Question

HW: 1-1-1 The beam AB is support at A by L shaped clips. Find the average resulting shear stress in the 20mm diameter bolt at A if the bolt is in (a) single shear, (b) double shear.

Answer 1

Part a: The single shear is 17.0 MPa

Part b: The double shear is 8.5 MPa

Step-by-step explanation:

The free body diagram is given as attached.

From the FBD, the

value of θ is given as

`\theta =tan^(-1) (0.78)/(2.3)\\\theta =18.73`

Taking moment about A gives

`\sum M_A=0`

`T_x * 0.82 +T_y * 2.3 -3500 * 2.3=0\\0.82T cos (18.73) +2.3Tsin(18.73)-8050=0\\T=5313.09\, N`

Now applying balance of Forces

`\sum F_x =0 \, ,\, \sum F_y=0`

For Forces along y axis

`A_y-3500+5313.09sin(18.73)=0\\A_y=1794 N`

Similarly Forces along x axis

`A_x-5313.09 cos(18.73) =0\\A_x=5031.72 N`

Now A is calculated as

`A=√(A_x^2+A_y^2)\\A=√((5031.72)^2+(1794)^2)\\A=5341.97 N`

On basis of this value of A, the single shear is given as

`J=(A)/((\pi d^2)/(4))\\J=(5341.97)/((\pi (20 * 10^(-3))^2)/(4))\\J=17.0 \, MPa \\`

The single shear is 17.0 MPa

On basis of this value of A, the single shear is given as

`J=(A)/((\pi d^2)/(4))\\J=(5341.97)/((\pi (20 * 10^(-3))^2)/(2))\\J=8.5\, MPa \\`

The double shear is 8.5 MPa