Question

A nonlinear structure has the following strain energy function: U=

L
AE
​
(L
2
+2Lx+L
x
2
+L
2

​
+a
2
) measured in joules, where x( m) is the displacement, AE=10
5
N is the section axial stiffness, and L=1.33 m. What is the force in kilonewtons when x=0.45 m ?

Answer 1

The force in a nonlinear structure when the displacement x is 0.45 m can be found by differentiating the given strain energy function U with respect to x and then substituting the given values to calculate the force, which is 356 kilonewtons (kN).

The question pertains to calculating the force in a nonlinear structure when a certain displacement is applied. The given strain energy function can be differentiated with respect to the displacement to find the force. This force is based on the concept that in a deformed system described by Hooke's law, the elastic potential energy (PEel) is given by the formula PEel = (1/2)kx² where x is the displacement, and k is the force constant, akin to the axial stiffness AE in the given problem.

Given:
Strain energy function, U = (AE/L)(L² + 2Lx + Lx² + a²)
AE = 10⁵ N (axial stiffness)
L = 1.33 m
x = 0.45 m (displacement)

To find the force, we differentiate the strain energy function with respect to displacement x:

dU/dx = AE(2L + 2x)
Force, F = dU/dx = 10⁵ N (2(1.33) + 2(0.45))

Substituting the values, F = 10⁵ N * (2.66 + 0.90)
F = 10⁵ N * 3.56
F = 356,000 N

To convert the force to kilonewtons (kN), divide by 1000:

F = 356 kN