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  1. Determine the diameter of a short rod to carry a compressive load of 2800 N, such that the percentage contraction and compressive stress do not exceed 0.01 and 30 N/mm2 . Given E = 200 GN/m2. Find the followings:

  1.  decrease in length if the original length is 10 cm.

  2.  changed length.

  3.  resilience?

  4.  stress if the same load is applied suddenly.

  5.  stress if the same load is applied as an impact from a height of 1 m.

  6.  proof load if the yield stress is 280 N/mm2.

  7.  decrease in diameter if the Poisson’s ratio is 0.25.

  8.  changed diameter.

  9.  shear modulus.

  10.  bulk modulus.

  11.  % change in area

  12.  % change in volume

  13.  ductility.

[ANS: Diameter = 14.93 mm]

  1. A 6 mm by 75 mm plate 600 mm long has a circular hole 25 mm located in the center. Find the axial tensile stress that may be applied to this plate in the longitudinal direction without exceeding the allowable stress of 220 M Pa. [ANS 66 k N]

  2. A mild steel bar 5 cm in diameter is compressed 0.006 cm by a compressive load of 200 k N. Determine its elongation under a pull of 100 k N. Take Young’s modulus in compression and tension as equal. [ANS 0.003 cm]

  3. Calculate the force required to punch a hole of 10 mm diameter through a mild plate 3,75 mm thick. Take the maximum shear strength of the material as 300 N/mm 2. Also find the compressive stress on the punch. [ANS: 35340 N, 450 N/mm 2].

Question on Stress Strain Curve

5. The following data were recorded during the tensile test of a 14-mm-diameter mild steel rod. The Gage length was 50 mm.

Load (N)

Elongation (mm)

Load (N)

Elongation (mm)



46 200


6 310


52 400


12 600


58 500


18 800


68 000


25 100


59 000


31 300


67 800


37 900


65 000


40 100


65 500


41 600


Plot the stress-strain diagram and determine the following mechanical properties: (a) proportional limits; (b) modulus of elasticity; (c) yield point; (d) ultimate strength; and (e) rupture strength.

Riveted Joint

  1. Determine the required number of rivets 20 mm in diameter for a double riveted lap joint joining two plates, each of thickness 8 mm. The tensile force on the joint is 200 k N. The allowable shear stress is 140 N/mm 2 and the allowable bearing stress is 320 N/m 2. ANS :5

  2. A rope is used to hoist a heavy load. When unloaded, the rope is exactly 12 m long. However, when the hoisting process begins and raises the load, the upper end of the rope is 12.36 m above the load. Calculate the % elongation. ANS : 3 %

Different Types of Loading

  1. Find the stresses on a bar of 20 mm diameter when a load of 5 k N is applied alternately as gradually applied load, suddenly applied load and as impact load. Length of the bar is 200 mm and the impact load is falling from a height of 150 mm. Given E = 200 G N/m 2.

  2. A cast iron column of hollow circular cross section with an external diameter of 30 cm and is subjected to a compressive load of 200 k N. Determine the necessary wall thickness if the allowable compressive stress is 80 N/mm 2. ANS: 2.7 mm


  1. At a point in an elastic material under strain, the stresses on three mutually perpendicular planes are follows: A normal tensile stress of 60 N/mm2 and a shear stress of 40 N/mm2 on one plane, a normal compressive stress of 40 N/mm2 and a complementary shear stress of 40 N/mm2 on the other plane, and

(a)  Principal stresses and principal planes

(b) The maximum shear stress and its plane

(c) The position of the plane on which there is no normal stress

ANS(+ 74 N/mm2, –54 N/mm2, 190, 1090, 64 N/mm2, -690)

2. The stresses at a particular point in a piece of material act upon the three planes whose relative angular positions are given by a triangle ABC, in which B is a right angle and angle C is 300. The normal stresses on these planes are 70 N/mm2 tension on AB, 30 N/mm2 compression on BC and 40 N/mm2 tension on AC. Determine the magnitude and direction of the shearing stresses on the given planes and the magnitude of the greatest direct stress and the greatest shearing stress at the point. ANS [ –52 N/mm2, + 69.3 N/mm2, 92.1 N/mm2 & 721 N/mm2].

3.Direct stress of 120 N/mm2 tension and 90 N/mmcompression are applied on a elastic material at a certain point on planes at right angles. The greater principal stress is limited to 150 N/mm2. What shearing stress may be applied on the given planes and what will be maximum shearing stress at the point? [135 N/mm2]

4. The principal stresses at a point in a member subjected to two dimensional stress are 120 N/mm2  and 50 N/mm2  both tensile. Find the plane on which the resultant stress has maximum obliquity and the magnitude of this obliquity.           [ 570 9’, 24018’]

5. At a point in a stressed body, the normal stresses are 90 N/mm2   tension on a vertical plane and 30 N/mm2 compression on a horizontal plane. A negative (clockwise) shearing stress of 50 N/mm2 acts on the vertical plane at that point. Determine and show on a sketch the principal and maximum shearing stresses. Also calculate and show normal stress on the plane of maximum shear stress.


  1. A 4 m span wooden beam  of rectangular section 120 mm deep and 60 mm wide is bent into an arc of radius 100 m. Determine the following quantities:

  1. Maximum tensile stress developed

  2. Maximum compressive stress developed

  3. Bending moment applied.

  4. Point load at the center if the beam is simply supported.

  5. UDL over the entire span if the beam is simply supported.

  6. Point load at the free end if the beam is cantilever.

  7. UDL over the entire length if the beam is a cantilever.

  8. Moment of resistance if the ultimate tensile stress is 30 N/mm2 .

Take factor of safety as 3 and E = 10 GN/m2.

SOLUTION: Given b=60 mm, d= 120 mm, R = 100 x 1000 mm, L= 4000 mm, σ = 30 N/mm2,FOS =3

  1. TENSILE = σt max = Ey/R=10 x 1000 x 60/100 x 1000 = 6 N/mm2.    (as y=d/2)

  2. Compressive = σc max = Ey/R=10 x 1000 x 60/100 x 1000 = 6 N/mm2.

  3. M = σ I/ymax= 6 x (1/12) 60 x 1203 / 60 = 0.864 kNm

  4. For simply supported beam with ‘W’ at center, M = WL/4, W = 0.864 kN

  5. For a simply supported beam with UDL, M = wL2/4, w =( 4M/L)5 = 0.93 kN/m

  6. Point load on a cantilever at the free end, M=WL, 0.864=W x 4, W = 0.216 kN

  7. UDL over entire length of cantilever, M=w2L/2, 0.864=w x 16/2,w = 0.108 kN/m

  8. σall = σult/FOS = 30/3 = 10 , MR = σall I/ymax = 10 (1/12) 60 1203/60 = 1.44 kN/mm

2. A simply supported beam of span 6 meters has the cross section 100 mm x 250 mm. If the permissible stress is 8 MPa, find the maximum intensity of the uniformly distributed load it can carry and the maximum concentrated load ‘W’ applied at 2 m from one end it can carry. (PTU DEC 2014)


M =wL2/4, I =(1/12)100 x 2503, y=125 mm, σ=8, Using M/I =σ/y,

we get w=(8/125)x(1/12) 100 x 2503x4/6000 x 6000=0.87 N/mm

(B) RA =(2/3)W, M=(4/3)W

M =8 x 125x(1/12) 100 x 2503

W= (8/125)x(1/12) 100 x 2503x(3/4)/1000= 8.04 k N

 3. A C.I. beam simply supported carries vertical loads only. The allowable stresses in tension and compression are 150 and 500 N/mm2. Refer fig, for cross section. Find the followings:1. Moment of resistance,2. The actual skin stresses, 3. Show graphically the actual stresses.

4. A C.I. pipe 250 mm external diameter having wall thickness of 10 mm rests on two supports 12 m apart on the ends. It is filled with water. What are the maximum normal stresses in the pipe if the specific gravity of C.I. is 7.8? Also show the stress and strain variations in the pipe. ANS 37.5 N/mm2 each in compression in the top fiber and tension in the bottom fiber.


Q1. Find the slope and deflection under

a concentrated load of 20 k N acting at

the free end of a cantilever of span 10 m.

Q2. Find the slope and deflection of the

free end of a cantilever of span 10 m.

Find slope and deflection under the

load of 20 k N acting at 8 m from the

fixed end.

Q3. The deflection of a simply supported

beam of span of 6 m under a

concentrated load of 2k N is 7.2 mm

under the load. Determine the position

of the load. Find the slope at the end

nearer to the load, magnitude of

maximum deflection, distance of point of

maximum deflection from the center of

the beam.

Take EI = 1000 kNm2

Q4. Determine the slope and deflection

at the free end of the cantilever beam of

span 10 m and having loads of 20 k N, 40

k N at 3m and 7m from the fixed end


Using the Macaulay’s Method.

Take EI = 1000 kNm2

4. Define slope and deflection.

5. Derive the approximate differential equation of a bent beam.

6. Describe double integration method

with an example.

7. State Macaulay’s Method with an example.

8. Discuss the moment area method of

finding slope and deflection.


  1. A torque of 10 k N m acts on a 100 mm diameter steel shaft. Determine the maximum shearing stress in the shaft. Also calculate the stresses and strains at 20 mm, 40 mm, 60 mm and 80 mm diameters. Show the stress variation and strain variation. Given G = 80 G N/m 2.

        ANS: (101.93, 20.387, 40.77, 61.16 and 81.54 N/m2)

  1. A propeller shaft 45 m long transmits 10 MW at 80 RPM. The internal and external diameters of the shaft are 24 and 57 cm respectively. Assuming the maximum torque to be 1.19 times the average (mean) torque, find the maximum shearing stress produced. Find the relative angular movement of the ends of the shaft when transmitting average torque. Given G = 80 G N/m 2. [ 40.49 N/m m2, 4.580]

  2. A hollow shaft of diameter ratio 3/5 transmits 600 kW at 110 RPM. The maximum torque being 12 % greater than the mean torque. The shear stress is not to exceed 60 N/m2 .  The twist in a length of 3 m not to exceed 10. Calculate the minimum external diameter satisfying these conditions.  80 G N/m 2.   [261 mm]

  3. A hole equal to half the diameter of a shaft is drilled throughout its length. BY what percentage is its weight reduction? What is % reduction in torsional strengths? [ 25 %, and 6.3 %]

  4. A flange coupling is required for a shaft transmitting 220 kW at 240 r.p.m. The bolts are 16 mm in diameter and are to be arranged on a pitch circle diameter of 250 mm. The working shear stress in both (shaft as well as bolt) is not to exceed 60 N/m m 2. Determine the number of bolts required and the actual stress realized. [ 6 and 58 N/m m 2]

  5. A steel bar of 19 mm diameter is closely fitted in a brass tube of 32 mm external diameter. It is  securely fixed at both ends. The compound bar (Shaft) is subjected to a torque of 520 N m.  The angle of twist measured on a Gage length of 250 mm is found to be 11.80 . If G for steel is 80 G N/m 2, calculate G for brass. Also find maximum stress in both the materials. [ 34.37 G N/m 2, 95.44 N/m m 2 in steel and 69.48 N/m m 2]

  6. A solid shaft transmits 560 kW at 300 r.p.m. with a maximum shear stress of 60 N/m m 2. What is the shaft diameter?

What would be the diameters of a hollow shaft of the same material (diameters ratio 2). These transmit the same power at the same speed and same stress?  Compare the stiffness of equal lengths of these shafts.

[114 mm, 117 and 57.5 mm, 1.06 solid vs hollow]

  1. A solid shaft of diameter ‘d’. It is subjected to a bending moment of 15 k N m. This is subjected to   a turning moment of 25 k N m. What is the minimum diameter of the shaft?  The maximum shear stress is not to exceed 160 N/m m 2 . The direct stress is not to exceed 200 N/m m 2? [95.77 mm]

  2. A ship has a propeller shaft of 0.5 m diameter which supports a propeller mass of 10 tons. The propeller may be considered as a concentrated load at the end of a cantilever of 3 m length. At a speed of 30 km/h of the ship, the propeller revolves at 100 rpm. If the engine is developing 20 MW, calculate the principal stresses and maximum shear stress in the shaft. The shaft propulsion efficiency is 80 %. [ –165.51 , + 35.91 and 100.71 N/m m 2]



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