JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III STRUCTURAL ANALYSIS - II Previous Question Paper (September - 2021 - R18)

For a two-hinged parabolic arch with span L = 10 m and rise h = 2.5 m, the equation of the arch is y = 4hx(L-x)/L² = 0.1x(10-x). The horizontal thrust H is determined by the condition of zero displacement at the support: H = (∫ M₀y dx) / (∫ y² dx), where M₀ is the bending moment of a simply supported beam.

For a UDL of w = 2 kN/m over the left half (0 ≤ x ≤ 5):

1. The bending moment M₀ = (wLx/4) - (wx²/2) for 0 ≤ x ≤ 5, and M₀ = (wL/2)(L-x)/2 for 5 ≤ x ≤ 10.
2. Calculating the integrals: ∫ M₀y dx = ∫[0,5] (5x - x²)(0.1x(10-x)) dx + ∫[5,10] (25 - 2.5x)(0.1x(10-x)) dx = 13.02 + 13.02 = 26.04.
3. The denominator ∫[0,10] y² dx = ∫[0,10] (0.1x(10-x))² dx = 0.01 ∫[0,10] (100x² - 20x³ + x⁴) dx = 0.01(10000/3 - 5000 + 10000/5) = 16.67.
4. Horizontal thrust H = 26.04 / 16.67 = 1.56 kN.

Vertical reactions at supports A (left) and B (right):

The bending moment at any section is M_x = M₀ - Hy. The maximum positive bending moment occurs in the loaded portion (x < 5), and the maximum negative moment occurs in the unloaded portion (x > 5). Calculating M_x at x = 2.5 m gives M = 3.75(2.5) - 2(2.5)²/2 - 1.56(0.1*2.5*7.5) = 9.375 - 6.25 - 2.925 = 0.2 kN-m. At x = 7.5 m, M = 1.25(2.5) - 1.56(0.1*7.5*2.5) = 3.125 - 2.925 = 0.2 kN-m. The maximum bending moment is 0.2 kN-m.

To analyze a portal frame using the Moment Distribution Method (MDM), follow these systematic steps:

1. Calculate Fixed End Moments (FEMs): Determine the fixed end moments for each member based on the applied loads. For a member of length L with a point load P at center, FEM = ±PL/8; for a UDL of w, FEM = ±wL²/12.
2. Determine Stiffness and Distribution Factors (DF): Calculate the relative stiffness (K) for each member. For a member fixed at one end and pinned/hinged at the other, K = 3EI/L. For a member fixed at both ends, K = 4EI/L. The distribution factor for a joint is calculated as DF = K / ΣK, where ΣK is the sum of stiffnesses of all members meeting at that joint.
3. Account for Sidesway: Since portal frames are often unsymmetrical or subjected to lateral loads, they undergo sidesway. Perform two separate analyses: 1) A frame analysis with sidesway prevented (by applying an artificial support) to find the internal moments and the resulting horizontal reaction (H₁) at the support. 2) A frame analysis with an arbitrary sway displacement (Δ) to find the moments and the corresponding restoring force (H₂).
4. Superposition: Calculate the actual sway force (H) acting on the frame. Determine the correction factor (λ = H / H₂) and adjust the moments obtained in the sway analysis by multiplying them by λ. The final moments are obtained by adding the moments from the non-sway analysis to the adjusted sway moments: M_final = M_non-sway + λ × M_sway.
5. Moment Distribution Table: Construct a table listing joints and members. Perform successive cycles of balancing the joints (distributing unbalanced moments) and carrying over the moments to the far ends until the values converge to near zero.

Sum the moments at each joint to obtain the final bending moments, which are then used to draw the Bending Moment Diagram (BMD).