Deflection of Bar MCQ Quiz in मल्याळम - Objective Question with Answer for Deflection of Bar - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 15, 2025

നേടുക Deflection of Bar ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Deflection of Bar MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Deflection of Bar MCQ Objective Questions

Top Deflection of Bar MCQ Objective Questions

Deflection of Bar Question 1:

When a circular rod of uniform cross-section A, Length L, and material Young’s modulus E, is subjected to a tensile force of P, then an increase in the length of the rod is given by the equation______

  1. PLAE
  2. P2LAE
  3. PL2AE
  4. PLA2E

Answer (Detailed Solution Below)

Option 1 : PLAE

Deflection of Bar Question 1 Detailed Solution

Concept:

If a rod is under tensile force, there will be an increase or decrease in length due to the developed stress and strain in it.

Stress is calculated by Force (F) divided by the cross-section area (A) on which force is applied.

Stress=FA

The strain is the change in length of rod (ΔL) divided by the original length (L) of the rod.

Strain=ΔLL

Hook's Law of elasticity

 Youngs modulus=stressstrain

F1 J.K 2.7.20 Pallavi D2

Calculation:

Given:

Tensile force = P,  Increase in length of rod = ΔL, Original length of rod = L, young’s modulus = E, cross-section area of rod = A.

Stress=PA

Strain=ΔLL

Youngs modulus=stressstrain

E=PAΔLL

E=PLΔLA

ΔL=PLEA

Deflection of Bar Question 2:

When a rod of circular cross-section is fixed at one end and subjected to an axial load of 500 N, the deflection under the load is found to be 2.4 mm, what will be the amount of deflection under the same load if the dia of rod is doubled & length is reduced to half of the original length?

  1. 1.2 mm
  2. 0.6 mm
  3. 0.3 mm
  4.  0.15 mm

Answer (Detailed Solution Below)

Option 3 : 0.3 mm

Deflection of Bar Question 2 Detailed Solution

Concept:

Deflection of circular bar under axial loading

δ=4PLπd2E

where P = load, L = length, d = diameter of cross sectional area, E = Young's modulus

Calculation:

Given:

For case-1: P1 = 500 N, δ1 = 2.4 mm, L1 = L, d1 = d, δ1 = 2.4 mm

For case-2: P2 = 500 N, δ2 = 2.4 mm, L2 = L2, d2 = 2d

let E be the Young's modulus for both the cases, hence ratio of deflecction in both the case is;

δ1δ2=4PL1πd12E4PL2πd22E

δ1δ2=Ld2L2(2d)2

δ1δ2=8

δ2δ1=18

δ2=18×2.4

δ2 = 0.3 mm

Hence deflection under the same load for second case is 0.3 mm

Deflection of Bar Question 3:

A cantilever beam AB of length L and uniform flexural rigidity E has a bracket AC attached to its free end shown in figure. Vertical load is applied to free end C of the bracket. In order that the deflection of point A to be zero the ratio a/L should be

Assignment 35 Jyoti ISRO ME 2010 13Qs.docx   1

  1. ½
  2. 1/3
  3. ¼
  4. 2/3

Answer (Detailed Solution Below)

Option 4 : 2/3

Deflection of Bar Question 3 Detailed Solution

Concept:

So, given below is the equivalent diagram showing Force P and moment (P.a) acting on the free end A of the cantilever beam.

Diagram

Downward deflection due to load P at the free end A = PL33EI

Upward deflection due to moment at free end A = ML22EI

Since moment at free end M = P × a

∴ Upward deflection due to moment at free end A = (P×a)L22EI

Therefore, for zero deflection at A:

Upward deflection = downward deflection

 (P.a)L23EI=PL33EI

 aL=23

Deflection of Bar Question 4:

In case of a composite bar having equal length, the total load on a composite bar is equal to

  1. sum of the loads carried by each different materials
  2. inverse of sum of the loads carried by each different materials
  3. square of sum of the loads carried by each different materials
  4. square root of sum of the loads carried by each different materials

Answer (Detailed Solution Below)

Option 1 : sum of the loads carried by each different materials

Deflection of Bar Question 4 Detailed Solution

Concept:

Composite bar:

  • A composite bar (bars in parallel) is made of different materials and they are rigidly fixed together so that both bars strain together under external load.
  • The extension or contraction of the bar is equal.
  • The total external load on the bar is equal to the sum of the loads carried by different materials.

For a composite bar, the strain in both the bars will be the same i.e.

ϵ=PAE=constant

Total load (P) = P1 + P2

qImage10379

 

Deflection of Bar Question 5:

A 0.75 meter aluminium bar 25 × 10-4 m2 in cross-sectional are attached to a 0.50 meter steel bar 15 × 10-4 m2 in cross-section area, as shown in the figure. Take E (Young’s modulus) value of 200 GPa for steel and 70 GPa for aluminum. Total shortening due to an axial compressive force of 175 kN is

quesOptionImage250

  1. 157168 mm
  2. 175168 mm
  3. 175186 mm
  4. 157186 mm

Answer (Detailed Solution Below)

Option 2 : 175168 mm

Deflection of Bar Question 5 Detailed Solution

Concept:

The FBD of both bars are shown below.

quesImage1393

The compression of both bars due to compressive load is given as, 

δTotal = δsteel + δAl

The elongation or compression of bar is, δ=PLAE

where, P = load, L = length of a bar, A = cross-section area, E = modulus of elasticity

Calculation:

Given:

For steel: 

P = 175 kN, L = 0.50 m , A = 15 × 10-4 m2, E = 200 GPa = 200 × 109 Pa

For Aluminium:

P = 175 kN, L = 0.75 m, A = 25 × 10-4 m2, E = 70 GPa = 70 × 109 Pa  

Deformation in Steel bar:

⇒ δsteel=175 × 103 × 0.5015 × 104 × 200 × 109=724000 m

Deformation in Aluminium bar:

 δAl=175 × 103 × 0.7525 × 104 × 70 × 109=34000 m

Therefore, δTotal=724000+34000=1960 m=1960×1000=2524 mm

From option 175168=2524 mm

Deflection of Bar Question 6:

A simply supported steel beam when loaded with a force of 100 kg at A. is found to deflect 10 mm at B. what force at B would cause a deflection of 2.5 mm at A

  1. 400 kg

  2. 100 kg

  3. 50 kg

  4. 25 kg

Answer (Detailed Solution Below)

Option 4 :

25 kg

Deflection of Bar Question 6 Detailed Solution

Concept:

Maxwell reciprocal theorem:

Maxwell's reciprocal theorem says that the deflection at D due to a unit load at C is the same as the deflection at C if a unit load was applied at D.

PcYc = PdYd

Calculation:

Given:

PA = 100 kg,   YB = 10 mm

Find PB = ? when YA = 2.5 mm

We Know that 

PA × YA PB × YB 

100 × 2.5 = PB × 10

PB = 25 kg.

Deflection of Bar Question 7:

A steel wire hangs vertically under its own weight. If its density is 10000 kg/m3 and allowable stress is 3000 kg/cm2 then how much length it can sustain.

  1. 2500 m
  2. 1250 m
  3. 3000 m
  4. 5000 m 

Answer (Detailed Solution Below)

Option 3 : 3000 m

Deflection of Bar Question 7 Detailed Solution

Concept:

Stress=Force appliedArea

σ = FA = mgA

σg = mA  ( ∵ unit of stress is kg/cm2 )

For a steel wire which is hanged vertically,

Force applied = Weight

Let A be the cross-sectional area of the wire, γ be the density of wire, and L is its length.

So, the weight of barg = γ AL

Stress due to self-weight =λALA=γL

Now, for bar to be safe, γL = σperm ...(i)

Calculation:

Given:

γ = 10000 kg/m3, σperm = 3000 kg/cm2 = 3000 × 104 kg/m2

Using Equation (i)

γL = σperm 

10000 × L = 3000 × 104

L = 3000 m

Deflection of Bar Question 8:

A solid conical bar of circular cross-section is suspended vertically as shown in the figure. The diameter of the bar at the base, D, equals 100 mm and its length, L is 0.5 m. If E = 200 GN/m2 and its weight per unit volume is 80 kN/m3,the elongation of the bar under self-weight is

F1 A.M 25.6.20 Pallavi D 6

  1. 1.50 × 10-6 mm
  2. 1.67 × 10-5 mm
  3. 1.71 × 10-6 mm
  4. 1.87 × 10-5 mm

Answer (Detailed Solution Below)

Option 2 : 1.67 × 10-5 mm

Deflection of Bar Question 8 Detailed Solution

Concept:

Elongation of the elemental strip is given by,

δ=PxdxAxE

Here, Px = load at x from bottom, dx = length of elemental strip, Ax = area of c-s at x and E = modulus of elasticity

Calculation:

F1 A.M Madhu 11.07.20 D5

Px=weightdensity×volumebelowaxisconsidered=γ×(13πrx2×x)

δ=PxdxAxE=γ×(13πrx2×x)×dxπrx2×E=γxdx3E

Elongation of the entire bar,

=0lγxdx3E=γl26E

Given,

L = 0.50 m, γ = 80 kN/m3 and E = 200 GN/m2

=γl26E=(80×1000)×0.526×(200×109)=16.67×109m1.67×105mm

Deflection of Bar Question 9:

A rigid body is very slowly dropped on another body and a deflection δ occurs in the second body. If the body be placed suddenly, the value of the impact factor will be

  1. 0
  2. 1
  3. 2

Answer (Detailed Solution Below)

Option 4 : 2

Deflection of Bar Question 9 Detailed Solution

Concept:

Strain energy (U)

U = Strain energy per unit volume (u) × Volume of the Member (V)

U = Area under load-deformation curve

Strain energy per unit volume ⇒ u=12×stress(σ)×strain(ϵ)=12×σ×σE=σ22E

Also,

Load deformation curve for gradual and sudden loading is shown below:

04.11.2017.029

In gradual loading, the loading starts from zero and increases gradually till the body is fully loaded, while in sudden loading, the load is suddenly applied to the body.

Calculation:

1. For Gradual Loading:

Strain Energy (U):

U=12×P×δL=12×P×σ×LE      ......(i)

U=u×V=σ22E×A×L      ......(ii)

From equation (i) and (ii)

12×P×σ×LE=σ22E×A×L

σgradual=PA

2. For sudden loading

Strain Energy (U):

U=P×δL=P×σ×LE      ......(i)

U=u×V=σ22E×A×L      ......(ii)

From equation (i) and (ii)

P×σ×LE=σ22E×A×L

σsudden=2PA

∴ Stress-induced in sudden loading is double that of normal loading.

Deflection of Bar Question 10:

A tapered circular rod of diameter varying from 20 mm to 10 mm is connected to another uniform circular rod of diameter 10 mm as shown in the following figure. Both bars are made of same material with the modulus of elasticity, E = 2x105 MPa. When subjected to a load P = 30π kN, the deflection at point is _______________ mm.

14

Answer (Detailed Solution Below) 14.5 - 15.5

Deflection of Bar Question 10 Detailed Solution

Concept:

The change in length of a prismatic bar with axial force P is given by,

δl=PLAE

The change in length of a tapered bar with diameter d1 at one end and diameter d2 at other end is given by,

δl=PLπ4(d1.d2)E

Where,

δl = change in length of the bar

P = Axial force on the bar

A = cross-sectional area of a bar

E = modulus of elasticity

Calculation:

P = 30π kN, d1 = 20 mm, d2 = 10 mm

L1 = 2 m, L2 = 1.5 m

15

Total elongation

δLTotal = δLAB + δLBC

δLTotal=PLAE+4PLπd1d2EδLTotal=30π×1.5π×(0.01)24×2×105+30π×2π×(0.01)×(0.02)4×2×105

δLTotal = 9 + 6 = 15 mm

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