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Total elongation of stepped-load steel bar

Short answer

Determine axial forces: AB 90 kN, BC 70 kN, CD 40 kN. Sum segment elongations Δ = NL/AE using E = 200 GPa and A = 600 mm². Total elongation = 1.396 mm.

Find the total elongation in the bar. Take E for the material as the 200 GPa. A Steel bar of 600 mm2600\ \text{mm}^2 cross-sectional area is carrying loads as shown in the figure given below. [Figure: bar A-B-C-D; 90 kN acting to the left at A, 20 kN acting to the right at B, 30 kN acting to the right at C, 40 kN acting to the right at D; A-B 750 mm, B-C 1000 mm, C-D 750 mm.]

Loads: 90 kN (←) at A, 20 kN (→) at B, 30 kN (→) at C, 40 kN (→) at D. Check: 20 + 30 + 40 = 90 kN, so the bar is in equilibrium.

Axial force in each portion (tension +ve)

 90 kN <--A-----750-----B--------1000--------C----750----D--> 40 kN
                        |                    |
                     20 kN -->            30 kN -->

Using the section method from the left end:

PortionLength (mm)Axial force
AB750+90+90 kN (tension)
BC100090−20=+7090 - 20 = +70 kN (tension)
CD75070−30=+4070 - 30 = +40 kN (tension)

Elongation

δ=1AE∑NiLi=90×103(750)+70×103(1000)+40×103(750)600×200×103=167500000120000000=1.3958 mm\begin{aligned} \delta &= \frac{1}{AE}\sum N_iL_i = \frac{90\times10^3(750)+70\times10^3(1000)+40\times10^3(750)}{600\times200\times10^3} = \frac{167500000}{120000000} = 1.3958\ \text{mm} \end{aligned}

Portion by portion: δAB=0.5625\delta_{AB} = 0.5625 mm, δBC=0.5833\delta_{BC} = 0.5833 mm, δCD=0.2500\delta_{CD} = 0.2500 mm, total =1.3958= 1.3958 mm.

Answer: total elongation = 1.396 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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