Exam Details

Subject structural analysis - i
Paper
Exam / Course pddc
Department
Organization Gujarat Technological University
Position
Exam Date January, 2019
City, State gujarat, ahmedabad


Question Paper

1
Seat No.: Enrolment
GUJARAT TECHNOLOGICAL UNIVERSITY
PDDC SEMESTER-II EXAMINATION WINTER 2018
Subject Code:X20603 Date: 09/01/2019
Subject Name: Structural Analysis I
Time: 02:30 PM TO 05:00 PM Total Marks: 70
Instructions:
1. Attempt all questions.
2. Make suitable assumptions wherever necessary.
3. Figures to the right indicate full marks.
Q.1

A solid circular shaft of 60 mm diameter transmitting 50 kW power at 150 RPM is to be replaced by a hollow circular shaft of the same material. The hollow circular is having external diameter 1.2 times internal diameter. Calculate the size of hollow circular shaft and also saving in material.
07

A hollow shaft of diameter ration 2/3 to transmit 350 kW at 200 r.p.m. If allowable shear stress is 50 N/mm2 and max torque is 30 greater then mean. If hollow shaft is to be replaced by a solid shaft of the same material, what should be the diameter of solid shaft?
07
Q.2

A thin cylindrical shell of 600 mm diameter is 1500 mm lonf and 10 mm thick. It is subjected to an internal pressure of 2 N/mm2 calculate the change in diameter, length and volume. Take E 200 GPa, 1/m =0.27
07

Distinguish between plane frame and gird plane truss and space truss.
07
OR

Draw shear force, bending moment and axial force diagrams for a frame shown in Figure-(1).
07
Q.3

A simply beam of span 10 m is loaded at shown in Figure-(2). Determine the position and amount of maximum deflection by Macaulay's method. Take E 200 GPa, I 6.95 x 108 mm4.
07

Find slope and deflection at free end for a cantilever beam shown in Figure-(3) by moment area method. E 2 x 105 N/mm2, I 5 x 108 mm4.
OR
Q.3

Calculate slope at point A,B and D deflection at point C for a beam loaded as shown in Figure-(4) by conjugate beam method. Take E 2 x 105 N/mm2, I 3 x 108 mm4.
07

Calculate vertical and horizontal deflection of the joint C of the pin jointed plane frame shown in Figure-(5). The cross sectional are of AB is 100 mm2 and of AC and BC 150 mm2. each E 2 x 105 N/mm2
07
Q.4

A cable is used to support five equal and equidistant loads over a span of 30 m. Find the length of the cable required ad its sectional area if the safe tensile stress is 140 N/mm2. The central dip of the cable is 2.5 m and loads are 5 kN each. Also find tensions each part.
07

A bar 54 mm in diameter is 4 m long. An axial load of 180 kN is suddenly applied to it. Find The maximum instantaneous stress The maximum instantaneous elongation The work stored in the bar at the instant of maximum elongation. Take E 2 x 105 N/mm2
07
OR
2
Q.4

A beam ACB 7 m long is fixed at A and is simply supported at and is provided with an internal hinge at 4 m from A. Draw influence line diagrams for the followings, RA, RB MA B.M. at the middle AC B.M. at 2 m from B.
07

A cylindrical chimney, 25 m high, of uniform circular section is 5 m external diameter and 2 m internal diameter. It is subjected to a horizontal wind pressure of 1400 N/mm2 . If the coefficient of wind pressure is 0.6 and unit weiht of masonry is 22 kN/m3 find the maximum and minimum stresses at the base of the section.
07
Q.5

A round steel rod of 15 mm diameter and 2 m length is subjected to a gradually increasing axial compressive load. Using Euler's formula find the buckling load. Find also the maximum lateral deflection corresponding to the buckling condition. Both ends of the rod may be taken as hinged. Take E 2.1 x 105 N/mm2 and yield stress of steel is 250 E 2 x 105 N/mm2.
07

Explain with the help of sketches effective length of column for different end conditions.
07
OR
Q.5

Derive Euler's formula for crippling load for a strut fixed at both the ends.
07

A steel column of hollow circular section 65 mm external diameter and 50 mm internal diameter is 2.5 m long and hinged at ends. The line of action of the load is parallel to the axis but is eccentric. Find maximum eccentricity for a crippling load equal to 75 of the Eulerian axial load. The yield stress of steel is equal to 310 N/mm2 and E 2.06 x 105 N/mm2.
07


A
B
C
D
4
3
2
20k
45
OR Q-2 Figure-(1)
40k
20k
60k
2
3
4
1
A
B
C
D
E
Q-3 Figure-(2)
4.5kN N
A
B
C
D
0.15
0.15
0.15

(2I
Q-3 Figure-(3)
A
60k
B
C
D
20kN/m
3m m
1
2
OR Q-3 Figure-(4)
3k
3k
6k
A
B
C

4
4
OR Q-3 Figure-(5)


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