Answer to Question #290500 in Mechanical Engineering for Alan Enrico V Tuib

Question

The steel rod is stress-free before the

axial loads P1 = 150 kN and P2 = 90 kN

are applied to the rod. Assuming that the

walls are rigid, calculate the axial force in

each segment after the loads are applied.

Use E = 200 GPa.

Answer 1

Solution;

The figure is as shown below;

From the diagrams; Section AB is under compression and Section BC and CD are under tension.

Let $R_1$ be the reaction at point A and $R_2$ be the reaction at point D.

$R_2=150+90-R_1$

The axial forces on the sections can be expressed as;

$P_{AB}=R_1$

$P_{BC}=150-R_1$

$P_{CD}=240-R_1$

The strain in a member is given as;

$\delta=\frac{PL}{AE}$

Since the walls are rigid;

$\delta_{AB}=\delta_{BC}+\delta_{CD}$

But E is the same for all sections;

$\frac{R_1(500)}{900}=\frac{(150-R_1)250}{2000}+\frac{(240-R_1)350}{1200}$

Simplify;

$\frac59R_1=\frac18(150-R_1)+\frac7{24}(240-R_1)$

$R_1=91.285kN$

The axial force in each segments are;

$P_{AB}=R_1=91.285kN$

$P_{BC}=150-R_1=58.715kN$

$P_{CD}=240-R_1=148.715kN$