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# Strength of Materials MCQs Introduction to Shear Stress

At ________ the shearing stress in a beam are maximum.

a) Extreme fibres

b) Modulus of section

c) Neutral axis

d) Along the cross-sectional area

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Answer: cExplanation: Shearing stress in a beam is maximum at the neutral axis. Shearing stress is defined as the resistance offered by the internal stress to the shear force.

Determine the shear stress at the level of neutral axis, if a beam has a triangle cross section having base “b” and altitude “h”. Let the shear force be subjected is F.

a) 3F/8bh

b) 4F/3bh

c) 8F/3bh

d) 3F/6bh

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Answer: cExplanation: For a triangular section subjected to shear force the shear stress in neutral axis is Shear stress at NA = 4/3 [Average shear stress]. = 4/3 [F/0.5 bh] = 8F / 3bh.

The maximum shear stress is ______ times the average shear stress [For rectangular beams].

a) 2.5

b) 3

c) 1.2

d) 1.5

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Answer: dExplanation: The maximum shear stress occurs at neutral axis. Then y = 0. Max shear stress = 3F/2bd = 3/2 [F/bd]. = 1.5 Average shear stress.

Shear stress in a beam is zero at ______

a) Neutral axis

b) Extreme fibres

c) Cross section

d) Junctions

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Answer: bExplanation: The resistance offered by the internal stress to shear is known as shearing stress. Shearing stress is zero at extreme fibres of the beam. The bending stresses are maximum at extreme fibres of the beam cross section.

Shear stress distribution over rectangular section will be _________

a) parabolic

b) elliptical

c) triangular

d) trapezoida

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Answer: a<br/>

Explanation:

Maximum shear stress is 1.5 times that of average shear stress.

The shear stress distribution is parabolic.

A round Steel rod of 100 mm diameter is bent into an arc of radius 100m. What is the maximum stress in the rod? Take E = 2×105 N/mm2.

a) 150 N/mm2

b) 200 N/mm2

c) 100 N/mm2

d) 300 N/mm2

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Answer: cExplanation: D = 100m y = 50 mm R = 10 × 103mm By equation of flexure; E/R = f/y f=E/R ×y = 2×105/ 100 × 103× 50 = 100 N/mm2.

For circular section, the maximum shear stress is equal to ____________ times of average shear stress.

a) 2/3

b) 3/2

c) 4/3

d) 3/4

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Answer: cExplanation: Maximum shear stress occurs at neutral axis & y = 0. Max. Shear stress = 4/3 [ F/A ]. F/A is average shear stress. The maximum shear stress distribution is 33% more than average shear stress.

A steel beam is 200 mm wide and 300 mm deep. The beam is simply supported and carries a concentrated load w. If the maximum stress are 2 N/mm2. What will be the corresponding load?

a) 50 kN

b) 80 kN

c) 40 kN

d) 85 kN

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Answer: bExplanation: For a rectangular cross section Max. Shear stress = 3/2 [F/A ] 2 = 3/2 [w/200 × 300]. w = 80 kN.

Maximum shear stress in thin cylindrical shell be ___________

a) pr/2t

b) pr/3t

c) pr/4t

d) pr/ 5t

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Answer: cExplanation: Hoop stress P(h) = pr/t Longitudinal stress P(l) = pr/2t Thus, hoop stress is twice the longitudinal stress Max. Shear stress = P(h) – P(l) / 2 = pr/4t.

Circumferential stress is same as of _________

a) Hoop stress

b) Longitudinal stress

c) Transverse stress

d) Phreatic stress

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Answer: aExplanation: In a thin cylindrical shell of internal radius r thickness t when subjected to internal fluid pressure P, the stress developed in the internal walls can be termed as circumferential stress or hoop stress. P(h) = pr/t.

A beam has a triangular cross-section, having altitude ”h” and base “b”. If the section is being subjected to a shear force “F”. Calculate the shear stress at the level of neutral axis in the cross section.

a) 4F/5bh

b) 4F/3bh

c) 8F/3bh

d) 3F/4bh

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Answer: cExplanation: For a triangular section subjected to a shear force, the shear stress at neutral axis is = 4/3 × average shear stress = 4/3 × F/A/2 ; A = bh = 8F/3bh.

The maximum shear stress in the rectangular section is ______________ times the average shear stress.

a) 3/4

b) 3/7

c) 5/3

d) 3/2

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Answer: dExplanation: The maximum shear stress occurs at the neutral axis. So, y = 0. Maximum shear stress = 3/2 × F/bd (• Average shear stress = F/bd ). = 3/2 × average shear stress.

The modular ratio for M20 grade concrete is _____________

a) 16

b) 13

c) 11

d) 07

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Answer: bExplanation: According to Indian Standards 456-2000 The modular ratio (m) = 280/3× cbc, For M20; compressive bearing capacity in concrete = 7 N/mm2 & tensile strength = 330 N/mm2. Modular ratio (m) = 280/21 = 13.33.

In doubly reinforced beam, the maximum shear stress occurs ______________

a) along the centroid

b) along the neutral axis

c) on the planes between neutral axis and tensile reinforcement

d) on the planes between neutral axis and compressive reinforcement

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Answer: dExplanation: In continuous beam the moments developed at the supports are greater than a moment’s developed at the mid span, show the maximum bending moment occurs at the supports. For continuous beams, the maximum shear stress occurs at the planes intersecting the compressive reinforcement and the neutral axis.

A cylindrical section having no joint is known as _______________

a) Proof section

b) Seamless section

c) Target section

d) Mown section

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Answer: bExplanation: A cylindrical section having no joint is known as seamless section. Built up section is not that strong as a seamless section of the same thickness.

The efficiency of cylindrical section is the ratio of the strength of joint to the strength of _______________

a) Solid plate

b) Boilerplate

c) Circumferential plate

d) Longitudinal plate

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Answer: aExplanation: The strength of plate or strength of rivet whichever is less is called the strength of joint. The ratio of the strength of joint to the strength of steel plate is called the efficiency of the cylinder.

Calculate the modulus of section for a hollow circular column of external diameter 60 mm and 10 mm thickness.

a) 170 m

b) 190 m

c) 250 m

d) 300 m

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Answer: a<br/>

Explanation: Given data :

D = 60 mm ; t = 10 mm & d = 60 – 2×10 = 40 mm

For hollow circular section, modulus of section( Z ) = 3.14 × D4 – d4 / 32 D.

= 17016.3 mm = 170 m.

Determine the modulus of a section for an I section, given the distance from neutral axis is 50 mm and moment of inertia is 2.8×106 mm4.

a) 59m

b) 51m

c) 58m

d) 63m

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Answer: cExplanation: The modulus of section is the ratio of the moment of inertia to the distance of the neutral axis. Given y = 50 mm I = 2.8×106 mm4. & Z = I/y = 2.8×106 / 50 = 57.76 ×103 mm = 57.7 ~ 58 m.

A circular Beam of 0.25 m diameter is subjected to you shear force of 10 kN. Calculate the value of maximum shear stress. [Take area = 176 m2].

a) 0.75 N/mm2

b) 0.58 N/mm2

c) 0.73 N/mm2

d) 0.65 N/mm2

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Answer: aExplanation: Given diameter = 0.25 m Area (A) = 176 m2 Shear Force (F) = 10 kN ~ 10000 N. For circular cross section the maximum shear stress is equal to 4/3 times of average shear stress Maximum shear force = 4/3 × F/A = 4/3 × 10000/176 = 0.75 N/mm2.

The maximum shear stress distribution is _____________ percentage more than average shear stress in circular section.

a) 54 %

b) 60 %

c) 33 %

d) 50 %

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Answer: cExplanation: Maximum shear stress occurs at neutral axis; y =0 Maximum shear stress = 16/3 × average shear stress But 4F / A is the average shear stress. So, the maximum shear stress = 4/3 times the average shear stress. Hence the maximum shear stress is 33% more than the average shear stress in a circular section.

### Strength of Materials MCQs Maximum Shear Stress – 1

Shear stress at top most fibre of rectangular section is _____________

a) Maximum

b) Minimum

c) Zero

d) Uniform through out

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Answer: cExplanation: In rectangular section, The shear stress at a distance “y “ from NA = 6F/bd3 × u (u = d2/4-y) The maximum shear stress occurs at a neutral axis, in the above equation when y is equal to zero. q is max. Hence the shear stress topmost fibre of rectangular section is zero.

1 GPA = ____________ pa.

a) 105

b) 106

c) 108

d) 109

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Answer: dExplanation: 1 Giga Pascal is equal to 109N/m2(Pascal) In the same way 1 kilo Pascal equal to 103 pascals 1 mega Pascal is equal to 106 pascals.

The maximum shear stress in an I section is __________

a) F/8I ×[B/b (D2-d2)+d2]

b) F/6I ×[B/b (D2-d2)+d2]

c) F/8I ×[B/b (D3-d3)+d2]

d) F/4I ×[B/b (D2-d2)+d2]

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Answer: aExplanation: Shear stress at top flange of the I section is zero. Shear stress at the junction of web and flange= B/b ×F/8I (D2-d2). Shear stress at bottom of the flange = F/8I (D2-d2). And shear force is maximum at neutral axis i.e F/8I ×[B/b (D2-d2)+d2].

Find the modulus of section of square beam of size 150 × 150 mm?<br/>

a) 654.5m^{3<br/>}

b) 550.85m^{3<br/>}

c) 562.5m^{3<br/>}

d) 586.9m^{3}

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Answer: cExplanation: Here, a = side of a square section = 150 mm. Moment of inertia for square section = a4/12; y=a/2. Section modulus Z = I/y = a3/6 = 1503/6 = 562.5 ×103 mm3.

In steel sections, the junction between a flange and web is known as ________

a) Edge

b) Fillet

c) Corner

d) Lug

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Answer: bExplanation: In a steel section, the junction between the flange and the web is known as fillet. The connections solve issues of complex geometry for joining the members of a central hub while they provide the standard connection through out. They are not readily available.

The percentage of carbon in structural steel is __________

a) 0.2 – 0.27 %

b) 0.6 – 0.85 %

c) 0.7 – 1.23 %

d) 1.23 – 1.45%

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Answer: aExplanation: The percentage of carbon in structural steel is 0. 2 to 0.27. Percentage of the carbon in steel increases the ductility of the Steel decreases.

The minimum percentage elongation for mild steel is __________

a) 6%

b) 13%

c) 23%

d) 34%

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Answer: cExplanation: The minimum percentage elongation for mild steel is 23% and the tensile strength of steel is usually taken as 42 to 54 kg/mm2.

GOST standards are used in _________

a) Italy

b) Poland

c) Russia

d) Pakistan

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Answer: cExplanation: GOST is an acronym for gosudastvennyy standard used in Russia. It usually carries two part number, one indicates serial number and other indicates the year of issue For example; GOST 155-70.

The allowable tensile stresses in steel structures is taken as 1500 kg /cm2 to ______

a) 1765 kg /cm2

b) 1900 kg /cm2

c) 2125 kg /cm2

d) 2455 kg/cm2

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Answer: cExplanation: Steel structures are available in various sections such as rolled I beams, channels, angle iron, bars, flat plates etc. The allowable tensile stress in steel structures is 1500 kg /cm2 to 2125 kg /cm2.

As per IS:800, the minimum thickness of web should not be less than ______

a) d/250

b) d/300

c) d/350

d) d/125

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Answer: aExplanation: As per IS: 800, the minimum thickness of web should not be less than d/250; [Where d = clear distance between Flange angles]. In case of unstiffened web, the minimum thickness of web plate should not be less than d/85.

The failing of a very long column is initially by ___________

a) Crushing

b) Collapsing

c) Buckling

d) Twisting

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Answer: cExplanation: The members considerably long in comparison of lateral dimensions are called Long columns. The members essentially fail by buckling (or) crippling to bending. According to Euler’s formula the long column can be determined.

What is the allowable stress in cast iron?

a) 3200 N/mm2

b) 2400 N/mm2

c) 3400 N/mm2

d) 5500 N/mm2

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Answer: dExplanation: The allowable stress in cast iron is 5500 N/mm2. Position Stress (N/mm2) Rankine’s Constant Mild steel 3200 1/7500 Wrought iron 2500 1/9000 Cast iron 5500 1/1600

Modulus of resilience is defined as __________

a) Resilience at ultimate stress

b) Resilience per unit volume

c) Resilience at proportional limit

d) Resilience at elastic limit

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Answer: bExplanation: The resilience per unit volume is defined as modulus of resilience. It is a property of the material. The Modulus of resilience is equal to 1Mpa for Steel with the proportionality limit of 200 Mpa.

A spring used to absorb shocks and vibrations is called as _______

a) Conical spring

b) Leaf spring

c) Disc spring

d) Torsion spring

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Answer: bExplanation: A leaf spring used to absorb shocks and vibrations and the springs in brakes and clutches are invariably used in order to apply forces.

A rectangular beam of 500 mm wide is subjected to maximum shear force of 250kN, the corresponding maximum shear stress been 3 N/mm2. The depth of the beam is equal to ______<br/>

a) 200mm<br/>

b) 250mm<br/>

c) 300mm<br/>

d) 350mm

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Answer: bExplanation: The maximum shear force in a rectangular section is 3N/mm2. In rectangular sections; Maximum shear force = 3/2 ×[F/bd] & 3 = 3/2 ×[250 ×103/ 500 × d] d = 250mm.