Engineering Questions with Answers - Multiple Choice Questions

Home » MCQs » Agriculture Engineering » Strength of Materials MCQs Maximum Shear Stress – 2

# Strength of Materials MCQs Maximum Shear Stress – 2

Calculate the maximum shear force for square beam of side is 320 mm. If the shear force is 94kN.

a) 1.37N/mm2

b) 2.36N/mm2

c) 5.21N/mm2

d) 4.32N/mm2

**
View Answer**

Answer: a

Explanation: Maximum shear force is 3/2 × F/a×a ( a = side of square)

= 3/2 × 94×103/320×320

= 1.3769 N/mm2.

A simply supported beam of span 8 metres carries a udl of 16 kN/m at a point out of 60 kN acting at it’s centre. Calculate the maximum shear force.

a) 87kN

b) 45kN

c) 78kN

d) 94kN

**
View Answer**

Answer: d<br/>

Explanation: Maximum shear force is w×l/2

= 60+16×8 / 2

= 94 kN.

The ratio of creep strain to elastic strain is known as ___________

a) Creep factor

b) Creep postulate

c) Creep coefficient

d) Creep variable

**
View Answer**

Answer: c

Explanation: Creep is defined as plastic deformation under a constant load or stress the creep Coefficient which is defined as the ratio of ultimate creep strain to the elastic strain at various ages of loadings.

Poisson’s ratio for high strength concrete is __________

a) 0.049

b) 0.095

c) 0.1

d) 0.1111

**
View Answer**

Answer: c

Explanation: Poisons ratio varies between 0.1 for high strength concrete and 0.2 for weak concrete. Usually it is taken as 0.15 for strength design and 0.2 for serviceability conditions.

Partial safety factor for concrete is taken as ____________

a) 1.3

b) 1.2

c) 1.5

d) 1.6

**
View Answer**

Answer: c

Explanation: A higher value of partial safety factor for concrete 1.5 has been adopted because there are greater chances of variation of the strength of concrete due to improper compaction, inadequate curing, improper batching and mixing.

The design compressive strength of concrete is ___________ times of characteristic compressive strength of concrete.

a) 0.313

b) 0.253

c) 0.466

d) 0.411

**
View Answer**

Answer: c

Explanation: The compressive strength of concrete in the structure is assumed to be 0.67 times the characteristic strength of concrete. The partial safety factor equal to 1.5 is applied to the strength of concrete in addition to it therefore the design compressive strength of concrete is 0.67 fck / 1.5 equal to 0.446 fck. [fck = characteristic compressive strength].

In cantilever beams, the steel bars are placed at ___________

a) Bottom of the beam

b) Top of the Beam

c) Midspan of the Beam

d) Near supports

**
View Answer**

Answer: b

Explanation: In cantilever beams, steel bars are placed near the top of the beam to resist the tensile stresses developed in top layers due to the negative bending moment that is hogging bending moment.

Calculate the level arm factor of a section of M20 grade and if Fe 415 Steel. [Take critical neutral axis factor as 0.289].

a) 0.78

b) 0.9

c) 0.58

d) 0.73

**
View Answer**

Answer: b

Explanation: Lever arm factor (j) = 1-k/3

Where k= 0.289

j = 1-0.289/3

= 0.904~0.9.

Working stress method is based on elastic theory assumptions.

a) True

b) False

**
View Answer**

Answer: a

Explanation: Working stress method is based on elastic theory assuming reinforced concrete as elastic material. The stress strain curve of concrete is assumed as linear from zero at the neutral axis to a maximum value at the extreme fibre. In the working stress method, members are designed for working loads such that the stresses developed are within the allowable stress.

Modular ratio method is also known as ______

a) Ultimate stress method

b) Limit state method

c) Working stress method

d) Stress and strain method

**
View Answer**

Answer: c

Explanation: The stress in steel is linearly related to the stresses in adjoining concrete by constant factor called modular ratio (defined as the ratio of modulus of elasticity of steel to that of concrete) Working stress method is therefore also known as modular ratio method.

Find the moment of inertia about centroid axis of a triangular section are having base 100 mm and height 150 mm.<br/>

a) 9.21×106mm4<br/>

b) 9.45×106mm4<br/>

c) 9.37×106 mm4<br/>

d) 8.51×106 mm4

**
View Answer**

Answer: c

Explanation: b = 100mm

h = 150 mm

Moment of inertia about centroid Axis = bh3 / 36.

= 100 ×1503/ 36

= 9.37×106mm4.

The stress corresponding to ______ of strain in the stress-strain curve of mild steel is known as proof stress.

a) 0.2%

b) 0.32%

c) 0.5%

d) 0.6%

**
View Answer**

Answer: a

Explanation: The stress corresponding to 0.2% of strain in the stress-strain curve of mild steel is known as proof stress. This is also taken as yield stress. The maximum stress is generally taken as yield stress.

__________ is the device used for measuring normal stresses on the surface of a stressed object.

a) Nephelometer

b) Straining appurtenances

c) Resistance strain gauge

d) Volt-Hypsometer gauge

**
View Answer**

Answer: c

Explanation: An electrical resistance strain gauge is a device for measuring normal strains on the surface of a stressed object. The gauges are small (less than half inch) made of wires that are bonded on surface of the object. We can use the transformation equations for plane strain to calculate the strains in various directions.

The compressive strength of brittle materials is ______ its tensile strength.

a) Less than

b) Greater than

c) Equal to

d) Depends on material

**
View Answer**

Answer: b

Explanation: The compressive strength of brittle materials is always greater than its tensile strength. In the same way, the tensile strength of ductile materials is greater than its compressive strength.

The breaking stress is _______ the ultimate stress.

a) Less than

b) Greater than

c) Depends on time

d) Equal to

**
View Answer**

Answer: a

Explanation: The stress corresponding to the ultimate load is known as ultimate stress and the stress corresponding to breaking point is known as breaking stress. In the stress strain curve, the ultimate stress is above the breaking stress. Hence the ultimate stress is greater than breaking stress.

### Strength of Materials MCQs Combined Stress

Bond stress for M20 grade concrete in tension is ____________

a) 1.4

b) 1.2

c) 1.5

d) 1.8

**
View Answer**

Answer: b

Explanation: Bond stress is the shear stress acting parallel to the bar on the interface between the reinforcing bar and the surrounding concrete. Hence it is the stress developed between the contact surface of Steel and concrete to keep them together. The value of M20 designs Bond stress is 1.2 in tension.

The formation of diagonal cracks at junctions is due to ________

a) Shear stress

b) Bond stress

c) Temperature stress

d) Lateral stress

**
View Answer**

Answer: a

Explanation: Bending is usually accompanied by shear. The combination of shear and bending stresses produces the principle stresses which causes diagonal tension in the beam section. This should be resisted by providing shear reinforcement in the form of vertical stirrups (or) bent up bars along with stirrups.

Calculate the factored bending moment of a rectangular reinforced concrete beam of effective span 4300 mm and load imposed 37.5 kN/m.

a) 100kNm

b) 127kNm

c) 130kNm

d) 145kNm

**
View Answer**

Answer: c

Explanation: Factored load (w) = 1.5×37.5 = 56.25 kN/m.

Factored bending moment for simply supported beam (M) = wl2/ 8. = 56.25×(4.3)2/ 8 = 130kNm.

Determine the limiting percentage of steel for singly reinforced sections of M20 grade & Fe415.

a) 0.68

b) 0.79

c) 0.96

d) 1.76

**
View Answer**

Answer: c

Explanation: The limiting percentage of steel for singly reinforced sections of M20 grade & Fe415 is 0.96.

Grade of concrete Limiting percentage of tensile Steel for a Fe415

M15 0.72

M20 0.96

M25 1.19

Calculate the limiting depth of the neutral axis for mild steel of effective depth 400 mm.

a) 318mm

b) 212mm

c) 455mm

d) 656mm

**
View Answer**

Answer: b

Explanation: The limiting depth of neutral axis Fe 250 steel is

Xu (max) = 0.53 × d ( for Fe250)

= 0.53 × 400

= 212mm.

Lap splices should not be used for bars larger than _____ mm.

a) 45mm

b) 54mm

c) 36mm

d) 72mm

**
View Answer**

Answer: c

Explanation: Splices are provided when the length of the bar is less than that required. The splicing of reinforcement is provided either by lap joint or mechanical joint or welded Joint. Lap splices should not be used for bars larger than 36 mm for larger diameter, bars may be welded.

Anchorage value for “U” hook is ________

a) 16 × diameter of bar

b) 12 × diameter of bar

c) 10 × diameter of bar

d) 8 × diameter of bar

**
View Answer**

Answer: a

Explanation: Anchorage value for “U” hook is 16 × diameter of bar.

Type of Hook / Bend in degrees Anchorage Value

U hook 16 × diameter of bar

45 bend 4 × diameter of bar

90 bend 8 × diameter of bar

135 bend 12 × diameter of bar

The standard __________ are provided in deformed bars.

a) Anglets

b) Bends

c) Fillets

d) Lugs

**
View Answer**

Answer: b

Explanation: In situations, where straight anchorage length cannot be provided due to lack of space. To improve the anchorage of bars, standard bends are provided in deformed bars.

Transverse bars are also called as _________

a) Main bars

b) Anchor bars

c) Distribution bars

d) Stirrups

**
View Answer**

Answer: c

Explanation: In addition to main bars, along the shorter direction provided at the bottom, minimum reinforcement along the longer span and are also provided on top of the main bars and at right angles to them. These are called distribution bars are transverse bars.

A slab supporting only in two edges opposite to each other is ______

a) Two way slab

b) One way slab

c) Continuous slab

d) Cantilever slab

**
View Answer**

Answer: b

Explanation: If the ratio of the longest span the shorter span is greater than 2 or A slab supporting only in two edges (opposite to each other) is called one way slab. This slab spans across shorter span practically.

Torsion reinforcement is provided in ___________ slab

a) One way slab

b) Two way slab

c) Simply supported slab

d) Cantilever slab

**
View Answer**

Answer: b

Explanation: A slab supporting on all four edges is known as two way slab. In this slab, the ratio of longest span to the shorter span is less than 2. It requires torsional reinforcement because there’s a chance of twisting at corners.

Generally in residential buildings, the width of stay is kept as ____________

a) 2m

b) 1m

c) 5m

d) 4m

**
View Answer**

Answer: b

Explanation: The stair consists of series of steps with landings at appropriate intervals. The width of stair depends upon the type of building in which it is provided. Generally, in residential buildings, the width of stair is 1 m.

As per IS 456:2000; the slope or pitch of stairs should be in between 25 ° to ___________

a) 45°

b) 90°

c) 40°

d) 120°

**
View Answer**

Answer: c

Explanation: Each step has one tread and one rise. As per IRC, the tread is in between 250mm to 300 mm. The slope or pitch of the stairs should be in between 25° to 40°.

When space is less, the ___________ staircases is much preferred.

a) Open well

b) Dog legged

c) Spiral stair

d) Circular

**
View Answer**

Answer: b

Explanation: The most common type of Stairs arranged with two adjacent flights running parallel with mid landing. Where the space is less, dog legged staircase is generally provided resulting in economical utilisation of available place.

The ______________ of a column is the distance between the points of zero bending moments.

a) Slenderness ratio

b) Eccentricity

c) Radius of gyration

d) Effective length

**
View Answer**

Answer: d

Explanation: Effective length of a column is the distance between the points of zero bending moments (point of contra flexure) of a buckled column the effective length of the column depends upon the unsupported length and the end conditions.