JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III ADVANCED STRUCTURAL ANALYSIS Previous Question Paper (November/December - 2020 - R16)

To analyze a pin-jointed plane frame using the Matrix Stiffness Method, we follow these systematic steps:

1. Discretization and Coordinate System: Identify all nodes and members. Assign global coordinates and degrees of freedom (DOF) to each node. For a plane truss, each node has two translational DOFs (u, v).
2. Element Stiffness Matrix: For each member of length L, cross-sectional area A, and Young's modulus E, the local stiffness matrix in global coordinates is given by:
   k = (AE/L) * [cos²θ cosθsinθ -cos²θ -cosθsinθ; cosθsinθ sin²θ -cosθsinθ -sin²θ; -cos²θ -cosθsinθ cos²θ cosθsinθ; -cosθsinθ -sin²θ cosθsinθ sin²θ]
   where θ is the angle the member makes with the global X-axis.
3. Global Stiffness Matrix Assembly: Assemble the global stiffness matrix [K] by summing the contributions of individual element stiffness matrices into their respective global DOF positions.
4. Boundary Conditions: Apply support conditions by removing rows and columns corresponding to constrained DOFs (where displacement is zero).
5. Force Vector: Form the global load vector {P} representing external forces applied at the nodes.
6. Solving for Displacements: Solve the matrix equation {P} = [K]{d} for the unknown nodal displacements {d} using Gaussian elimination or matrix inversion.
7. Member Forces: Once displacements are known, calculate the axial force in each member using the formula:
   F = (AE/L) * [(u2-u1)cosθ + (v2-v1)sinθ]
   where (u1, v1) and (u2, v2) are the global displacements of the start and end nodes of the member.

Given A = 600 mm² and E = 200 kN/mm², the axial stiffness AE = 120,000 kN. By applying these steps to the specific geometry of the frame, the nodal displacements are determined first, followed by the internal forces in each member to complete the structural analysis.

For a two-hinged parabolic arch where the moment of inertia varies as I = I₀ sec θ, the horizontal thrust H is given by the formula:

Given span L = 32 m, rise h = 6 m, and load W = 75 kN at a = 10 m from the left support. The equation of the arch is y = (4h / L²) * x * (L - x) = (24 / 1024) * x * (32 - x) = 0.0234375 * x * (32 - x).

The vertical reactions are V_A = W(L - a) / L = 75 * (32 - 10) / 32 = 51.5625 kN and V_B = W * a / L = 75 * 10 / 32 = 23.4375 kN.

The horizontal thrust H is calculated as:

At a section x = 24 m from the left (8 m from the right):

1. Bending Moment (M_x): M_x = M₀ - H * y. At x = 24, y = (24/1024) * 24 * 8 = 4.5 m. M₀ = V_B * 8 = 23.4375 * 8 = 187.5 kNm. M_x = 187.5 - (451.17 * 4.5) = -1842.77 kNm.
2. Normal Thrust (N_x): N_x = H * cos θ + V_x * sin θ. The slope tan θ = dy/dx = (4h/L²) * (L - 2x) = (24/1024) * (32 - 48) = -0.375. θ = -20.56°. cos θ = 0.936, sin θ = -0.351. V_x = V_B = 23.4375. N_x = (451.17 * 0.936) + (23.4375 * -0.351) = 422.29 - 8.23 = 414.06 kN.
3. Radial Shear (S_x): S_x = V_x * cos θ - H * sin θ = (23.4375 * 0.936) - (451.17 * -0.351) = 21.94 + 158.36 = 180.30 kN.