The aircraft member is made of aluminum (6 = 27.0 GPa) and have a length of 1.00

Question

The aircraft member is made of aluminum (6 = 27.0 GPa) and have a length of 1.00 m. Assume that the stress concentrations at corners are negligible. A torque T = 9.50 kN-m is applied to the torsion member whose cross section is shown in Figure below. Where does the maximum shear stress occur in the member? Explain why. Determine the maximum shear stress in the member. Determine the shear stress (tau_AB, tau_BC, tau_CD, and tau_DA) in the each segment (AB, BC, CD, and DA) of the cross section. Determine the rate of twist (theta, the angle of twist per unit length) and the angle of twist (beta).

Explanation / Answer

solution:

1)here for given alluminium section subjected torsion then governing equation is

T/J=G*angle/L=t/R

2)here polar moment inertia around z axis is

max dimension 64*83 and minimum 77*53

J=bh/12(b^2+h^2)

J=(64*83/12)(64^2+83^2)-(77*56/12)(77^2+56^2)=1605336.667 mm4

2)here maximum shear stress would be at farthest distance of 83 mm/2=41.5 mm

hence maximum stress is

t=T*R/J=9500*41.5/1605336.667=.2455 MPa

3)where stresses in member Ab and CD are

tab=tcd=T*R/J=9500*40/J=.2367 MPa

4)where stresses in member BC and AD are

tad=tbc=9500*30/J=.1775 MPa

5)here rate of angle of twist and angle turned is givenby

angle/L=t/R*G

for BC and AD are

angle/L=.1775/27000*30=2.1919*10^-7 rad/m

angle=2.1917*10^-7 rad

for memebr

AD and BC are

angle/L=.2367/27000*40=2.1916*10^-7 rad/m

angle=2.1916*10^-7 rad

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