Question

Problem 4.2: The mechanism below is in equilibrium for an applied load of P = 20 kN. Specifications for the mechanism limit the shear stress in the steel [G = 80 GPa) shaft BC to 70 MPa, the shear stress in bolt A to 100 MPa, and the vertical deflection of joint D to a maximum value of 25 mm. Assume that the bearings allow the shaft to rotate freely. Using L = 1,200 mm, a = 110 mm, and b= 210 mm, calculate: (a) the minimum diameter required for shaft BC. (b) the minimum diameter required for bolt A. Bolt Bearings

the modulus of elasticity is not given

Answer 1

Answer:

Given that,

Load, P = 20 kN

Shear stress, G = 70 MPa

Shear stress in bolt A = 100 MPa

Vertical deflection = 25 mm

L = 1200 m

a = 110 mm

b = 210 mm

a) > Displacement Diagram of- shaft DC. - b a) The angle at point a gind = 1 d = sin ) = sin(0.005) = sint (0 : 1190) = 6.9370 Free body diagram of lever DC . 2) The Distance x 2 = Cose gezocoso

=) The torque in the shaft, T= Px - Precos! = Pacos T = (0000) X. Co. 21) eos (6.3879 = 20000 x 0.21*0.9937 2.2170.13 Nm :) The polar moment of Inertia of the shaft BC J = 177ch here, ca oradius of the shaft Be. asistenta - Tablet Talu = 27 Tau) here, Tau is allowable shear stress of the shaft Bc. 2 X 1170.13713 Lax70x1067 (8340-26 7 / 2 19.918106

C=0.03359 m 5. que - RTL GAO - Lose - T2TL 14 CAGO here, Gts shear modulus, c 2. % 1170.13 X 1-2 ÄX80x109 10008.312 3880X10% (0.1103) c= 0:02m035m From the two dradiuses. ... The maximum radius should be consides, Because it satisfies both maximum shear and angle of twist contoria. max = 0.0339am - The diameter of the shaft Be d-RCmay

d = 2 8003359 = 0.06718 = 67.18mm ". The minimum diameter required for shaft Beis 67:18mm b) The force acting on the bolt. T= Fxa F = 1 - B170.13 0.110 = 37910.27N =) The diameter of the bolt a Toolt = F Abolt Id bott abolt = .- Тюнхт 37910. 2784 dbolt Oox7 M82.68 - M82.68 abolt 21.97mm ... The minimum diameterol the