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# MCQs on Stresses in Members-1

1 - Question

**MCQs on Normal & Shear Stress**

In the given figure a stepped column carries loads. What will be the maximum normal stress in the column at B in the larger diameter column if the ratio of P/A here is unity?

a) 1/1.5

b) 1

c) 2/1.5

d) 2

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Answer: cExplanation: Normal stress at B = Total load acting at B / Area of a cross-section at B = (P + P) / 1.5 A = 2P/ 1.5A = 2/1.5.

2 - Question

The stress which acts in a direction perpendicular to the area is called ____________

a) Shear stress

b) Normal stress

c) Thermal stress

d) None of the mentioned

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Answer: bExplanation: Normal stress acts in a direction perpendicular to the area. Normal stress is of two types tensile and compressive stress.

3 - Question

Which of these are types of normal stresses?

a) Tensile and compressive stresses

b) Tensile and thermal stresses

c) Shear and bending

d) Compressive and plane stresses

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Answer: aExplanation: The normal stress is divided into tensile stress and compressive stress. advertisement

4 - Question

In a body loaded under plane stress conditions, what is the number of independent stress components?

a) 1

b) 2

c) 3

d) 6

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Answer: cExplanation: In a body loaded under plane stress conditions, the number of independent stress components is 3 I.e. two normal components and one shear component.

5 - Question

If a bar of large length when held vertically and subjected to a load at its lower end, its won-weight produces additional stress. The maximum stress will be ____________

a) At the lower cross-section

b) At the built-in upper cross-section

c) At the central cross-section

d) At every point of the bar

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Answer: bExplanation: The stress is the load per unit area. After the addition of weight in the bar due to its loading on the lower end the force will increase in the upper cross-section resulting in the maximum stress at the built-in upper cross-section.

6 - Question

Which type of stress does in a reinforcement bar is taken by the concrete?

a) Tensile stress

b) Compressive stress

c) Shear stress

d) Bending stress

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Answer: bExplanation: Concrete has the property of taking a good amount of compressive stress. So, In the reinforcement bar, the compressive stress is taken by the concrete.

7 - Question

A material has a Poisson’s ratio of 0.5. If uniform pressure of 300GPa is applied to that material, What will be the volumetric strain of it?

a) 0.50

b) 0.20

c) 0.25

d) Zero

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Answer: dExplanation: As volumetric strain = (1-2μ)σ/E Here the value of μ is 0.5 so 1 – 2 * 0.5 becomes zero. Therefore whatever be the stress the value of volumetric strain will be zero.

8 - Question

A diagram which shows the variations of the axial load for all sections of the pan of a beam is called ____________

a) Bending moment diagram

b) Shear force diagram

c) Thrust diagram

d) Stress diagram

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Answer: dExplanation: The stress diagram shows the variation of the axial load for all sections of the pan. The bending moment diagram shows the variation of moment in a beam. The shear force diagram shows the variation in the shear force due to loading in the beam.

9 - Question

The stress induced in a body, when subjected to two equal and opposite forces which are acting tangentially across the resisting section resulting the shearing of the body across its section is called ____________

a) Bending stress

b) Compressive stress

c) Shear strain

d) Shear stress

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Answer: dExplanation: Shear stress makes the body to shear off across the section. It is tangential to the area over which it acts. The corresponding strain is the shear strain.

10 - Question

What is the formula for shear stress?

a) Shear resistance/shear area

b) Force/unit area

c) Bending strain/area

d) Shear stress/length

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Answer: aExplanation: When force is applied, the twisting divides the body. The resistance is known as shear resistance and shear resistance per unit area is known as shear stress.

11 - Question

Which of the following stresses are associated with the tightening of a nut on a bolt? P. Crushing and shear stress in threads Q. Bending stress due to the bending of bolt R. Torsional shear stress due to frictional resistance between the nut and the bolt Select the correct answer using the codes given below:

a) P and Q

b) P and R

c) Only P

d) Only R

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Answer: aExplanation: Bending stress comes when there is some kind of eccentric load. Torsional stress will come when the nut is rotating. Shear stress will come in tightening of a nut on bolt.

12 - Question

The transverse shear stress acting in a beam of rectangular cross-section, subjected to a transverse shear load, is ____________

a) variable with maximum at the bottom of the beam

b) Variable with maximum at the top of the beam

c) Uniform

d) Variable with maximum on the neutral axis

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Answer: dExplanation: Maximum value of shear stress at neutral axis is τ = 3/2 τmean So, transverse shear stress is variable with a maximum in the neutral axis.

13 - Question

A block 100mm x 100mm base and 10mm height. What will the direct shear stress in the element when a tangential force of 10kN is applied to the upper edge to a displacement 1mm relative to lower face?

a) 1Pa

b) 1MPa

c) 10MPa

d) 100Pa

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Answer: bExplanation: Shear stress = 10kN / 100mmx100mm = 1 N/mm2 = 1MPa.

14 - Question

**MCQs on Bending Stress**

A beam is said to be of uniform strength, if ____________

a) B.M. is same throughout the beam

b) Shear stress is the same through the beam

c) Deflection is the same throughout the beam

d) Bending stress is the same at every section along its longitudinal axis

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Answer: dExplanation: Beam is said to be uniform strength if at every section along its longitudinal axis, the bending stress is same.

15 - Question

Stress in a beam due to simple bending is ____________

a) Directly proportional

b) Inversely proportional

c) Curvilinearly related

d) None of the mentioned

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Answer: aExplanation: The stress is directly proportional to the load and here the load is in terms of bending. So the stress is directly proportional to bending.

16 - Question

Which stress comes when there is an eccentric load applied?

a) Shear stress

b) Bending stress

c) Tensile stress

d) Thermal stress

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Answer: bExplanation: When there is an eccentric load it means that the load is at some distance from the axis. This causes compression in one side and tension on the other. This causes bending stress. advertisement

17 - Question

What is the expression of the bending equation?

a) M/I = σ/y = E/R

b) M/R = σ/y = E/I

c) M/y = σ/R = E/I

d) M/I = σ/R = E/y

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Answer: aExplanation: The bending equation is given by M/I = σ/y = E/R where M is the bending moment I is the moment of inertia y is the distance from neutral axis E is the modulus of elasticity R is the radius.

18 - Question

On bending of a beam, which is the layer which is neither elongated nor shortened?

a) Axis of load

b) Neutral axis

c) Center of gravity

d) None of the mentioned

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Answer: bExplanation: When a beam is in bending the layer in the direction of bending will be In compression and the other will be in tension. One side of the neutral axis will be shortened and the other will be elongated.

19 - Question

The bending stress is ____________

a) Directly proportional to the distance of layer from the neutral layer

b) Inversely proportional to the distance of layer from the neutral layer

c) Directly proportional to the neutral layer

d) Does not depend on the distance of layer from the neutral layer

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Answer: aExplanation: From the bending equation M/I = σ/y = E/R Here stress is directly proportional to the distance of layer from the neutral layer.

20 - Question

Consider a 250mmx15mmx10mm steel bar which is free to expand is heated from 15C to 40C. what will be developed?

a) Compressive stress

b) Tensile stress

c) Shear stress

d) No stress

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Answer: dExplanation: If we resist to expand then only stress will develop. Here the bar is free to expand so there will be no stress.

21 - Question

The safe stress for a hollow steel column which carries an axial load of 2100 kN is 125 MN/m2. if the external diameter of the column is 30cm, what will be the internal diameter?

a) 25 cm

b) 26.19cm

c) 30.14 cm

d) 27.9 cm

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Answer: bExplanation: Area of the cross section of column = π/4 (0.302 – d2) m2 Area = load / stress. So, π/4 ( 0.302 – d2) m2 = 21 / 125 d = 26.19cm.

22 - Question

**MCQs on Tensile Stress**

During a tensile test on a ductile material ____________

a) Nominal stress at fracture is higher than the ultimate stress

b) True stress at fracture is higher than the ultimate stress

c) True stress a fracture is the same as the ultimate stress

d) None of the mentioned

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Answer: bExplanation: In a ductile material, the true stress at fracture will be higher the ultimate stress.

23 - Question

When equal and opposite forces applied to a body, tend to elongate it, the stress so produced, is called ____________

a) Shear stress

b) Compressive stress

c) Tensile stress

d) Transverse stress

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Answer: cExplanation: When subjected to two equal and opposite pulls as a result of which there is an increase in length. This produces tensile stress.

24 - Question

Which of the following stresses are associated with the tightening of a nut on a bolt? P. Tensile stress due to the streching of bolt Q. Bending stress due to the bending of bolt R. Torsional shear stress due to frictional resistance between the nut and the bolt Select the correct answer using the codes given below.

a) P and Q

b) P and R

c) Only p

d) R and Q

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Answer: aExplanation: Bending stress comes when there is some kind of eccentric load. When nut is tightened, the bolt will pull itself and stretching will be there resulting in the tensile stress. Torsional stress will come when the nut is rotating. advertisement

25 - Question

In a tensile test, near the elastic limit zone ____________

a) Tensile stress increases in linear proportion to the stress

b) Tensile stress increases at a faster rate

c) Tensile stress decreases at a faster rate

d) None of the mentioned

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Answer: cExplanation: The stress first decreases and then decreases before the strain hardening occurs. The decreases in the stress is due to the attraction between carbon molecules.

26 - Question

5. Match the following and give the correct code given in options: List 1 List 2 A. Tensile test on CI 1. Plain fracture on a transverse plane B. Tensile test on MS 2. Granular helecoidal fracture C. Torsion test on CI 3. Cup and cone 4. Granular fracture in a transverse plane

a) A – 1 B – 2 C – 4

b) A – 1 B – 4 C – 2

c) A – 3 B – 1 C – 2

d) A – 3 B – 4 C – 1

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Answer: dExplanation: Tensile test on CI is done on cup and cone. Torsion test on MS is on plain fracture on a traverse plane.

27 - Question

The phenomenon of slow growth of strain under a steady tensile stress is called ____________

a) Yielding

b) Creeping

c) Breaking

d) None of the mentioned

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Answer: bExplanation: Creeping is the phenomenon of slow growing strain under a stress for a period of time.

28 - Question

A rod 150cm long and of diameter 2cm is subjected to an axial pull of 20kN. What will be the stress?

a) 60 N/mm2

b) 65 N/mm2

c) 63.6 N/mm2

d) 71.2 N/mm2

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Answer: cExplanation: The stress = load / area Load = 20,000N Area = π/4 (20)2 = 100π mm2.

29 - Question

The stress in a rod is 70 N/mm2 and the modulus of elasticity is 2 x 105 N/mm2. what will be the strain in the rod?

a) 0.00052

b) 0.00035

c) 0.00030

d) 0.00047

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Answer: bExplanation: As E = σ/e Here, E = 2 * 105 N/mm2 And, σ = 70 N/mm2 e = 70/2*105 = 0.00035.

30 - Question

What will be the minimum diameter of a steel wire, which is used to raise a load of 4000N if the stress in the rod is not to exceed 95 MN/m2?

a) 6mm

b) 6.4mm

c) 7mm

d) 7.3mm

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Answer: dExplanation: As stress = load / area Area = load/stress Also, area is π/4 D2 so π/4 D2 = 4000 / 95 And D = 7.32.

31 - Question

A tensile test was conducted on mild steel bar. The load at elastic limit was 250kN and the diameter of the steel bar was 3cm. What will be the value of stress?

a) 35368 x 104 N/m2

b) 32463 x 104 N/m2

c) 35625 x 104 N/m2

d) 37562 x 104 N/m2

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Answer: aExplanation: The stress = load / area Load = 150 x 1000N Area = π/4 (0.03)2 m2.

32 - Question

**MCQs on Compressive Stress**

For keeping the stress wholly compressive the load may be applied on a circular column anywhere within a concentric circle of diameter _____________

a) D/2

b) D/3

c) D/4

d) D/8

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Answer: cExplanation: The load application on a circular column affects stress. If it is under D/4 the stress will be wholly compressive.

33 - Question

Consider two bars A and B of same material tightly secured between two unyielding walls. Coefficient of thermal expansion of bar A is more than that of B. What are the stresses induced on increasing the temperature?

a) Tension in both the materials

b) Tension in material A and compression in material B

c) Compression in material A and tension in material B

d) Compression in both the materials

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Answer:dExplanation: Since both the supports are fixed and both bars will try to expand, so rise in temperature will cause compressive stresses in the bars.

34 - Question

What will be the unit of compressive stress?

a) N

b) N/mm

c) N/mm2

d) Nmm

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Answer: cExplanation: As the stress is the ratio of force to the area, so it will be N/mm2. Here mm is normally used in its calculation most of the time. advertisement

35 - Question

A cast iron T section beam is subjected to pure bending. For maximum compressive stress to be 3 times the maximum tensile stress, centre of gravity of the section from flange side is ____________

a) h/2

b) H/3

c) H/4

d) 2/3h

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Answer: cExplanation: H/4 when the applied moment is sagging. Otherwise, I.e. if the applied moment is hogging it is H/4. as in the options both are not given means we have to take hogging.

36 - Question

A solid circular shaft of diameter d is subjected to a torque T. the maximum normal stress induced in the shaft is ____________

a) Zero

b) 16T/πd3

c) 32T/πd3

d) None of the mentioned

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Answer: bExplanation: The maximum torque transmitted by a circular solid shaft is obtained from the maximum shear stress induced at the outer surface of the solid shaft and given by T = πD3/16 x normal stress, So, normal stress = 16T/πd3.

37 - Question

When a rectangular beam is loaded transversely, the maximum compressive stress develops on ____________

a) Bottom fibre

b) Top fibre

c) Neutral axis

d) Every cross-section

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Answer: bExplanation: Loaded means loaded downwards. In that case, upper fibres will be compressed while lower will be expanded. Hence maximum compressive stress will be developed in top layer.

38 - Question

An axial residual compressive stress due to a manufacturing process is present on the outer surface of a rotating shaft subjected to bending. Under a given bending load, the fatigue of the shaft in the presence of the residual compressive stress is ____________

a) Decreased

b) Increases or decreased, depending on the external bending load

c) Neither decreased nor increased

d) Increases

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Answer: dExplanation: From the Gerber’s parabola that is the characteristic curve of the fatigue life of the shaft in the presence of the residual compressive stress. The fatigue life of the material is effectively increased by the introduction of compressive mean stress, whether applied or residual.

39 - Question

A steel bar of 40mm x 40mm square cross-section is subjected to an axial compressive load of 200kN. If the length of the bar is 2m and E=200GPa, the elongation of the bar well be ____________

a) 1.25mm

b) 2.70mm

c) 4.05mm

d) 5.40mm

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Answer: aExplanation: Elongation of the bar = Pl/AE = -200x103x 2000 / ( 1600 x 200 x 103) = -1.25 The minus sign here shows that the stress here is compressive.

40 - Question

**MCQs on Thermal Stress**

The length, Young’s modulus and coefficient of thermal expansion of bar P are twice that of bar Q. what will be the ration of stress developed in bar P to that in bar Q if the temperature of both bars is increased by the same amount?

a) 2

b) 8

c) 4

d) 16

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Answer: cExplanation: Temperature Stress = EαδT Stress in bar P / Stress in bar Q = (EP / EQ) x (αP / αQ) = 2×2 = 4.

41 - Question

A steel bar 600mm long and having 30mm diameter, is turned down to 25mm diameter for one fourth of its length. It is heated at 30 C above room temperature, clamped at both ends and then allowed to cool to room temperature. If the distance between the clamps is unchanged, the maximum stress in the bar ( α = 12.5 x 10-6 per C and E = 200 GN/m2) is

a) 25 MN/m2

b) 40 MN/m2

c) 50 MN/m2

d) 75 MN/m2

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Answer: dExplanation: As temperature stress do not depend upon properties of cross section like length and area. They only depends upon properties of the material. Therefore, σ=αEδT = 12.5 x 10-6 x 200 x 103 x 30 = 75 MN/m2.

42 - Question

A cube having each side of length p, is constrained in all directions and is heated unigormly so that the temperature is raised to T.C. What will be the stress developed in the cube?

a) δET / γ

b) δTE / (1 – 2γ)

c) δTE / 2 γ

d) δTE / (1 + 2γ)

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Answer: bExplanation: δV/V = P / K = a3 (1 + aT)3 – a3) / a3 Or P / (E / 3(1-2γ)) = 3αT. advertisement

43 - Question

A steel rod 10mm in diameter and 1m long is heated from 20 to 100 degree celcius, E = 200 GPa and coefficient of thermal expansion is 12 x10-6 per degree celcius. Calculate the thermal stress developed?

a) 192 MPa(tensile)

b) 212 MPa(tensile)

c) 192MPa(compressive)

d) 212 MPa(compressive)

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Answer: cExplanation: αEδT = (12 x 10-6) ( 200 x 103) (100-20) = 192MPa.

44 - Question

A cube with a side length of 1m is heated uniformly a degree celcius above the room temperature and all the sides are free to expand. What will be the increase in the volume of the cube? Consider the coefficient of thermal expansion as unity.

a) Zero

b) 1 m3

c) 2 m3

d) 3 m3

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Answer: dExplanation: Coefficient of thermal expansion = 3 x coefficient of volume expansion.

45 - Question

The thermal stress is a function of _____________ P. Coefficient of linear expansion Q. Modulus of elasticity R. Temperature rise

a) P and Q

b) Q and R

c) Only P

d) Only R

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Answer: dExplanation: Stress in the rod is only due to temperature rise.

46 - Question

A steel rod is heated from 25 to 250 degree celcius. Its coefficient of thermal expansion is 10-5 and E = 100 GN/m2. if the rod is free to expand, the thermal stress developed in it is:

a) 100 kN/m2

b) 240 kN/m2

c) Zero

d) Infinity

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Answer: cExplanation: Thermal stress will only develop if the body is restricted.

47 - Question

Which one of the following pairs is NOT correctly matched?

a) Temperature strain with permitted expansion – ( αTl – δ)

b) Temperature thrust – ( αTE)

c) Temperature stress – (αTEA)

d) Temperature stress with permitted expansion – E(αTl – δ) / l

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Answer: aExplanation: Dimension analysis gives Temperature strain with permitted expansion – ( αTl – δ)is wrong. In other options the dimensions are correctly matched.

48 - Question

A steel rod of length L and diameter D, fixed at both ends, is uniformly heated to a temperature rise of δT. The Youngs modulus is E and the coefficient of linear expansion is unity. The thermal stress in the rod is ____________

a) Zero

b) T

c) EδT

d) EδTL

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Answer: cExplanation: As α = δl / l δT So, δl = l x 1 x δT And temperature strain = δl / l = δT As E = stress / strain Stress = E δT.

49 - Question

A uniform, slender cylindrical rod is made of a homogeneous and isotropic material. The rod rests on a frictionless surface. The rod is heated uniformly. If the radial and longitudinal thermal stress are represented by σx and σz, then ___________

a) σx = 0, σy = 0

b) σx not equal to 0, σy = 0

c) σx = 0, σy not equal to 0

d) σx not equal to 0, σy not equal to 0

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Answer: aExplanation: We know that due to temperature changes, dimensions of the material change. If these changes in the dimensions are prevented partially or fully, stresses are generated in the material and if the changes in the dimensions are not prevented, there will be no stress set up. (Zero stresses). Hence cylindrical rod Is allowed to expand or contract freely. So, σx = 0 and σy = 0.

50 - Question

which one of the following are true for the thermal expansion coefficient?

a) αaluminium > αbrass> αcopper > αsteel

b) αbrass > αaluminium > αcopper > αsteel

c) αcopper > αsteel > αaluminium > αbrass

d) αsteel > αaluminium > αbrass > αcopper

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Answer: aExplanation: Aluminium has the largest value of thermal expansion coefficient, then brass and then copper. Steel among them has lowest value of thermal expansion coefficient.

51 - Question

The length, coefficient of thermal expansion and Youngs modulus of bar A are twice of bar B. If the temperature of both bars is increased by the same amount while preventing any expansion, then the ratio of stress developed in bar A to that in bar B will be ___________

a) 2

b) 4

c) 8

d) 16

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Answer: bExplanation: Temperature Stress = EαδT So σ1 / σ2 = E1α1δT1/E2α2δT2 From question, α and E of bar A are double that of bar B.

52 - Question

**MCQs on Stress due to Materials Used and Their Applications**

Which test is conducted to measure the ability of a material to resist scratching, abrasion, deformation and indentation?

a) Creep test

b) Fatigue test

c) Hardness test

d) Compression test

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Answer: cExplanation: The ability of a material to resist scratching, abrasion, deformation and indentation is called hardness. So to measure this hardness test is used. it is generally expressed by Brinell, Rockwell or Vickers hardness numbers.

53 - Question

Which test is conducted to measure the endurance limit of the material?

a) Creep test

b) Fatigue test

c) Compression test

d) Hardness test

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Answer: bExplanation: The fatigue test is used to design components subjected to varying load. It experimentally determines the endurance limit of the material.

54 - Question

What is the process in which the metal is cooled rapidly in water after heating the metal above the lower critical temperature to increase the hardness of the material?

a) Quenching

b) Tampering

c) Hardening

d) Annealing

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Answer: aExplanation: Quenching is the process in which the metal is cooled rapidly in water after heating the metal above the lower critical temperature to increase the hardness of the material. Hardness is achieved during the quenching process depends on the amount of carbon content and cooling rate. advertisement

55 - Question

What is the process of heating the metal in the furnance to a temperature slightly above the upper critical temperature and cooling slowly In the furnance.

a) Quenching

b) Tampering

c) Annealing

d) Normalizing

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Answer: cExplanation: Annealing is the process of heating the metal in the furnance to a temperature slightly above the upper critical temperature and cooling slowly In the furnance. It produces an even grain structure, reduces hardness and increases ductility usually at a reduction of strength.

56 - Question

Photo stress method is ___________

a) Stress analysis method

b) Creep test

c) Ultra violet test

d) None of the mentioned

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Answer: aExplanation: Photo stress is a widely used full field technique for accurately measuring surface strains to determine the stresses in a part or structure during static of dynamic testing.

57 - Question

What is the factor of safety?

a) The ratio of total stress to the permissible stress

b) The ratio of ultimate stress to the permissible stress

c) The ratio of ultimate stress to the applied stress

d) The ratio of ultimate stress to the modulus of elasticity

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Answer: bExplanation: The ratio of ultimate stress to the permissible or working stress is called the factor of safety. This factor of safety is kept in mind in designing any structure.

58 - Question

Which one of the following has the largest value of thermal coefficient?

a) Brass

b) Copper

c) Steel

d) Aluminium

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Answer: dExplanation: Aluminium has the large value of thermal coefficient among them of value 24 x 10-6. whereas brass and copper has 19×10-6 and 17×10.

59 - Question

Identify which factor may cause a lowered body temperature:

a) Infection

b) Stress

c) Shock

d) Exercise

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Answer: cExplanation: Shock can cause the body temperature to drop, and so the cause of shock must be found. Other factors that can cause a lowered body temperature include: very young/old, serious haemorrhage, recovery from anesthesis and poisons.