Question

Question 2 The single-cell thin-walled beam of Figure 2 with all dimensions in mm has a constant torque T-3150 Nm applied to it. The shear modulus for all skin section is 27.5 GPa. Determine (a) the shear flow distribution, (b) the maximum shear stress, and (c) the rate of twist. 13 225 75 3150 N.m 400 Fig. 2

Answer 1

Explanation:

Area enclosed by wall midline is given by A_E= 1/2(bxh)

                                                                =

                                                                 =45000mm²

a)The shear flow distribution

Since the Torque acts at the exact middle of the cross section,

$q=\frac{T}{2A_{E}}$

$q=\frac{3150}{2*0.045}$

    =35000 Ns/m

b) The maximum shear stress,

Since the Torque acts at the exact middle of the cross section, and thickness of the wall is constant throughout the section,

$\tau =\frac{q}{t}$

$\tau =\frac{35000}{0.00135}$

Maximum shear stress $\tau =$ 25.926 Mpa.

c) Rate of twist

Angle of twist is given by,

$\phi =\frac{T*L}{G*J}$

$rate\, of\, twist=\frac{\phi}{L} =\frac{T}{G*J}$

T=3150 Nm

G=27.5 Gpa

$J=\frac{4*A_{E}^{2}*t}{p}$

P(perimeter)=406.97+427.2+225

                  =1059.17 mm

$J=\frac{4*0.045^{2}*1.35}{1059.17*10^{-3}}$

J= 1.0324 x 10^-5 m⁴

$rate\, of\, twist=\frac{\phi}{L} =\frac{T}{G*J}$

                                     $=\frac{3150}{27.5*10^{9}*1.0324*10^{-5}}$

                                      =0.01109 rad/m.