Answer to Question #230028 in Mechanical Engineering for jubair

Question

The bent steel bar shown in figure is 201mm square. Determine the normal stresses at A and B.

Answer 1

$Cos\theta=\frac{a}{500}=\frac{4}{5}$

a=400mm

$Sec\theta=\frac{b}{400}=\frac{5}{4}$

b=500mm

$\sum M_1=0$

(a+b)R₂=(500+400+100)450

900R₂=1000*450

R₂=500kN

$Cos\theta =\frac{c}{200}=\frac{4}{5}$

c=160mm

M₃=500(201)-450(c+100)

=100500-450(160+100)

=-16500kNmm

M₃=16500kNmm clockwise

F_a=450 $Sin \theta=450(\frac{3}{5})$

=270kN

$\sigma_a=\frac{P}{A}=\frac{270(1000)}{201*201}$

=6.683MPa

$\sigma_f=\frac{6M}{bd^{2}}=\frac{6*16500*1000}{201*201^{2}}$

=12.19MPa

$\sigma_A=-\sigma_a+\sigma_f$

=-6.683+12.19

=5.507MPa

$\sigma_B=-\sigma_a-\sigma_f$

=-6.683-12.19

=-18.873MPa