How are the axial forces in the segments found?

Divide each load by the number of segments

Add all loads and use the same force everywhere

How is the total change in length obtained for the stepped bar?

Sum of F L / (A E) for the segments, with signs

Total load times total length / (smallest A x E)

Largest segment's elongation only

Average stress times total length

What is the reaction at A and why?

40 kN, from overall equilibrium (and equal to the force in AB)

20 kN, equal to the force in CD

60 kN, the largest applied load

120 kN, the sum of the magnitudes of all loads

Practice · 8 marks

Elongation of a stepped steel bar

Short answer

By equilibrium, axial forces: AB = +40 kN, BC = -20 kN, CD = +20 kN. Stresses: AB 56.6 MPa (T), BC -40.7 MPa (C), CD 63.7 MPa (T). Total elongation = Σ(FL/AE) = 0.311 mm. Reaction at A = 40 kN.

A steel bar ABCD is fixed at end A and free at end D. The segments are AB: 1.0 m long, 30 mm diameter; BC: 0.8 m long, 25 mm diameter; CD: 0.6 m long, 20 mm diameter. Axial loads act as follows: 60 kN pulling to the right at B, 40 kN pushing to the left at C, and 20 kN pulling to the right at D. Take E = 200 GPa. Find (a) the axial force and stress in each segment, (b) the total change in length of the bar, (c) the reaction at A.

Free-body and axial forces

 |A-----30mm-----B----25mm----C---20mm---D
 |<--- 1.0 m --->|<-- 0.8 m ->|<- 0.6 m ->
 R_A <---        60kN->   <-40kN     20kN->

Take forces to the right as positive. Cut each segment and balance the loads on the free side (right-hand side):

Segment	Axial force $F$	Nature
CD	$+20$ kN	Tension
BC	$+20 - 40 = -20$ kN	Compression
AB	$-20 + 60 = +40$ kN	Tension

(a) Stresses

$A_{AB} = \frac{\pi}{4}(30)^2 = 706.9\ \text{mm}^2$ , $A_{BC} = 490.9\ \text{mm}^2$ , $A_{CD} = 314.2\ \text{mm}^2$ .

Segment	$\sigma = F/A$ (MPa)
AB	$40000/706.9 = +56.6$
BC	$-20000/490.9 = -40.7$
CD	$20000/314.2 = +63.7$

(b) Total change in length

\delta = \sum \frac{F L}{A E}

\begin{aligned} \delta_{AB} &= \frac{40000\times1000}{706.9\times200000} = +0.283\ \text{mm}\\ \delta_{BC} &= \frac{-20000\times800}{490.9\times200000} = -0.163\ \text{mm}\\ \delta_{CD} &= \frac{20000\times600}{314.2\times200000} = +0.191\ \text{mm} \end{aligned}

\delta = 0.283 - 0.163 + 0.191 = +0.311\ \text{mm (elongation)}

(c) Reaction at A

Equilibrium of the whole bar: $-R_A + 60 - 40 + 20 = 0$ , so $R_A = 40$ kN acting to the left (equal to the force in AB).

Answer: forces $+40$ , $-20$ , $+20$ kN; stresses $56.6$ , $-40.7$ , $63.7$ MPa; total elongation $= 0.311$ mm; $R_A = 40$ kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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All Strength of Materials chapter 2 questions →

Elongation of a stepped steel bar

Free-body and axial forces

(a) Stresses

(b) Total change in length

(c) Reaction at A

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