structural and stress analysis second edition pdf

www.EngineeringEBooksPdf.com Structural and Stress Analysis Second Edition by Dr T.H.G Megson www.EngineeringEBooksPdf.com www.EngineeringEBooksPdf.com Solutions Manual Solutions to Chapter Problems S.2.1 (a) Vectors representing the 10 and 15 kN forces are drawn to a suitable scale as shown in Fig S.2.1 Parallel vectors AC and BC are then drawn to intersect at C The resultant is the vector OC which is 21.8 kN at an angle of 23.4◦ to the 15 kN force B C 10 kN R 60° u O 15 kN A FIGURE S.2.1 (b) From Eq (2.1) and Fig S.2.1 R2 = 152 + 102 + × 15 × 10 cos 60◦ which gives R = 21.8 kN Also, from Eq (2.2) tan θ = 10 sin 60◦ 15 + 10 cos 60◦ so that θ = 23.4◦ www.EngineeringEBooksPdf.com • Solutions Manual S.2.2 (a) The vectors not have to be drawn in any particular order Fig S.2.2 shows the vector diagram with the vector representing the 10 kN force drawn first 12 kN kN 20 kN R u 10 kN FIGURE S.2.2 The resultant R is then equal to 8.6 kN and makes an angle of 23.9◦ to the negative direction of the 10 kN force (b) Resolving forces in the positive x direction Fx = 10 + cos 60◦ − 12 cos 30◦ − 20 cos 55◦ = −7.9 kN Then, resolving forces in the positive y direction Fy = cos 30◦ + 12 cos 60◦ − 20 cos 35◦ = −3.5 kN The resultant R is given by R2 = (−7.9)2 + (−3.5)2 so that R = 8.6 kN Also tan θ = 3.5 7.9 which gives θ = 23.9◦ www.EngineeringEBooksPdf.com Solutions to Chapter Problems • S.2.3 Initially the forces are resolved into vertical and horizontal components as shown in Fig S.2.3 y (1.0, 1.6) x 20.0 kN 69.3 kN 30° 80 kN 40 kN (Ϫ1, 1.25) 40 kN 50 kN 35.4 kN 30° Rx 34.6 kN y 45° (0, 0.5) 35.4 kN Ry (1.25, 0.25) x O 60 kN FIGURE S.2.3 Then Rx = 69.3 + 35.4 − 20.0 = 84.7 kN Now taking moments about the x axis Rx y¯ = 35.4 × 0.5 − 20.0 × 1.25 + 69.3 × 1.6 which gives y¯ = 1.22 m Also, from Fig S.2.3 Ry = 60 + 40 + 34.6 − 35.4 = 99.2 kN Now taking moments about the y axis Ry x¯ = 40.0 × 1.0 + 60.0 × 1.25 − 34.6 × 1.0 so that x¯ = 0.81 m The resultant R is then given by R2 = 99.22 + 84.72 from which R = 130.4 kN Finally θ = tan−1 99.2 = 49.5◦ 84.7 www.EngineeringEBooksPdf.com • Solutions Manual S.2.4 (a) In Fig S.2.4(a) the inclined loads have been resolved into vertical and horizontal components The vertical loads will generate vertical reactions at the supports A and B while the horizontal components of the loads will produce a horizontal reaction at A only since B is a roller support kN kN kN 6.1 kN 60° A RA,H 5.7 kN 45° 3.5 kN B 5.7 kN RA,V RB 4m 6m 5m 5m FIGURE S.2.4(a) Taking moments about B RA,V × 20 − × 16 − 6.1 × 10 − 5.7 × = which gives RA,V = 6.9 kN Now resolving vertically RB,V + RA,V − − 6.1 − 5.7 = so that RB,V = 7.9 kN Finally, resolving horizontally RA,H − 3.5 − 5.7 = so that RA,H = 9.2 kN Note that all reactions are positive in sign so that their directions are those indicated in Fig S.2.4(a) (b) The loads on the cantilever beam will produce a vertical reaction and a moment reaction at A as shown in Fig S.2.4(b) Resolving vertically RA − 15 − × 10 = which gives RA = 65 kN www.EngineeringEBooksPdf.com Solutions to Chapter Problems • kN/m MA A B 15 kN RA 10 m FIGURE S.2.4(b) Taking moments about A MA − 15 × 10 − × 10 × = from which MA = 400 kN m Again the signs of the reactions are positive so that they are in the directions shown (c) In Fig S.2.4(c) there are horizontal and vertical reactions at A and a vertical reaction at B 20 kN 5m 10 kN 15 kN kN/m A RA,H B RA,V RB 2m 4m 2m 2m FIGURE S.2.4(c) By inspection (or by resolving horizontally) RA,H = 20 kN Taking moments about A RB × + 20 × − × × − 15 × − 10 × = which gives RB = 12.5 kN Finally, resolving vertically RA,V + RB − 10 − 15 ì = www.EngineeringEBooksPdf.com ã Solutions Manual so that RA,V = 22.5 kN (d) The loading on the beam will produce vertical reactions only at the supports as shown in Fig S.2.4(d) 75 kN/m kN/m A B RA RB 3m 9m FIGURE S.2.4(d) Taking moments about B RA × 12 + 75 − × 12 × = Hence RA = 41.8 kN Now resolving vertically RB + RA − × 12 = so that RB = 54.2 kN S.2.5 (a) The loading on the truss shown in Fig P.2.5(a) produces only vertical reactions at the support points A and B; suppose these reactions are RA and RB respectively and that they act vertically upwards Then, taking moments about B RA × 10 − × 16 − 10 × 14 − 15 × 12 − 15 × 10 − × + × = which gives RA = 57 kN (upwards) Now resolving vertically RB + RA − − 10 − 15 − 15 − − = from which RB = −2 kN (downwards) www.EngineeringEBooksPdf.com Solutions to Chapter Problems • (b) The angle of the truss is tan−1 (4/10) = 21.8◦ The loads on the rafters are symmetrically arranged and may be replaced by single loads as shown in Fig S.2.5 These, in turn, may be resolved into horizontal and vertical components and will produce vertical reactions at A and B and a horizontal reaction at A 7427.9 N 2000 N 8000 N 21.8° 21.8° RA,H RA,V 21.8° 4m 2970.9 N 742.7 N RB 1857.0 N 20 m FIGURE S.2.5 Taking moments about B RA,V × 20 + 742.7 × − 1857.0 × 15 + 2970.9 × + 7427.9 × = which gives RA,V = −835.6 N (downwards) Now resolving vertically RB + RA,V − 1857.0 + 7427.9 = from which RB = −4735.3 N (downwards) Finally, resolving horizontally RA,H − 742.7 − 2970.9 = so that RA,H = 3713.6 N www.EngineeringEBooksPdf.com 252 • Solutions Manual The maximum positive shear force at these sections is then: With the head of the load at b, S(max) = 1.2 × 0.1 × = 0.12 c, S(max) = 1.2 × 0.3 × = 0.36 d, S(max) = 1.2 × 0.5 × = 0.60 e, S(max) = 1.2 × 0.7 × = 0.84 f, S(max) = 1.2 × 0.9 × = 1.08 y, S(max) = 1.2 × 0.05 × 0.5 = 0.03 The diagram of maximum positive shear force is then plotted to scale from these values; the diagram of maximum negative shear force follows, both are shown in Fig S.20.9(b) Superimposed on the diagrams of maximum shear force is the dead load shear inverted Then, from Fig S.20.9(b) shear reversal will occur over the central 1.3 m of the beam 1.25 1.3 m Shear reversal Dead load shear 1.25 FIGURE S.20.9(b) S.20.10 The force in the member CD is given by −MF /4 (using the method of sections) The influence line for FCD is shown in Fig S.20.10 A C D E 0.625 0.625 1.25 FIGURE S.20.10 Then, with the loads in the given positions FCD = −(20 × 0.625 + 10 × 1.25 + × 0.625) = −28.1 kN www.EngineeringEBooksPdf.com Solutions to Chapter 20 Problems • 253 S.20.11 From the method of sections FCG = MD /1.5, FHD = −MC /1.5, FFE sin α = RA when unit load is to the right of F and FFE sin α = RB when unit load is to the left of E; sin α = 1.5/2.5 = 0.6 The influence lines for the forces in CG, HD and FE are then as shown in Figs S.20.11(a), (b) and (c), respectively 2m 3m 3m 1.67 1.67 8ϫ8 ϭ 2.67 16 ϫ 1.5 (a) 1.67 1.75 ϫ 10 ϭ 2.5 16 ϫ 1.5 (b) 0.21 0.29 Ϫve 1.25 1.67 (c) FIGURE S.20.11 For the maximum force in CG place the right hand 70 kN load over the maximum ordinate in the IL; the ordinates under the other loads are found by similar triangles Then FCG (max) = 15 2.67 × 16 + 40 × + 70 × 1.67 + 70 × 2.67 + 60 × 1.67 = 763 kN For the maximum force in HD place the left hand 70 kN load over the maximum ordinate in the IL Then FHD (max) = − 15 2.5 × 16 + 40 × 1.67 + 70 × 2.5 + 70 × 1.75 + 60 × = −724 kN For the maximum force in FE place the left hand 70 kN load over the 1.25 ordinate Then FFE (max) = 15 0.21 × 2.29 − 15 1.25 × 13.71 + 40 × 0.21 − 70 × 1.25 − 70 × 0.94 − 60 × 0.625 i.e FFE (max) = −307 kN www.EngineeringEBooksPdf.com 254 • Solutions Manual S.20.12 When unit load is to the right of E, FCE may be found by taking moments about the intersection of GC and ED produced and is then equal to 0.95RA When unit load is to the left of D, FCE may be found by taking moments about the intersection of GH and DE produced Then, FCE = −2.085RB The IL is then as shown in Fig S.20.12(a) Further, FDE = MC /3 and the IL is as shown in Fig S.20.12(b) 2.38 m 1.62 m 0.95 D,C 0.475 E 0.695 2.085 (a) ϫ 16 24 ϫ ϩve ϭ 1.78 (b) FIGURE S.20.12 Then FCE (max pos) = −30 0.695 × 10.38 + 45 0.475 × 10.38 + 30 0.475 × FDE = 45 × 24 × 1.78 = 961.2 kN FCE (max neg) = −45 0.695 × 13.62 13.62 = 37.3 kN = −65.3 kN S.20.13 The family of influence lines for the shear force in each panel of the truss is drawn and is shown in Fig S.20.13 r a 1.0 c q p n2 n1 d o n3 e n m l k 1.0 n4 f b g h i j The central panel will require counterbracing www.EngineeringEBooksPdf.com FIGURE S.20.13 Solutions to Chapter 20 Problems For panel S = 2.8 × area n3 pa − 1.2(area n3 fb − area n3 pa) = × 3.375 − 1.2 × 5.625 = 0.38 Therefore panel (and 6) require counterbracing For panel S = × 2.25 9 − 1.2 × 6.75 = −1.7 Therefore panel and the remaining panels not require counterbracing S.20.14 The influence lines are shown in Figs S.20.14(a), (b), (c) and (d) l l C A B D A (b) RB (a) RA l A B B C C D E E D A FIGURE S.20.14 (c) SE B (d) ME C D l S.20.15 Release the beam at A and apply a load W as shown in Fig S.20.15 w B C A x RB 2m 2m FIGURE S.20.15 Then, RB = 2W www.EngineeringEBooksPdf.com • 255 256 • Solutions Manual Using Macauley’s method EI EI d2 v dx2 = Wx − 2W [x − 2] dv dx = EIv = Wx2 − W [x − 2]2 + C1 Wx3 W [x − 2]3 − + C1 x + C2 The boundary conditions are: v = when x = m and m These give C1 = −10W 16W , C2 = 3 so that v= x3 W EI [x − 2]3 − − 10x + 16 For the RA IL, v = when x = so that 1= W EI 16 and W = EI 16 Then RA = x3 16 − [x − 2]3 − 10x + 16 When a udl of 30 kN/m covers the span AB RA = 30 16 x3 − i.e RA = 26.25 kN www.EngineeringEBooksPdf.com 10x + 16 dx Solutions to Chapter 21 Problems • 257 Solutions to Chapter 21 Problems S.21.1 With the spring in position the forces acting on the column in its buckled state are as shown in Fig S.21.1 d 4P kd ␷ x L y FIGURE S.21.1 Then, from Eq (21.1) EI d2 v dx2 = 4P(δ − v) − kδ(L − x) (i) The solution of Eq (i) is v = A cos µx + B sin µx + δ[4P + k(x − L)] where µ2 = When x = 0, v = which gives (ii) 4P EI δ(kL − 4P) 4P Also, when x = 0, dv/dx = from which −δk B= 4Pµ Eq (ii) then becomes A= v= δ (kL − 4P) cos µx − + 4P + k(x − L) 4P When x = L, v = δ, then δ= k( sin µx) µ δ (kL − 4P) cos µL − k( sin µL) µ 4P from which k= 4Pµ (µL − tan µL) www.EngineeringEBooksPdf.com + 4P (iii) 258 • Solutions Manual S.21.2 Suppose that the buckling load of the column is P Then from Eq (21.1) and referring to Fig S.21.2, P D L EI C x 4EI ␷ L B L EI y A P FIGURE S.21.2 in AB EI d2 v dx2 = −Pv (i) 4EI d2 v dx2 = −Pv (ii) and in BC The solutions of Eqs (i) and (ii) are, respectively vAB = A cos µx + B sin µx µx µx vBC = C cos + D sin 2 (iii) (iv) where P EI When x = 0, vAB = so that, from Eq (iii), A = Therefore µ2 = vAB = B sin µx (v) When x = L/2, (dv/dx)BC = Then, from Eq (iv) D = C tan µL and vBC = C cos µx µL µx + tan sin (vi) When x = L/4, vAB = vBC so that, from Eqs (v) and (vi) B sin µL = C sec µL µL cos www.EngineeringEBooksPdf.com (vii) Solutions to Chapter 21 Problems • 259 When x = L/4, (dv/dx)AB = (dv/dx)BC so that again, from Eqs (v) and (vi) Bcos µL C µL µL sec sin = (viii) Dividing Eq (vii) by Eq (viii) gives tan µL µL tan =2 or tan2 − tan2 µL µL =2 from which tan i.e µL = 0.707 µL = 0.615 i.e P EI L = 0.615 so that P= 24.2EI L2 S.21.3 The compressive load P will cause the column to be displaced from its initial curved position to that shown in Fig S.21.3 P ␷ x L ␷0 y P FIGURE S.21.3 www.EngineeringEBooksPdf.com 260 • Solutions Manual From Eq (21.1) and noting that the bending moment at any point in the column is proportional to the change in curvature produced EI d2 v dx2 − EI But v0 = a Then d2 v0 dx2 4x L2 = −Pv (i) (L − x) d2 v0 −8a = 2 dx L and Eq (i) becomes d2 v + dx2 P EI v= −8a L2 (ii) The solution of Eq (ii) is v = A cos µx + B sin µx − where µ2 = 8a (µL)2 (iii) P EI When x = 0, v = so that A= 8a (µL)2 Also, when x = L/2, (dv/dx) = i.e B= 8a tan µL (µL)2 Eq (iii) then becomes v= 8a µL [ cos µx + tan sin µx − 1] 2 (µL) The maximum bending moment occurs when v is a maximum at x = L/2, i.e Mmax = −Pvmax Therefore, from Eq (iv) Mmax = −8aP[sec µL (µL)2 www.EngineeringEBooksPdf.com − 1] (iv) Solutions to Chapter 21 Problems • 261 S.21.4 The solution may be obtained directly from Eq (21.40) Then σmax =    P   1+ π dt  − PPCR δd πd3 t 8πdt    i.e P π dt σmax = 1+ (1 − α) 4δ d S.21.5 (a) The Euler buckling load for a pin-ended column is given by Eq (21.5) The second moment of area of a circular section column is π D4 /64 which, in this case, is given by I = π × 12.54 /64 = 1198.4 mm4 The area of cross section is A=π× 12.52 = 122.72 mm2 (i) P = π × 200 000 × 1198.4 (500)2 i.e P = 9462.2 N Therefore the test result conforms to the Euler theory (ii) P = π × 200 000 × 1198.4 (200)2 i.e P = 59 137.5 N Therefore the test result does not conform to the Euler theory (b) From Eq (21.27) and considering the first test result 9800 = 122.72σs 1+ kL2 1198.4 122.72 which simplifies to 79.86 + 2.05 ì 106 k = s www.EngineeringEBooksPdf.com 262 ã Solutions Manual Similarly, from the second test result 215.12 + 0.88 × 106 k = σs Solving gives k = 1.16 × 10−4 , σs = 317.2 N/mm2 S.21.6 The second moment of area of the column is given by π D4 − I= 7D = 0.0203D4 64 The area of cross section is given by π D2 − A= 7D = 0.1841D2 Then r2 = 0.0203D4 = 0.11D2 0.1841D2 Substituting in Eq (21.27) 300 × × 103 = 0.1841D2 1+ 7500 (2.5×103 )2 (0.11D2 ) which simplifies to D4 − 14.85 × 103 D2 − 0.125 × 103 = Solving D = 122 mm Say D = 128 mm, t = mm S.21.7 The column will buckle about an axis parallel to its web The second moment of area of the column is then given by I= × × 1303 184 × 63 + = 2.93 × 106 mm4 12 12 The area of cross section is A = × 130 × + 184 × = 3184 mm2 www.EngineeringEBooksPdf.com Solutions to Chapter 21 Problems • 263 Then r = 2.93 × 106 = 920.2 3184 so that r = 30.3 mm Then L 2.5 × 103 = = 82.5 r 30.3 Also σCR = π × 200 000 82.52 (see Eq (21.25)) i.e σCR = 290.0 N/mm2 Substituting these values etc in Eq (21.46) gives σ = 180.9 N/mm2 Then the maximum load the column can withstand is given by P = 180.9 × 3184 × 10−3 = 576 kN The Euler buckling load is PCR = 290 × 3184 × 10−3 = 923.4 kN Then P 576 = = 0.62 PCR 923.4 S.21.8 The bending moment at any section of the column is given by M = PCR v = PCR kx(L − x) Also dv = k(L − 2x) dx Substituting in Eq (21.65) U +V = k2 PCR 2E I1 a (Lx − x2 )2 dx + I1 L L−a I2 L−a a (Lx − x2 )2 dx − www.EngineeringEBooksPdf.com (Lx − x2 )2 dx+ PCR k2 L (L − 2x)2 dx 264 • Solutions Manual i.e U +V = PCR k2 2EI2 I2 −1 I1 + L2 a3 La4 a5 L2 (L − a)3 − + − L(L − a)4 (L − a)5 − + I2 I1 L5 30 − PCR k2 L3 From the principle of the stationary value of the total potential energy ∂(U + V ) =0 ∂k Then, since I2 = 1.6I1 and a = 0.2L this gives PCR = 14.96EI1 L2 Without the reinforcement π EI1 L2 PCR = The ratio of the two is 14.96/π = 1.52 Therefore the increase in strength is 52% S.21.9 Assume the equation of the deflected centre line of the column is v= 4δ L2 x2 in which δ is the deflection of the ends of the column relative to its centre and the origin for x is at the centre of the column The second moment of area varies in accordance with the relationship x I = I1 − 1.6 L The bending moment at any section of the column is given by M = PCR (δ − v) = PCR δ − x2 L2 Also, from the above dv = dx 8δ L2 x Substituting in Eq (21.65) U +V = PCR δ EI1 L3 L (L2 − 4x2 )2 dx − L − 1.6x www.EngineeringEBooksPdf.com 64PCR δ L4 L x2 dx Solutions to Chapter 21 Problems • 265 i.e U +V = δ2 L 0.3803PCR 8PCR δ − EI1 3L From the principle of the stationary value of the total potential energy δL 0.7606PCR ∂(U + V ) 16PCR δ = − =0 ∂δ EI1 3L which gives 7.01EI1 L2 For a column of constant thickness and second moment of area I2 PCR = PCR = π EI2 L2 For the columns to have the same buckling load π EI2 7.01EI1 = L L2 so that I2 = 0.7I1 Therefore, since the radii of gyration are the same, A2 = 0.7A1 The weight of the constant thickness column is equal to ρA2 L = 0.7ρA1 L The weight of the tapered column is equal to ρ × average thickness × L = ρ × 0.6A1 L so that the saving in weight is 0.1ρA1 L, i.e saving in weight = 0.1 ρA1 L = 0.143 0.7ρA1 L i.e about 15% www.EngineeringEBooksPdf.com www.EngineeringEBooksPdf.com .. .Structural and Stress Analysis Second Edition by Dr T.H.G Megson www.EngineeringEBooksPdf.com www.EngineeringEBooksPdf.com Solutions Manual Solutions to... FIGURE S.3.14 Shear force Between A and B the shear force is constant and equal to −60 kN Between B and C the shear force is equal to −60 + 50 = −10 kN and between C and D the shear force is equal... www.EngineeringEBooksPdf.com 12 • Solutions Manual At any section between A and B, SAB = −RA,V = −6.9 kN At any section between B and C, SBC = −6.9 + = −3.9 kN At any section between C and D, SCD = −6.9