Q. 2.24

A bar ABC consists of two parts AB and BC each of 1 m length. Both the portions are square in section. Portion AB is 15 mm square while portion BC is 20 mm square. The bar is suspended vertically from A with a clearance of 1 mm between the lower end c and rigid horizontal support as shown in Figure 2.37. A load of 100 kN is applied at B. Find the stress in AB and BC and the reactions at supports. E = 200 × 10³ kN/mm².

Verified Solution

When the load of 100 kN is applied at B, AB will elongate. Initially, there would be no stress in BC. But as C moves forward, it will push against the lower support. Then, BC goes for compression. Let stress in AB be $\sigma_{ AB }$ MPa (tensile) and stress in BC be $\sigma_{ BC }$ MPa (compression).

Area of section AB = 15 × 15 = 225 mm²
Area of section BC = 20 × 20 = 400 mm²

Force in AB + Force in BC = 100 × 10³

\begin{aligned} & \sigma_{ AB } A_{ AB }+\sigma_{ BC } A_{ BC }=100 \times 10^3 \\ & 225 \sigma_{ AB }+400 \sigma_{ BC }=100 \times 10^3 \end{aligned}

$\sigma_{ AB }+1.8 \sigma_{ BC }=444.44$            (1)

Also,

Elongation of AB – Compression of BC = 1 mm

(Note: if there were to be no gap, then, elongation of AB = Compression of BC)

∴              $\frac{\sigma_{ AB } \times 1000}{200 \times 10^3}-\frac{\sigma_{ BC } \times 1000}{200 \times 10^3}=1$

$\sigma_{ AB }-\sigma_{ BC }=200$            (2)

Equation (1) – (2) gives, $2.8 \sigma_{ BC }=244.44, \quad \therefore \quad \sigma_{ BC }=87.3 MPa$

Substituting $\sigma_{ BC } \text { in Eq. (2), we get } \sigma_{ AB }=287.3 MPa$

Reaction at upper support = Force in portion AB

$=\sigma_{ AB } \times A_{ AB }=287.3 \times 225=64642.5 N =64.64 kN \text { (Pull) }$