Answer:
a). α = 26.57
b). Maximum load is 50 .kN
Explanation:
a).
The normal force is given by
N = σ A cosec β
where, σ is the normal stress
A is the cross sectional area
Similarly, shear force is given by
S= τ A cosec β
where, τ is the shearing stress
Now from the figure,
tan β = S/N
= τ/σ
Therefore, [tex]$\beta = \tan^{-1}(2)$[/tex] = 63.43
α = 90 - β = 26.57
b).
The normal force is given by
[tex]$N=(100\times 10^6)(400\times 10^{-6}) \text{ cosec}\ 63.43$[/tex]
[tex]$N=44.78\times 10^3$[/tex] N
We have
[tex]$\Sigma F_y=0$[/tex]
∴ N - F sin β = 0
⇒ F = N / sin β
= [tex]$\frac{44.72\times 10^3}{\sin(63.43)} = 50\times 10^3 N$[/tex]
Similarly,
The shear force is given by
S = τ A cosec β
= [tex]$(50\times 10^6)(400\times 10^{-6}) \text{ cosec}\ 63.43 = 22.36\times 10^3 N$[/tex]
[tex]$\Sigma F_x=0$[/tex]
∴ S - F cos β = 0
⇒ F = S / cos β
[tex]$\frac{22.36\times 10^3}{\cos(63.43)} = 49.99\times 10^3 N$[/tex]
Therefore, force is 50 kN.
