Test: Slope & Deflection Level - 3 - Mechanical Engineering MCQ

# Test: Slope & Deflection Level - 3 - Mechanical Engineering MCQ

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## 15 Questions MCQ Test Strength of Materials (SOM) - Test: Slope & Deflection Level - 3

Test: Slope & Deflection Level - 3 for Mechanical Engineering 2024 is part of Strength of Materials (SOM) preparation. The Test: Slope & Deflection Level - 3 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Slope & Deflection Level - 3 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Slope & Deflection Level - 3 below.
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Test: Slope & Deflection Level - 3 - Question 1

### A simply supported beam of span 6 m carries two concentrated loads of 2N each at 1 m distance from each of the supports. If the flexural rigidity of the beam is constant. the slope at the support is

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 1

Slope at B = due to symmetry

=

=

=

Test: Slope & Deflection Level - 3 - Question 2

### A prismatic beam of length L is simply supported at its ends and subjected to a total UDL of W spread over its entire span. It is then propped at its Centre to neutralize the deflection. The net B.M. at its Centre will be

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 2

The propped reaction neutralize Downward deflection due to udl = upward Deflection due to support at C

∴ moment at C Mc =

RA + RB + RC = W

∴ Rc =

RA + RB = W -

Due to symmetry RA = RB =

∴ Mc =

=

=

= (Clockwise)

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Test: Slope & Deflection Level - 3 - Question 3

### A cantilever beam of uniform EI has a span equal to ‘L’. An upward force W acts at the midpoint of the beam and a downward force P acts at the free end. In order that the deflection at the free end is zero, the relation between P and W should be

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 3

Upward deflection at B due to W

=

=

=

Downward deflection due to P =

W=

Test: Slope & Deflection Level - 3 - Question 4

A point load ‘W’ is acting at the mid-span of a cantilever of length ‘L’. If the free end is supported on a rigid prop, the reaction of the prop is

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 4

yb =

ya =

Equating upward and downward deflection

VB =

Test: Slope & Deflection Level - 3 - Question 5

A cantilever AB, 3 m long carries a uniformly distributed load of 2 kN/m throughout its length, rests over a similar cantilever of same cross-section and same material, 2 m long and shown below. What is reaction C?

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 5

Let the reactions at C is R. Deflection at C in cantilever CD,

δC =

=

= δB (in cantilever AB) ⋯ ①

YB1 =

YB2 =

YB = YB1 + YB2 =

From equation ① and ②

R = 1.736 kN

Test: Slope & Deflection Level - 3 - Question 6

A beam ABC of length 3L has one support at the left end and the other support at a distance 2L from the left end. The beam carries a point load W at the right end. The deflection at the right end is

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 6

solving RA =

RB = 3/2W

Deflection at C,

δc =

Test: Slope & Deflection Level - 3 - Question 7

A cantilever of Length ‘L’ carries a uniformly distributed load of w per unit run for a distance ‘a’ from the fixed end. The deflection at free end is

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 7

BM at B in conjugate beam = deflection at B in original beam

δn =

=

=

=

Method 2:

Deflection B = deflection at C + θc × (L − a)

=

=

Test: Slope & Deflection Level - 3 - Question 8

A uniform beam of length ‘L’ is simply and symmetrically supported to a span ‘l’ find the ratio L/l so that the upward deflection at each end equals the downward deflection at mid span, due to a concentrated point load at mid span.

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 8

Deflection at midpoint = deflection at free end

Test: Slope & Deflection Level - 3 - Question 9

A prismatic simply supported beam carries a point load at mid-span section due to which a slope of 2° is produced at support sections. The slope at quarter span sections is

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 9

=

=

= 2 - 0.5 = 1.5°

Test: Slope & Deflection Level - 3 - Question 10

A beam of span 6m and of uniform flexural rigidity EI = 40000 kNm2 is subjected to a clockwise couple of 300 kNm at a distance of 4m from the left end. The deflection at the point of application of the couple is _______ mm

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 10

Taking moments about A, Vb × 6 = 300

Vb = 50 kN ↑

∴ Va = 50 kN ↓

The B.M at any section distant x from A is given by,

Integrating,

Integrating !gain,

EIy =

AT x = 0, t = 0

∴ C2 = 0

At x = 6, y = 0

∴ 0 =

∴ C1 = 200

Deflection at C. Putting x = 4m, in the deflection equation,

EIyc =

= 266.67

Max. deflection. This will occur in the larger segment. Equating the slope to zero, we get,

∴ x = 2√2 m

Putting x = 2√2 m in the deflection equation

EIymax =

ymax =

= 9.43 mm

Test: Slope & Deflection Level - 3 - Question 11

A cantilever 3m long, and of symmetrical section 250 mm deep carries a uniformly distributed load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end. The deflection at the free end is __________ mm (Take E = 200 GPa, I = 54000 cm4)

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 11

AB fixed at end B, Free at end A carrying a point load W at L1 from B end UDL of W throughout to length EI = flexural rigidity δ = deflection at free end

=

where, L1 = 1.2m, L2 = 1.8m, L = 3m,

W = 80 kN and W = 30 kN/m

EI = 200 × 106 kN/m2 × 54000 × 10−8 m4

= 108000 kNm2

δ =

= 0.42666 × 10−3 + 0.96 × 10−3 + 2.8125 × 10−3

= 4.2 × 10−3m = 4.2 mm

Test: Slope & Deflection Level - 3 - Question 12

A cantilever of length 6 m is loaded as shown in the figure below. The maximum deflection in the beam is

E = 200 MPa, I = 1 × 10−4 m−4, W = 1 kN.

P1 = 3W P2 = 2W P3 = 1W

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 12
EI = 200 × 106 × 1 × 10−4 = 200,00 kNm2

Slope at free end

I =

=

=

Deflection at free end

y =

=

=

=

= 0.00893 m = 8.93 mm

Test: Slope & Deflection Level - 3 - Question 13

A horizontal steel rod ABC of diameter 50 mm and length 400 mm carrying a uniformly distributed load of 1.8 kN/m is simply supported at its ends A and C and is also supported at its mid-point B by its connection to a hanging vertical steel wire of 5 mm diameter and length 360 mm as shown in figure below. The deflection at point B is _________ mm (Take E = 2 × 105 N/mm2).

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 13

Let T = Tension in the wire BD Deflection of B = Extension of the wire

l1 = 400 mm, l2 = 360 mm,

W = 1.8 kN/m = 1.8 N/mm

D = 50 mm, I = = 306796.875 mm

d = 5 mm, A = = 19.635 mm2

T = = 86.228 N

∴ Deflection of B

= = 0.0079 mm

Test: Slope & Deflection Level - 3 - Question 14

A horizontal beam (I = 8.616 × 107mm4) carries a uniformly distributed load of 50 kN over its length of 3m. The beam is supported by three vertical steel tie rods, each 1.80m long. One at each end and one at in the middle, the end rods having a diameter of 25mm and the centre rod a diameter of 30mm. The deflection of the centre of the beam is _________mm (Take E = 2 × 105 N/mm2)

(Take E = 2 × 105 N/mm2)

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 14

Area of each rod

= A1 = = 490.87 mm2

Area of the middle rod

= A2 = = 706.86 mm2

Let, P1 = Tension in each end rod

P2 = Tension in the middle rod

f1 = Stress in the rods

f2 = Stress in the middle rod

δ1 = Extension of each end rod

δ2 = Extension of the middle rod

Deflection of the midpoint of the beam

δ2 − δ1 =

EI(δ2 − δ1) =

Deflection of beams

8.616 × 107 × 1800(f2 − f1)

=

2205.696(f2 − f1) = 250 − 5654.88f2

2205.696f2 − 2205.696f1 = 250 − 5654.88f2

7860.576f2 − 2205.696f1 = 250 ⋯ ①

2f1A1 + f2A2 = 50

2f1(490.87) + f2 (706.86) = 50

981.74f, + 706.86f, = 50 ⋯ ②

From equation ① and ②,

7860.576f2 − 2205.696f1 = 5(981.74f,1 + 706.86f2)

7860.576f2 − 2205.696f1 = 4908.70f1 + 3534.3f2

4326.76f2 = 7114.396f1

∴ f2 = 1.644f1

Substituting in equation ②

981.74f1 + 706.86 × 1.644f1 = 50

2143.8178f1 = 50

f1 = 0.02332 kN/mm2 = 23.32 N/mm2

f2 = 1.644 × 23.32 = 38.33 N/mm2

Deflection of the midpoint of the beam

δ2 − δ1 =

=

=0.135 mm

Test: Slope & Deflection Level - 3 - Question 15

determine the slope and deflection at a point in a beam is suitable for beams subjected to concentrated loads and can be extended to uniformly distributed loads Reason(R): Macaulay’s method is based upon the modification of moment of area method. This is applicable to a simple beam carrying a single concentrated load but by superposition, this method can be extended to cover any kind of loading.

Detailed Solution for Test: Slope & Deflection Level - 3 - Question 15
Macaulay’s method is based on singularity function. It is applicable for prismatic beams only. While Mohr’s moment area method can be used for prismatic and non-prismatic beams.

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