A steel bar 10 m long and 10 cm * 2.5 cm in section. It is subjected to axial pull of 250 KN. Determine the intensities of normal and tangential stresses on a plane section at 60 degree to the longitudinal axis.
To determine the intensities of normal and tangential stresses on a plane section at a specific angle to the longitudinal axis, we can use the equations for normal and tangential stresses due to an axial load.
The normal stress, also known as the axial stress, can be calculated using the formula:
σ = F / A
where:
σ = normal stress (intensity of stress)
F = axial force (pull) acting on the bar
A = cross-sectional area of the bar
The tangential stress can be calculated using the formula:
τ = F / A * tan(2θ)
where:
τ = tangential stress (intensity of stress)
F = axial force (pull) acting on the bar
A = cross-sectional area of the bar
θ = angle between the plane section and the longitudinal axis
Given information:
Length of the bar (L) = 10 m
Cross-sectional dimensions of the bar:
Width (W) = 10 cm = 0.1 m
Height (H) = 2.5 cm = 0.025 m
Axial force (F) = 250 kN = 250,000 N
Angle (θ) = 60 degrees
First, calculate the cross-sectional area (A):
A = W * H
A = 0.1 m * 0.025 m
A = 0.0025 m²
Next, calculate the normal stress (σ):
σ = F / A
σ = 250,000 N / 0.0025 m²
σ = 100,000,000 N/m² or Pa
Finally, calculate the tangential stress (τ):
Convert the angle from degrees to radians:
θ_radians = θ * π / 180
θ_radians = 60 degrees * π / 180
θ_radians = π / 3 radians
τ = F / A * tan(2θ_radians)
τ = 250,000 N / 0.0025 m² * tan(2 * π / 3)
τ ≈ 108,253.98 N/m² or Pa
Therefore, the intensities of normal and tangential stresses on the plane section at 60 degrees to the longitudinal axis are approximately:
Normal stress (σ) = 100,000,000 N/m² or Pa
Tangential stress (τ) ≈ 108,253.98 N/m² or Pa
To determine the intensities of normal and tangential stresses on a plane section at a 60 degree angle to the longitudinal axis of the steel bar, we can use the equations for normal and tangential stress.
The normal stress, denoted by σ, is the force divided by the cross-sectional area:
σ = F / A
where F is the axial pull force and A is the cross-sectional area of the bar.
Given:
Length of the bar (L) = 10 m
Width of the bar (w) = 10 cm = 0.1 m
Height of the bar (h) = 2.5 cm = 0.025 m
Axial pull force (F) = 250 KN = 250,000 N
First, we need to calculate the cross-sectional area (A):
A = w * h = 0.1 m * 0.025 m = 0.0025 m²
Next, we can substitute the values into the equation for normal stress:
σ = F / A = 250,000 N / 0.0025 m² = 100,000,000 N/m² = 100 MPa
So, the intensity of the normal stress on the plane section is 100 MPa.
To calculate the tangential stress, denoted by τ, we can use the formula:
τ = σ * sin(2θ)
where θ is the angle between the plane section and the longitudinal axis. In this case, θ = 60 degrees.
Substituting the values into the equation for tangential stress:
τ = 100 MPa * sin(2 * 60) = 100 MPa * sin(120) ≈ 100 MPa * 0.866 = 86.6 MPa
Therefore, the intensity of the tangential stress on the plane section is approximately 86.6 MPa.