two pistons so that the circumferential stress

in the cylinder does not exceed 3 MPa Each piston has a radius of 45 mm and the cylinder has a wall thickness

cover plate along the rivet line a–a, and (c) the shear

stress in the rivets.

8–1. A spherical gas tank has an inner radius of r = 1.5 m

If it is subjected to an internal pressure of p = 300 kPa,

determine its required thickness if the maximum normal

stress is not to exceed 12 MPa.

8–2. A pressurized spherical tank is to be made of

12-mm-thick steel If it is subjected to an internal pressure

of p = 1.4 MPa, determine its outer radius if the

maximum normal stress is not to exceed 105 MPa.

8–3. The thin-walled cylinder can be supported in one of

two ways as shown Determine the state of stress in the wall

of the cylinder for both cases if the piston P causes the

internal pressure to be 0.5 MPa The wall has a thickness of

6 mm and the inner diameter of the cylinder is 200 mm.

*8–4. The tank of the air compressor is subjected to an

internal pressure of 0.63 MPa If the internal diameter of the

tank is 550 mm, and the wall thickness is 6 mm, determine

the stress components acting at point A Draw a volume

element of the material at this point, and show the results

on the element.

A

Prob 8–4

8–11. The staves or vertical members of the wooden tank are held together using semicircular hoops having a thickness of 12 mm and a width of 50 mm Determine the

normal stress in hoop AB if the tank is subjected to an

internal gauge pressure of 14 kPa and this loading is transmitted directly to the hoops Also, if 6-mm-diameter bolts are used to connect each hoop together, determine the

tensile stress in each bolt at A and B Assume hoop AB

supports the pressure loading within a 300-mm length of the tank as shown.

*8–8. The gas storage tank is fabricated by bolting together

two half cylindrical thin shells and two hemispherical shells as

shown If the tank is designed to withstand a pressure of 3 MPa,

determine the required minimum thickness of the cylindrical

and hemispherical shells and the minimum required number of

longitudinal bolts per meter length at each side of the cylindrical

shell The tank and the 25 mm diameter bolts are made from

material having an allowable normal stress of 150 MPa and 250

MPa, respectively The tank has an inner diameter of 4 m.

8–9. The gas storage tank is fabricated by bolting together

two half cylindrical thin shells and two hemispherical shells

as shown If the tank is designed to withstand a pressure of

3  MPa, determine the required minimum thickness of the

cylindrical and hemispherical shells and the minimum

required number of bolts for each hemispherical cap The

tank and the 25 mm diameter bolts are made from material

having an allowable normal stress of 150 MPa and 250 MPa,

respectively The tank has an inner diameter of 4 m.

Probs 8–8/9

8–10. A wood pipe having an inner diameter of 0.9 m is

bound together using steel hoops each having a

cross-sectional area of 125 mm 2 If the allowable stress for the

hoops is sallow = 84 MPa, determine their maximum

spacing s along the section of pipe so that the pipe can resist

an internal gauge pressure of 28 kPa Assume each hoop

supports the pressure loading acting along the length s of

Prob 8–11

*8–12. A pressure-vessel head is fabricated by welding the circular plate to the end of the vessel as shown If the vessel sustains an internal pressure of 450 kPa, determine the average shear stress in the weld and the state of stress in the wall of the vessel.

8–13. The 304 stainless steel band initially fits snugly around

the smooth rigid cylinder If the band is then subjected to a

nonlinear temperature drop of ∆T = 12 sin 2 u °C, where u is

in radians, determine the circumferential stress in the band.

*8–16. The cylindrical tank is fabricated by welding a strip

of thin plate helically, making an angle u with the

longitudinal axis of the tank If the strip has a width w and thickness t, and the gas within the tank of diameter d is pressured to p, show that the normal stress developed along

the strip is given by s u = (pd>8t) (3 - cos 2u).

8–14. The ring, having the dimensions shown, is placed

over a flexible membrane which is pumped up with a

pressure p Determine the change in the inner radius of the

ring after this pressure is applied The modulus of elasticity

for the ring is E.

8–15. The inner ring A has an inner radius r1 and outer

radius r2 The outer ring B has an inner radius r3 and an outer

radius r4, and r2 7 r3 If the outer ring is heated and then

fitted over the inner ring, determine the pressure between

the two rings when ring B reaches the temperature of the

inner ring The material has a modulus of elasticity of E and

a coefficient of thermal expansion of a.

pretension in the filament is T and the vessel is subjected to

an internal pressure p, determine the hoop stresses in the

filament and in the wall of the vessel Use the free-body diagram shown, and assume the filament winding has a

thickness t′ and width w for a corresponding length L of