Answer:
SHEAR FORCE = 257.11 N
Explanation:
given data:
depth d = 7 cm
height h = 3 cm
P = 400 N
Area of cross section normal to P = 30*70 = 2100 mm2
Area of cross section (inclined portion) = A'
[tex]sin\theta =\frac{A}{A^{'}}[/tex]
[tex]A^{'}= \frac{A}{sin\theta}[/tex]
[tex]A^{'}= \frac{2100}{sin50}[/tex]
A' =2741.35 mm2
Taking summation of vertical force
[tex]V-Pcos\theta = 0[/tex]
v = 400*cos50° = 257.11 N
Average shear force = [tex]\frac{v}{A'}[/tex]
Average shear force = [tex]\frac{257.11}{2741.35} = 0.093 MPa[/tex]
