Question #290500

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.

1
Expert's answer
2022-01-26T01:37:02-0500

Solution;

The figure is as shown below;



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

Let R1R_1 be the reaction at point A and R2R_2 be the reaction at point D.

R2=150+90R1R_2=150+90-R_1

The axial forces on the sections can be expressed as;

PAB=R1P_{AB}=R_1

PBC=150R1P_{BC}=150-R_1

PCD=240R1P_{CD}=240-R_1

The strain in a member is given as;

δ=PLAE\delta=\frac{PL}{AE}

Since the walls are rigid;

δAB=δBC+δCD\delta_{AB}=\delta_{BC}+\delta_{CD}

But E is the same for all sections;

R1(500)900=(150R1)2502000+(240R1)3501200\frac{R_1(500)}{900}=\frac{(150-R_1)250}{2000}+\frac{(240-R_1)350}{1200}

Simplify;

59R1=18(150R1)+724(240R1)\frac59R_1=\frac18(150-R_1)+\frac7{24}(240-R_1)

R1=91.285kNR_1=91.285kN

The axial force in each segments are;

PAB=R1=91.285kNP_{AB}=R_1=91.285kN

PBC=150R1=58.715kNP_{BC}=150-R_1=58.715kN

PCD=240R1=148.715kNP_{CD}=240-R_1=148.715kN





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