Lecture-3-1.pptx

Lecture-3
Mechanics of Materials-I
Waqas Asghar
waqas.asghar@uettaxila.edu.pk
Lecturer, Mechanical Dept.
UET Taxila

SAINT-VENANT’S PRINCIPLE
• The principle was first noticed by the French scientist
Barré de Saint-Venant in 1855.
• Notice the stress distribution acting at sections
a–a, b–b, and c–c
• At c–c , stress distribution is uniform & is located at
a distance ≥ width of bar.
• At b–b , stress distribution is non-uniform.
• At a–a , stress distribution is highly non-uniform.
In a loaded material, “if the cross section is taken away from the point of load application or
support, stress & strain distribution on that cross-section will be independent of applied load”.

SAINT-VENANT’S PRINCIPLE
If we convert the point load to uniformly distributed load, resultant stress and strain distribution
will become uniform.

• A bar having variable cross-sectional area along its length L, is shown in figure.
• Bar is subjected to concentrated loads at its end.
• 𝛿 =? = Relative displacement of one end w.r.t other, caused by loading.
• Neglect the localized deformation caused at the point of concentrated loading and where the
cross section changes suddenly,
• Take an element of length dx, area A(x) from bar, at any position x.
• Stress and strain in element will be given by:
𝝈 =
𝑷(𝒙)
𝑨(𝒙)
and 𝜺 =
d𝜹
𝒅𝒙
Elastic Deformation Of Axially Loaded Member

Within elastic limit, Hooke’s law is applicable; i.e.
𝝈 = 𝑬. 𝜺
Putting the values of 𝝈 & 𝜺 from previous slide, Hooke’s law becomes
𝑷(𝒙)
𝑨(𝒙)
= 𝑬.
d𝜹
𝒅𝒙
d𝜹 =
𝑷 𝒙 dx
𝑨 𝒙 𝑬
For entire length of bar
𝜹 = 𝟎
𝑳 𝑷 𝒙 dx
𝑨 𝒙 𝑬
where
𝜹 = displacement of one point on the bar relative to the other point
L = original length of bar
P(x) = Internal axial force at the section (in terms of x), determined by method of sections
A(x) = cross-sectional area of the bar, expressed as a function of x
E = modulus of elasticity for the material

𝜹 =
𝑷𝑳
𝑨𝑬
Elastic Deformation of Bar Subjected To Constant Axial Force,
Having Constant Cross –Sectional Area & Constant ‘E’
𝜹 = ∑
𝑷𝑳
𝑨𝑬
Elastic Deformation of Bar Subjected To Several Different Axial
Forces, Having Variable Cross –Sectional Areas & Variable ‘E’

Sign Convention
Force P and displacement 𝛿 will be +ve if they cause tension or elongation and vice versa.

Example 4.1: A-36 steel bar ( E=29× 103 𝑘𝑠𝑖) is made from two segments
having cross-sectional areas of AAB = 1 in2 and ABD = 2 in2. Find 𝛿𝐴 the vertical
displacement (𝜹) of end A and the displacement (𝜹) of B relative to C.
Solution:
Find internal force in each member. Due to the application of variable
external loadings, the internal axial forces in regions AB, BC, and CD
will all be different & can be obtained by applying method of sections .
Section AB:
+↑ ∑𝑃𝑦 = 0
15 kip – PAB = 0
PAB = 15 kip (T)

Section AC:
+↑ ∑𝑃𝑦 = 0
15 – (4 ×2 ) – PBC = 0
PBC = 7 kip (T)
Section AD:
+↑ ∑𝑃𝑦 = 0
15 – (4×2 ) – (8×2) – PCD = 0
PCD = – 9 kip (T)
Displacement of A relative to the fixed support D:
+ve sign shows elongation, so the displacement at A is upward.

Displacement of B relative to C:
+ve sign shows elongation, as B moves away from C.

Example 4.3: Rigid beam AB rests on the two short posts. Post AC is made of
steel having 20 mm dia and Est = 200 G. Post BD is made of aluminum having 40
mm dia and Eal = 70 G. Determine the displacement (𝜹) of point F.
Solution:
Find the internal forces, by using force and moment
equilibrium equations, as:
+↑ ∑𝐹𝑦 = 0
− 90 + FA + FB = 0
FA + FB = 90 kN
+↺ ∑𝑀𝐴 = 0
− 90(200) + (600)FB = 0
FB = 30 kN
Putting FB into above eq, we have : FA = 60 kN
Support Reactions at Point A & B

𝜹 of point F can’t be found directly. For this purpose, initially
we have to found the deflection of point A & B.
Post AC:
Post BD:
Compressive forces acting on post AC & BD are equal in
magnitude to the internal forces, calculated previously (FA
= 60 kN, FA = 30 kN) but will act in opposite direction to
maintain equilibrium.
Displacement of Point A & B
Compressive Forces Acting on Post AC & BD

Now from trigonometry taking proportion of sides of two blue shaded triangles Δ𝑧𝑛𝑦 𝑎𝑛𝑑 Δ𝑧𝑚𝑥
𝑧𝑛
𝑧𝑚
=
𝑛𝑦
𝑚𝑥
⟹
400 𝑚𝑚
600 𝑚𝑚
=
𝑛𝑦
0.184
𝐧𝐲 = 𝟎. 𝟏𝟐𝟐𝟕 𝒎𝒎
Diagram Showing Displacements of Points A, B & F
We have already calculated the
𝜹𝑨 & 𝜹𝑩. So From figure
𝜹𝑨 = 𝜹𝑩 +mx
0.286 𝑚𝑚 = 0.102 𝑚𝑚 + mx
𝒎𝒙 = 𝟎. 𝟏𝟖𝟒 𝒎𝒎
Displacement of Point F
As, 𝜹𝑭 = 𝜹𝑩 +ny
𝜹𝑭 = 0.102 𝑚𝑚 + 0.1227 mm
𝜹𝑭 = 𝟎. 𝟐𝟐𝟓 𝒎𝒎 ↓

The Principle of Superposition
In a linearly elastic structure, combined effect of several loads acting simultaneously on a
member is equal to the summation of the loads acting separately.
• This Principle simplifies the stress (𝝈), displacement (𝜹) or deflection problems having
complicated or multiple loadings.
• This is done by breaking the member down as many times as necessary for each force acting
on it.
• Once all the stresses or deflections are found, they can then be added to get a final answer.
• Necessary conditions required to apply principle of superposition are:
 The loading must be linearly related to the stress or displacement, that is to be determined.
OR Hooke’s law remains applicable.
 The loading must not significantly change the original geometry of the member.

Forces in all segment are of tensile nature
Start simplifying from R.H.S
By Principle of superposition
Problem: A steel bar of 25 mm dia, is acted upon by forces as shown in figure.
What is the total elongation of bar when E = 190 GPa.
Solution:
Net Elongation = Sum of Individual Elongations
ΔL = ΔL1 + ΔL2 + ΔL3
As E =
𝜎
𝜀
=
P
A
Δ𝐿
𝐿
=
𝑃𝐿
𝐴∆𝐿
Elongation or Extension or Deformation of whole bar, ΔL =
PL
𝐴𝐸
ΔL =
P1
L1
𝐴𝐸
+
P2
L2
𝐴𝐸
+
P3
L3
𝐴𝐸
=
1
𝐴𝐸
P1L1+ P2L2 + P3L3

A =
𝜋
4
0.025 2 = 4.908 × 10
_4 m2
ΔL =
1
𝐴𝐸
P1L1+ P2L2 + P3L3
ΔL =
1
4.908 × 10−4 190 ×109 (60000 × 2)+(80000 × 1)+(50000 × 3)
= 3.75 × 10
_3 m = 3.75 mm
Elongation of whole bar, 𝜟L
𝑨 =
𝝅
𝟒
𝒅 𝟐

ΔL = ΔL1 + ΔL2 + ΔL3
ΔL =
1
𝐸
[
P1
L1
𝐴1
−
P2
L2
𝐴2
+
P3
L3
𝐴3
]
E = 205 × 109
Pa = 205 × 109 𝑁
𝑚2 =
205×109 𝑁
1000𝑚𝑚 2
= 205 × 103 𝑁
𝑚𝑚2
ΔL =
1
205×103 [
50000×150
176.71
−
250000 ×200
706.86
+
80000 ×250
314.16
] = ≈ 0.173 mm
Problem: A circular bar (E = 205 GPa) has three segments, as shown in figure.
Find: (a) Total elongation in bar. (b) Length of middle segment to have zero
elongation of bar. (c) Dia of last segment to have zero elongation of bar.
Solution:
(a) Total Elongation in the Bar:
𝐴 =
𝜋
4
𝑑 2
𝐴1 = 176.71 𝑚𝑚2
𝐴2 = 706.86 𝑚𝑚2
𝐴3 = 314.16 𝑚𝑚2
Net Elongation =Sum of Individual Elongations

Putting ΔL = 0
0 =
1
205×103 [
50000×150
176.71
−
250000 ×L2
706.86
+
80000 ×250
314.16
]
L2 = 300 mm
(b) Length of Middle Segment (L2) at Zero Elongation of Bar:
Putting ΔL = 0
1
205×103 [
50000×150
176.71
− (
250000 ×200
706.86
)+(
80000 ×250
𝜋
4
𝑑3
2 )] = 0
𝑑3 = 30 mm
(c) Dia of Last Segment (D3) to have Zero Elongation of Bar:

Statically Indeterminate Systems
The systems in which equilibrium equations are not sufficient
to find all reaction on a member.
• For this purpose, deformation equation of material (ΔL =
PL
𝐴𝐸
) is also taken into account.
• Statically indeterminate systems may consist of ≥ 2
members or of different materials
• Example: Equilibrium equation of bar shown is given
below: Simple Bar
Free
Body
Diagram
+↑ 𝑭 = 𝟎, ⇨ 𝑭𝑨 +𝑭𝑩 − 𝑷 = 𝟎

Simple Bar
Free
Body
Diagram
Kinematic or Compatibility condition.
• Additional equation required for the solution of statically
indeterminate systems, is referred as Compatibility condition.
• In Compatibility condition, it is necessary to consider how
points on the bar displace w.r.t. each other
• In compatibility condition of the shown bar, displacement of
one end of the bar w.r.t other end becomes zero ( 𝜹𝑨
𝑩
= 𝟎),
since the end supports are fixed.
• Hence, the compatibility condition becomes:

As bar remains horizontal than deformation in brass and steel
will also be same i.e. (means 𝜟Lb = 𝜟Ls).
Problem: Three equally spaced circular rods in same vertical plane supports a rigid bar
AB. Two outer rods are of brass each 600 mm long and 25mm diameter. The central
steel rod is of steel that is 800mm long and 30mm diameter. Determine forces in the
bars due to applied load of 120 KN through the mid point of the bar. Bar remains
horizontal after application of load.Take Es/EB = 2
Solution:
Put Pb into above equation
2(
25
54
Ps) + Ps = 120
⇨ Ps =
𝟏𝟐𝟎×𝟓𝟒
𝟏𝟎𝟒
= 𝟔𝟐. 𝟑 𝒌𝑵
PbLb
𝐴𝑏𝐸𝑏
=
PsLs
𝐴𝑠𝐸𝑠
𝐏𝐛 = Ps [
Ls
Lb
×
𝐴𝑏
𝐴𝑠
×
𝐸𝑏
𝐸𝑠
] = Ps [
800
600
×
𝜋
4
25 2
𝜋
4
30 2
×
1
2
]
FBD
⇨ 𝐏𝐛 =
𝟐𝟓
𝟓𝟒
𝐏𝐬
⇨ 2Pb+ Ps = 120
+↑ ∑𝐹𝑦 = 0 ⇨ Pb+ Ps + Pb – 120 = 0
⇨ Pb = 𝟐𝟖. 𝟖 𝒌𝑵

Problem: Circular bar is loaded as shown. Find (a) Reaction at lower support. (b)
Stresses in bars. Take E = 205 Gpa (205×103 N/mm2)
Solution:
From the figure
ΔLnet = ΔL1 + ΔL2 =0.8mm
From equilibrium eq
R1 + R2 – 40 = 0 ⇨ R1 = 40kN – R2
As we know
ΔL1 – ΔL2 = 0.8 𝑚𝑚
Given: A1 = 80 mm2 , A2 = 160 mm2
1
205×103 [
40000−R2
80 𝑚𝑚2 −
2400 ×R2
160 𝑚𝑚2 ] = 0.8 mm ⇨ R2 = 14533.3 N
P1
L1
𝐴1
𝐸
–
P2
L2
𝐴2
𝐸
= 0.8 mm
⇨ R1 = 25466.7 N

(b) For stresses
𝜎1 =
𝑃1
𝐴1
=
𝑅1
𝐴1
=
25466.67
80
𝝈𝟏 = 318.3 N/mm2
𝜎2 =
𝑃2
𝐴2
=
𝑅2
𝐴2
=
1453.3
160
𝝈𝟐= 90.83 N/mm2

Problem: A load of 2MN is applied on short concrete column ( 500×500 )mm.
Column is reinforced by four steel bars of 10mm dia. One in each corner. Find
stresses in the concrete and steel bars. Est = 2.1×105 N/mm2, Econ = 1.4 ×104 N/mm2
Solution:
P = 2×106 N
Area of steel bar =
𝜋
4
× 102 = 25 𝜋 mm2,
Est = 2.1×105 N/mm2 Econ = 1.4 ×104 N/mm2
Steel bars dia = 10 mm, Quantity of Steel bars = 4
Concrete cross section = 500×500 mm,
Given:
Area of concrete = ( 500×500 ) – 4 (25𝜋) = 249685.8 mm2
As load P is shared by column (Steel bar + concrete)
So,
P = 4Psteel + Pconcrete
2×106 = 4 (𝜎s × 𝐴s) + (𝜎c × 𝐴c)
2×106 = 4 (𝜎s × 25𝜋) + (𝜎c ×249685.8) ………….. (1)
⇨ 𝝈 =
𝑷
𝑨
or P = 𝝈A

Compression in any steel bar = Compression in concrete bar
𝑃s
Ls
𝐴𝑠
𝐸𝑠
=
𝑃c
Lc
𝐴𝑐
𝐸𝑐
𝜎s
Ls
𝐸𝑠
= 𝜎c
Lc
𝐸𝑐
𝜎s
𝜎c
=
𝐸𝑠
𝐸𝑐
𝜎s =
𝐸𝑠
𝜎c
𝐸𝑐
=
2.1×105
1.4×104 𝜎c
𝜎s = 15 𝜎c …….(2)
Putting Equation 2 in equation 1
∴ 𝐋𝐬 = 𝐋𝐜
⇨ 𝝈𝐬 = 117.92 N/mm2 ⇨ 𝝈𝐜 = 7.86 N/mm2

Problem: Two vertical rods (steel and copper) each are rigidly fixed at the top and
50 cm apart. Dia and lengths of each rod are 2cm and 4m respectively. A cross bar
fixed to the rods at the lower ends carries a load of 5000N such that the cross bar
remains horizontal even after loading. Find the stress in each rod and the position
of the load on the bar. E for steel = 2×105 N/mm2 and E for Copper = 1×105 N/mm2
Solution:
Given:
(a) As bar remains horizontal, so
Extension in steel = Extension copper
∆ Lsteel = ∆Lcu
𝑃s
Ls
𝐴𝑠
𝐸𝑠
=
𝑃c
Lc
𝐴𝑐
𝐸𝑐
𝜎s
Ls
𝐸𝑠
= 𝜎c
Lc
𝐸𝑐
ds = dc = 2 cm, , Ls = Lc = 4 cm
Esteel = 2×105 N/mm2 , ECopper = 1×105 N/mm2
∴ 𝐋𝐬 = 𝐋𝐜
⇨ 𝝈𝐬 =
𝑬𝒔
𝝈𝐜
𝑬𝒄
⇨
𝜎s
𝜎c
=
𝐸𝑠
𝐸𝑐
By putting values of Es & Ec
𝜎s = 2 𝜎c ….(1)

Taking +↑ ∑𝐹𝑦 = 0 ⇨ Ps + Pc – 5000 = 0
Ps + Pc = 5000
𝜎𝑠 As + 𝜎c Ac = 5000
(𝜎𝑠 + 𝜎c)A = 5000
𝜎𝑠 + 𝜎c = 15.92 …..(2)
By solving eq (1) and eq (2)
⇨ 𝝈 =
𝑷
𝑨
or P = 𝝈A
∴ 𝑨𝒔 𝒅𝐬 = 𝒅𝐜 , So 𝐀𝐬 = 𝐀𝐜
𝝈𝐬 = 10.61 N/mm2 𝝈𝐬 = 5.3 N/mm2
(b) Position of the load on the bar.
Also, +↺ ∑𝑀𝐵 = 0
– Ps (500mm) + 5000x = 0
x =
Ps
10
=
𝜎𝑠
As
10
=
10.61×
𝜋
4
×(20)2
10
= 333.3 mm

31
Practice Questions of Ch. # 4 (Hibbeler 8th ed.)
• Problems = 4- 31, 35, 38, 39, 41, 42, 44, 45, 46, 62

4.31,35,38,39,41,42,44,45,46,62
Numerical from Hibler
4.65,69,70,71,72,73,75,77,78,79,84,85,86
Thermal Stresses

FORCE METHOD OF ANALYSIS FOR
AXIALLY LOADED MEMBERS
Statically indeterminate problems can be solved by writing the
compatibility equation using the principle of superposition.
Take the pt. B and temporarily remove its effect on bar
• P causes B to be displaced downward by an amount 𝛿𝑝
• reaction must displace end B of the bar upward by an amount
𝛿𝐵
So net displacement occurs at B = 0. So compatibility equation
for displacements at point B is
0 = 𝛿𝑝 - 𝛿𝐵

Thermal Stresses
Case-I (Free
Expansion)
• Temperature
increases from t1 to
t2
T = t2 - t1
∆𝐿 = α T L
Where
α = linear coefficient
of thermal expansion
• If material is
allowed to expand
freely no thermal
stress is induced
Case-II (No Free Expansion)
T = Rise in temperature
 Support are preventing the
expansion
Strain prevented = ∆𝑙 = α T L
Strain =
𝐸𝑥𝑛𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑝𝑟𝑒𝑣𝑒𝑛𝑡𝑒𝑑
𝑂𝑟𝑖𝑔𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
𝜀 =
αT L
𝐿
𝜀 = αT
Thermal or compressive strain
From Hook’s law
E =
σ
𝜀
𝜀 =
σ
𝐸
αT =
σ
𝐸
σ = EαT
Thermal stress compressive
nature

If temperature decreases bar will be contract. So strain and stress
developed will be tensile on nature.
∆𝐿 = α T L In case of free expansion
𝜀 = αT Thermal or compressive stress No allowance for
expansion
σ = EαT Thermal stress is present
When allowance for expansion is present
𝜀 =
αT L − ∆𝑙
𝐿
σ = 𝜀 E = (
αT L − ∆𝑙
𝐿
)E
Where
α = linear coefficient of thermal expansion
T = Change in temperature
L = Original length
∆𝑙 = Change in length

Given:
t1 = 24℃ L = 32m
a) Stress in rails at 80℃ when there is no allowance for expansion
σ = αTE
σ = 11×10-6 /c × (80-24) × 205×109 Pa
= 126280×103 Pa
σ = 126.8 MPa
Rails are laid down such that there is no stress in them at 24℃. If rails length = 32 m, find
a) Stress in rails at 80℃ when there is no allowance for expansion
b) Stress in rails at 80℃ when there is an allowance for expansion of 8m/rail.
c) Expansion allowance for no stress in rails at 80℃
d) Max. temperature for no stress in the rails when expansion allowance is 8mm
α = 11×10-6 /c E = 205 GPa
Problem#
Solution

b) Stress in rails at 80℃ when there is an allowance for expansion of
8m/rail
σ = 𝜀 E = (
αT L − ∆𝑙
𝐿
)E
σ = (
11×10−6 /c × (56×32)−(8 ×10
_
3
)
32
) × 205×103
σ = 7503.000 Nm2
σ = 75 MPa
c) Expansion allowance for no stress in rails at 80℃
𝜀 = αT
∆𝑙
𝐿
= αT
∆𝑙 = αTL
∆𝑙 = 8 × 10
_3/c × (80-24)℃ × 32m = 0.01971m = 19.7 mm
d) Max. temperature for no stress in the rails when expansion allowance
is 8mm
∆𝑙 = 8 mm T = t2 – t1 = t2 – 24
∆𝑙 = αTL
8 ×10-3 m = 8 × 10
_3 × (t2 – 24) × 3L
t2 = 46.72 ℃

Thermal stress in composite bars
Brass and steel bars rigidly fixed together
Pics Remaining
αbrass > αsteel
Compressive stress in brass
Material having high α will elongate less than expected. So there
will be compressive strain and stress in member and vice versa.
Overall expansion in brass = Overall expansion in steel
α𝑏TL –
α𝑏
L
𝐸𝑏
= α𝑠TL –
α𝑠
L
𝐸𝑠
Free Expansion
From equilibrium condition
Tensile force = Compressive force
Ptensile = Pcomp
Psteel = Pbrass
σsAs = σbAb

Steel Tube Cu Rod
do = 30 mm d = 15 mm
di = 20 mm
t1 = 10℃ (no stress)
t2 = 200℃
T = (200-10)℃ = 190℃
αCu (Undergo Compressive stress) > αsteel (Undergo Tensile stress)
Steel tube of 30 mm out. dia and 20mm internal dia, encloses a Cu rod of 15m dia to
which it is rigidly joined at end. If at a temp. of 10 there is no stress. Find stress in rod and
tube when temp is raised to 200℃. Esteel = 2.1×105 N/mm2 ECu = 105 N/mm2,
α𝑠𝑡𝑒𝑒𝑙 = 11×10-6/℃ α𝐶𝑢 = 18×10-6/℃
Problem#
Solution

Overall expansion in steel = Overall expansion in Cu
α𝑠TLs +
α𝑠
Ls
𝐸𝑠
= α𝐶𝑢TLs –
α𝐶𝑢
LCu
𝐸𝐶𝑢
Where Ls = LCu
α𝑠T +
α𝑠
𝐸𝑠
= α𝐶𝑢T –
α𝐶𝑢
𝐸𝐶𝑢
……………(1)
From equilibrium
Force in steel = Force in Cu
Psteel = Pcu
σsAs = σCuACu
σs
𝜋
4
(302 − 202) = σCu
𝜋
4
(152)
σs =
9
20
σcu……………………………..(2)
Putting eq (2) into eq (1)
α𝑠T +
α𝑠
𝐸𝑠
= α𝐶𝑢T –
α𝐶𝑢
𝐸𝐶𝑢
(11×10-6 × 190)(
9σ𝐶𝑢
20 ×2.1 ×105) = (18×10-6 × 190) - (
σ𝐶𝑢
105)

Simplifying
σCu = 109.5 N/mm2
σs = 9/20 × 10.75 N/mm2 = 49.2 N/mm2

The Principle of Superposition
In a loaded material, resultant stress (𝝈) or displacement (𝜹) at a point can be determined by
algebraically summing the stress or displacement caused by each load component, applied
separately to the member.
• Necessary conditions required to apply principle of superposition are:
 The loading must be linearly related to the stress or displacement, that is to be determined.
OR Hooke’s law remains applicable.
 The loading must not significantly change the original geometry of the member.
• This Principle simplifies the stress and displacement problems having complicated loadings.
• This is done by subdividing the loading into components, then algebraically adding the results
• In a linearly elastic structure, combined effect of several loads acting simultaneously on a
member is equal to the summation of the loads acting separately.
• This Principle is used to solve complex stress (𝝈), displacement (𝜹) or deflection problems with
multiple loads and/or reactions acting on the member.
• Superposition helps us solve these problems by breaking the member down as many times as
necessary for each force acting on it.
• Once all the stresses displacements or deflections for the point of interest are found, they can then be
added all together to get a final answer.

Statically Indeterminate Systems
The systems in which equilibrium equations are not sufficient to find all
reaction on a member.
• For this purpose deformation of equation of material (ΔL =
PL
𝐴𝐸
) is also
taken into account.
• If system consist of two members of different materials it become
necessary to take deformation into account.
• An additional equation required for the solution of statically
indeterminate systems, is referred as Compatibility condition.
• In Compatibility condition, it is necessary to consider how points on
the bar displace w.r.t. each other
• An additional equation is required for solution, which should consider
how points on the bar displace wrt each other. Specifically, an
equation that specifies the conditions for displacement is denoted to
as a In this case, a suitable compatibility condition would require the
Simple Bar
Free
Body
Diagram
ndition.

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