A hollow steel member has an inside diameter of 50mm and a wall thickness of 5mm at a length
of 1500mm and a modulus of elasticity of 210GN/M^2. If the member is subjected to a tensile
load of 5KN determine each of the following changes in dimension 1. Elongation 2. change in
thickness. Taking poissons ratio to be 0.4 show all steps and calculations
To determine the changes in dimension of the hollow steel member, we can use the following
formulas:
Elongation (δ) = PL/AE
Change in thickness (Δt) = -νδt
Where P is the tensile load, L is the length of the member, A is the cross-sectional area, E is the
modulus of elasticity, ν is Poisson's ratio, and t is the original thickness of the member.
Given data:
Inside diameter of the hollow steel member = 50mm
Wall thickness of the hollow steel member = 5mm
Length of the hollow steel member = 1500mm
Modulus of elasticity of the steel member = 210GN/m^2
Tensile load = 5KN
Poisson's ratio (ν) = 0.4
We need to calculate the cross-sectional area and original thickness of the member before
calculating the elongation and change in thickness.
Cross-sectional area:
The outside diameter of the member can be calculated as follows: Outside diameter = Inside
diameter + 2 × Wall thickness
= 50mm + 2 × 5mm= 60mm
The cross-sectional area can be calculated using the formula for the area of a hollow cylinder:
A = π/4 × (D^2 - d^2)
where D is the outside diameter and d is the inside diameter.
A = π/4 × (60^2 - 50^2)
= 942.48 mm^2
Original thickness:
The original thickness (t) of the member is given as 5mm.
Now we can substitute the given data into the formulas to calculate the elongation and change in
thickness:
Elongation:
δ = PL/AE
= (5 × 10^3 N) × (1500mm) / (942.48mm^2 × 210 × 10^9 N/m^2)
= 0.0000447mm
Therefore, the elongation of the hollow steel member is 0.0000447mm.
Change in thickness:
Δt = -νδt
= -0.4 × (0.0000447mm) × (5mm)
= -0.00000089mm
Therefore, the change in thickness of the hollow steel member is -0.00000089mm. Note that the
negative sign indicates that the thickness of the member has decreased under the tensile load.