Question

4. (36 pts) A certain force F applied at D near the end of the lever shown in the figure results in certain stresses in the cantilever. The cantilever is of AISI 1035 steel, forged and heat-treated and it has yield strength of 81 kpsi. At normal temperature the material is a ductile material; hence, the stress concentration at shoulder A need not be considered. Neglect the transverse shear stress. We want to determine the maximum F that you can applied such that the cantilever OABC does not yield. в с 12 1" D. Cantilever shoulder 15" (a) (12 pts) Draw a free body diagram of cantilever OABC. At C, please indicate what kind of loads is applied to the cantilever and express these loads in terms of "F" (b) (3x2 pts) For a small element E at A on the top surface, what is the bending stress and shear stress due to F? (e) (3x2 pts) For a small element F at 0 on the top surface, what is the bending stress and shear stress due to F? Lever

Answer 1

F 6 Torque = Fx15" ra) E due to F (6) Bending stress at & shear stres - Bending - I = second moment of area. At section E =M 7 I o= 14x FX 2 X 3 At section a = 14 x 32x F T - 142.67F. ① Bending at F o=16*F* 2 8 x64 0 - 48.3137 F

② Shear stres at E T= 15FX 12 x 32 TX 14 . 7. = 76.433 F © Bened shear stres at F 7= 155 * ² x 32 = 22.64F * Both shear stress and Bending stres is maximum E at Point For Using calculation of F. maximum distortion Energy theory (It) permissible = / 6x² + 372 81 kpsi = √ (142.67 F)² + 3 (76.4394 F =. 173.2 lb.