Question

75 mm Question 1.7 Find the distance d and the shear force V of the object shown, if the shear strain is y=0.008 and G-45 GPa.. 100 mm 50 mm

Answer 1

Shear strain = 0.008

Shear stress/ shear strain = G

Shear stress / 0.008=45 X 10⁹N/m2

=45KN/mm^​2​​​​

Therefor shear stress = 0.008 X 45KN/mm^{​​​​​2​​​​.}

     And shear force = 0.008 X 45KN/mm^​​​2​​​​ X area of crossection

= 0.008X45KN/mm^{​​​​​2}​​​​ X 75 X 100

= 2700 KN

Now ,

d/50=0.008

d= 0.008 X 50 =0.4 mm