A cantilever of length 6 m carries a point load of 48 kN at its centre. The cantilever is propped rigidity at the free end. Determine the reaction at the rigid prop.

Question

Question

This question was previously asked in

JKSSB JE CE 2021 Official Paper Shift 1 (Held on 27 Oct 2021)

Answer 1

Explanation:

Here the prop end will resist the deflection of the B end, hence there will generate a reaction, let it be RB.

The load is at midspan,

Hence, L1 = L2 = L/2

F1 Shraddha Chandramaouli 17.06.2021 D8

Deflection at the free end due to load on midspan = δ1 + δ2

δ1 = \(\frac{1}{3}× \frac{WL_1^3 }{EI}\)

= \(\frac{1}{3}× \frac{W(\frac{L}{2})^2 }{EI}\) [Putting L1 = L/2]

= \(\frac{1}{24}× \frac{WL^3 }{EI}\)

δ2 = Slope due load at midspan × L/2

= \(\frac{1}{2}× \frac{WL_2^2 }{EI}\) × L/2

= \(\frac{1}{16}× \frac{WL^3 }{EI}\)[Putting L2 = L/2]

Now, the prop installed won't allow the end to sink down.

Let, R_B is the reaction at end B.

∴ \(\frac{R_B \times L^3}{3EI} = \frac{W \times L^3}{24EI} + \frac{W\times L^3}{16EI}\)

⇒ R_B = \(\frac{5W}{16}\)

Calculation:

Given:

W = 48 kN

\(R_B=\frac{5W}{16} \Rightarrow R_B=\frac{5 \times 48}{16} = 15 ~kN\)