Case (a): Ans Ans. Case (b): ; s 1 = 65(4) Ans. s 1 = pr t. = 1.04 ksi. Ans. s 2 = pr 2t ; s 2 = 65(4) = 520 psi

Transcription

1 8 3. 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) () s 1 = 1.04 ksi, s 2 = 0, (b) s 1 = 1.04 ksi, s 2 = 520 psi 734

2 *8 12. A pressure-vessel hed is fbricted by gluing the circulr plte to the end of the vessel s shown. If the vessel sustins n internl pressure of 450 kp, determine the verge sher stress in the glue nd the stte of stress in the wll of the vessel. 20 mm 450 mm 10 mm + c F y = 0; p(0.225) 2 450(10 3 ) - t vg (2p)(0.225)(0.01) = 0; t vg = 5.06 MP s 1 = p r t = 450(103 )(0.225) 0.02 = 5.06 MP s 2 = p r 2 t = 450(103 )(0.225) 2(0.02) = 2.53 MP 743

3 8 23. The clmp is mde from members A nd AC, which re pin connected t A. 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 A E 180 N There is no moment in this problem. Therefore, the compressive stress is produced by xil force only. s const = P A = 240 = 1.07 MP (0.015)(0.015) s const = 1.07 MP 755

4 8 30. The rib-joint pliers re used to grip the smooth pipe C. If the force of 100 N is pplied to the hndles, determine the stte of stress t points A nd on the cross section of the jw t section -. Indicte the results on n element t ech point. 25 mm 25 mm C mm 100 N A 10 mm Support Rections: Referring to the free-body digrm of the hndle shown in Fig., 20 mm 7.5 mm Section 100 N + M D = 0; 100(0.25) - F C (0.05) = 0 F C = 500 N Internl Lodings: Consider the equilibrium of the free-body digrm of the segment shown in Fig. b, F y = 0; V = 0 + MC = 0; M - 500(0.025) = 0 V = 500 N M = 12.5 N # m Section Properties: The moment of inerti of the cross section bout the centroidl xis is I = 1 12 (0.0075)(0.023 ) = 5(10-9 ) m 4 Referring to Fig. c, Q A nd Q re Q A = 0 Q = y A =0.005(0.01)(0.0075) = 0.375(10-6 ) m 3 Norml Stress: The norml stress is contributed by bending stress only. Thus s = My I For point A, y = 0.01 m. Then 12.5(0.01) s A = - 5(10-9 = -25 MP = 25 MP (C) ) For point, y = 0. Then s = 0 Sher Stress: The sher stress is contributed by trnsverse sher stress only. Thus, t A = VQ A It = 0 t = VQ It = 500[0.375(10-6 )] 5(10-9 )(0.0075) = 5 MP The stte of stress of points A nd re represented by the elements shown in Figs. d nd e respectively. 762

5 8 30. Continued s A = 25 MP (C), s = 0, t A = 0, t = 5 MP 763

6 8 37. 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, in bck, 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 A 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 A = pa = 25pA10-6 m 2 I z = p 4 A = pA10-9 m 4 J = p 2 A = pA10-9 m 4 Referring to Fig. b, Q is Q = y A = 4(0.005) 3p c p 2 A d = A10-9 m 3 Norml Stress: The norml stress is combintion of xil nd bending stress. Thus, For point, y = 0. Then s = N A - M zy I z s = pA = MP = 1.53 MP (C) 771

7 8 37. Continued Sher Stress: The trnsverse sher stress developed t point is c At xy V d = V yq = I z t 90c83.333A10-9 d pA10-9 (0.01) = MP The torsionl sher stress developed t point is c At xy T d = Tc J = 20(0.005) pA10-9 = MP Thus, At xy = 0 At xy = c At xy T d - c At xy V d = = MP = 100 MP The stte of stress t point is represented on the element shown in Fig. d. s = 1.53 MP (C), t = 100 MP 772

8 8 57. The sign is subjected to the uniform wind loding. Determine the stress components t points A nd on the 100-mm-dimeter supporting post. Show the results on volume element locted t ech of these points. z 2 m 1 m 3 m 1.5 kp C A D 2 m y Point A: x s A = Mc (T) I = 10.5(103 )(0.05) p = 107 MP 4 (0.05)4 t A = Tc J = 3(103 )(0.05) p = (10 6 ) = 15.3 MP 4 (0.05)4 Point : s = 0 t = Tc J - VQ It = (10 6 ) (4(0.05)/3p))(1 2 )(p)(0.05)2 p 4 (0.05)4 (0.1) t = 14.8 MP s A = 107 MP (T), t A = 15.3 MP, s = 0, t = 14.8 MP 800

9 8 66. Determine the stte of stress t point on the cross section of the pipe t section in. A 50 lb x y z 1 in. Section 60 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 V y = lb V z = 25 lb 10 in. 12 in. Mx = 0; T + 50 sin 60 (12) = 0 T = lb # in My = 0; M y - 50 cos 60 (10) = 0 M y = 250 lb # in Mz = 0; M z + 50 sin 60 (10) = 0 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 4 A = in 4 J = p 2 A = in 4 Referring to Fig. b, AQ z = 0 AQ y = y 1 œ A 1 œ - y 2 œ A 2 œ = 4(1) 3p c p 2 A12 d - 4(0.75) 3p c p 2 A0.752 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 At xy T d = Tr C J = (1) = psi

10 8 66. Continued The trnsverse sher stress developed t point is c At xz V d = 0 c At xy V d = V yaq y = 43.30( ) = psi I z t (2-1.5) Combining these two sher stress components, At xy = c At xy T d - c At xy V d = = 422 psi At xz = 0 s = 466 psi (C), t = 422 psi 813