Post on 28-Mar-2018
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.
Strain energy density
• 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.
Strain energy density
• 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
Strain energy density
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
Strain energy density
• (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
Elastic Strain Energy for Normal Stresses
• 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?
Elastic Strain Energy due to axial force
P P b B
t
𝑏𝑥 = b +𝐵 − 𝑏
𝐿𝑥
= 𝑏 + 𝑘𝑥
P P b B
t L
𝑥
Elastic Strain Energy due to axial force
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.
Elastic Strain Energy due to axial force
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.)
Elastic Strain Energy due to axial force
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
Elastic Strain Energy due to axial force
• What is the expression for strain energy of a prismatic bar under its own weight.
Elastic Strain Energy due to axial force
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
Elastic Strain Energy due to axial force
L 𝛿𝑥
• (b) Due to transverse load on beams:
Elastic Strain Energy for Normal Stresses
• 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∴
Elastic Strain Energy for Normal Stresses
• 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
Elastic Strain Energy for Normal Stresses
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
6𝐸𝐼2ℎ + 3𝑏
• Let the horizontal movement of D be 𝛿
• External work done, 𝑊𝑒 =1
2Pδ
• We know, 𝑊𝑒 = U
• ∴ 1
2Pδ =
𝑃2ℎ2
6𝐸𝐼2ℎ + 3𝑏
• ∴ 𝛿 =𝑃ℎ2
3𝐸𝐼2ℎ + 3𝑏
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
Resolving vertically,
𝑃𝐷𝐵=20 kN.
20 kN
40 𝑃𝐷𝐸
𝑃𝐷𝐵
D
• For the equilibrium of joint B,
Deflection of Trusses Problems
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
Resolving vertically,
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
• For the equilibrium of joint B,
Deflection of Trusses Problems
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
Resolving horizontally,
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
Deflection of Trusses Problems
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
Castigliano’s First theorem derivation
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
Castigliano’s First theorem derivation
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,
Castigliano’s First theorem derivation
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
Castigliano’s First theorem derivation
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),
Castigliano’s First theorem derivation
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.
Deflection of Trusses Problems
Deflection of Trusses Problems
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.
• For the equilibrium of joint C,
Resolving vertically,
𝑃𝐶𝐵 sin 𝜃=P ∴ 𝑃𝐶𝐵 = P 5 kN.
Resolving horizontally,
𝑃𝐶𝐷= 𝑃𝐶𝐵 cos 𝜃
= P 5 ×2
5 = 2P kN.
∴ 𝑃𝐶𝐷 = 2𝑃 𝑘𝑁.
Deflection of Trusses Problems
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,
Deflection of Trusses Problems
A
B
C
D E
20 kN P kN
2 m
2 m 2 m
1m 2𝑃
P 5
Resolving horizontally, 𝑃𝐷𝐸= 2𝑃
Resolving vertically,
𝑃𝐷𝐵=20 kN.
20 kN
2P 𝑃𝐷𝐸
𝑃𝐷𝐵
D
• For the equilibrium of joint B,
Deflection of Trusses Problems
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
𝑃𝐵𝐸 𝑃𝐵𝐴
𝜃 𝜃
Resolving vertically,
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
• For the equilibrium of joint B,
Deflection of Trusses Problems
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
𝑃𝐵𝐸 𝑃𝐵𝐴
𝜃 𝜃
Resolving horizontally,
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
• Strain energy, 𝑈 = 𝑃2𝐿
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.
Deflection of Trusses Problems
• Horizontal displacement of Joint C:
• Now assume a horizontal load of Q kN at C.
Deflection of Trusses Problems
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.
• For the equilibrium of joint C,
Deflection of Trusses Problems
A
B
C
D E
20 kN 20 kN
2 m
2 m 2 m
1m
Q
𝜃
20
C
𝑷𝑪𝑩
𝑸 𝑷𝑪𝑫
Resolving vertically,
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
Deflection of Trusses Problems
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
Resolving vertically,
𝑃𝐷𝐵=20 kN.
Resolving horizontally, 𝑃𝐷𝐸= 𝑄 − 40
Deflection of Trusses Problems
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
𝑃𝐵𝐸 𝑃𝐵𝐴
𝜃 𝜃
For the equilibrium of joint B,
𝜃
Resolving vertically,
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)
Deflection of Trusses Problems
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
𝑃𝐵𝐸 𝑃𝐵𝐴
𝜃 𝜃
𝑃𝐵𝐴 = 40 5 − 𝑃𝐵𝐸------(1)
Resolving horizontally,
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
Deflection of Trusses Problems
20 5 10 5
30 5
• Strain energy, 𝑈 = 𝑃2𝐿
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 of Trusses Problems
• Deflection at joint C, 𝛿𝐶 = 𝛿𝑉 𝐶2+ 𝛿𝐻 𝐶
2
= 5.494 2 + −1 2
∴ 𝛿𝐶 =5.59 mm
Deflection of Trusses Problems
• 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
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).
A.U. Question paper problems
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)
A.U. Question paper problems
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.U. Question paper problems
• 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.U. Question paper problems
• 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)
A.U. Question paper problems
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
Williot diagram - Problem
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
Williot diagram - Problem
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.
Williot diagram - Problem
(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)
A.U. Question paper problems
3 cm 50 cm 50 kN 50 kN
Compressive 100 N/mm2
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)
A.U. Question paper problems
3 cm 50 cm 50 kN 50 kN
Compressive 100 N/mm2
Compressive 100 N/mm2
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
A.U. Question paper problems
• Assume 𝜇 = 0.3, 𝐸 = 2 × 105 N/mm2
A.U. Question paper problems
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
𝑑𝑥 + 𝑀𝑦2
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
𝑑𝑥 + 𝑀𝑦2
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.
A.U. Question paper problems
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)
A.U. Question paper problems
2m
3m 3 m
40 kN
For the truss shown in Figure find the total strain energy stored.
A.U. Question paper problems
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’.
A.U. Question paper problems
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 .
A.U. Question paper problems
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.U. Question paper problems
• 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)
A.U. Question paper problems
1 m 4 m 3 m
A B C D
40 kN/m