Question

A solid shaft with 100-mm-diameter is made of cast iron and is subjected to the distributed and
concentrated loadings as shown in Figure 9. The shear modulus of cast iron is 39GPa.
(A) Determine the angle of twist at the free end A of the shaft due to these loadings. [9 marks]
(B) Determine the maximum shear stress in this shaft. [4 marks]
(C) Cast iron is a brittle material with an ultimate strength of 260MPa. Determine whether this
structure will fail or not. [2 marks]

A solid shaft with 100-mm-diameter is made of cast iron and is subjected to the distributed and concentrated loadings as shown in Figure 9. The shear modulus of cast iron is 39GPa. (A) Determine the angle of twist at the free end A of the shaft due to these loadings. [9 marks] (B) Determine the maximum shear stress in this shaft. [4 marks] (C) Cast iron is a brittle material with an ultimate strength of 260MPa. Determine whether this structure will fail or not. [2 marks] 2 kN.m/m 600 N·m B 400 N·m 0.8 m 0.6 m Figure 9.

Answer 1

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Soln Givena Clockwise torque =) +ve Anti clockwise torque =) -ve d= 100mm G= 39 GPa 1. = 39x10²MPa J = Td99.82x10 mmy 32 KNX 400NM GOON 0.8m 0.6m Torque Diagram TA 1400Nmt 400Nm ----+-200Nm Section AB: Torque at any section in AB = 400Nm (clockwise). At point B, there is concentrated moment of cooNm in

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ant clock wise direction. Therefore, there will be a shift in torque diagram vertically downward and by an amount of GooNm and reaches -200 Nm Section BC Consider a section xx from free end.. caule in Bc at a distance x² 2000 Nm 7400Nm Heeeeeece orm for T 4 R6oONM Then, torque at section xx is given by , Txe = +400-600+ 2000 (-0.6) inclined = 2000 (x-0.6) - 200 (straight line) At x = 1.4m (fixed end), T= 2000(14-0.6)-200 = 1400 Nm (clockcoise direction) torque at c will be 1400 Nm in anticlock- Therefore, reaction wise direction.

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(A) Angle of twist at free end A is given by, GJ/Авар – 400 * 0.6x1066-267x164 yad. IGJ AB 39X10°x 9.82x106 = GJBC. GJ = $(2000 (2-066) –200) dz G 0.6 Baklodą.semob (2000 (4-062)-200x** 3941039.82716 [O- (-480 5 7 - Nm2. = 480X106 39x10'49.82x106 = 1.2533x10"rad 04 = 6.267x104 + 1.2533x10-3 = 1.88x16" rad

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(B) Maximum shear stress will occur in the section at which there is maximum torque. In this case, maximum torque occur at fixed and c (1400 nm). Therefore, Maximum shear stress, Tmax = 16T T23 = 16x 1400X103 TX 1003 = 7.13 MPa 0 Ultimate strength of cast iron, Tu = 260MPa. Therefore, ultimate shear strength, Tour Cų - 260 = 130 MPa Since the maximum shear stress Tmax = 713MPa is less than 130MPa, the structure will not fail.