JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year THEORY OF STRUCTURES Previous Question Paper (March - 2022 - R18)
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Main
A steel bar 25 mm diameter and 500 mm long is encased in a brass tube of same length with an external diameter of 40 mm and internal diameter of 32 mm. The composite bar is heated through 50℃. Calculate the stresses induced in steel bar and brass tube. The coefficient of thermal expansion of steel and brass are 12×10-6/℃ and 2.1×10-5/℃ respectively. E = 2×105 N/mm2 for steel and 0.9×105 N/mm2 for brass. [15]
Answer
To determine the thermal stresses induced in the composite bar, we analyze the equilibrium and compatibility conditions. Let the subscripts s denote steel and b denote brass.
- Geometric Properties:
Area of steel (As) = π/4 × 252 = 490.87 mm2
Area of brass (Ab) = π/4 × (402 - 322) = 452.39 mm2 - Given Data:
ΔT = 50°C
αs = 12 × 10-6/°C, αb = 21 × 10-6/°C
Es = 2 × 105 N/mm2, Eb = 0.9 × 105 N/mm2 - Compatibility Condition:
The final length of both materials must be equal. The free expansion of brass is greater than steel, so brass is in compression and steel is in tension. Let σs and σb be the stresses.
(αb ΔT - σb/Eb) = (αs ΔT + σs/Es) - Equilibrium Condition:
The internal forces must balance: σsAs = σbAb
σb = σs × (490.87 / 452.39) = 1.085 σs - Calculation:
Substituting into the compatibility equation:
(21 × 10-6 × 50) - (1.085 σs / 0.9 × 105) = (12 × 10-6 × 50) + (σs / 2 × 105)
0.00105 - 1.205 × 10-5 σs = 0.0006 + 0.5 × 10-5 σs
0.00045 = 1.705 × 10-5 σs
σs = 26.39 N/mm2 (Tensile) - Brass Stress:
σb = 1.085 × 26.39 = 28.63 N/mm2 (Compressive)
Draw the shear force and bending moment diagrams for the beam supported and loaded as shown in Figure 1. [15]
Answer
To analyze the beam, we first determine the support reactions. Assuming a simply supported beam of length L with a point load P at distance a from the left support (A) and b from the right support (B):
- Equilibrium Equations: Summing moments about A (ΣMA = 0) gives RB = (P × a) / L. Summing vertical forces (ΣV = 0) gives RA = P - RB = (P × b) / L.
- Shear Force Diagram (SFD): The shear force is constant at +RA from A to the load point. At the load point, the shear force drops by magnitude P to a value of -RB, remaining constant until support B.
- Bending Moment Diagram (BMD): The bending moment increases linearly from 0 at support A to a maximum value of (RA × a) at the point of load application. From the load point, it decreases linearly back to 0 at support B.
For a beam with a Uniformly Distributed Load (UDL) of intensity w over the entire span L:
- Support Reactions: By symmetry, RA = RB = (w × L) / 2.
- SFD: The shear force varies linearly from +wL/2 at A to -wL/2 at B, crossing zero at the mid-span (L/2).
- BMD: The bending moment follows a parabolic curve, starting at 0 at the supports and reaching a maximum value of Mmax = (w × L2) / 8 at the mid-span.
Note: For specific numerical values, substitute the given span lengths and load magnitudes into these standard equations. Ensure that the SFD is drawn with units of Force (kN) and the BMD with units of Moment (kNm).
A simply supported beam of 5 m span has an I-section, top flange 100 mm×12 mm, web 10 mm thick and bottom flange 150 mm×12 mm. The depth of the section is 250 mm. It carries uniformly distributed load of 100 kN/m over its entire span and 125 kN concentrated load at its mid-span. Draw the variation of maximum bending stresses across the depth of the section. Also find the compressive and tensile force developed at mid-span section. [15]
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A simply supported steel beam of span 4.5 m and depth 300 mm is subjected to a concentrated load of 240 kN at its mid-span. The beam has an I-section with the top flange 150 mm×16 mm, web 12 mm thick and bottom flange 200 mm×20 mm. Draw the maximum shear stress distribution across the depth of the section. [15]
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A simply supported beam of span 6 m is subjected to the loading as shown in Figure 2, find the maximum deflection and the slope at the supports. Use moment-area method. [15]
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Using moment distribution method, analyse the frame shown in Figure 3. Assume the flexural rigidity is constant. Draw BMD. [15]
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The state of stress at a point of a strained member is shown in Figure 4. Determine the following from the first principles: (a) Magnitude of stresses acting on a plane 1-1, (b) Magnitude of principal stresses and direction of principal planes and (c) Planes carrying maximum shear stress and accompanied normal stresses. Also verify the above results using Mohr's circle. [15]
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