Answer:
Part a: The single shear is 17.0 MPa
Part b: The double shear is 8.5 MPa
Explanation:
The free body diagram is given as attached.
From the FBD, the
value of θ is given as
[tex]\theta =tan^{-1} \frac{0.78}{2.3}\\\theta =18.73[/tex]
Taking moment about A gives
[tex]\sum M_A=0[/tex]
[tex]T_x \times 0.82 +T_y \times 2.3 -3500 \times 2.3=0\\0.82T cos (18.73) +2.3Tsin(18.73)-8050=0\\T=5313.09\, N[/tex]
Now applying balance of Forces
[tex]\sum F_x =0 \, ,\, \sum F_y=0[/tex]
For Forces along y axis
[tex]A_y-3500+5313.09sin(18.73)=0\\A_y=1794 N[/tex]
Similarly Forces along x axis
[tex]A_x-5313.09 cos(18.73) =0\\A_x=5031.72 N[/tex]
Now A is calculated as
[tex]A=\sqrt{A_x^2+A_y^2}\\A=\sqrt{(5031.72)^2+(1794)^2}\\A=5341.97 N[/tex]
On basis of this value of A, the single shear is given as
[tex]J=\frac{A}{\frac{\pi d^2}{4}}\\J=\frac{5341.97}{\frac{\pi (20 \times 10^{-3})^2}{4}}\\J=17.0 \, MPa \\[/tex]
The single shear is 17.0 MPa
On basis of this value of A, the single shear is given as
[tex]J=\frac{A}{\frac{\pi d^2}{4}}\\J=\frac{5341.97}{\frac{\pi (20 \times 10^{-3})^2}{2}}\\J=8.5\, MPa \\[/tex]
The double shear is 8.5 MPa
