UNIT I ENERGY PRINCIPLES - Sri Venkateswara College of ... of... · • Strain energy due to shear...

UNIT I ENERGY PRINCIPLES Contents: • Strain energy and strain energy density

• Strain energy due to axial load

• Strain energy due to shear

• Strain energy due to flexure

• Strain energy due to torsion

• Castigliano’s theorems

• Maxwell’s reciprocal theorem

• Principle of virtual work

• Application of energy theorems for computing deflections in beams and trusses

• Williot Mohr's Diagram

UNIT I ENERGY PRINCIPLES

• References:

Ferdinand P. Beer, E. Russell Johnston Jr., John T. Dewolf, “Mechanics of Materials” McGraw-Hill, New York, 2006.

Punmia B.C.,"Theory of Structures" (SMTS) Vol II, Laxmi Publishing Pvt Ltd, New Delhi 2004.

Rattan.S.S., "Strength of Materials", Tata McGraw Hill Education Pvt. Ltd., New Delhi, 2011.

Rajput R.K., "Strength of Materials (Mechanics of Solids)", S.Chand & company Ltd., New Delhi, 2010.

Strain energy

• Strain Energy of the elastic body is defined as the internal work done by the external load in deforming or straining the body.

• Strain energy is always scalar quantity.

• When an elastic body is deformed under external loading, work is done. The used energy is stored in the body as strain energy and is recovered without loss on removal of the load.

• Example: Clockwork device stores strain energy and then gives it up.

• Strain energy is associated with different forms of stress encountered in structures.

• Different forms of stress are direct stress, shear stress, bending stress and torsion stress.

• Strain energy concept is useful to calculate deflection of structures.

• Strain energy is usually denoted by symbol U.

• Strain energy, U= 𝑓2

2𝐸 × volume

Strain energy

Strain energy due to axial load • A uniform cross sectional rod is subjected to a gradually

increasing axial load.

• The elementary work done by the load P as the rod elongates by a small amount 𝑑𝑥 is

which is equal to the area of width dx under the load-deformation diagram.

workelementarydxPdU

Strain energy due to axial load • The total work done by the load P for a

deformation x1 results in an increase of

strain energy

energystrainworktotaldxPU

x

1

0

• In the case of a linear elastic deformation,

11212

121

0

1

xPkxdxkxU

x

Strain energy density

• To eliminate the effects of size,

evaluate the strain- energy per

unit volume,

densityenergy straindu

L

dx

A

P

V

U

x

x

1

1

0

0

• The total strain energy density resulting from the

deformation is equal to the area under the curve to 1.


• As the material is unloaded, the

stress returns to zero but there

is a permanent deformation.

• Only the strain energy

represented by the triangular

area is recovered.

• Remainder of the energy spent in deforming the

material is dissipated as heat.


• The strain energy density resulting from setting 휀1 = 휀R is the modulus of toughness.

• The energy per unit volume required to cause the material to rupture is related to its ductility as well as its ultimate strength.

• If the stress remains within the proportional limit,

E

EdEu x

22

21

21

0

1

1


resilience of modulusE

u YY

2

2

• The strain energy density resulting from setting σ1 = σY is the modulus of resilience.

• Strain energy per unit volume of the material is known as strain energy density or resilience.

• Resilience, u= 𝑈

𝑉

• When the stress 𝑓 is equal to proof stress, 𝑓𝑝 at the

elastic limit, the corresponding resilience is known as

proof resilience, up = 𝑓𝑝2

2𝐸

• The proof resilience is known as modulus of resilience. It is the property of the material. It’s unit is N-m/m3= N/m2


• (a) Due to axial force:

Elastic Strain Energy for Normal Stresses

• In an element with a nonuniform stress

distribution,

energystrain totallim0

dVuUdV

dU

V

Uu

V

• For values of u < uY , i.e., below the proportional limit,

energy strainelasticdVE

U x 2

2

Strain energy due to axial force

• A member subjected to an external load W.

• Let the extension of the member be ‘δ’.

• Since the load is applied gradually, the magnitude

of the load is increased gradually from zero to the

value ‘W’ and the member also has gradually

extended.

• External work done, 𝑊𝑒= Avg.load × displacement

= 0+𝑊

2 × δ =

1

2 Wδ.

• Let the energy stored by the member be 𝑊𝑖.

• We have 𝑊𝑒=𝑊𝑖,

• Let the tension in the member be ‘S’.

• For the equilibrium of the member, S = W.

w

w

δ

A

• Tensile stress, 𝑓 =𝑆

𝐴

• Tensile strain, e = 𝑓

𝐸=𝑆

𝐴𝐸

• Where E is the young’s modulus of the material.

• Change in length, δ= e× 𝑙 = 𝑆𝑙

𝐴𝐸

• Strain energy stored = work done = 1

2 Wδ

• Strain energy stored per unit volume = 𝑠2𝑙

2𝐴𝐸 ÷ 𝐴𝑙 =

𝑓2

2𝐸

Strain energy due to axial force (contd…)

𝑈𝐴=1

2× 𝑆 ×

𝑆𝑙

𝐴𝐸= 𝑠2𝑙

2𝐴𝐸

∴ 𝑈𝐴= 𝑠2𝑙

2𝐴𝐸

w

w

δ

A


• Under axial loading, dxAdVAPx

L

dxAE

PU

0

2

2

energy strainelasticdVE

U x 2

2

• For a rod of uniform cross-section,

AE

LPU

2

2

Problems

Problem 1:

A steel bar 15 mm in diameter is pulled axially by a load of 10 KN. If the bar is 250mm long. Calculate the strain energy stored by the bar. Take E = 2 x 105 N/mm2.

Problems

2A 176.71mm4

1 Cross sectional area of the bar,

Strain energy stored by the bar,

Solution:

d = 15 mm, P=10 kN, l=250 mm and E= 2× 105N/mm2

AE

LPU

2

2

𝑈 = 10×1000 2×250 2×176.71×2×105

=353.69 N-mm

• What is the strain energy in a bar of conical section subjected to axial load P?

Elastic Strain Energy due to axial force

P P

d

D

• 𝑑𝑥 = d +𝐷−𝑑

𝐿𝑥

= 𝑑 + 𝑘𝑥

Strain energy of small length = 𝑃2𝛿𝑥

2𝐴𝑥𝐸

Strain energy of whole bar = 𝑃2

2𝜋

4 𝑑+𝑘𝑥 2𝐸

𝐿

0 𝑑𝑥

∴ U = 2𝑃2𝐿

𝜋𝐸𝑑𝐷

P P

d

D

Elastic Strain Energy due to axial force L

𝛿𝑥 𝑥

• What is the strain energy in a bar of trapezoidal section subjected to axial load P?


P P b B

t

𝑏𝑥 = b +𝐵 − 𝑏

𝐿𝑥

= 𝑏 + 𝑘𝑥

P P b B

t L

𝑥


Strain energy of small length = 𝑃2𝛿𝑥

2𝐴𝑥𝐸

Strain energy of whole bar = 𝑃2

2 𝑏+𝑘𝑥 𝑡𝐸

𝐿

0 𝑑𝑥

∴ 𝑈 =𝑃2𝐿

2 𝐵 − 𝑏 𝑡𝐸log𝑒𝐵

𝑏

• A steel bar is acted upon by forces as shown in the following Figure. Determine the strain energy stored in the bar if A is the area of cross section of the bar and E is the modulus of elasticity.


P 2P 5P

A B C D

L/3 L/3 L/3

• For segment CD, force at D=force at C= P

• For segment BC, force at C=force at B= 3P

• For segment AB, force at B=force at A= 2P (Comp.)


P P C D

3P 3P C D

2P 2P

C D

P 2P 5P

A B C D

L/3 L/3 L/3

• Total strain energy = 𝑃2𝐿

2𝐴𝐸

=𝐿/3

2𝐴𝐸𝑃2 + 3𝑃 2 + −2𝑃 2

• ∴ 𝑈 =7𝑃2𝐿

3𝐴𝐸


• What is the expression for strain energy of a prismatic bar under its own weight.


L

Weight per unit volume of the bar = w Weight of the bar below the small section, 𝑤𝑥 = 𝑤𝐴𝑥

Strain energy of the element =𝑤𝐴𝑥 2𝑑𝑥

2𝐴𝐸

Total strain energy = 𝑤𝐴𝑥 2𝑑𝑥

2𝐴𝐸

𝐿

0

= 𝑤2𝐴

2𝐸 𝑥2𝑑𝑥𝐿

0

∴ 𝑈 =𝑤2𝐴𝐿3

6𝐸


L 𝛿𝑥

• (b) Due to transverse load on beams:


• For a beam subjected to a transverse load (i.e., beam is under bending),

dVEI

yMdV

EU x

2

222

22

I

yMx

Setting dV = dA dx,

dxdAyEI

MdxdA

EI

yMU

L

A

L

A

0

2

2

2

0

2

22

22

dxEI

MU

L

0

2

2∴


• For an end-loaded cantilever beam find the strain

energy and deflection at the free end of the beam.

dxEI

MU

L

0

2

2

PxM

EI

LPU

6

32

∴

EI

LPdx

EI

xPL

62

32

0

22

• Work done by the external load, 𝑊𝑒 = 1

2Pδ

• We know 𝑊𝑒=U

• ∴1

2Pδ =

𝑃2𝐿3

6𝐸𝐼

• and hence, 𝛿 =𝑃𝐿3

3𝐸𝐼


Problems

Taking into account only the normal stresses due to

bending, determine the strain energy of the beam for the

loading shown.

Problem 1:

• Solution:

• Determine the reactions at A

and B from a free-body

diagram of the complete

beam.

Problems

• Develop a diagram of

the bending moment

distribution.

L

PaR

L

PbR BA

vL

PaMx

L

PbM 21

Problems

vL

PaM

xL

PbM

2

1

BD,portion Over the

AD,portion Over the

• Integrate over the volume of the

beam to find the strain energy.

baEIL

baPbaab

L

P

EI

dxxL

Pa

EIdxx

L

Pb

EI

dvEI

Mdx

EI

MU

ba

ba

2

2223232

2

2

0

2

0

2

0

22

0

21

6332

1

2

1

2

1

22

EIL

baPU

6

222

Strain energy for shearing stress

• For a material subjected to plane shearing stresses,

xy

xyxy du

0

• For values of xy within the proportional

limit,

GGu

xyxyxyxy

2

2

212

21

• The total strain energy is found from

dVG

dVuU

xy

2

2

Strain energy for shearing stress

J

Txy

dVGJ

TdV

GU

xy

2

222

22

• For a shaft subjected to a torsional load,

• Setting dV = dA dx,

L

L

A

L

A

dxGJ

T

dxdAGJ

TdxdA

GJ

TU

0

2

0

22

2

02

22

2

22

• In the case of a uniform shaft,

GJ

LTU

2

2

• A portal frame ABCD has its end A hinged and end D is placed on-rollers. A horizontal force P is applied on the end D as shown in the following Figure. Determine the horizontal movement of D. Assume all members have the same flexural rigidity.

Problems

P

h

b

A

B C

D

• Solution:

There will be a horizontal reaction P at A.

Strain energy stored by the frame, U

= Strain energy stored by the columns

+ Strain energy stored by the beam

∴ 𝑈 = 𝑀2

2𝐸𝐼

𝐿

0dx

= 2 𝑃2𝑦2𝑑𝑦

2𝐸𝐼

ℎ

0

+ 𝑃2ℎ2𝑑𝑥

2𝐸𝐼

𝑏

0

= 2 ×𝑃2ℎ3

6𝐸𝐼+𝑃2ℎ2𝑏

2𝐸𝐼=𝑃2ℎ3

3𝐸𝐼+𝑃2ℎ2𝑏

2𝐸𝐼

=𝑃2ℎ2

6𝐸𝐼2ℎ + 3𝑏

Problems

P

h

b

A

B C

D Y Y

𝑥

P

• U =𝑃2ℎ2


• Let the horizontal movement of D be 𝛿

• External work done, 𝑊𝑒 =1

2Pδ

• We know, 𝑊𝑒 = U

• ∴ 1

2Pδ =

𝑃2ℎ2


• ∴ 𝛿 =𝑃ℎ2


Problems

P

h

b

A

B C

D Y Y

𝑥

P

𝛿

P

h

b B C

D Y Y

𝑥

Problems

• Find the strain energy stored in the cantilever beam subjected to u.d.l. of w/m for whole span.

• 𝑀𝑥 =𝑊𝑥2

2

W kN/m

‘L’ m

𝑈 = 𝑀𝑥2

2𝐸𝐼

𝐿

0dx

𝑥

∴ 𝑈 =𝑊2𝐿5

40𝐸𝐼

• Problem : • A cantilever truss ABCDE is hinged at two points A and E. E is 2 m

below A. EDC is the horizontal bottom chord, C being the free end. ED=DC=2m. AB=BC. BD and BE are the vertical and diagonal members. The truss is loaded with 20 kN loads at C and D. Cross sectional area of each member is 8 cm2. Find the strain energy stored in the truss. E=2× 105 MPa.

Problems - Strain energy stored in a Truss

20 kN 20 kN

A

E

2 m

D

C

2 m 2 m

B

1m

Strain energy stored in a Truss

A cantilever truss ABCDE is hinged at two points A and E. E is 2 m below A.

EDC is the horizontal bottom chord, C being the free end. ED=DC= 2m

AB=BC. BD and BE are the vertical and diagonal members.

The truss is loaded with 20 kN loads at C and D.

Cross sectional area of each member is 8 cm2. Find the strain energy stored in the truss. E=2× 105 MPa.

• Solution: Let us first determine the forces in the various members.

• For the equilibrium of joint C,

Resolving vertically,

𝑃𝐶𝐵 sin 𝜃=20 ∴ 𝑃𝐶𝐵 = 20 5 kN.

Resolving horizontally,

𝑃𝐶𝐷= 𝑃𝐶𝐵 cos 𝜃

= 20 5 ×2

5 = 40 kN.

∴ 𝑃𝐶𝐷 = 40 𝑘𝑁.

A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m 𝜃

sin 𝜃 =1

5

cos 𝜃 =2

5

𝜃

20 kN

C

𝑷𝑪𝑩

𝑷𝑪𝑫

Strain energy stored in a Truss

• For the equilibrium of joint D,

Deflection of Trusses Problems

A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m 40

20 5

Resolving horizontally, 𝑃𝐷𝐸= 40 𝑘𝑁


𝑃𝐷𝐵=20 kN.

20 kN

40 𝑃𝐷𝐸

𝑃𝐷𝐵

D

• For the equilibrium of joint B,


A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m 20 5

20 𝜃

20 kN

20 5 B

𝑃𝐵𝐸 𝑃𝐵𝐴

𝜃 𝜃


𝑃𝐵𝐴 sin 𝜃 + 𝑃𝐵𝐸 sin 𝜃 = 20 + 20 5 sin 𝜃

𝑃𝐵𝐴 ×1

5+ 𝑃𝐵𝐸 ×

1

5= 20 + 20 5 ×

1

5

𝑃𝐵𝐴 + 𝑃𝐵𝐸 = 20 + 20 5

𝑃𝐵𝐴 = 40 5 − 𝑃𝐵𝐸 -------(1)

sin 𝜃 =1

5



A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m 20 5

20 𝜃

20 kN

20 5 B


𝜃 𝜃


𝑃𝐵𝐴 cos 𝜃 = 𝑃𝐵𝐸 cos 𝜃 + 20 5 cos 𝜃

𝑃𝐵𝐴 = 𝑃𝐵𝐸 + 20 5 Substitute eq.(1) in the above eq.

40 5 − 𝑃𝐵𝐸 = 𝑃𝐵𝐸 + 20 5

𝑃𝐵𝐸 = 10 5 kN

𝜃

𝑃𝐵𝐴 = 40 5 − 𝑃𝐵𝐸------(1)

∴ 𝑃𝐵𝐴 = 30 5 kN

• Strain energy, 𝑈 = 𝑃2𝐿

2𝐴𝐸

• U=1

2×8×10−4×2×108

( 20 52× 5) + (402× 2) + (202× 1)

+ 402 × 2 + ( 10 52× 5) + ( 30 5

2× 5)

∴ 𝑈 =0.07016 kNm


A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m 20 5

40 40

30 5

10 5 20

Problems

A simply supported beam is loaded as shown in Figure. Determine the strain energy stored due to bending and deflection at centre of the beam.

W kN

L/2 m L/2 m

A B C

X m

• Solution:

Problems

W kN

L/2 m L/2 m

A B C X m

𝑀𝑥 =𝑊𝑥

2

𝑊

2

𝑊

2

𝑈 = 𝑀𝑥2

2𝐸𝐼

𝐿

0dx

∴ 𝑈 =𝑊2𝐿3

96𝐸𝐼

𝛿 =𝜕𝑈

𝜕𝑊=𝑊𝐿3

48𝐸𝐼

Problems

• Problem :

A simply supported beam is loaded as shown in Figure. Determine the strain energy stored due to bending E= 210 GN/m2.

15 kN

3 m 1.5 m

A B C

X m 6 cm

9 cm

• Solution:

Calculation of reactions:

𝑅𝐴 × 4.5 = 15 × 1.5

∴ 𝑅𝐴 = 5 kN.

Problems

X m

15 kN

3 m 1.5 m

A B

C

5 kN

Consider the section X X at a distance X from A,

𝑀𝑥 = 5𝑥, → limits x= 0 to 3 m

𝑀𝑥 = 5𝑥 − 15(𝑥 − 3), → limits 𝑥 = 3 𝑡𝑜 4.5 𝑚

Strain energy, 𝑈 = 𝑀𝑥2

2 𝐸𝐼

𝑙

0

𝑑𝑥

• ∴ 𝑈 =1

2𝐸𝐼 25𝑥23

0𝑑𝑥 + 5𝑥 − 15(𝑥 − 3) 2

4.5

3𝑑𝑥

=1

2𝐸𝐼 25𝑥23

0

𝑑𝑥 + 100𝑥2 + 2025 − 900𝑥 4.5

3

𝑑𝑥

=1

2𝐸𝐼25 ×𝑥3

30

3

+ 100 ×𝑥3

3+ 2025𝑥 − 900

𝑥2

23

4.5

=1

2𝐸𝐼225 + 2137.5 − 5062.5 + 3037.5

= 337.5×106

2×210×109× 6×93

12×10−4

= 220.46 Nm

Problems

• Find the strain energy stored in the simply supported beam subjected to u.d.l. of w/m for whole span.

Problems

w kN/m

‘L’ m

x

wL/2 wL/2

Castigliano’s theorems

• Castigliano’s first theorem:

• For linearly elastic structure, the Castigliano’s first theorem may be defined as the first partial derivative of the strain energy of the structure with respect to any particular force gives the displacement of the point of application of that force in the direction of its line of action.

Castigliano’s First theorem derivation

Consider an elastic beam AB subjected to loads W1 and W2, acting at points 1 and 2 respectively


22212

12111

11111 WIf

where ∆11= deflection at 1 due to a unit load at 1 and

21121 W

with ∆21 = deflection at 2 due to a unit load at 1


22222 W , with ∆22 = deflection at 2 due to a unit load at 2 &

21212 W , with ∆12 = deflection at 1 due to a unit load at 2.

Then

(I) 212111

12111

WW

Similarly,

(II) 222121

22212

WW

(III) ----- 2

1

2

1

2

1

2

1

2

1

2

1

2112

2

222

2

111

212122221111

121222111

WWWW

WWWWWW

WWW

Considering the work done = U

• Now applying W2 at Point 2 first and then applying W1 at Point 1,


11121211121 WW

12122221222 WW

Strain energy, U

(IV) ------- 2

1

2

1

2

1

2

1

2

1

2

1

2

1112121

2

222

111112122222

111212222

WWWW

WWWWWW

WWW

Similarly,

• Considering equation (III) and (IV), and equating them, it can be shown that


2

1112121

2

222

2112

2

222

2

111

2

1

2

1

2

1

2

1

WWWW

WWWWU

2112 This is called Betti – Maxwell’s reciprocal theorem

Deflection at point 1 due to a unit load at point 2 is equal to the deflection at point 2 due to a unit load at point 1.

• From Eqn. (III),


1212111

1

WW

W

U

From Eqn. (IV),

2121222

2

WW

W

U

iWi

U

∴ This is Castigliano’s first theorem.

2

1

2

12112

2

222

2

111 WWWWU

2

1112121

2

2222

1

2

1WWWWU

Castigliano’s second theorem

Similarly the energy U can be express in terms of spring

stiffnesses k11, k12 (or k21), & k22 and deflections δ1 and δ2;

then it can be shown that

2

2

1

1

WU

WU

This is Castigliano’s second theorem.

i

iM

U

When rotations are to be determined,

Method of least work

The partial derivative of internal energy with respect to a load applied at a point where the deflection is zero, Then

𝜕𝑈

𝜕𝑊= 0.

Maxwell’s reciprocal theorem

• In any beam (or) truss, the deflection at any point C due to load W at any point B is the same as the deflection at B due to the same load W applied at C.

𝛿𝐶 = 𝛿𝐵

A B C D

𝛿𝐶

W

A B C D

𝛿𝐵

W =

• Problem : • A cantilever truss ABCDE is hinged at two points A and E. E is 2

m below A. EDC is the horizontal bottom chord, C being the free end. ED=DC=2m. AB=BC. BD and BE are the vertical and diagonal members. The truss is loaded with 20 kN loads at C and D. Cross sectional area of each member is 8 cm2. Find the deflection of the free end C of the truss. E=2× 105 MPa.



A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m

• Solution: To find the vertical displacement at joint C, take the vertical downward load P at C and find the forces in all members.



𝑃𝐶𝐵 sin 𝜃=P ∴ 𝑃𝐶𝐵 = P 5 kN.


𝑃𝐶𝐷= 𝑃𝐶𝐵 cos 𝜃

= P 5 ×2

5 = 2P kN.

∴ 𝑃𝐶𝐷 = 2𝑃 𝑘𝑁.


A

B

C

D E

20 kN P kN

2 m

2 m 2 m

1m 𝜃

sin 𝜃 =1

5

cos 𝜃 =2

5

𝜃

P kN

C

𝑷𝑪𝑩

𝑷𝑪𝑫

• For the equilibrium of joint D,


A

B

C

D E

20 kN P kN

2 m

2 m 2 m

1m 2𝑃

P 5

Resolving horizontally, 𝑃𝐷𝐸= 2𝑃


𝑃𝐷𝐵=20 kN.

20 kN

2P 𝑃𝐷𝐸

𝑃𝐷𝐵

D



A

B

C

D E

20 kN P kN

2 m

2 m 2 m

1m P 5

20 𝜃

20 kN

P 5 B


𝜃 𝜃


𝑃𝐵𝐴 sin 𝜃 + 𝑃𝐵𝐸 sin 𝜃 = 20 + 𝑃 5 sin 𝜃

𝑃𝐵𝐴 ×1

5+ 𝑃𝐵𝐸 ×

1

5= 20 + 𝑃 5 ×

1

5

𝑃𝐵𝐴 + 𝑃𝐵𝐸 = 20 + 𝑃 5

𝑃𝐵𝐴 = 20 + 𝑃 5 − 𝑃𝐵𝐸

sin 𝜃 =1

5



A

B

C

D E

20 kN P kN

2 m

2 m 2 m

1m P 5

20 𝜃

20 kN

P 5 B


𝜃 𝜃


𝑃𝐵𝐴 cos 𝜃 = 𝑃𝐵𝐸 cos 𝜃 + 𝑃 5 cos 𝜃

𝑃𝐵𝐴 = 𝑃𝐵𝐸 + P 5 Substitute eq.(1) in the above eq.

20 + 𝑃 5 − 𝑃𝐵𝐸 = 𝑃𝐵𝐸 + P 5

𝑃𝐵𝐸 = 10 5 kN

𝜃

𝑃𝐵𝐴 = 20 + 𝑃 5 − 𝑃𝐵𝐸------(1)

∴ 𝑃𝐵𝐴 = 10 + 𝑃 5


2𝐴𝐸

• U=

1

2𝐴𝐸

𝑃 52× 5 + 2𝑃 2 × 2 + 20 2 × 1

+ 2𝑝 2 × 2 + 10 52× 5 + 𝑃 + 10 2 5

2× 5

• Vertical downward deflection at C, 𝛿𝑐 𝑉 =𝜕𝑈

𝜕𝑃

•𝜕𝑈

𝜕𝑃=1

2𝐴𝐸2𝑃 × 5 5 + 16𝑃 + 16𝑃 + 5 5 2𝑃 + 20

• By substituting P=20 kN,

•𝜕𝑈

𝜕𝑃=1

2𝐴𝐸2 × 20 × 5 5 + 32 × 20 + 5 5 40 + 20

∴ 𝛿𝑉 𝐶 =1758.03

2 × 8 × 10−4 × 2 × 108= 0.005494 m = 5.494 mm.


• Horizontal displacement of Joint C:

• Now assume a horizontal load of Q kN at C.


A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m

Q

• Let us first determine the forces in the various members.



A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m

Q

𝜃

20

C

𝑷𝑪𝑩

𝑸 𝑷𝑪𝑫


𝑃𝐶𝐵 sin 𝜃=20 ∴ 𝑃𝐶𝐵 = 20 5 kN. Resolving horizontally,

𝑃𝐶𝐷+20 5 cos 𝜃=Q

𝑃𝐶𝐷+ 20 5 ×2

5 = Q.

∴ 𝑃𝐶𝐷 = (𝑄 − 40) 𝑘𝑁.

sin 𝜃 =1

5

cos 𝜃 =2

5


A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m

Q

20 5

𝑄 − 40

For the equilibrium of joint D,

20 kN

𝑄 − 40

𝑃𝐷𝐵

D 𝑃𝐷𝐸


𝑃𝐷𝐵=20 kN.

Resolving horizontally, 𝑃𝐷𝐸= 𝑄 − 40


20 5

A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m

Q 𝑄 − 40 𝑄 − 40

20

20 kN

20 5 B


𝜃 𝜃

For the equilibrium of joint B,

𝜃


𝑃𝐵𝐴 sin 𝜃 + 𝑃𝐵𝐸 sin 𝜃 = 20 + 20 5 sin 𝜃

𝑃𝐵𝐴 ×1

5+ 𝑃𝐵𝐸 ×

1

5= 20 + 20 5 ×

1

5

𝑃𝐵𝐴 + 𝑃𝐵𝐸 = 40 5

𝑃𝐵𝐴 = 40 5 − 𝑃𝐵𝐸 ----------(1)


A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m

Q 𝑄 − 40 𝑄 − 40

20

20 5

For the equilibrium of joint B,

20 kN

20 5 B


𝜃 𝜃

𝑃𝐵𝐴 = 40 5 − 𝑃𝐵𝐸------(1)


𝑃𝐵𝐴 cos 𝜃 = 𝑃𝐵𝐸 cos 𝜃 + 20 5 cos 𝜃

𝑃𝐵𝐴 = 𝑃𝐵𝐸 + 20 5 Substitute eq.(1) in the above eq.

40 5 − 𝑃𝐵𝐸 = 𝑃𝐵𝐸 + 20 5

𝑃𝐵𝐸 = 10 5 kN ∴ 𝑃𝐵𝐴 = 30 5

𝑃𝐵𝐴 = 40 5 − 10 5

From equation (1),

A

B

C

D E

20 kN 20 kN

2 m

2 m 2 m

1m

Q

𝑄 − 40 𝑄 − 40

20


20 5 10 5

30 5


2𝐴𝐸

• U=

1

2𝐴𝐸

20 52× 5 + 𝑄 − 40 2 × 2 + 20 2 × 1

+ 𝑄 − 40 2 × 2 + 10 52× 5 + 30 5

2× 5

• Horizontal deflection at C, 𝛿𝐻 𝐶 =𝜕𝑈

𝜕𝑄

•𝜕𝑈

𝜕𝑄=1

2𝐴𝐸(2𝑄 − 80) × 2 + 2𝑄 − 80 × 2

• By substituting Q=0 kN,

•𝜕𝑈

𝜕𝑃=1

2𝐴𝐸−320

∴ 𝛿𝐻 𝐶 =−320

2 × 8 × 10−4 × 2 × 108= −0.001 m = 1 mm.


• Deflection at joint C, 𝛿𝐶 = 𝛿𝑉 𝐶2+ 𝛿𝐻 𝐶

2

= 5.494 2 + −1 2

∴ 𝛿𝐶 =5.59 mm


• A simply supported beam of span 3 m is carrying a point load of 20 kN at 1m from left support in addition to a u.d.l. of 10 kN/m spread over the right half span. Using castigliano’s theorem determine the deflection under the point load. Take EI is constant throughout. (May/June 2012)

A.U. Question paper problems


10 kN/m 3 m

20 kN

1.5 m 1 m A B C D

𝛿𝑐 =𝜕𝑈

𝜕𝑊𝑐=𝜕𝑈

𝜕𝑊

3 m

W kN

1.5 m 1 m A B C D

0.67W+3.75 0.33W+11.25

𝑀𝑥 = 0.67W+ 3.75 x; x=0 to 1

𝑀𝑦 = 0.67W+ 3.75 y −W(y − 1); y=1 to 1.5

𝑀𝑧 = 0.33W+ 11.25 z − 10 ×𝑧2

2; z=0 to 1.5

𝑈 = 𝑀𝑥2

2 𝐸𝐼

1

0

𝑑𝑥 + 𝑀𝑦2

2 𝐸𝐼

1.5

1

𝑑𝑦 + 𝑀𝑧2

2 𝐸𝐼

1.5

0

𝑑𝑧

X

Y z

10 kN/m

• A simply supported beam of span 8 m carries two concentrated loads of 32 kN and 48 kN at 3m and 6 m from left support. Calculate the deflection at the centre by strain energy principle (Nov/Dec 2007).


Problems

48+0.5 W

Calculation of support reactions: 𝑅𝐵 × 8 = 32 × 3 + 𝑊 × 4 + 48 × 6 𝑅𝐵 = 48 + 0.5 𝑊 𝑅𝐴 = 32 + 0.5 𝑊

Solution:

8 m

32 kN 48 kN

2 m 3 m

W kN

A B C D E

4 m

𝛿𝐸=?

Problems

• 𝑈 = 𝑀𝑥2

2 𝐸𝐼

𝑙

0𝑑𝑥 =

𝑀𝐴𝐶2

2 𝐸𝐼

3

0𝑑𝑎+

𝑀𝐶𝐸2

2 𝐸𝐼

4

3𝑑𝑏 +

𝑀𝐵𝐷2

2 𝐸𝐼

2

0𝑑𝑥 +

𝑀𝐷𝐸2

2 𝐸𝐼

4

2𝑑𝑦

• 𝑀𝐴𝐶=(32+0.5W)𝑎; 𝑙𝑖𝑚𝑖𝑡𝑠 0 𝑡𝑜 3

• 𝑀𝐶𝐸=(32+0.5W)𝑏 − 32 𝑏 − 3 ; 𝑙𝑖𝑚𝑖𝑡𝑠 3 𝑡𝑜 4

• 𝑀𝐵𝐷=(48+0.5W)𝑥; 𝑙𝑖𝑚𝑖𝑡𝑠 0 𝑡𝑜 2

• 𝑀𝐷𝐸=(48+0.5W)𝑦 − 48 𝑦 − 2 ; 𝑙𝑖𝑚𝑖𝑡𝑠 2 𝑡𝑜 4

8 m

32 kN 48 kN

2 m 3 m

W kN

A B C D E

4 m 48+0.5 W

a y 𝛿𝐸 =

𝜕𝑈

𝜕𝑊𝐸=𝜕𝑈

𝜕𝑊

b

Y

x

• Find the strain energy stored in the simply supported beam subjected to u.d.l. of w/m for whole span.

Problems

w kN/m

‘L’ m

x

wL/2 wL/2

• For the beam shown in the following Figure find the slope and deflection at C. (Nov/Dec 2011)


4 kN

4 m 1 m

6 kN/m

(2EI) (EI)

A B C

A beam 4m in length is simply supported at the ends and carries a uniformly distributed load of 6 kN/m length. Determine the strain energy stored in the beam. Take E = 200 GPa and I = 1440 cm4. (April/May 2011)


• A beam simply supported over a span of 3m carries a UDL of 20 kN/m over the entire span. The flexural rigidity EI = 2.25 MNm2 Using Castigliano’s theorem, determine the deflection at the centre of the beam. (April/May 2011)


• A cantilever of rectangular section breadth b, depth d and of length l carries uniformly distributed load spread from free end to the mid section of the cantilever. Using Castigliano’s theorem find: Slope and deflection due to bending at the free end. (Nov/Dec 2010)


Williot diagram

Williot diagram: To determine deflection of truss joints graphically.

• Due to external loads acting at joints of a truss, the truss members are subjected to axial forces (Compressive or tensile).

• Hence the members undergo changes in their lengths.

• Due to the changes in lengths of the members, deflections of joints take place.

• If the extensions and contractions of the members are known, it is possible to determine the displacements of the joints graphically.

• A crane consists of a jib 7.5 m long, of 15 cm2 cross sectional area and a horizontal tie 6m long of 10 cm2 sectional area. Determine the vertical and horizontal displacements of crane head, when a load of 10 tons is suspended from it.

𝐸 = 2 × 106 kg/cm2.

Williot diagram - Problem

• Graphical method – Williot diagram


AB = 7.52 − 62 = 4.5 m.

tan 𝜃 =4.5

6

sin 𝜃 =4.5

7.5

cos 𝜃 =6

7.5

Resolving the forces at c vertically, we have PCBsin 𝜃=10

PCB=10

sin 𝜃=10

4.5× 7.5 = 16.67 𝑡 = 16.67 × 103 kg.

Resolving horizontally at C,

𝑃𝐶𝐴=𝑃𝐶𝐵 cos 𝜃 = 16.67 ×6

7.5= 13.34 t = 13.34 × 103kg.

A

B

C

10 t

𝜃

6 m

10 t

𝜃

PCB

PCA C

ABC=15 cm2

AAC=10 cm2

𝐸 = 2 × 106 kg/cm2

• Decrease in length of BC =𝑃𝐿

𝐴𝐸 𝑩𝑐=16.67×103×750

15×2.0×106=0.417 cm=4.2mm

• Increase in length of AC =𝑃𝐿

𝐴𝐸 𝑨𝑪=13.34×103×600

10×2.0×106=0.40 cm=4.0mm


A

B

C

10 t

𝜃 13.34 × 103kg.

• Plot a point and designate it as (a,b)

• Measure ac1= 4.0 mm parallel to AC

• Draw bc2 = 4.2 mm parallel to CB

• Draw C1C and C2C perpendicular to

aC1and bc2 respectively and obtain

the point C.

• The actual deflection of the joint C is from

(a,b) to c.

• By scaling from the diagram,

Vertical deflection of C= 𝜹𝑽 = 𝟏𝟐. 𝟖 𝒎𝒎.

Horizontal deflection of C = 𝜹𝒉 = 𝟒𝒎𝒎.


(a,b) c1

c2

c

𝛿ℎ = 4

𝛿𝑉 =12.8 mm

1. Find the strain energy stored in a steel bar 50 cm long and 3 cm× 1cm in cross section , shown in the following figure when it is subjected simultaneously to an axial pull of 50 kN and compressive stress of 100 N/mm2 on its narrow edge. (May/June 2013)


3 cm 50 cm 50 kN 50 kN

Compressive 100 N/mm2


• 𝜎𝑥 =50×103

30×10= 166.67 N/𝑚𝑚2

• 𝜎𝑦 = 100 N/𝑚𝑚2 , 𝜎𝑧 = 0

• Strain in x direction, 휀𝑥 =𝜎𝑥

𝐸− μ −

𝜎𝑦

𝐸

Strain in Y direction, 휀𝑦 = −𝜎𝑦

𝐸− μ

𝜎𝑥

𝐸

∴ 휀𝑦 = −(𝜎𝑦

𝐸+ μ

𝜎𝑥

𝐸)


3 cm 50 cm 50 kN 50 kN



x

y

Z

• Strain Energy, 𝑈 = 1

2

𝐿

0𝜎𝑥휀𝑥dv +

1

2

𝐿

0𝜎𝑦휀𝑦dv

• 𝑈 = 1

2

𝐿

0𝜎𝑥휀𝑥dA dx +

1

2

𝐿

0𝜎𝑦휀𝑦 dA dx

𝑈 = 1

2

500

0

𝜎𝑥𝜎𝑥𝐸+ μ𝜎𝑦𝐸(30 × 10) dx

+ 1

2

30

0

(−𝜎𝑦) −(𝜎𝑦𝐸+ 𝜇𝜎𝑥𝐸

) (500 × 10) dy

𝑈 = 1

2𝐸

500

0

𝜎𝑥2 + μ 𝜎𝑥𝜎𝑦 30 × 10 dx

+ 1

2𝐸

30

0

𝜎𝑦2 + 𝜇𝜎𝑥𝜎𝑦 (500 × 10) dy


• Assume 𝜇 = 0.3, 𝐸 = 2 × 105 N/mm2


𝑈 = 1

2𝐸

500

0

166.672 + 0.3 166.67 × 100 (30 × 10) dx

+ 1

2𝐸

30

0

1002 + 0.3 166.67 × 100 (500 × 10) dy

𝑈 =1

2 × 2 × 105166.672 + 0.3 166.67 × 100 (30 × 10) × 500

+ 1

2 × 2 × 1051002 + 0.3 166.67 × 100 (500 × 10) × 30

𝑈 = 12, 291.67 + 5625.03 ∴ 𝑈 = 17,916.71 𝑁𝑚𝑚

• For beams:

𝜹 = 𝑴𝒎 𝒅𝒙

𝑬𝑰

𝒍

𝟎

• For trusses:

𝜹 = 𝑷𝒌𝒍

𝑨𝑬

Principle of virtual work

• Find the slope at the centre of a cantilever beam subjected to u.d.l. of w kN/m for the whole span using energy principle.

Problems

W kN/m

‘L’ m

Problems

𝜃𝐶

Slope at C, 𝜃𝐶 =𝜕𝑈

𝜕𝑀𝐶

Since there is no external moment at C, assume a moment M at C.

Moment at any section XX between B and C from free end is given by,

𝑀𝑥 = −(w𝑥2

2); limits: 0 to L/2

𝜕𝑀𝑥𝜕𝑀= 0

C

W kN/m

‘L’ m

A B

M

𝑥 X

X

Y

𝑀𝑦 = −(M +w𝑦2

2); Limits: L/2 to L

𝜕𝑀𝑦𝜕𝑀= −1

Moment at any section YY between C and A from free end is given by,

𝑈 = 𝑀𝑥2

2 𝐸𝐼

𝐿/2

0


2 𝐸𝐼

𝐿

𝐿/2

𝑑𝑦

Problems

C

W kN/m

‘L’ m

A B

M

𝑥 X

X

Y

𝜃𝐶 𝜃𝑐 =𝜕𝑈

𝜕𝑀𝑐=𝜕𝑈

𝜕𝑀

=𝜕

𝜕𝑀 𝑀𝑥2

2 𝐸𝐼

𝑙/2

0


2𝐸𝐼

𝐿

𝐿/2

𝑑𝑦

= 2𝑀𝑥2𝐸𝐼

𝐿/2

0

𝜕𝑀𝑥𝜕𝑀𝑑𝑥 +

2𝑀𝑦

2𝐸𝐼

𝐿

𝐿/2

𝜕𝑀𝑦

𝜕𝑀𝑑𝑦

∴ 𝜃𝑐 =1

𝐸𝐼 𝑀𝑥

𝐿/2

0

𝜕𝑀𝑥𝜕𝑀𝑑𝑥 +1

𝐸𝐼 𝑀𝑦

𝐿

𝐿/2

𝜕𝑀𝑦𝜕𝑀𝑑𝑦

Problems

substitute, 𝑀𝑥 = −(w𝑥2

2);

𝜕𝑀𝑥𝜕𝑀= 0

𝑀𝑦 = −(M +w𝑦2

2);

𝜕𝑀𝑦

𝜕𝑀= −1

∴ 𝜃𝑐 =1

𝐸𝐼 −(w𝑥2

2) × 0

𝐿/2

0

𝑑𝑥 +1

𝐸𝐼 −(M +

w𝑦2

2) ×

𝐿

𝐿/2

−1 𝑑𝑦

∴ 𝜽𝒄 = 0 +1

𝐸𝐼 w𝑦2

2

𝐿

𝐿/2

𝑑𝑦 =1

2𝐸𝐼×𝑤𝑦3

3 =𝑤

6𝐸𝐼𝐿3 −𝐿

8

𝟑

=𝟕𝒘𝑳𝟑

𝟒𝟖𝑬𝑰

Substitute M=0

L/2

L

Problems

∴ 𝜃𝐵 =1

𝐸𝐼 𝑀𝑥

𝑙

0

𝜕𝑀𝑥𝜕𝑀𝑑𝑥

𝑀𝑥 = −(M +w𝑥2

2)

𝜕𝑀𝑥𝜕𝑀= −1

∴ 𝜃𝐵 =1

𝐸𝐼 −(M +

w𝑥2

2) (−1)

𝑙

0

𝑑𝑥

Substitute M=0 in the above equation, we get

𝜃𝐵 =1

𝐸𝐼 w𝑥2

2

𝑙

0

𝑑𝑥 =𝑤

2𝐸𝐼

𝑥3

3

0

L

∴ 𝜃𝐵 =𝑤𝐿3

6𝐸𝐼

• Find the slope at the free end of a cantilever beam subjected to u.d.l. of w kN/m for the whole span using energy principle.

Problems

W kN/m

‘L’ m

𝑀𝑥 = −(M +w𝑥2

2)

𝑈 = 𝑀𝑥2

2 𝐸𝐼

𝑙

0

𝑑𝑥

Problems

W kN/m

‘L’ m

A B

𝜃𝐵

Slope at B, 𝜃𝐵=𝜕𝑈

𝜕𝑀𝐵

M

𝜃𝐵 =𝜕𝑈

𝜕𝑀𝐵=𝜕𝑈

𝜕𝑀=𝜕

𝜕𝑀 𝑀𝑥2

2 𝐸𝐼

𝑙

0

𝑑𝑥 = 2𝑀𝑥2𝐸𝐼

𝑙

0

𝜕𝑀𝑥𝜕𝑀𝑑𝑥

Since there is no external moment at B, assume a moment M at B.

Moment at any section XX from free end is given by,

𝑥 X

X

• For the beam shown in the Figure, find the deflection at C and slope at D.

I= 40 ×107 mm4

E = 200 GPa.


2 m

40 kN

2 m 2 m

30 kN

A B C D

• Determine the vertical deflection at the free end of the cantilever truss shown in the following Figure. Take cross sectional area of compression members as 850 mm2 and tension members as 1000 mm2. Modulus of elasticity, E = 210 Gpa for all the members. (May/June 2012)


2m

3m 3 m

40 kN

For the truss shown in Figure find the total strain energy stored.


1 kN

4 m 4 m

3 m

A

B

C

E : 2 × 105 N/mm2

Area : AB : 100 mm2

BC : 100 mm2

AC : 80 mm2

(Nov/Dec 2011)

For the truss shown in Figure find the vertical deflection at ‘C’.


B

3 m

C

4 m A D

5 kN

Cross sectional area of all the members : 100 mm2

E = 2 × 105 N/mm2

(Nov/Dec 2011)

• For the truss shown in Figure, find the horizontal movement of the roller at D. AB, BC, CD area = 8 cm2

AD and AC = 16 cm2

E = 2 ×105 N/mm2 .


B

4 m

C

3 m A D

5 kN

• A bar of uniform cross section A and length L hangs vertically, subjected to its own weight. Prove that the strain energy stored

within the bar is given by 𝑢 =𝐴𝑥𝜌3𝑥𝐿3

6𝐸. (Nov/Dec 2014)


• A simply supported beam having 8 m span and carries UDL of 40 KN/m as shown in fig.(a). Determine the deflection of the beam at its midpoint and also the position of maximum deflection and maximum deflection. Take E=2x105 N/mm2 and I=4.3 x108 mm4.

(Nov/Dec 2014)


1 m 4 m 3 m

A B C D

40 kN/m

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UNIT I ENERGY PRINCIPLES - Sri Venkateswara College of ... of... · • Strain energy due to shear...

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