JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II THEORY OF STRUCTURES Previous Question Paper (March - 2021 - R18)

B.Tech | II | II-I | R18 | March - 2021

Answer PaperPaper Code: 153BXMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

1

A structural member supports loads, which produce, at a particular point, a direct tensile stress of 80 N/mm² and a shear stress of 45 N/mm² on the same plane. Calculate the values and directions of the principal stresses at the point and the maximum stress, stating on which planes this will act.

15 marks

Answer

Given: Direct tensile stress (σx) = 80 N/mm², Shear stress (τxy) = 45 N/mm², and σy = 0 N/mm² (as it is not specified).

The principal stresses (σ1, σ2) are calculated using the formula:

σ1,2 = (σx + σy)/2 ± √(((σx - σy)/2)² + τxy²)

Substituting the values:

σ1,2 = (80 + 0)/2 ± √(((80 - 0)/2)² + 45²)
σ1,2 = 40 ± √(40² + 45²)
σ1,2 = 40 ± √(1600 + 2025) = 40 ± 60.21

The principal stresses are:

σ1 = 100.21 N/mm² (Major Principal Stress)
σ2 = -20.21 N/mm² (Minor Principal Stress)

The directions of the principal planes (θp) are given by:

tan(2θp) = 2τxy / (σx - σy) = (2 × 45) / (80 - 0) = 90 / 80 = 1.125
2θp = 48.37° or 228.37°
θp1 = 24.18° and θp2 = 114.18°

The maximum shear stress (τmax) is calculated as:

τmax = (σ1 - σ2) / 2 = (100.21 - (-20.21)) / 2 = 60.21 N/mm²

The planes of maximum shear stress act at 45° to the principal planes, specifically at θs = θp1 - 45° = -20.82° and θs2 = θp1 + 45° = 69.18°.

2

A bending moment of 3,000 Nm is applied to the purlin in a plane at an angle of 30° to the vertical y axis. If the sense of the bending moment is such that its components Mx and My both produce tension in the positive xy quadrant, calculate the maximum direct stress in the purlin stating clearly the point at which it acts.

15 marks

Answer

To calculate the maximum direct stress, we first resolve the applied bending moment M = 3,000 Nm into its components along the principal axes x and y. Given the angle θ = 30° to the vertical y-axis, the components are:

Mx = M cos(30°) = 3,000 × 0.866 = 2,598 Nm
My = M sin(30°) = 3,000 × 0.5 = 1,500 Nm

The direct stress σ at any point (x, y) in the cross-section is given by the formula:

σ = (My × x) / Iy + (Mx × y) / Ix

Since the problem states that both components produce tension in the positive xy quadrant, the maximum tensile stress occurs at the corner where both x and y coordinates are positive (the extreme fiber in the first quadrant). Assuming a rectangular section with width b and depth d, where Ix = (bd3)/12 and Iy = (db3)/12, the maximum stress is:

σmax = (My × (b/2)) / ((db3)/12) + (Mx × (d/2)) / ((bd3)/12)
σmax = (6 × My) / (d × b2) + (6 × Mx) / (b × d2)

The maximum direct stress acts at the extreme fiber located at the corner (x = b/2, y = d/2) in the first quadrant, where the tensile effects of both bending moments are additive.

3

A rectangular block of material is subjected to a tensile stress of 100 N/mm² on one plane and a tensile stress of 60 N/mm² on a plane at right angle, together with shear stresses of 45 N/mm² on the same plane. Find:

15 marks
a
The direction of principal planes
5 marks

Answer

Sign in to view the complete answer paper.

b
The magnitude of the principal stress
5 marks

Answer

Sign in to view the complete answer paper.

c
The magnitude of the greatest shear stress.
5 marks

Answer

Sign in to view the complete answer paper.

4

A beam on simple supports is subjected to a parabolically distributed load of intensity

q = 4xq0(L-x)/L2
where q0 is the maximum intensity of the load. Derive the equation of the deflection curve and determine the maximum deflection
δmax
.

15 marks

Answer

Sign in to view the complete answer paper.

5

What are the basic assumptions in theory of elasticity and derive the equations of equilibrium and compatibility equations?

15 marks

Answer

Sign in to view the complete answer paper.

6

A hollow cylindrical cast iron column is 4 m long with both ends fixed. Determine the minimum diameter of the column if it has to carry a safe load of 250 kN with a factor of safety of 5. Take the internal diameter as 0.8 times the external diameter. Take ultimate compressive stress for the column material is 550 N/mm² and a = 1/1600 in Rankine’s formula.

15 marks

Answer

Sign in to view the complete answer paper.

7

Calculate the vertical deflection of the joint B and the horizontal movement of the support D in the truss shown. The cross-sectional area of each member is 1800 mm² and Young’s modulus, E, for the material of the members is 200,000 N/mm².

15 marks

Answer

Sign in to view the complete answer paper.

815 marks
a
How will you apply Clapeyron’s theorem of three moments to a continuous beam with simply supported ends and continuous beam with fixed end supports?
7 marks

Answer

Sign in to view the complete answer paper.

b
A continuous beam ABCD of length 15 m rests on four supports covering 3 equal spans and carries a uniformly distributed load of 1.5 kN/m length. Calculate the moments at the supports.
8 marks

Answer

Sign in to view the complete answer paper.

About this paper

This page contains the question paper of THEORY OF STRUCTURES from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech AE II II-I, March - 2021, R18.

For the solved question paper with answers, switch to the Answer Paper tab.

Other papers in THEORY OF STRUCTURES

Loading Question Papers

Fetching data from database...