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QUESTION: 1

A simply supported beam of span ‘l’ and flexural rigidity EI carries a unit load at its mid-span section; the strain energy of the beam is

Solution:

U1 =

=

=

U1 =

U = 2U1 = 2 ×

QUESTION: 2

A simply supported beam of span l carries a uniformly distributed load w/m over the entire span; the strain energy of the beam is given by

Solution:

M_{x} =

U =

=

=

=

=

=

=

QUESTION: 3

The strain energy of stored due to bending for cantilever beam of span L, modulus of elasticity E, moment of inertia I and a point load P at the free end is

Solution:

U =

=

=

QUESTION: 4

Total strain energy stored in a simply supported beam of span, 'L' and flexural rigidity 'EI 'subjected to a concentrated load 'W' at the centre is equal to

Solution:
Strain energy

=

=

=

Alternative method: In a funny way you may use Castigliano’s theorem,

δ =

We know that δ = for simply supported beam in concentrated load at mid span.

Then δ = or

U = partially integrating

with respect to W we ge U =

QUESTION: 5

A steel specimen 150 mm^{2} in cross-section stretches by 0 · 05 mm over a 50 mm gauge length under an axial load of 30 kN. What is the strain energy stored in the specimen? (Take E = 200 GPa)

Solution:
Strain Energy stored in the specimen

=

=

= 0.75 N-m

QUESTION: 6

A cantilever beam, 2 m in length, is subjected to a uniformly distributed load of 5 kN/m. If E = 200 GPa and I = 1000 cm^{4}, the strain energy stored in the beam will be

Solution:

U =

=

= 10Nm

QUESTION: 7

A truck weighing 150 kN and travelling at 2 m/sec impacts with a buffer spring which compresses 1.25 cm per 10 kN. The maximum compression of the spring is

Solution:
Kinetic energy of the truck = strain energy of the string

x =

= 0.2766m = 27.66cm

QUESTION: 8

A member formed by connecting a steel bar to an aluminium bar is shown in figure. Assuming that the bars are prevented from bucking sidewise, calculate the magnitude of the force P that will cause the total length of the member to decrease by 0.25 mm. The values of the elastic modulus for steel and aluminium are 2.1 × 105 N/mm^{2} and 0.7 × 105 N/mm^{2 }respectively. What is the total work done by the force P?

(A) 27.5

(B) 28.5

Solution:
Area of the steel bar = A_{s} = 50 × 50 = 2500 mm^{2}

A!rea of aluminium bar = A_{a} = 100 × 100 = 10000 mm^{2}

Total change in length = δ

= δ

= 0.25mm

P × 0.11143 × 10^{−5} = 0.25

∴ P = 224356 N

Total work done = 1/2 × load × deformation

=

= 28044 Nmm

= 28.044 Nm = 28.044 Joule

Question_Type: 5

QUESTION: 9

A 10 mm diameter mild steel bar of length 1.50 metres is stressed by a weight of 120 N dropping freely through 20 mm before commencing to stretch the bar. Find the maximum instantaneous stress and the elongation produced in the bar. (Take E = 2 × 105 N/mm^{2})

Solution:
!rea of the bar = π / 4× (10^{2}) = 78.54 mm^{2}

Let the maximum instantaneous stress be p N/mm^{2}

∴ Maximum elongation = δl =

Equating the loss of potential energy to the strain energy stored by the member, we have

P(h + δl) =

∴ 2400 + 120

∴ p^{2} − 240 p/A=

∴ p^{2} −

∴ p^{2} − 3.056 p = 8148.714

∴ (p − 1.528)^{2} = 1848.714 + 2.334

∴ (p − 1.528)2 = 8151.049

∴ p − 1.528 = 90.283

p = 91.811 N/mm^{2}

Alternatively, p =

But, = 1.528 N/mm^{2}

= 1.5282 = 2.335

∴ p = 1.528 +

= 91.811 N/mm^{2}

∴ Maximum elongation

= δl =

= 0.6885 mm

QUESTION: 10

The L-shaped bar shown in the figure is of uniform cross-section 60 mm × 120 mm. Calculate the total strain energy. (Take E = 2 × 10^{5} MPa, G = 0.8 × 10^{5} MPa)

Solution:

Total strain energy U = UB + US + UA = Strain energy due to bending of BC + Strain energy due to shear of BC + Strain energy due to axial load W

=

Where, W = 10,000 N, L1 = 2000 mm,

L2 = 1000 mm, E = 2 × 105 MPa,

G = 0.8 × 105 MPa

I = =8.64 × 10^{6} mm^{4}

b = 60 mm, d = 120 mm

A = b × d = 60 × 120 = 7200 mm^{2}

Putting the values Strain energy,

U = + +

= 77160.5 + 208.33 + 357.143 Nmm

= 77725.967 Nmm = 77.725 Nm

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