Using strain energy method, the vertical deflection at O is (take modulus of elasticity, E = 2 × 10^{5} N/mm^{2}, cross-sectional area of wire, A = 100 mm^{2}).

This question was previously asked in

MPPSC AE CE 2016 Official Paper - II

Option 1 : 2.12 mm

CT 3: Building Materials

3014

10 Questions
20 Marks
12 Mins

**Concept:**

The strain energy stored in truss,

U = U_{OA} + U_{OB}

\(U = \frac{{P_1^2L}}{{2AE}} + \frac{{P_2^2L}}{{2AE}}\)

The deflection at O will be, \(\Delta = \frac{{\partial U}}{{\partial P}}\)

**Calculation:**

At joint O

ΣF_{x} = 0, F_{OA} = F_{OB}..............(i)

ΣF_{y} = 0, F_{OA }× cosθ + F_{OB} × cosθ = P ..........(ii)

2F_{OA} × cosθ = P

Here, cosθ = 3/4.24 and P = 10 kN

\({F_{OA}} = \frac{P}{{2\cos \theta }} = \frac{P}{{2 \times \frac{3}{{4.24}}}} = 0.7067P\)

The strain energy stored in truss,

U = UOA + U_{OB}

\(U = \frac{{F_{OA}^2L}}{{2AE}} + \frac{{F_{OB}^2L}}{{2AE}} = \frac{{2F_{OA}^2L}}{{2AE}} = \frac{{F_{OA}^2L}}{{AE}}\)

⇒ \(U = \frac{{{{(0.7067P)}^2}L}}{{AE}} = \frac{{0.5{P^2}L}}{{AE}}\)

The deflection at O will be, \(\Delta = \frac{{\partial U}}{{\partial P}}\)

\(\Delta = \frac{{0.5 \times 2PL}}{{AE}} = \frac{{PL}}{{AE}}\)

\(\Delta = \frac{{10 \times {{10}^3} \times 4.24 \times {{10}^3}}}{{100 \times 2 \times {{10}^5}}} = 2.12\ mm\)