Ans. Ans. Ans. Ans. Ans. Ans.

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1 08 Solutions /28/10 8:34 M Pge sphericl gs tnk hs n inner rdius of r = 1.5 m. If it is subjected to n internl pressure of p = 300 kp, determine its required thickness if the mximum norml stress is not to exceed 12 MP. s llow = p r 2 t ; 12(106 ) = 300(103 )(1.5) 2 t t = m = 18.8 mm 8 2. pressurized sphericl tnk is to be mde of 0.5-in.-thick steel. If it is subjected to n internl pressure of p = 200 psi, determine its outer rdius if the mximum norml stress is not to exceed 15 ksi. s llow = p r 2 t ; 15(103 ) = 200 r i 2(0.5) r i = 75 in. r o = 75 in in. = 75.5 in The thin-wlled cylinder cn be supported in one of two wys s shown. Determine the stte of stress in the wll of the cylinder for both cses if the piston P cuses the internl pressure to be 65 psi. The wll hs thickness of 0.25 in. nd the inner dimeter of the cylinder is 8 in. P 8 in. 8 in. P () (b) Cse (): s 1 = pr t s 2 = 0 ; s 1 = 65(4) 0.25 = 1.04 ksi Cse (b): s 1 = pr t ; s 1 = 65(4) 0.25 = 1.04 ksi s 2 = pr 2t ; s 2 = 65(4) = 520 psi 2(0.25) 532

2 08 Solutions /28/10 8:34 M Pge 533 *8 4. The tnk of the ir compressor is subjected to n internl pressure of 90 psi. If the internl dimeter of the tnk is 22 in., nd the wll thickness is 0.25 in., determine the stress components cting t point. Drw volume element of the mteril t this point, nd show the results on the element. Hoop Stress for Cylindricl Vessels: Since nlysis cn be used. pplying Eq. 8 1 r t = = , then thin wll s 1 = pr t = 90(11) 0.25 = 3960 psi = 3.96 ksi Longitudinl Stress for Cylindricl Vessels: pplying Eq. 8 2 s 2 = pr 2t = 90(11) = 1980 psi = 1.98 ksi 2(0.25) 8 5. The sphericl gs tnk is fbricted by bolting together two hemisphericl thin shells of thickness 30 mm. If the gs contined in the tnk is under guge pressure of 2 MP, determine the norml stress developed in the wll of the tnk nd in ech of the bolts.the tnk hs n inner dimeter of 8 m nd is seled with 900 bolts ech 25 mm in dimeter. Norml Stress: Since sphericl tnk s wll, r t = = , thin-wll nlysis is vlid. For the s = pr 2t = 2(4) = 133 MP 2(0.03) Referring to the free-body digrm shown in Fig., P = p = c p. Thus, 4 82 d = 32p10 6 N + c F y = 0; 32p P b - 450P b = 0 P b = p N The norml stress developed in ech bolt is then s b = P b = p = 228 MP b p

3 08 Solutions /28/10 8:34 M Pge The sphericl gs tnk is fbricted by bolting together two hemisphericl thin shells. If the 8-m inner dimeter tnk is to be designed to withstnd guge pressure of 2 MP, determine the minimum wll thickness of the tnk nd the minimum number of 25-mm dimeter bolts tht must be used to sel it.the tnk nd the bolts re mde from mteril hving n llowble norml stress of 150 MP nd 250 MP, respectively. Norml Stress: For the sphericl tnk s wll, s llow = pr 2t = 2106 (4) 2t t = m = 26.7 mm Since r t = = , thin-wll nlysis is vlid. Referring to the free-body digrm shown in Fig., P = p = c p. Thus, 4 82 d = 32p10 6 N + c F y = 0; 32p n 2 (P b) llow - n 2 (P b) llow = 0 n = 32p106 (P b ) llow (1) The llowble tensile force for ech bolt is (P b ) llow = s llow b = c p d = pn Substituting this result into Eq. (1), 32p10 6 n = p10 3 = =

4 08 Solutions /28/10 8:34 M Pge boiler is constructed of 8-mm thick steel pltes tht re fstened together t their ends using butt joint consisting of two 8-mm cover pltes nd rivets hving dimeter of 10 mm nd spced 50 mm prt s shown. If the stem pressure in the boiler is 1.35 MP, determine () the circumferentil stress in the boiler s plte prt from the sem,(b) the circumferentil stress in the outer cover plte long the rivet line, nd (c) the sher stress in the rivets. 8 mm 50 mm 0.75 m ) s 1 = pr = 1.35(106 )(0.75) = (10 6 ) = 127 MP t b) (10 6 )(0.05)(0.008) = s 1 (2)(0.04)(0.008) s 1 =79.1 MP c) From FD() + c F y = 0; F b (10 6 )[(0.008)(0.04)] = 0 F b = 25.3 kn (t vg ) b = F b p = 322 MP (0.01)

5 08 Solutions /28/10 8:34 M Pge 536 *8 8. The gs storge tnk is fbricted by bolting together two hlf cylindricl thin shells nd two hemisphericl shells s shown. If the tnk is designed to withstnd pressure of 3 MP, determine the required minimum thickness of the cylindricl nd hemisphericl shells nd the minimum required number of longitudinl bolts per meter length t ech side of the cylindricl shell. The tnk nd the 25 mm dimeter bolts re mde from mteril hving n llowble norml stress of 150 MP nd 250 MP, respectively. The tnk hs n inner dimeter of 4 m. Norml Stress: For the cylindricl portion of the tnk, the hoop stress is twice s lrge s the longitudinl stress. s llow = pr t ; = 3106 (2) t c t c = 0.04 m = 40 mm For the hemisphericl cp, s llow = pr t ; = 3106 (2) 2t s t s = 0.02 m = 20 mm r Since, thin-wll nlysis is vlid. t 6 10 Referring to the free-body digrm of the per meter length of the cylindricl portion, Fig., where P = p = [4(1)] = N, we hve + c F y = 0; n c (P b ) llow - n c (P b ) llow = 0 n c = 6106 (P b ) llow (1) The llowble tensile force for ech bolt is (P b ) llow = s llow b = c p d = N Substituting this result into Eq. (1), n c = = 49 bolts>meter 536

6 08 Solutions /28/10 8:34 M Pge The gs storge tnk is fbricted by bolting together two hlf cylindricl thin shells nd two hemisphericl shells s shown. If the tnk is designed to withstnd pressure of 3 MP, determine the required minimum thickness of the cylindricl nd hemisphericl shells nd the minimum required number of bolts for ech hemisphericl cp. The tnk nd the 25 mm dimeter bolts re mde from mteril hving n llowble norml stress of 150 MP nd 250 MP, respectively. The tnk hs n inner dimeter of 4 m. Norml Stress: For the cylindricl portion of the tnk, the hoop stress is twice s lrge s the longitudinl stress. s llow = pr t ; = 3106 (2) t c t c = 0.04 m = 40 mm For the hemisphericl cp, s llow = pr t ; = 3106 (2) 2t s t s = 0.02 m = 20 mm r Since, thin-wll nlysis is vlid. t 6 10 The llowble tensile force for ech bolt is (P b ) llow = s llow b = c p d = N Referring to the free-body digrm of the hemisphericl cp, Fig. b, where P = p = c p, 4 42 d = 12p10 6 N : + F x = 0; 12p n s 2 (P b) llow - n s 2 (P b) llow = 0 n s = 12p106 (P b ) llow (1) Substituting this result into Eq. (1), n s = = 308 bolts 537

7 08 Solutions /28/10 8:34 M Pge wood pipe hving n inner dimeter of 3 ft is bound together using steel hoops ech hving crosssectionl re of 0.2 in 2. If the llowble stress for the hoops is s llow = 12 ksi, determine their mximum spcing s long the section of pipe so tht the pipe cn resist n internl guge pressure of 4 psi. ssume ech hoop supports the pressure loding cting long the length s of the pipe. 4 psi s 4 psi s s Equilibrium for the steel Hoop: From the FD : + F x = 0; 2P - 4(36s) = 0 P = 72.0s Hoop Stress for the Steel Hoop: s 1 = s llow = P 12(10 3 ) = 72.0s 0.2 s = 33.3 in The stves or verticl members of the wooden tnk re held together using semicirculr hoops hving thickness of 0.5 in. nd width of 2 in. Determine the norml stress in hoop if the tnk is subjected to n internl guge pressure of 2 psi nd this loding is trnsmitted directly to the hoops. lso, if 0.25-in.-dimeter bolts re used to connect ech hoop together, determine the tensile stress in ech bolt t nd. ssume hoop supports the pressure loding within 12-in. length of the tnk s shown. 6 in. 6 in. 18 in. 12 in. 12 in. F R = 2(36)(12) = 864 lb F = 0; 864-2F = 0; F = 432 lb s h = F = 432 = 432 psi h 0.5(2) s b = F 432 = = 8801 psi = 8.80 ksi b p 4 (0.25)2 538

8 08 Solutions /28/10 8:34 M Pge 539 *8 12. Two hemispheres hving n inner rdius of 2 ft nd wll thickness of 0.25 in. re fitted together, nd the inside guge pressure is reduced to -10 psi. If the coefficient of sttic friction is m s = 0.5 between the hemispheres, determine () the torque T needed to initite the rottion of the top hemisphere reltive to the bottom one, (b) the verticl force needed to pull the top hemisphere off the bottom one, nd (c) the horizontl force needed to slide the top hemisphere off the bottom one in. 2 ft Norml Pressure: Verticl force equilibrium for FD(). + c F y = 0; 10Cp(24 2 )D - N = 0 N = 5760p lb The Friction Force: pplying friction formul F f = m s N = 0.5(5760p) = 2880p lb ) The Required Torque: In order to initite rottion of the two hemispheres reltive to ech other, the torque must overcome the moment produced by the friction force bout the center of the sphere. T = F f r = 2880p( >12) = lb # ft = 18.2 kip # ft b) The Required Verticl Force: In order to just pull the two hemispheres prt, the verticl force P must overcome the norml force. P = N = 5760p = lb = 18.1 kip c) The Required Horizontl Force: In order to just cuse the two hemispheres to slide reltive to ech other, the horizontl force F must overcome the friction force. F = F f = 2880p = 9048 lb = 9.05 kip The 304 stinless steel bnd initilly fits snugly round the smooth rigid cylinder. If the bnd is then subjected to nonliner temperture drop of T = 20 sin 2 u F, where u is in rdins, determine the circumferentil stress in the bnd. Comptibility: Since the bnd is fixed to rigid cylinder (it does not deform under lod), then d F - d T = 0 10 in. u 1 in in. P(2pr) E 2p - Trdu = 0 L0 2pr E P b = 20r L0 2p 2p E s c = 10 (1 - cos 2u)du L s c = 10E 0 2p sin 2 udu however, P = s c = 10(9.60) = 2.69 ksi 539

9 08 Solutions /28/10 8:34 M Pge The ring, hving the dimensions shown, is plced over flexible membrne which is pumped up with pressure p. Determine the chnge in the internl rdius of the ring fter this pressure is pplied. The modulus of elsticity for the ring is E. p r o r i w Equilibrium for the Ring: Form the FD : + F x = 0; 2P - 2pr i w = 0 P = pr i w Hoop Stress nd Strin for the Ring: s 1 = P = pr i w (r s - r i )w = pr i r s - r i Using Hooke s Lw e 1 = s 1 E = pr i E(r s - r i ) [1] However, e 1 = 2p(r i) 1-2pr i = (r i) 1 - r i = dr i. 2pr Then, from Eq. [1] dr i r i = r i pr i E(r s - r i ) r i dr i = pr i 2 E(r s - r i ) 540

10 08 Solutions /28/10 8:34 M Pge The inner ring hs n inner rdius r 1 nd outer rdius r 2. efore heting, the outer ring hs n inner rdius r 3 nd n outer rdius r 4, nd r 2 7 r 3. If the outer ring is heted nd then fitted over the inner ring, determine the pressure between the two rings when ring reches the temperture of the inner ring.the mteril hs modulus of elsticity of E nd coefficient of therml expnsion of. r 2 r 1 r 4 r 3 Equilibrium for the Ring: From the FD Hoop Stress nd Strin for the Ring: Using Hooke s lw : + F x = 0; 2P - 2pr i w = 0 P = pr i w s 1 = P = pr i w (r o - r i )w = pr i r o - r i e 1 = s 1 E = pr i E(r o - r i ) [1] However, e 1 = 2p(r i) 1-2pr i = (r i) 1 - r i = dr i. 2pr Then, from Eq. [1] dr i r i = r i pr i E(r o - r i ) r i dr i = pr i 2 E(r o - r i ) Comptibility: The pressure between the rings requires dr 2 + dr 3 = r 2 - r 3 [2] From the result obtined bove dr 2 = Substitute into Eq. [2] pr 2 2 E(r 2 - r 1 ) dr 3 = pr 2 2 E(r 2 - r 1 ) + pr 3 E(r 4 - r 3 ) = r 2 - r 3 2 pr 3 2 E(r 4 - r 3 ) p = E(r 2 - r 3 ) r 2 2 r 2 - r 1 + r 3 2 r 4 - r 3 541

11 08 Solutions /28/10 8:34 M Pge 542 *8 16. The cylindricl tnk is fbricted by welding strip of thin plte heliclly, mking n ngle u with the longitudinl xis of the tnk. If the strip hs width w nd thickness t, nd the gs within the tnk of dimeter d is pressured to p, show tht the norml stress developed long the strip is given by s u = (pd>8t)(3 - cos 2u). u w Norml Stress: s h = s 1 = pr t = p(d>2) t = pd 2t s l = s 2 = pr 2t = p(d>2) = pd 2t 4t Equilibrium: We will consider the tringulr element cut from the strip shown in Fig.. Here, h = (w sin u)t nd l = (w cos u)t. Thus, F h = s h h = pd pwd (w sin u)t = sin u nd 2t 2 F l = s l l = pd pwd (w cos u)t = cos u. 4t 4 Writing the force eqution of equilibrium long the x xis, F x = 0; c pwd 2 However, sin 2 u + cos 2 u = 1. This eqution becomes lso, sin 2 u = 1 (1 - cos 2u), so tht 2 N u = pwd (3 - cos 2u) 8 Since u = wt, then sin u d sin u + c pwd 4 N u = pwd 4 N u = pwd 4 s u = N u u = cos u d cos u - N u = u 2 sin 2 u + cos 2 u sin 2 u + 1 pwd (3 - cos 2u) 8 wt s u = pd (3 - cos 2u) 8t (Q.E.D.) 542

12 08 Solutions /28/10 8:34 M Pge In order to increse the strength of the pressure vessel, filment winding of the sme mteril is wrpped round the circumference of the vessel s shown. If the pretension in the filment is T nd the vessel is subjected to n internl pressure p, determine the hoop stresses in the filment nd in the wll of the vessel. Use the free-body digrm shown, nd ssume the filment winding hs thickness t nd width w for corresponding length of the vessel. p t L w s 1 T t s 1 Norml Stress in the Wll nd Filment efore the Internl Pressure is pplied: The entire length w of wll is subjected to pretension filment force T. Hence, from equilibrium, the norml stress in the wll t this stte is T 2T - (s l ) w (2wt) = 0 (s l ) w = T wt nd for the filment the norml stress is (s l ) fil = T wt Norml Stress in the Wll nd Filment fter the Internl Pressure is pplied: The stress in the filment becomes s fil = s l + (s l ) fil = pr (t + t ) + T wt nd for the wll, s w = s l - (s l ) w = p r (t + t ) - T wt The verticl force P cts on the bottom of the plte hving negligible weight. Determine the shortest distnce d to the edge of the plte t which it cn be pplied so tht it produces no compressive stresses on the plte t section. The plte hs thickness of 10 mm nd P cts long the center line of this thickness. s = 0 = s - s b 200 mm 300 mm 500 mm 0 = P - M c I 0 = P (0.2)(0.01) - P(0.1 - d)(0.1) 1 12 (0.01)(0.23 ) P( d) = 0 P d d = m = 66.7 mm 543

13 08 Solutions /28/10 8:34 M Pge Determine the mximum nd minimum norml stress in the brcket t section when the lod is pplied t x = mm 15 mm 100 kn x 200 mm 150 mm Consider the equilibrium of the FD of the top cut segment in Fig., + c F y = 0; N = 0 N = 100 kn + M C = 0; 100(0.1) - M = 0 M = 10 kn # m = 0.2(0.03) = m 2 I = 1 12 (0.03)(0.23 ) = 20.0(10-6 ) m 4 The norml stress developed is the combintion of xil nd bending stress. Thus, s = N ; M y I For the left edge fiber, y = C = 0.1 m. Then s L = - 100(103 ) (103 )(0.1) 20.0(10-6 ) = (10 6 ) P = 66.7 MP (C) (Mx) For the right edge fiber, y = 0.1 m. Then s R = (103 ) (103 )(0.1) 20.0(10-6 ) = 33.3 MP (T) 544

14 08 Solutions /28/10 8:34 M Pge 545 *8 20. Determine the mximum nd minimum norml stress in the brcket t section when the lod is pplied t x = 300 mm. 15 mm 15 mm 100 kn x 200 mm 150 mm Consider the equilibrium of the FD of the top cut segment in Fig., + c F y = 0; N = 0 N = 100 kn + M C = 0; M - 100(0.2) = 0 M = 20 kn # m = 0.2 (0.03) = m 2 I = 1 12 (0.03)(0.23 ) = 20.0(10-6 ) m 4 The norml stress developed is the combintion of xil nd bending stress. Thus, s = N ; M y I For the left edge fiber, y = C = 0.1 m. Then s C = - 100(103 ) (103 )(0.1) 20.0(10-6 ) = 83.33(10 6 ) P = 83.3 MP (T)(Min) For the right edge fiber, y = C = 0.1 m. Thus s R = - 100(103 ) (103 )(0.1) 20.0(10-6 ) = 117 MP 545

15 08 Solutions /28/10 8:34 M Pge The coping sw hs n djustble blde tht is tightened with tension of 40 N. Determine the stte of stress in the frme t points nd. 75 mm 3 mm 8 mm 8 mm 100 mm 3 mm 50 mm s = - P + Mc I = - 40 (0.008)(0.003) + 4(0.004) = 123 MP 1 (0.003)(0.008)3 12 s = Mc I = 2(0.004) = 62.5 MP 1 (0.003)(0.008) The clmp is mde from members nd C, which re pin connected t. If it exerts compressive force t C nd of 180 N, determine the mximum compressive stress in the clmp t section.the screw EF is subjected only to tensile force long its xis. 15 mm 30 mm 40 mm F C 180 N 15 mm Section E 180 N There is no moment in this problem. Therefore, the compressive stress is produced by xil force only. s mx = P = 240 = 1.07 MP (0.015)(0.015) 546

16 08 Solutions /28/10 8:34 M Pge The clmp is mde from members nd C, which re pin connected t. If it exerts compressive force t C nd of 180 N, sketch the stress distribution cting over section. The screw EF is subjected only to tensile force long its xis. 15 mm 30 mm 40 mm F C 180 N 15 mm Section E 180 N There is moment in this problem. Therefore, the compressive stress is produced by xil force only. s mx = P = 240 = 1.07 MP (0.015)(0.015) *8 24. The bering pin supports the lod of 700 lb. Determine the stress components in the support member t point. The support is 0.5 in. thick in in. 0.5 in. 3 in. F x = 0; N cos 30 = 0; N = lb 1.25 in. F y = 0; V sin 30 = 0; V = 350 lb 700 lb + M = 0; M - 700( sin 30 ) = 0; M = 175 lb # in. s = N - Mc I = (0.75)(0.5) - (175)(0.375) 1 12 (0.5)(0.75)3 s = ksi t = 0 (since Q = 0) 547

17 08 Solutions /28/10 8:34 M Pge The bering pin supports the lod of 700 lb. Determine the stress components in the support member t point. The support is 0.5 in. thick in in. 0.5 in. 3 in. F x = 0; N cos 30 = 0; N = lb F y = 0; V sin 30 = 0; V = 350 lb + M = 0; M - 700( sin 30 ) = 0; M = 175 lb # in. s = N + Mc I = (0.75)(0.5) + 175(0.375) 1 12 (0.5)(0.75) in. 700 lb s = 5.35 ksi t = 0 (since Q = 0) The offset link supports the loding of P = 30 kn. Determine its required width w if the llowble norml stress is s llow = 73 MP. The link hs thickness of 40 mm. P s due to xil force: s due to bending: s = P = 30(103 ) (w)(0.04) = 750(103 ) w s b = Mc I = 30(103 w )( )(w 2 ) 1 12 (0.04)(w)3 = 4500 (103 )( w 2 ) w 2 s mx = s llow = s + s b 50 mm P w 73(10 6 ) = 750(103 ) w (103 )( w 2 ) w 2 73 w 2 = 0.75 w w 73 w 2-3 w = 0 w = m = 79.7 mm 548

18 08 Solutions /28/10 8:34 M Pge The offset link hs width of w = 200 mm nd thickness of 40 mm. If the llowble norml stress is s llow = 75 MP, determine the mximum lod P tht cn be pplied to the cbles. P = 0.2(0.04) = m 2 I = 1 12 (0.04)(0.2)3 = (10-6 ) m 4 s = P + Mc I 75(10 6 ) = P P(0.1) (10-6 ) 50 mm w P = 109 kn P *8 28. The joint is subjected to force of P 80 lb nd F 0. Sketch the norml-stress distribution cting over section if the member hs rectngulr cross-sectionl re of width 2 in. nd thickness 0.5 in. s due to xil force: s = P = 80 = 80 psi (0.5)(2) 0.5 in. 2 in. s due to bending: F s = Mc I = 100(0.25) = 1200 psi 1 (2)(0.5)3 12 (s mx ) t = = 1280 psi = 1.28 ksi (s mx ) c = = 1120 psi = 1.12 ksi P 1.25 in. y 1.28 = (0.5 - y) 1.12 y = in. 549

19 08 Solutions /28/10 8:34 M Pge The joint is subjected to force of P = 200 lb nd F = 150 lb. Determine the stte of stress t points nd nd sketch the results on differentil elements locted t these points. The member hs rectngulr cross-sectionl re of width 0.75 in. nd thickness 0.5 in. 0.5 in. 2 in. F 1.25 in. P = 0.5(0.75) = in 2 Q = y œ =0.125(0.75)(0.25) = in 3 ; Q = 0 I = 1 12 (0.75)(0.53 ) = in 4 Norml Stress: s = N ; My I s = = 533 psi (T) s = (0.25) = psi = 1067 psi (C) Sher stress: t = VQ I t t = 150( ) = 600 psi ( )(0.75) t = 0 550

20 08 Solutions /28/10 8:34 M Pge If the 75-kg mn stnds in the position shown, determine the stte of stress t point on the cross section of the plnk t section. The center of grvity of the mn is t G. ssume tht the contct point t C is smooth. G C 600 mm mm 600 mm 1.5 m 12.5 mm Section nd b b 50 mm Support Rections: Referring to the free-body digrm of the entire plnk, Fig., + M = 0; F C sin 30 (2.4) - 75(9.81) cos 30 (0.9) = 0 F C = N F x = 0; x - 75(9.81) sin cos 30 = 0 x = N F y = 0; y sin 30-75(9.81) cos 30 = 0 y = N Internl Lodings: Consider the equilibrium of the free-body digrm of the plnk s lower segment, Fig. b, F x = 0; N = 0 N = N F y = 0; V = 0 V = N + MO = 0; M (0.6) = 0 M = N # m Section Properties: The cross-sectionl re nd the moment of inerti bout the centroidl xis of the plnk s cross section re Referring to Fig. c, Q is = 0.6(0.05) = 0.03 m 2 Q = y = (0.0125)(0.6) = m 3 Norml Stress: The norml stress is the combintion of xil nd bending stress. Thus, For point, y = m. Then I = 1 12 (0.6)0.053 = m 4 s = N ; My I s = (0.0125) = kp = 504 kp (C) 551

21 08 Solutions /28/10 8:34 M Pge Continued Sher Stress: The sher stress is contributed by trnsverse sher stress. Thus, t = VQ c d = = 14.9 kp It (0.6) The stte of stress t point is represented on the element shown in Fig. d. 552

22 08 Solutions /28/10 8:34 M Pge Determine the smllest distnce d to the edge of the plte t which the force P cn be pplied so tht it produces no compressive stresses in the plte t section. The plte hs thickness of 20 mm nd P cts long the centerline of this thickness. P d 200 mm 300 mm Consider the equilibrium of the FD of the left cut segment in Fig., : + F x = 0; N - P = 0 N = P + M C = 0; M - P(0.1 - d) = 0 M = P(0.1 - d) = 0.2 (0.02) = m 4 I = 1 12 (0.02)(0.23 ) = (10-6 ) m 4 The norml stress developed is the combintion of xil nd bending stress. Thus s = N ; My I Since no compressive stress is desired, the norml stress t the top edge fiber must be equl to zero. Thus, 0 = P P(0.1 - d)(0.1) ; (10-6 ) 0 = 250 P P (0.1 - d) d = m = 66.7 mm 553

23 08 Solutions /28/10 8:34 M Pge 554 *8 32. The horizontl force of P = 80 kn cts t the end of the plte. The plte hs thickness of 10 mm nd P cts long the centerline of this thickness such tht d = 50 mm. Plot the distribution of norml stress cting long section. Consider the equilibrium of the FD of the left cut segment in Fig., P d 200 mm 300 mm : + F x = 0; N - 80 = 0 N = 80 kn + M C = 0; M - 80(0.05) = 0 M = 4.00 kn # m = 0.01(0.2) = m 2 I = 1 12 (0.01)(0.23 ) = 6.667(10-6 ) m 4 The norml stress developed is the combintion of xil nd bending stress. Thus, s = N ; My I t point, y = 0.1 m. Then s = 80(103 ) (103 )(0.1) 6.667(10-6 ) = -20.0(10 6 ) P = 20.0 Mp (C) t point, y = 0.1 m. Then s = 80(103 ) (103 )(0.1) 6.667(10-6 ) = 100 (10 6 ) P = 100 MP (T) The loction of neutrl xis cn be determined using the similr tringles. 554

24 08 Solutions /28/10 8:34 M Pge The pliers re mde from two steel prts pinned together t. If smooth bolt is held in the jws nd gripping force of 10 lb is pplied t the hndles, determine the stte of stress developed in the pliers t points nd C. Here the cross section is rectngulr, hving the dimensions shown in the figure in. D 0.2 in. 0.1 in. E D E 30 C 3 in. 10 lb 0.2 in. 0.2 in. C 0.2 in in. + Q F x = 0; N - 10 sin 30 = 0; N = 5.0 lb 2.5 in. 4 in. 10 lb + F y = 0; V - 10 cos 30 = 0; V = lb + M C = 0; M - 10(3) = 0 M = 30 lb # in. = 0.2(0.4) = 0.08 in 2 I = 1 12 (0.2)(0.43 ) = (10-3 ) in 4 Q = 0 Q C = y =0.1(0.2)(0.2) = 4(10-3 ) in 3 Point : s = N + My I = (0.2) (10-3 = 5.56 ksi(t) ) t = VQ I t = 0 Point C: Sher Stress : s C = N + My I = = psi = 62.5 psi(c) t C = VQ I t = 8.660(4)(10-3 ) (10-3 = 162 psi )(0.2) 555

25 08 Solutions /28/10 8:34 M Pge Solve Prob for points D nd E in. D 0.2 in. 0.1 in. E D E 30 C 3 in. 10 lb 0.2 in. 0.2 in. C 0.2 in in. + M = 0; -F(2.5) + 4(10) = 0; F = 16 lb 2.5 in. 4 in. 10 lb Point D: s D = 0 t D = VQ It = 16(0.05)(0.1)(0.18) [ 1 = 667 psi 12 (0.18)(0.2)3 ](0.18) Point E: s E = My I 28(0.1) = = 23.3 ksi (T) 1 12 (0.18)(0.2)3 t E = 0 556

26 08 Solutions /28/10 8:34 M Pge The wide-flnge bem is subjected to the loding shown. Determine the stress components t points nd nd show the results on volume element t ech of these points. Use the sher formul to compute the sher stress. 500 lb 2500 lb 3000 lb 2 ft 2 ft 2 ft 4 ft 6 ft 0.5 in. I = 1 12 (4)(73 ) (3.5)(63 ) = in in. 4 in. 2 in. 4 in. 0.5 in. = 2(0.5)(4) + 6(0.5) = 7 in 2 Q = y =3.25(4)(0.5) + 2(2)(0.5) = 8.5 in 3 Q = 0 s = -Mc I = (12)(3.5) = ksi t = 0 s = My I t = VQ I t = 11500(12)(1) = 2.69 ksi = 2625(8.5) = ksi 51.33(0.5) 557

27 08 Solutions /28/10 8:34 M Pge 558 *8 36. The drill is jmmed in the wll nd is subjected to the torque nd force shown. Determine the stte of stress t point on the cross section of drill bit t section. y 400 mm x 20 N m 125 mm y 5 mm z 150 N Section Internl Lodings: Consider the equilibrium of the free-body digrm of the drill s right cut segment, Fig., F x = 0; N b = 0 N = 120 N F y = 0; b - V y = 0 V y = 90 N M x = 0; 20 - T = 0 T = 20 N # m M z = 0; b(0.4) b(0.125) + M z = 0 M z = 21 N # m Section Properties: The cross-sectionl re, the moment of inerti bout the z xis, nd the polr moment of inerti of the drill s cross section re = p = 25p10-6 m 2 I z = p = p10-9 m 4 J = p = p10-9 m 4 Referring to Fig. b, Q is Q = 0 Norml Stress: The norml stress is combintion of xil nd bending stress. Thus, s = N - M zy I z For point, y = m. Then s = p (0.005) p10-9 = MP = 215 MP (C) 558

28 08 Solutions /28/10 8:34 M Pge Continued Sher Stress: The trnsverse sher stress developed t point is c t xy V d = V yq = 0 I z t The torsionl sher stress developed t point is Thus, C(t xz ) T D = Tc J = 20(0.005) p10-9 = MP t xy = 0 t xz = c t xz T d = 102 MP The stte of stress t point is represented on the element shown in Fig c. 559

29 08 Solutions /28/10 8:34 M Pge The drill is jmmed in the wll nd is subjected to the torque nd force shown. Determine the stte of stress t point on the cross section of drill bit t section. y 400 mm x 20 N m 125 mm Internl Lodings: Consider the equilibrium of the free-body digrm of the drill s right cut segment, Fig., F x = 0; N b = 0 N = 120 N z y 5 mm N F y = 0; b - V y = 0 V y = 90 N Section M x = 0; 20 - T = 0 T = 20 N # m M z = 0; b(0.4) b(0.125) + M z = 0 M z = 21 N # m Section Properties: The cross-sectionl re, the moment of inerti bout the z xis, nd the polr moment of inerti of the drill s cross section re = p = 25p10-6 m 2 I z = p = p10-9 m 4 J = p = p10-9 m 4 Referring to Fig. b, Q is Q = y = 4(0.005) 3p c p d = m 3 Norml Stress: The norml stress is combintion of xil nd bending stress. Thus, For point, y = 0. Then s = N - M zy I z s = p = MP = 1.53 MP(C) 560

30 08 Solutions /28/10 8:34 M Pge Continued Sher Stress: The trnsverse sher stress developed t point is c t xy V d = V yq = I z t 90c d p10-9 (0.01) The torsionl sher stress developed t point is = MP c t xy T d = T C J = 20(0.005) p10-9 = MP Thus, t C = 0 t xy = c t xy T d - c t xy V d = = MP = 100 MP The stte of stress t point is represented on the element shown in Fig. d. 561

31 08 Solutions /28/10 8:34 M Pge Since concrete cn support little or no tension, this problem cn be voided by using wires or rods to prestress the concrete once it is formed. Consider the simply supported bem shown, which hs rectngulr cross section of 18 in. by 12 in. If concrete hs specific weight of 150 lb>ft 3, determine the required tension in rod, which runs through the bem so tht no tensile stress is developed in the concrete t its center section. Neglect the size of the rod nd ny deflection of the bem. 4 ft 4 ft 16 in. 18 in. 2 in. 6 in. 6 in. Support Rections: s shown on FD. Internl Force nd Moment: : + F x = 0; T - N = 0 N = T + M o = 0; M + T(7) - 900(24) = 0 M = T Section Properties: = 18(12) = 216 in 2 I = 1 12 (12)183 = 5832 in 4 Norml Stress: Requires s = 0 s = 0 = N + Mc I 0 = -T ( T)(9) T = 2160 lb = 2.16 kip 562

32 08 Solutions /28/10 8:34 M Pge Solve Prob if the rod hs dimeter of 0.5 in. Use the trnsformed re method discussed in Sec E E c = st = ksi, 2 ksi. 4 ft 4 ft 16 in. 18 in. 2 in. 6 in. 6 in. Support Rections: s shown on FD. Section Properties: n = E st = 29(103 ) E con 3.6(10 3 ) = con = (n - 1) t = ( ) p 4 b 0.52 = in 2 = 18(12) = in 2 y = y = 9(18)(12) + 16(1.3854) = in. I = 1 12 (12) (18)( ) ( ) 2 = in 4 Internl Force nd Moment: : + F x = 0; T - N = 0 N = T + M o = 0; M + T(6.9554) - 900(24) = 0 M = T Norml Stress: Requires s = 0 s = 0 = N + Mc I 0 = -T ( T)(8.9554) T = lb = 2.16 kip 563

33 08 Solutions /28/10 8:34 M Pge 564 *8 40. Determine the stte of stress t point when the bem is subjected to the cble force of 4 kn. Indicte the result s differentil volume element. 4 kn G 250 mm 375 mm D 2 m C 0.75 m 1 m 20 mm Support Rections: + M D = 0; 4(0.625) - C y (3.75) = mm 200 mm 15 mm 20 mm 150 mm C y = kn : + F x = 0; C x - 4 = 0 C x = 4.00 kn Internl Forces nd Moment: : + F x = 0; N = 0 N = 4.00 kn + c F y = 0; V = 0 V = kn + M o = 0; M (1) = 0 M = kn # m Section Properties: = 0.24(0.15) - 0.2(0.135) = m 2 I = 1 12 (0.15) (0.135)0.23 = m 4 Q = y =0.11(0.15)(0.02) (0.1)(0.015) = m 3 Norml Stress: s = N ; My I s = 4.00(103 ) 9.00(10-3 ) (103 )(0) 82.8(10-6 ) = MP (T) Sher Stress: pplying sher formul. t = VQ It = (103 )C0.405(10-3 )D 82.8(10-6 )(0.015) = MP 564

34 08 Solutions /28/10 8:34 M Pge Determine the stte of stress t point when the bem is subjected to the cble force of 4 kn. Indicte the result s differentil volume element. 4 kn G 250 mm 375 mm D 2 m C 0.75 m 1 m 20 mm Support Rections: + M D = 0; 4(0.625) - C y (3.75) = mm 200 mm 15 mm 20 mm 150 mm C y = kn : + F x = 0; C x - 4 = 0 C x = 4.00 kn Internl Forces nd Moment: : + F x = 0; N = 0 N = 4.00 kn + c F y = 0; V = 0 V = kn + M o = 0; M (1) = 0 M = kn # m Section Properties: = 0.24(0.15) - 0.2(0.135) = m 2 I = 1 12 (0.15) (0.135)0.23 = m Q = 0 Norml Stress: s = N ; My I s = 4.00(103 ) 9.00(10-3 ) (103 )(0.12) 82.8(10-6 ) = MP = MP (C) Sher Stress: Since Q = 0, then t = 0 565

35 08 Solutions /28/10 8:34 M Pge The br hs dimeter of 80 mm. Determine the stress components tht ct t point nd show the results on volume element locted t this point. 200 mm 300 mm Consider the equilibrium of the FD of br s left cut segment shown in Fig., F y = 0; V y b = 0 V y = 3 kn kn F z = 0; V z b = 0 V z = -4 kn M y = 0; M y b(0.3) = 0 M y = -1.2 kn # m M z = 0; M z b(0.3) = 0 M z = -0.9 kn # m I y = I t = p 4 (0.044 ) = 0.64(10-6 )p m 4 Referring to Fig. b, (Q y ) = 0 (Q z ) = z = 4(0.04) 3p c p 2 (0.042 ) d = 42.67(10-6 ) m 3 The norml stress is contributed by bending stress only. Thus, s = - M zy I z + M yz I y For point, y = m nd z = 0. Then s = (103 )(-0.04) 0.64(10-6 )p + 0 = (10 6 )P = 17.9 MP (C) The trnsverse sher stress developed t point is t xy v = V y(q y ) I z t t xz v = V z(q z ) I y t = 0 = 4(103 )C42.67(10-6 )D 0.64(10-6 )p (0.08) = 1.061(10 6 ) P = 1.06 MP The stte of stress for point cn be represented by the volume element shown in Fig. c, 566

36 08 Solutions /28/10 8:34 M Pge Continued 567

37 08 Solutions /28/10 8:34 M Pge The br hs dimeter of 80 mm. Determine the stress components tht ct t point nd show the results on volume element locted t this point. 200 mm 300 mm Consider the equilibrium of the FD of br s left cut segment shown in Fig., F y = 0; V y b = 0 V y = 3 kn kn F z = 0; V z b = 0 V z = -4 kn M y = 0; M y b(0.3) = 0 M y = -1.2 kn # m M z = 0; M z b(0.3) = 0 M z = -0.9 kn # m I y = I z = p 4 (0.044 ) = 0.64(10-6 )p m 4 Referring to Fig. b, Q y = y =c 4(0.04) 3p d c p 2 (0.042 ) d = 42.67(10-6 ) m 3 Q z = 0 The norml stress is contributed by bending stress only. Thus, s = - M zy I z + M yz I y For point, y = 0 nd z = 0.04 m. Then s = (103 )(0.04) 0.64(10-6 )p = (10 6 ) P = 23.9 MP (C) The trnsverse sher stress developed t point is t xy v = V y(q y ) I z t = 3(103 )C42.67(10-6 )D 0.64(10-6 )p (0.08) = (10 6 ) MP = MP 568

38 08 Solutions /28/10 8:34 M Pge Continued t xz v = V z (Q z ) I y t = 0 The stte of stress for point cn be represented by the volume element shown in Fig. c 569

39 08 Solutions /28/10 8:34 M Pge 570 *8 44. Determine the norml stress developed t points nd. Neglect the weight of the block. 6 kip 3 in. 12 kip Referring to Fig., F x = (F R ) x ; = F F = kip M y = (M R ) y ; 6(1.5) - 12(1.5) = M y M y = kip # in 6 in. M z = (M R ) z ; 12(3) - 6(3) = M z M z = 18.0 kip # in The cross-sectionl re nd moment of inerti bout the y nd z xes of the crosssection re = 6(3) = 18 in 2 I y = 1 12 (6)(3)3 = 13.5 in 4 I z = 1 12 (3)(63 ) = 54.0 in 4 The norml stress developed is the combintion of xil nd bending stress. Thus, s = F - M z y I z + M y z I y For point, y = 3 in. nd z = -1.5 in. s = (3) (-1.5) = ksi = 1.00 ksi (C) For point, y = 3 in nd z = 1.5 in. s = (3) (1.5) = ksi = 3.00 ksi (C) 570

40 08 Solutions /28/10 8:34 M Pge Sketch the norml stress distribution cting over the cross section t section. Neglect the weight of the block. 6 kip 3 in. 12 kip 6 in. Referring to Fig., F x = (F R ) x ; = F F = kip M y = (M R ) y ; 6(1.5) - 12(1.5) = M y M y = kip # in M z = (M R ) z ; 12(3) - 6(3) = M z M z = 18.0 kip # in The cross-sectionl re nd the moment of inerti bout the y nd z xes of the cross-section re = 3 (6) = 18.0 in 2 I y = 1 12 (6)(33 ) = 13.5 in 4 I z = 1 12 (3)(63 ) = 54.0 in 4 The norml stress developed is the combintion of xil nd bending stress. Thus, s = F - M zy I z + M yz I y For point, y = 3 in. nd z = -1.5 in. s = (3) (-1.5) = ksi = 1.00 ksi (C) For point, y = 3 in. nd z = 1.5 in. s = (3) (1.5) = ksi = 3.00 ksi (C) 571

41 08 Solutions /28/10 8:34 M Pge Continued For point C, y = -3 in. nd z = 1.5 in. s C = (-3) (1.5) = ksi = 1.00 ksi (C) For point D, y = -3 in. nd z = -1.5 in. s D = (-3) (-1.5) = 1.00 ksi (T) The norml stress distribution over the cross-section is shown in Fig. b 572

42 08 Solutions /28/10 8:34 M Pge The support is subjected to the compressive lod P. Determine the bsolute mximum nd minimum norml stress cting in the mteril. 2 2 P 2 2 Section Properties: w = + x = ( + x) I = 1 12 () ( + x)3 = ( + x)3 12 Internl Forces nd Moment: s shown on FD. Norml Stress: s = N ; Mc I = -P ( + x) ; 0.5PxC1 2 ( + x)d ( + x)3 12 = P -1 + x ; 3x ( + x) 2 R s = - P 1 + x + = - P 4x + ( + x) 2 R s = P -1 + x + = P 2x - ( + x) 2 R 3x ( + x) 2 R 3x ( + x) 2 R [1] [2] ds In order to hve mximum norml stress,. dx = 0 ds dx = - P ( + x)2 (4) - (4x + )(2)( + x)(1) ( + x) 4 R = 0 - P (2-4x) = 0 3 ( + x) P Since, then ( + x) 3 Z 0 2-4x = 0 x =

43 08 Solutions /28/10 8:34 M Pge Continued Substituting the result into Eq. [1] yields s mx = - P 4(0.500) + ( + 0.5) 2 R = P 2 = 1.33P 2 (C) ds In order to hve minimum norml stress,. dx = 0 ds dx = P ( + x)2 (2) - (2x - )(2)( + x)(1) ( + x) 4 R = 0 P (4-2x) = 0 3 ( + x) P Since, then ( + x) 3 Z 0 4-2x = 0 x = 2 Substituting the result into Eq. [2] yields s min = P 2(2) - ( + 2) 2 R = P 3 2 (T) 574

44 08 Solutions /28/10 8:34 M Pge The support is subjected to the compressive lod P. Determine the mximum nd minimum norml stress cting in the mteril. ll horizontl cross sections re circulr. r P Section Properties: d =2r + x = p(r + 0.5x) 2 I = p (r + 0.5x)4 4 Internl Force nd Moment: s shown on FD. Norml Stress: s = N ; Mc I = P 0.5Px(r + 0.5x) ; 2 p p(r + 0.5x) (r + 0.5)4 4 = P p 1 (r + 0.5x) 2 ; 2x (r + 0.5x) 3 R s = - P p 1 (r + 0.5x) 2 + 2x (r + 0.5x) 3 R = - P p r + 2.5x (r + 0.5x) 3 R [1] s = P p 1 (r + 0.5x) 2 + 2x (r + 0.5x) 3 R = P p 1.5x - r (r + 0.5x) 3 R [2] ds In order to hve mximum norml stress,. dx = 0 ds dx = - P p (r + 0.5x)3 (2.5) - (r + 2.5x)(3)(r + 0.5x) 2 (0.5) (r + 0.5x) 6 R = 0 - P (r - 2.5x) = 0 4 p(r + 0.5x) P Since, then p(r + 0.5x) 4 Z 0 r - 2.5x = 0 x = 0.400r Substituting the result into Eq. [1] yields s mx = - P p r + 2.5(0.400r) [r + 0.5(0.400r)] 3 R = P r 2 = 0.368P r 2 (C) 575

45 08 Solutions /28/10 8:34 M Pge Continued ds In order to hve minimum norml stress,. dx = 0 ds dx = P p (r + 0.5x)3 (1.5) - (1.5x - r)(3)(r + 0.5x) 2 (0.5) (r + 0.5x) 6 R = 0 P Since, then p(r + 0.5x) 4 Z 0 P (3r - 1.5x) = 0 4 p(r + 0.5x) 3r - 1.5x = 0 x = 2.00r Substituting the result into Eq. [2] yields s min = P p 1.5(2.00r) - r [r + 0.5(2.00r)] 3 R = P r 2 (T) *8 48. The post hs circulr cross section of rdius c. Determine the mximum rdius e t which the lod cn be pplied so tht no prt of the post experiences tensile stress. Neglect the weight of the post. Require s = 0 s = 0 = P + Mc I ; 0 = -P p c 2 + (Pe)c p 4 c4 c e P e = c 4 576

46 08 Solutions /28/10 8:34 M Pge If the bby hs mss of 5 kg nd his center of mss is t G, determine the norml stress t points nd on the cross section of the rod t section. There re two rods, one on ech side of the crdle. 500 mm 15 G 75 mm 6 mm Section Section Properties: The loction of the neutrl surfce from the center of curvture of the rod, Fig., cn be determined from R = L d r where = p = 36p10-6 m 2 L d r = 2p r - 2r 2 - c 2 = 2p = m Thus, R = 36p = m Then e = r - R = = m Internl Lodings: Consider the equilibrium of the free-body digrm of the crdle s upper segment, Fig. b, + c F y = 0; -5(9.81) - 2N = 0 N = N + M O = 0; 5(9.81)( ) - 2M = 0 M = N # m Norml Stress: The norml stress is the combintion of xil nd bending stress. Thus, s = N + M(R - r) er Here, M = (negtive) since it tends to increse the curvture of the rod. For point, r = r = m. Then, s = p ( ) + 36p10-6 ( )10-3 (0.075) = MP = 89.1 MP (C) For point, r = r = m. Then, s = p10-6 = 79.3 kp (T) ( ) 36p10-6 ( )10-3 (0.087) L d r = 0.25 ln 5 4 =

47 08 Solutions /28/10 8:34 M Pge The C-clmp pplies compressive stress on the cylindricl block of 80 psi. Determine the mximum norml stress developed in the clmp. 4 in in. 1 in. 4.5 in in. R = 1 d r = 1(0.25) = P = s b = 80p (0.375) 2 = lb M = ( ) = lb # in. s = M(R - r) r(r - R) + P (s t ) mx = ( ) (1)(0.25)(4)( ) = 8.37 ksi (1)(0.25) (s c ) mx = ( ) 1(0.25)(5)( ) = ksi (1)(0.25) 578

48 08 Solutions /28/10 8:34 M Pge post hving the dimensions shown is subjected to the bering lod P. Specify the region to which this lod cn be pplied without cusing tensile stress to be developed t points,, C, nd D. x e y z P C D e z y Equivlent Force System: s shown on FD. Section Properties: = 2(2) (2)R = 62 2 I z = 1 12 (2)(2) (2) (2) + 3 b 2 R = 5 4 I y = 1 12 (2)(2) (2) (2) 3 b 2 R = Norml Stress: s = N - M zy I z + M y z I y = -P Pe yy Pe z z = P e y y + 18e z z t point where y = - nd z =, we require s P 30 4 C -52-6(-) e y + 18() e z D e y + 18e z 6e y + 18e z 6 5 When e z = 0, e y When e y = 0, e z Repet the sme procedures for point, C nd D. The region where P cn be pplied without creting tensile stress t points,, C nd D is shown shded in the digrm. 579

49 08 Solutions /28/10 8:34 M Pge 580 *8 52. The hook is used to lift the force of 600 lb. Determine the mximum tensile nd compressive stresses t section. The cross section is circulr nd hs dimeter of 1 in. Use the curved-bem formul to compute the bending stress. 300 lb 300 lb Section Properties: r = = 2.00 in. 2.5 in. 1.5 in. L d r = 2pr - 2r 2 - c 2 = 2p = in. 600 lb = p0.5 2 = 0.25p in 2 R = 1 d r = 0.25p = in Internl Force nd Moment: s shown on FD. The internl moment must be computed bout the neutrl xis. M = lb # in. is positive since it tends to increse the bem s rdius of curvture. Norml Stress: pplying the curved-bem formul. For tensile stress r - R = = in. (s t ) mx = N + M(R - r 1) r 1 (r - R) = ( ) p 0.25p(1.5)( ) = psi = 15.5 ksi (T) For compressive stress, (s c ) mx = N + M(R - r 2) r 2 (r - R) 600 = 0.25p ( ) 0.25p(2.5)( ) = psi = 9.31 ksi (C) 580

50 08 Solutions /28/10 8:34 M Pge The msonry pier is subjected to the 800-kN lod. Determine the eqution of the line y = f1x2 long which the lod cn be plced without cusing tensile stress in the pier. Neglect the weight of the pier m 2.25 m y x 800 kn 1.5 m y 1.5 m x C = 3(4.5) = 13.5 m 2 I x = 1 12 (3)(4.53 ) = m 4 I y = 1 12 (4.5)(33 ) = m 4 Norml Stress: Require s = 0 s = P + M xy I x + M yx I y 0 = -800(103 ) (103 )y(2.25) (103 )x(1.5) = 0.148x y y = x 581

51 08 Solutions /28/10 8:34 M Pge The msonry pier is subjected to the 800-kN lod. If x = 0.25 m nd y = 0.5 m, determine the norml stress t ech corner,, C, D (not shown) nd plot the stress distribution over the cross section. Neglect the weight of the pier. = 3(4.5) = 13.5 m m 2.25 m y x 800 kn 1.5 m y 1.5 m x I x = 1 12 (3)(4.53 ) = m 4 I y = 1 12 (4.5)(33 ) = m 4 s = P + M xy I x + M yx I y C s = -800(103 ) (103 )(2.25) (103 )(1.5) = 9.88 kp (T) s = -800(103 ) (103 )(2.25) (103 )(1.5) = kp = 49.4 kp (C) s C = -800(103 ) (103 )(2.25) (103 )(1.5) = -128 kp = 128 kp (C) s D = -800(103 ) (103 )(2.25) (103 )(1.5) = kp = 69.1 kp (C) 582

52 08 Solutions /28/10 8:34 M Pge The br hs dimeter of 40 mm. If it is subjected to the two force components t its end s shown, determine the stte of stress t point nd show the results on differentil volume element locted t this point. x 100 mm z 150 mm y 500 N 300 N Internl Forces nd Moment: F x = 0; N x = 0 F y = 0; V y = 0 V y = -300 N F z = 0; V z = 0 V z = 500 N M x = 0; T x = 0 M y = 0; M y - 500(0.15) = 0 M y = 75.0 N # m M z = 0; M z - 300(0.15) = 0 M z = 45.0 N # m Section Properties: = p = p m 2 I x = I y = p = p m 4 J = p = p m 4 Norml Stress: (Q ) z = 0 (Q ) y = 4(0.02) 3p c 1 2 p0.022 d = m 3 s = N - M zy I z + M yz I y s = (0) 40.0(10-9 )p (0.02) 40.0(10-9 )p = 11.9 MP (T) Sher Stress: The trnverse sher stress in the z nd y directions cn be obtined using the sher formul, t V = VQ. It (t xy ) = -t Vy = - 300C5.333(10-6 )D 40.0(10-9 )p (0.04) (t xz ) = t Vz = 0 = MP 583

53 08 Solutions /28/10 8:34 M Pge 584 *8 56. Solve Prob for point. x 100 mm z 150 mm y 500 N 300 N Internl Forces nd Moment: F x = 0; N x = 0 F y = 0; V y = 0 V y = -300 N F z = 0; V z = 0 V z = 500 N M x = 0; T x = 0 M y = 0; M y - 500(0.15) = 0 M y = 75.0 N # m M z = 0; M z - 300(0.15) = 0 M z = 45.0 N # m Section Properties: = p = p m 2 I x = I y = p = p m 4 J = p = p m 4 Norml Strees: (Q ) y = 0 (Q ) z = 4(0.02) 3p c 1 2 p0.022 d = m 3 s = N - M zy I z + M yz I y s = (0.02) 40.0(10-9 ) p (0) 40.0(10-9 ) p = MP = 7.16 MP (C) Sher Stress: The trnverse sher stress in the z nd y directions cn be obtined using the sher formul, t V = VQ. It (t xz ) = t Vz = 500C5.333(10-6 )D 40.0(10-9 ) p (0.04) (t xy ) = t Vy = 0 = MP 584

54 08 Solutions /28/10 8:34 M Pge The 2-in.-dimeter rod is subjected to the lods shown. Determine the stte of stress t point, nd show the results on differentil element locted t this point. x z 8 in. 600 lb y 12 in. Consider the equilibrium of the FD of the right cut segment, Fig., F y = 0 ; N y = 0 N y = -800 lb F z = 0 ; V z = 0 V z = -600 lb F x = 0 ; V x = 0 V x = 500 lb M y = 0 ; T y - 600(12) = 0 T y = 7200 lb # in M z = 0 ; M z + 800(12) + 500(8) = 0 M z = lb # in M x = 0 ; M x + 600(8) = 0 M x = lb # in 500 lb 800 lb I x = I z = p 4 (14 ) = p 4 in4 = p(1 2 ) = p in 2 J = p 2 (14 ) = p 2 in4 Referring to Fig. b, (Q x ) = 0 (Q z ) = y = 4(1) 3p c p 2 (12 ) d = in 3 The norml stress is contributed by xil nd bending stress. Thus, s = N + M xz I x - M zx I z For point, z = 0 nd x = 1 in. s = 800 p (0) (1) p>4 p>4 = 17.57(10 3 ) psi = 17.6 ksi (T) The torsionl sher stress developed t point is (t yz ) T = T yc J = 7200(1) = 4.584(10 3 ) psi = ksi T p>2 The trnsverse sher stress developed t point is. (t yz ) g = V z(q z ) I x t = 600(0.6667) p 4 (2) = psi = ksi T (t xy ) g = V x(q x ) I z t = 500(0) p 4 (2) = 0 585

55 08 Solutions /28/10 8:34 M Pge Continued Combining these two sher stress components, t yz = t yz T + t yz g = t xy = 0 = ksi = 4.84 ksi The stte of stress of point cn be represented by the volume element shown in Fig. c. 586

56 08 Solutions /28/10 8:34 M Pge The 2-in.-dimeter rod is subjected to the lods shown. Determine the stte of stress t point, nd show the results on differentil element locted t this point. z x 8 in. 600 lb y 12 in. Consider the equilibrium of the FD of the right cut segment, Fig., 500 lb F y = 0; N y = 0 N y = -800 lb 800 lb F z = 0; V z = 0 V z = -600 lb F x = 0; V x = 0 V x = 500 lb M y = 0; T y - 600(12) = 0 T y = 7200 lb # in M z = 0; M z + 800(12) + 500(8) = 0 M z = lb # in M x = 0; M x + 600(8) = 0 M x = lb # in. The cross-sectionl re the moment of inerti bout x nd Z xes nd polr moment of inerti of the rod re = p(1 2 ) = p in 2 I x = I z = p 4 (14 ) = p 4 in4 J = p 2 (14 ) = p 2 in4 Referring to Fig. b, (Q z ) = 0 (Q x ) = z = 4(1) 3p c p 2 (12 ) d = in 4 The norml stress is contributed by xil nd bending stress. Thus, s = N + M xz I x - M zx I z For point, x = 0 nd z = 1 in. s = 800 p (1) p> (0) p>4 = 5.86 ksi (C) The torsionl sher stress developed t point is (t xy ) T = T yc J = 7200(1) = 4.584(10 3 ) psi = ksi : p>2 The trnsverse sher stress developed t point is. (t xy ) v = V x(q x ) I z t = 500 (0.6667) p 4 (2) = psi = ksi : (t yz ) v = V z(q z ) I x t = 600 (0) p 4 (2) = 0 587

57 08 Solutions /28/10 8:34 M Pge Continued Combining these two sher stress components, t xy = (t xy ) T + (t xy ) v = = ksi = 4.80 ksi t yz = 0 The stte of stress of point cn be represented by the volume element shown in Fig. c. 588

58 08 Solutions /28/10 8:34 M Pge If P = 60 kn, determine the mximum norml stress developed on the cross section of the column. Equivlent Force System: Referring to Fig., 150 mm 150 mm 2P 15 mm 15 mm P + c F x = F R x ; = -F F = 180 kn 15 mm 75 mm 100 mm M y = (M R ) y ; -60(0.075) = -M y M y = 4.5kN # m M z = (M R ) z ; -120(0.25) = -M z M z = 30kN # m 100 mm 100 mm Section Properties: The cross-sectionl re nd the moment of inerti bout the y nd z xes of the cross section re = 0.2(0.3) (0.27) = m 2 I z = 1 12 (0.2) (0.185)0.273 = m 4 I y = 2c 1 12 (0.015)0.23 d (0.27) = m 4 Norml Stress: The norml stress is the combintion of xil nd bending stress. Here, F is negtive since it is compressive force. lso, M y nd M z re negtive since they re directed towrds the negtive sense of their respective xes. y inspection, point is subjected to mximum norml stress. Thus, s = N - M zy I z + M yz I y s mx = s = = MP = 71.0 MP(C) - C D(-0.15) C D(0.1)

59 08 Solutions /28/10 8:34 M Pge 590 *8 60. Determine the mximum llowble force P, if the column is mde from mteril hving n llowble norml stress of s llow = 100 MP. Equivlent Force System: Referring to Fig., 150 mm 150 mm 2P 15 mm 15 mm P + c F x = (F R ) x ; -P - 2P = -F 15 mm 75 mm 100 mm F = 3P 100 mm 100 mm M y = (M R ) y ; -P(0.075) = -M y M y = P M z = (M R ) z ; -2P(0.25) = -M z M z = 0.5P Section Properties: The cross-sectionl re nd the moment of inerti bout the y nd z xes of the cross section re = 0.2(0.3) (0.27) = m 2 I z = 1 12 (0.2) (0.185)0.273 = m 4 I y = 2c 1 12 (0.15)0.23 d (0.27) = m 4 Norml Stress: The norml stress is the combintion of xil nd bending stress. Here, F is negtive since it is compressive force. lso, M y nd M z re negtive since they re directed towrds the negtive sense of their respective xes. y inspection, point is subjected to mximum norml stress, which is in compression. Thus, s = N - M zy I z + M yz I y = - 3P (-0.5P)(-0.15) P(0.1) P = N = 84.5 kn 590

60 08 Solutions /28/10 8:34 M Pge The beveled ger is subjected to the lods shown. Determine the stress components cting on the shft t point, nd show the results on volume element locted t this point. The shft hs dimeter of 1 in. nd is fixed to the wll t C. 200 lb z C y 8 in. x F x = 0; V x = 0; V x = 125 lb F y = 0; 75 - N y = 0; N y = 75 lb F z = 0; V z = 0; V z = 200 lb M x = 0; 200(8) - M x = 0; M x = 1600 lb # in. M y = 0; 200(3) - T y = 0; T y = 600 lb # in. M z = 0; M z + 75(3) - 125(8) = 0; M z = 775 lb # in. = p(0.5 2 ) = in 2 3 in. 75 lb 125 lb J = p 2 (0.54 ) = in 4 I = p 4 (0.54 ) = in 4 (Q ) x = 0 (Q ) z = 4(0.5) 3p 1 2 b(p)(0.52 ) = in 3 (s ) y = - N y + M x c I = (0.5) = psi = 16.2 ksi (T) (t ) yx = (t ) V - (t ) twist = V x(q ) z It - T yc J = 125( ) (1) - 600(0.5) = psi = ksi (t ) yz = V z(q ) x It = 0 591

61 08 Solutions /28/10 8:34 M Pge The beveled ger is subjected to the lods shown. Determine the stress components cting on the shft t point, nd show the results on volume element locted t this point. The shft hs dimeter of 1 in. nd is fixed to the wll t C. 200 lb z C y 8 in. x 3 in. 75 lb 125 lb F x = 0; V x = 0; V x = 125 lb F y = 0; 75 - N y = 0; N y = 75 lb F z = 0; V z = 0; V z = 200 lb M x = 0; 200(8) - M x = 0; M x = 1600 lb # in. M y = 0; 200(3) - T y = 0; T y = 600 lb # in. M z = 0; M z + 75(3) - 125(8) = 0; M z = 775 lb # in. = p(0.5 2 ) = in 2 J = p 2 (0.54 ) = in 4 I = p 4 (0.54 ) = in 4 (Q ) z = 0 (Q ) x = 4(0.5) 3p 1 2 b(p)(0.52 ) = in 3 (s ) y = - N y + M z c I = (0.5) = 7.80 ksi (T) (t ) yz = (t ) V + (t ) twist = V z(q ) x It + T y c J = 200( ) (1) + 600(0.5) = 3395 psi = 3.40 ksi (t ) yx = V x (Q ) z It = 0 592

62 08 Solutions /28/10 8:34 M Pge The uniform sign hs weight of 1500 lb nd is supported by the pipe, which hs n inner rdius of 2.75 in. nd n outer rdius of 3.00 in. If the fce of the sign is subjected to uniform wind pressure of p = 150 lb>ft 2, determine the stte of stress t points C nd D. Show the results on differentil volume element locted t ech of these points. Neglect the thickness of the sign, nd ssume tht it is supported long the outside edge of the pipe. 6 ft 12 ft 150 lb/ft 2 3 ft E F C D z x y Section Properties: = p = p in 2 I y = I z = p = in 4 (Q C ) z = (Q D ) y = 0 (Q C ) y = (Q D ) z = 4(3) 3p c 1 2 (p)32 d - 4(2.75) 3p c 1 2 (p)2.752 d = in 3 J = p = in 4 Norml Stress: s = N - M z y I z + M y z I y s C = p - (-64.8)(12)(0) (12)(2.75) = 15.6 ksi (T) s D = p - (-64.8)(12)(3) (12)(0) = 124 ksi (T) Sher Stress: The trnverse sher stress in the z nd y directions nd the torsionl sher stress cn be obtined using the sher formul nd the torsion formul, t nd t twist = Tr V = VQ, respectively. It J (t xz ) D = t twist = 64.8(12)(3) (t xy ) D = t Vy = 0 = 62.4 ksi 593

63 08 Solutions /28/10 8:34 M Pge Continued (t xy ) C = t Vy - t twist = 10.8( ) (2)(0.25) (12)(2.75) = ksi (t xz ) C = t Vz = 0 Internl Forces nd Moments: s shown on FD. F x = 0; N x = 0 N x = kip F y = 0; V y = 0 V y = 10.8 kip F z = 0; V z = 0 Mx = 0; T x (6) = 0 T x = 64.8 kip # ft My = 0; M y (6) = 0 Mz = 0; 10.8(6) + M z = 0 M y = 9.00 kip # ft M z = kip # ft 594

64 08 Solutions /28/10 8:34 M Pge 595 *8 64. Solve Prob for points E nd F. 12 ft 6 ft 150 lb/ft 2 3 ft E F C D z x y Internl Forces nd Moments: s shown on FD. F x = 0; N x = 0 N x = kip F y = 0; V y = 0 V y = 10.8 kip F z = 0; V z = 0 Mx = 0; T x (6) = 0 T x = 64.8 kip # ft M y = 0; M y (6) = 0 M y = 9.00 kip # ft Mz = 0; 10.8(6) + M z = 0 M z = kip # ft Section Properties: = p = p in 2 I y = I z = p = in 4 (Q C ) z = (Q D ) y = 0 (Q C ) y = (Q D ) z = 4(3) 3p c 1 2 (p)32 d - 4(2.75) 3p c 1 2 (p)2.752 d = in 3 J = p = in 4 Norml Stress: s = N - M zy I z + M y z I y s F = p - (-64.8)(12)(0) (12)(-3) = ksi = 17.7 ksi (C) s E = p - (-64.8)(12)(-3) (12)(0) = -125 ksi = 125 ksi (C) 595

65 08 Solutions /28/10 8:34 M Pge Continued Sher Stress: The trnverse sher stress in the z nd y directions nd the torsionl sher stress cn be obtined using the sher formul nd the torsion formul, t nd t twist = Tr V = VQ, respectively. It J (t xz ) E = -t twist = (12)(3) (t xy ) E = t Vy = 0 = ksi (t xy ) F = t Vy + t twist = 10.8( ) (2)(0.25) (12)(3) = 67.2 ksi (t xy ) F = t Vy = 0 596

66 08 Solutions /28/10 8:34 M Pge Determine the stte of stress t point on the cross section of the pipe t section in. 50 lb x y z 1 in. Section in. 12 in. Internl Lodings: Referring to the free - body digrm of the pipe s right segment, Fig., F y = 0; V y - 50 sin 60 = 0 V y = lb F z = 0; V z - 50 cos 60 = 0 V z = 25 lb Mx = 0; T + 50 sin 60 (12) = 0 My = 0; M y - 50 cos 60 (10) = 0 Mz = 0; M z + 50 sin 60 (10) = 0 T = lb # in M y = 250 lb # in M z = lb # in Section Properties: The moment of inerti bout the y nd z xes nd the polr moment of inerti of the pipe re I y = I z = p = in 4 J = p = in 4 Referring to Fig. b, Q y = 0 Q z = y 1 œ 1 œ - y 2 œ 2 œ = 4(1) 3p c p 2 12 d - 4(0.75) 3p c p d = in 3 Norml Stress: The norml stress is contributed by bending stress only. Thus, s = - M zy I z + M yz I y For point, y = 0.75 in nd z = 0. Then s = (0.75) = psi = 605 psi (T) Sher Stress: The torsionl sher stress developed t point is c t xz T d = Tr J = (0.75) = psi

67 08 Solutions /28/10 8:34 M Pge Continued The trnsverse sher stress developed t point is c t xy V d = 0 c t xz V d = V zq z = I y t 25( ) (2-1.5) = psi Combining these two sher stress components, t xy = 0 t xz = c t xz T d - c t xz V d = = 327 psi 598

68 08 Solutions /28/10 8:34 M Pge Determine the stte of stress t point on the cross section of the pipe t section in. 50 lb x y z 1 in. Section in. 12 in. Internl Lodings: Referring to the free - body digrm of the pipe s right segment, Fig., F y = 0; V y - 50 sin 60 = 0 F z = 0; V z - 50 cos 60 = 0 Mx = 0; T + 50 sin 60 (12) = 0 My = 0; M y - 50 cos 60 (10) = 0 Mz = 0; M z + 50 sin 60 (10) = 0 V y = lb V z = 25 lb T = lb # in M y = 250 lb # in M z = lb # in Section Properties: The moment of inerti bout the y nd z xes nd the polr moment of inerti of the pipe re I y = I z = p = in 4 J = p = in 4 Referring to Fig. b, Q z = 0 Q y = y 1 œ 1 œ - y 2 œ 2 œ = 4(1) 3p c p 2 12 d - 4(0.75) 3p c p d = in 3 Norml Stress: The norml stress is contributed by bending stress only. Thus, s = - M zy I z + M yz I y For point, y = 0 nd z = -1. Then s = (1) = psi = 466 psi (C) Sher Stress: The torsionl sher stress developed t point is c t xy T d = Tr C J = (1) = psi

69 08 Solutions /28/10 8:34 M Pge Continued The trnsverse sher stress developed t point is c t xz V d = 0 c t xy V d = V yq y = 43.30( ) = psi I z t (2-1.5) Combining these two sher stress components, t xy = c t xy T d - c t xy V d = = 422 psi t xz = 0 600

70 08 Solutions /28/10 8:34 M Pge The eccentric force P is pplied t distnce from the centroid on the concrete support shown. Determine the rnge long the y xis where P cn be pplied on the cross section so tht no tensile stress is developed in the mteril. e y x P z 2 h 3 h 3 e y b 2 b 2 y Internl Lodings: s shown on the free - body digrm, Fig.. Section Properties: The cross-sectionl re nd moment of inerti bout the z xis of the tringulr concrete support re = 1 2 bh I z = 1 36 bh3 Norml Stress: The norml stress is the combintion of xil nd bending stress. Thus, s = N - M zy I z s = -P - Pe yy 1 2 bh 1 36 bh3 s = - 2P (1) bh 3 h2 + 18e y y Here, it is required tht s nd s. For point, y = h 0 0, Then. Eq. (1) gives 3 0 Ú - 2P bh 3 ch2 + 18e y h 3 bd 0 h 2 + 6he y e y Ú - h 6 For Point, y = - 2. Then. Eq. (1) gives 3 h 0 Ú - 2P bh 3 ch2 + 18e y hbd 0 h 2-12he y e y h 12 Thus, in order tht no tensile stress be developed in the concrete support, e y must be in the rnge of - h 6 e y h

71 08 Solutions /28/10 8:34 M Pge 602 *8 68. The br hs dimeter of 40 mm. If it is subjected to force of 800 N s shown,determine the stress components tht ct t point nd show the results on volume element locted t this point. 150 mm I = 1 4 p r4 = 1 4 (p)(0.024 ) = (10-6 ) m mm z = p r 2 = p( ) = (10-3 ) m 2 Q = y = 4 (0.02) 3p s = P + Mz I (0.02)2 bp b = (10-6 ) m 3 2 x 30 y 800 N = (10-3 ) + 0 = MP t = VQ I t = (5.3333)(10-6 ) = MP (10-6 )(0.04) Solve Prob for point. 150 mm 200 mm z x 30 y 800 N I = 1 4 p r4 = 1 4 (p)(0.024 ) = (10-6 ) m 4 = p r 2 = p( ) = (10-3 ) m 2 Q = 0 s = P - Mc I = (10-3 ) (0.02) = MP (10-6 ) t = 0 602

72 08 Solutions /28/10 8:34 M Pge The 4 -in.-dimeter shft is subjected to the loding shown. Determine the stress components t point. Sketch the results on volume element locted t this point. The journl bering t C cn exert only force components C y nd C z on the shft, nd the thrust bering t D cn exert force components D x, D y, nd on the shft. D z x C 125 lb 2 in. 8 in. 10 in. z 8 in. 20 in. 125 lb 2 in. D y 20 in. = p 4 (0.752 ) = in 2 I = p 4 ( ) = in 4 Q = 0 t = 0 s = M y c I = -1250(0.375) = ksi = 30.2 ksi (C) Solve Prob for the stress components t point. 125 lb z D x C 2 in. 10 in. 8 in. 8 in. 20 in. 125 lb 2 in. y 20 in. = p 4 (0.752 ) = in 2 I = p 4 ( ) = in 4 Q = y = 4(0.375) 3p 1 2 b(p)( ) = in 3 s = 0 t = V zq I t = 125( ) = ksi (0.75) 603

73 08 Solutions /28/10 8:34 M Pge 604 *8 72. The hook is subjected to the force of 80 lb. Determine the stte of stress t point t section. The cross section is circulr nd hs dimeter of 0.5 in. Use the curved-bem formul to compute the bending stress. 1.5 in lb The loction of the neutrl surfce from the center of curvture of the hook, Fig., cn be determined from R = L d r where = p( ) = p in 2 L d r = 2pr - 2r 2 - c 2 = 2p = in. Thus, R = p = in. Then e = r - R = = in. Referring to Fig. b, I nd Q re I = p 4 (0.254 ) = (10-3 )p in 4 Q = 0 Consider the equilibrium of the FD of the hook s cut segment, Fig. c, ; + F x = 0; N - 80 cos 45 = 0 N = lb + c F y = 0; 80 sin 45 - V = 0 V = lb + M o = 0; M - 80 cos 45 ( ) = 0 M = lb # in The norml stress developed is the combintion of xil nd bending stress. Thus, s = N + M(R - r) e r Here, M = lb # in since it tends to reduce the curvture of the hook. For point, r = 1.5 in. Then s = (98.49)( ) p p( )(1.5) = 9.269(10 3 ) psi = 9.27 ksi (T) The sher stress in contributed by the trnsverse sher stress only. Thus t = VQ It = 0 The stte of strees of point cn be represented by the element shown in Fig. d. 604

74 08 Solutions /28/10 8:34 M Pge The hook is subjected to the force of 80 lb. Determine the stte of stress t point t section.the cross section hs dimeter of 0.5 in. Use the curved-bem formul to compute the bending stress. 1.5 in lb The loction of the neutrl surfce from the center of curvture of the the hook, Fig., cn be determined from R = L d r Where = p( ) = p in 2 L d r = 2p r - 2r 2 - c 2 = 2p = in. Thus, R = p = in Then e = r - R = = in Referring to Fig. b, I nd Q re computed s I = p 4 (0.254 ) = (10-3 )p in 4 Q = y = 4(0.25) 3p c p 2 (0.252 ) d = in 3 Consider the equilibrium of the FD of the hook s cut segment, Fig. c, ; + F x = 0; N - 80 cos 45 = 0 N = lb + c F y = 0; 80 sin 45 - V = 0 V = lb + M o = 0; M - 80 cos 45 ( ) = 0 M = lb # in The norml stress developed is the combintion of xil nd bending stress. Thus, s = N + M(R - r) e r Here, M = lb # in since it tends to reduce. the curvture of the hook. For point, r = 1.75 in. Then s = (98.49)( ) p p ( )(1.75) = 1.62 psi (T) The sher stress is contributed by the trnsverse sher stress only. Thus, t = VQ It = ( ) (10-3 )p (0.5) = 3.84 psi The stte of stress of point cn be represented by the element shown in Fig. d. 605

75 08 Solutions /28/10 8:34 M Pge The block is subjected to the three xil lods shown. Determine the norml stress developed t points nd. Neglect the weight of the block. 100 lb 50 lb 5 in. 3 in. 2 in.4 in. 250 lb 2 in. 5 in. M x = -250(1.5) - 100(1.5) + 50(6.5) = -200 lb # in. M y = 250(4) + 50(2) - 100(4) = 700 lb # in. I x = 1 12 (4)(133 ) b(2)(33 ) = in 4 I y = 1 12 (3)(83 ) b(5)(43 ) = in 4 = 4(13) + 2(2)(3) = 64 in 2 s = P - M y x I y + M x y I x s = (4) -200 (-1.5) = psi s = (2) -200 (-6.5) = psi 606