As shown in figure 10 N force is applied at two ends of a rod. Calculate tensile stress and searing stress for section PR, Area of cross section PQ is 10 `cm^(2).`

Explanation

Suppose cross section PQ having area A and PR having A.

`therefore A. =(A)/(cos 30 ^(@)) = (10 xx 10 ^(-4))/( sqrt 3 // 2 ) `
`therefore A. = ( 2 xx 10 ^(-3))/( sqrt3) m ^(2) " "...(1)`
Here, angle between force `vecF` and ara of A. cross-section PR is `30^(@).` Force `vecF` has two component
`(i) F cos theta,` parallel component to A. which provides tension
`therefore F_(l) = F cos 30^(@)`
`=10 xx (sqrt3)/( 2)`
`= 5 sqrt3 N " "...(2)`
(ii) `F sin theta,` perpendicular component to A., it is tangential
`therefore F _(t) = F sin 30^(@)`
`=10 xx 1/2`
`therefore F_(t) = 5 N " "...(3)`
Tensile stress `sigma _(1) = ("Tensile force "(F _(t)))/(A.)`
`= ( 5 sqrt3 xxsqrt3)/(2xx10 ^(-3)`
(From equation (1) and (2))
`therefore sigma _(l ) = 7.5 xx 10 ^(+3) N //m ^(2)`
Tangential (shear) stress
` sigma _(t) = ("Tangential Force" (F _(t)))/(A.)`
`= ( 5 xx sqrt3)/( 2 xx 10 ^(3)) ` (From equation (1) and (3))
`therefore sigma _(t) = ( 5 sqrt3 xx 10 ^(3))/( 2) = 4.33 xx 10 ^(3) N//m ^(2)`