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JNTU B.Tech III STRUCTURAL ANALYSIS - II Previous Question Paper (March - 2021 - R18)
Answer any five questions. All questions carry equal marks.
Main
A two-hinged parabolic arch of span 24 m has central rise of 6 m and the moment of inertia of the cross-section of the arch varies as the secant of the slope of the arch axis. The arch is subjected to a concentrated load of 60 kN at a distance of 8 m from the right support. Find the support reactions and the maximum bending moment in the arch.
Answer
For a two-hinged parabolic arch where the moment of inertia varies as the secant of the slope (I = I0 sec θ), the horizontal thrust H is given by the formula:
Given span L = 24 m, rise h = 6 m, and load W = 60 kN at a = 8 m from the right support (x = 16 m from the left support). The equation of the arch is y = 4h/L2 * (Lx - x2) = 4(6)/242 * (24x - x2) = (24x - x2)/24.
The integration of y2 over the span is:
The bending moment M0 due to the load at x = 16 m is calculated as:
Reactions: RA = W(a)/L = 60(8)/24 = 20 kN; RB = W(L-a)/L = 60(16)/24 = 40 kN.
M0 for x < 16: M0 = RA * x = 20x. For x > 16: M0 = RB * (24-x) = 40(24-x).
Calculating ∫ M0 y dx:
Horizontal Thrust H = 6400 / 207.36 = 30.86 kN.
Support Reactions: VA = 20 kN, VB = 40 kN, H = 30.86 kN.
Maximum Bending Moment: The bending moment at any section x is Mx = M0 - H * y. At the load point (x = 16 m): y = 4(6)/242 * (24*16 - 162) = 5.33 m. M16 = 20(16) - 30.86(5.33) = 320 - 164.48 = 155.52 kNm.
Using the moment-distribution method, analyse the continuous beam supported and loaded as shown in Figure 1. Also draw the shear force and bending moment diagrams.
Answer
The moment distribution method is an iterative process used to determine the end moments of members in a continuous beam. The analysis involves calculating distribution factors, fixed-end moments, and performing successive cycles of distribution and carry-over until the moments converge.
1. Determine Stiffness and Distribution Factors: For each span, calculate the relative stiffness (k = I/L). At each interior support, calculate the distribution factor (DF) for the adjacent members using the formula DF = k / Σk.
2. Calculate Fixed-End Moments (FEM): Determine the fixed-end moments for each span based on the applied loading. For a point load P at the center, FEM = ±PL/8; for a uniformly distributed load w, FEM = ±wL2/12.
3. Moment Distribution Table: Set up a table listing all joints and members. Enter the FEMs and perform cycles of balancing the joints by distributing the unbalanced moment according to the DFs, followed by carrying over half of the distributed moment to the far end of the member.
4. Final Moments: Sum the moments at each joint to obtain the final support moments (MA, MB, MC, etc.).
5. Shear Force and Bending Moment Diagrams: Calculate the reactions at the supports by considering the equilibrium of each span under the applied loads and the calculated support moments. The shear force diagram is constructed by plotting the variation of shear along the beam, and the bending moment diagram is constructed by superimposing the free bending moment diagram (due to loads on simply supported spans) and the support moment diagram (due to continuity).
Using Kani’s method, analyse the frame shown in Figure 2 and draw the bending moment diagram.
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Analyse the frame shown in Figure 3, using portal method. Also draw the bending moment diagram. Assume the geometrical and material properties are the same for the elements of the frame.
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Using cantilever method, analyse the frame shown in figure 4. Draw the BM diagram. Assume the geometrical and material properties are the same for the elements of the frame.
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Using stiffness method, analyse the beam supported and loaded as shown in Figure 5. Assume EI is constant.
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Analyse the pin-jointed plane frame supported and loaded as shown in Figure 6. Use stiffness method.
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Draw the influence line diagrams for the reaction at the right support and the bending moment at the mid-point of the left span of a two-span continuous beam supported as shown in Figure 7.
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