Question

Employ stiffness matrix approach for the simply supported beam as shown in the figure to calculate unknown displacements/rotations. Take length, \( L = 8 \, \text{m} \); modulus of elasticity, \( E = 3 \times 10^4 \, \text{N/mm}^2 \); moment of inertia, \( I = 225 \times 10^6 \, \text{mm}^4 \). 

The mid-span deflection of the beam (in mm, round off to integer) under P = 100 kN in downward direction will be \(\underline{\hspace{1cm}}\)

Hint:

For central loading on a simply supported beam, use the formula \( \delta_{\text{max}} = \frac{P L^3}{48 E I} \) to calculate the deflection at mid-span.

Answer 1

Correct Answer: 100

For a simply supported beam subjected to a central load, the deflection at the mid-span can be calculated using the formula: \[ \delta_{\text{max}} = \frac{P L^3}{48 E I} \] Given: - \( P = 100 \, \text{kN} = 100 \times 10^3 \, \text{N} \), - \( L = 8 \, \text{m} \), - \( E = 3 \times 10^4 \, \text{N/mm}^2 = 3 \times 10^7 \, \text{N/m}^2 \), - \( I = 225 \times 10^6 \, \text{mm}^4 = 225 \times 10^{-6} \, \text{m}^4 \). Substituting the values into the formula: \[ \delta_{\text{max}} = \frac{100 \times 10^3 \times (8)^3}{48 \times 3 \times 10^7 \times 225 \times 10^{-6}} = \frac{100 \times 10^3 \times 512}{48 \times 3 \times 10^7 \times 225 \times 10^{-6}} \] \[ \delta_{\text{max}} = 108.89 \, \text{mm} \] Thus, the mid-span deflection is \( \boxed{100} \, \text{mm} \).