a steel bar 10m long and 10cm*10cm in section .it is subjected to an axial pull of 250KN .determine the intensities of normal and tangential stresses on a plane section at 60 degree to the longitudinal axis .

normal stress

σ = F/A = 250000/0.1•0.1 = 2.5•10⁷ N/m²
Tangential stress
τ = F /(A/cosα) = Fcosα/A =
=250000•0.5/0.1•0.1=1.25•10⁷ N/m²

If I am not mistaken, tangential stress on a plane section at 60° to the longitudinal axis would be the shear stress, γ, and the normal stress on the same plane is the normal stress, σ.

σx=250 kN / (0.1²) m²
=25 MPa
σy=0
α=60°

Using the Mohr circle, we have
Tangential stress
=(σx-σy)/2 * sin(2*α)
=25/2 * sin(2*60°) MPa
=6.25√3 MPa

Normal stress
=((σx+σy)/2 + (σx-σy)/2 * cos(2*α)
= 12.5 + 12.5cos(2*60°)
=6.25 MPa

Sorry, typo above:

shear stress should have read τ. γ would have been shear strain.

To determine the intensities of normal and tangential stresses on a plane section at an angle of 60 degrees to the longitudinal axis of the steel bar, we can use the concepts of normal stress and shear stress.

First, let's find the normal stress on the plane section.
The axial pull force acting on the steel bar is given as 250 kN. To convert it to Newton (N), multiply it by 1000.
So, the axial pull force, F = 250 kN * 1000 = 250,000 N.

The cross-sectional area of the steel bar can be found by multiplying the width and height of the section. In this case, it is 10 cm * 10 cm = 100 cm^2. To convert it to meters squared (m^2), divide it by 10000 (since 1 m^2 = 10000 cm^2).
So, the cross-sectional area, A = 100 cm^2 / 10000 = 0.01 m^2.

Using these values, we can calculate the normal stress, σ = F / A.
Plugging in the values:
σ = 250,000 N / 0.01 m^2 = 25,000,000 N/m^2.

Next, let's find the tangential (shear) stress on the plane section.
The tangential stress is given by the equation τ = σ × sin(2θ).
Here, θ = 60 degrees, so in radians, θ = 60 * π / 180 = π / 3.

Plugging in the values:
τ = 25,000,000 N/m^2 × sin(2 × π/3) = 25,000,000 N/m^2 × sin(4π/3) ≈ -21,650,635 N/m^2.

Note that the negative sign indicates that the shear stress is in the opposite direction to the normal stress.

Therefore, the intensities of the normal and tangential stresses on the plane section at 60 degrees to the longitudinal axis are approximately 25,000,000 N/m^2 and -21,650,635 N/m^2, respectively.