MECHANICS OF SOLIDS-I

MODULE-3
SIMPLE STRAINS
PROF. DR. MOHAMMAD ASHRAF
DEPARTMENT OF CIVIL ENGINEERING, UET PESHAWAR
2 CONTENTS
• Stress-Strain Relation
• Simple Strains
• Normal Strains
• Shearing Strains

• Poisson’s Ratio: Biaxial and Triaxial Deformations

• Analysis of Indeterminate Axially Loaded Members
3 STRESS-STRAIN CURVE
Let consider an axially loaded steel coupon:
A = Original cross sectional area of the coupon before application of load
A’ = Cross sectional area of the coupon after application of load
P = Axial load applied on coupon.
L = Original gauge length before application of load.
L’ = Gauge length after application of load
The stress in coupon is,
σ = P/A - - - - - - - - - - - - - (a)
The average strain in the gauge length is,
Є = ΔL/L = (L’-L)/L - - - - - (b)
4 STRESS-STRAIN CURVE
When the coupon is gripped between jaws of
Universal Testing Machine (UTM) and stretched
We apply strain and measure load.We increases the deformation and
gradually up to rupture. measure the load.

The load (P) resisted by the coupon and the

deformation (ΔL ) produced between the gauge
length are recorded.
Stresses and strains at various deformation/ load
levels are calculated from σ=P/A and Є=ΔL/L
A curve is drawn between the strain (as abscissa)
and stress (as ordinate) known as stress-strain
curve.
5 STRESS-STRAIN CURVE
Practically there may be very less difference
Salient Feature of Stress-Strain Curve of Mild Steel: b/w elatic and propotional limiit.Practically
OR Modulus of easticityy we consider it one.
Youngs Modulus: Initially, the stress-strain curve is
linear, i.e. the stress is directly proportional to
strains (Hook’s Law is applicable).
σ α Є => σ = EЄ
Where E is known as Young’s Modulus, or modulus
of elasticity which is the slope of the initial straight
line of σ-Є curve.
Proportional Limit: The maximum stress/strain up
to which stress in directly proportional to strain is
known as proportional limit.
6 STRESS-STRAIN CURVE
Salient Feature of Stress-Strain Curve of Mild Steel:
Elastic Limit: The point on the stress-strain curve
up to which if the coupon is unloaded, then it will
come back to its original position without any
permanent change in length.
Yield Point: The point at which there is appreciable
elongation (increase in strain) without any
corresponding increase in stress,; indeed, the
stress may actually decrease when yielding
occurs. At yield point the needle stop for some time of UTM machine
Starting point of yielding is called yeild point.It recover
This is the idealized version of
curve and they are shown as a one
some defromation on unloaading but some still remains..
Practically there is negligible difference between point on the curve.

the above three points.

after yeild point may be 1 to 2 % max decrease.

Hum propotional strength ki bat kam karte hai zida tar hm Yeild strength awr elastic strain ki bat karte hai.
Strenght at the elatic , or propional or yield is point is called coresponding strenght.This is strain softening
upto start of increase of stress again.sb/c strain increase without increase in stress.
7 STRESS-STRAIN CURVE
Salient Feature of Stress-Strain Curve of Mild Steel:
Ultimate Strength: The maximum strength on the
stress-strain curve is known as ultimate strength.
Rupture Strength: The strength at failure/ rupture of
the coupon is called rupture strength. Rupture
strength is less than the ultimate strength.
Strain Hardening: The gain in strength beyond the
yielding of material is called strain hardening.
While Reort of client contains yield and ultimate strength.There is
instrument available which will give you area in actaul situation
at differnt stages or do instrumentation.
 This neckang is not neccarily

8 STRESS-STRAIN CURVE
occur at mid but where is
imperfection neching will
strat from that ,Sometime
necking happen is jawas so
Salient Feature of Stress-Strain Curve of Mild Steel: we cannot measure elongation
so we took gauge legh aroun
point where necking occurs.
Necking: Up to the ultimate strength, the change in
cross sectional area of coupon is negligible. However,
when the coupon is stretched beyond the ultimate point,
the cross sectional area of the coupon decreases until
rupture occurs. This phenomenon is called necking.
Actual Rupture Strength: The rupture strength, when
calculated using the reduced cross sectional area, A’ is
called actual rupture strength.
σ’rupture= Prupture/A’
The actual rupture strength is greater than the rupture
strength.
9 STRESS-STRAIN CURVE When we combine that ho much load
it can take as well as deform

Salient Feature of Stress-Strain Curve of Mild Steel:

upto elastic point si it is modulus of
resileince and utpo rupture poin combination
is called modulus of toughness.
The area under stress-strain curve is equal to the work
done on the coupon per unit volume.
Modulus of resileince=1/2(Stress*strain)

Modulus of Resilience: The area under stress strain

curve from origin up to the yield is known as modulus
of resilience. Resilience is the ability of a material to
absorb energy when it is deformed elastically, and
release that energy upon unloading.
Modulus of Toughness: The total area under the stress-
strain curve from origin to the rupture point is known as
modulus of toughness. Toughness is the ability of a
material to absorb energy and plastically deform
without fracturing.
But in second figure the body does
not release the whole energy .some
stored in the body producing permanat
deformation.And if material obsorb more energy that would be good material
10 STRESS-STRAIN CURVE
Ductile and Brittle Materials Reinforcedc concrete if proper design are ductile

Ductile Material: Materials which gives warning

before its rupture are called ductile materials. These
materials, exhibiting large rupture strain, gives large
deformation before their failure. Mild steel,
aluminum, etc. are ductile materials.
Weaker ductile material
Brittle Material: Materials which fails abruptly are preffered than stronger
brittle material
without giving warning in the shape of large
deformation. Plain cement concrete, unreinforced
brick masonry, high carbon steel, etc. are brittle
materials.
11 NORMAL AND SHEAR DEFORMATIONS
Normal Strain: Let P is the axial load applied on a member of length L which produces an axial
deformation of Δ. The strain and stress produced in the member are:
Є = Δ/L - - - - - - (a)
σ = P/A - - - - - - (b)
From the Hook’s Law:
σ α Є => σ = EЄ - - - - - - (c)
Putting values of stress and strains from equations (a) and (b) respectively in equation (c),
P/A = E (Δ/L)
=> Δ = PL/AE - - - - - - (d)
AE is known as the axial rigidity of the member.
Example: Calculate axial deformation in a steel member (E = 29000 ksi) of L = 10 ft and A = 4 in2
subjected to a axial load P = 100 kips.
Δ = PL/AE = 100 x (10x12)/(4 x 29000) = 0.1034 in.
12 NORMAL AND SHEAR DEFORMATIONS
Shearing Strain: Shearing strain is the angular deformation “γ”. Let V is the shearing load applied on a
member of length L which produces an shearing deformation of Δ. The strain and stress produced in the
member are:
γ = Δ/L - - - - - - (a)
τ = V/Av - - - - - - (b)
From the Hook’s Law:
τ α γ => τ = G γ - - - - - - (c)
Where G is the modulus of rigidity. Putting values of stress and strains from equations (a) and (b)
respectively in equation (c),
V/Av = G (Δ/L)
=> Δ = VL/AvG - - - - - - (d)
Example: Calculate shear deformation in a steel member (G = 1100 ksi) of L = 1 in. and Av = 4 in2 subjected
to a shearing load V = 100 kips.
Δ = VL/AvG = 100 x 1/(4 x 1100) = 0.0227 in.
13 STRAIN PROBLEMS Deformati
onmm Load, N
Strain Stress
0.0000 0.0
0 0
0.01 6310 0.0002 41.0
0.02 12600 0.0004 81.9
Problem 203: The following data was recorded during a tensile 0.03 18800 0.0006 122.1
test of a 14 mm diameter mild steel rod. The gauge length was 50 0.04 25100 0.0008
0.0010
163.1
203.3
0.05 31300
mm. 0.06 37900 0.0012 246.2
0.063 40100 0.0013 260.5
Plot stress-strain curve and determine, a) Proportional limit, b) 0.433 41600 0.0087 270.2
1.25 46200 0.0250 300.1
modulus of elasticity, c) yield point, d) ultimate strength and e) 2.5 52400 0.0500 340.4
4.5 58500
rupture strength 7.5 68000
0.0900 380.0
0.1500 441.7
12.5 69000
Solution: A = πd2/4 = π x 142/4 = 153.94 mm2. 15.5 67800
0.2500
0.3100
448.2
440.4
20 65000
0.4000 422.2
Stress is calculated as P/A and strains as Δ/L, From the plot: 61500 399.5

Proportional Stress = 246.2 MPa Stress-Strain Curve

500.0
Young’s Modulus, E = 246.2/0.0012 = 205,169 MPa = 205 GPa
400.0
Yield strength = 270.2 MPa
300.0
Ultimate Strength = 448.2 MPa. Take offset at 0.002 and
darw line parralto original one
200.0
Rupture Strength = 399.5 MPa
100.0
0.002
0.0
0.0000 0.1000 0.2000 0.3000 0.4000
14 STRAIN PROBLEMS
Problem 205: A uniform bar of length L, cross sectional area A, and unit mass  is suspended
vertically from one end. Show that its total elongation is 𝛿 = 𝜌𝑔𝐿2 /2𝐸, if the total mass of the bar
is M, show also that 𝛿 = 𝑀𝑔𝐿/2𝐴𝐸
Solution: Let consider a differential element of depth dy at a vertical distance y from bottom as
shown. The differential element is subjected at a load, Px equal to the weight of portion below
the differential element.
Strain diagrm
𝑃𝑥 = 𝑦𝐴𝜌𝑔 Here load increases toward fix support
and strain also varying
Deformation in the differential element is,
𝑃𝑥 𝑑𝑦 𝑦𝜌𝑔
𝑑𝛿 = = 𝑑𝑦
𝐴𝐸 𝐸
The total elongation is obtained by integrating the above equation from 0 to L:
𝐿 𝐿
𝑦𝜌𝑔 𝜌𝑔 𝑦 2 𝜌𝑔𝐿2
𝛿=න 𝑑𝑦 = =
0 𝐸 𝐸 2 0
2𝐸

Now 𝑀 = 𝜌𝐴𝐿 ⇒ 𝜌 = 𝑀/𝐴𝐿

𝑀 2
𝜌𝑔𝐿2 (𝐴𝐿)𝑔𝐿 𝑀𝑔𝐿
𝛿= = =
2𝐸 2𝐸 2𝐴𝐸
15 STRAIN PROBLEMS
Problem 207: A steel wire is 30 ft long hanging vertically support a load of 500 lb. Neglecting the weight of
wire, determine the require diameter if the stress is not to exceed 20 ksi and the total elongation is not to
exceed 0.2 in. Take E = 29x106 psi.
Solution: Length of wire, L = 30 ft = 360 in.
Supported Load, P = 500 lb
Limiting stress in wire, σ = 20 ksi
Limiting elongation, 𝛿 = 0.2 in.
Young’s Modulus, E = 29x106 psi
Diameter of wire, d = ?
Limiting stress in the wire:
𝜎 = P/A = P/(πd2/4) => 20,000 = 500/(πd2/4) => d = 0.1784 in.
Limiting elongation in wire:
𝛿 = PL/AE => 0.2 = 500 x 360/(πd2/4 x 29x106) => d = 0.1988 in.
To limit the stress and elongation simultaneously, maximum value should be used, d = 0.1988 in.
16 STRAIN PROBLEMS
Problem 211: A bronze bar is fastened between a steel bar and an aluminum bar as shown in figure. Axial
loads are applied at the position indicated. Find the largest value of P that will not exceed an overall
deformation of 3.0 mm or the following stresses. 140 MPa in steel, 120 MPa in bronze, and 80 MPa in
aluminum. Assume that the assembly is suitably braced to prevent bucking.
Solution:
σst ≤ 140 MPa,
σbr ≤ 120 MPa
σal ≤ 80 Mpa
𝛿 overall ≤ 3.0 mm
First calculate internal axial forces in each segment
using equilibrium conditions and free body diagrams.
Pst = +P (tension)
Pbr = P – 3P = -2P (compression)
Pal = +2P (tension)
17 STRAIN PROBLEMS
Problem 211 (Cont..):
Now limiting stress in each segments one by one and calculate limiting value of P.
For steel, σst = P/A => P = σst x A = 140 x 480 = 67200 N = 67.2 kN
For Bronze, σbr = 2P/A => P = (σbr x A)/2 = (120 x 650)/2 = 39000 N = 39.0 kN
For Aluminum, σal = 2P/A => P = (σal x A)/2 = (80 x 320)/2 = 12800 N = 12.8 kN
Now limiting overall deformation in the compound bar,
𝛿 overall = 𝛿 st + 𝛿 br + 𝛿 al = (PL/AE) st + (PL/AE) br +(PL/AE)al
𝛿 overall = {Px1000/(480x200,000)} + {-2P x 2000/(650x83,000)} + {2Px1500/(320x70,000)}
𝛿 overall = 70.2025 x 10-6 P => 3.0 = 70.2025 x 10-6 P => P = 42733 N = 42.733 kN
Selecting least value, P = 12.8 kN
18 STRAIN PROBLEMS
Problem 214: The rigid bars AB and CD shown in figure are
supported by pins at A and C and the two rods. Determine the
maximum force P that can be applied as shown if its vertical
movement is limited to 5 mm. Neglect the weight of all members.
Only consider the rigid body deformation.
Solution:
Limiting movement at P, ΔP = 5 mm
Applying equilibrium conditions,
ΣMC = 0 => Pst x 6 = P x 3 => Pst = P/2
ΣMA = 0 => Pst x 6 = Pal x 3 => Pal = 2Pst = 2 (P/2) = P
Elongation is steel and aluminum rods are:
Δst = (PL/AE)st = (P/2) 2000/(300 x 200,000) = 16.67 x10-6 P
Δal = (PL/AE)al = (P) 2000/(500 x 70,000) = 57.14 x10-6 P
19 STRAIN PROBLEMS
Problem 214 (Cont..):
From the deform shape,
ΔP = ΔD /2 = (2 Δal + Δst)/2 = Δal + Δst/2
Putting values of Δal and Δst

ΔP = 57.14 x10-6 P + (16.67 x10-6 P) / 2 = 65.37 x10-6 P

=> P = ΔP / 65.37 x10-6 = 5 / 65.37 x10-6 = 76487 N
P = 76.5 kN
20 POISSON’S RATIO, BIAXIAL DEFORMATION
Poisson’s Ratio: When stress is applied to a body in a certain direction X, strains are
produced in all three directions of X, Y and Z. If tensile stress is applied in X direction, there
will be elongation in the X direction accompanied by shortening in the transverse
This is a material elati cncstant.like Elatic modulus.Wood , steel etc has theeir own poisnt
directions (Y and Z ). ratio.Poison ratio in 99.99% cases is poitive number.
The negative ratio of the lateral strain (ЄL) to the axial strain (Єa) is a material elastic
constant known as Poisson’s Ratio denoted by “”.
If elongation is in x-direction then htrere will be shortening in perpendiculr direction
 = - ЄL / Єa and shorting is neg to make it just positive this - ve sign is introduced.
Its range is 0<Poison ratio<0.5 and usual range is 0.2-0.3

Biaxial State of Stress: Let at a certain point in a body σX and σY are the normal stresses in X
and Y directions respectively as shown in figure.
From Principle of Superposition
ЄX = ЄX1 + ЄX2 = σX /E + (- σY /E ) = (σX - σY)/E
ЄY = ЄY1 + ЄY2 = -σX /E + σY /E ) = (σY - σX)/E
21 POISSON’S RATIO, BIAXIAL AND TRIAXIAL DEFORMATION
Similarly for Triaxial state of stress, stains in terms of stresses are:
1
∈𝑋 = 𝜎𝑋 − 𝜗 𝜎𝑌 + 𝜎𝑍 −−−− −(𝑎)
𝐸
1
∈𝑌 = 𝜎𝑌 − 𝜗 𝜎𝑋 + 𝜎𝑍 −−−− −(𝑏)
𝐸
1
∈𝑍 = 𝜎𝑍 − 𝜗 𝜎𝑋 + 𝜎𝑌 −−−− −(𝑏)
𝐸
Inverting equations (a) and (b) to write stresses in terms of strains
𝐸
𝜎𝑋 = ∈𝑋 +𝜗 ∈𝑌 +∈𝑍 −−−− −(𝑑)
1 − 𝜗2
𝐸
𝜎𝑌 = ∈𝑌 +𝜗 ∈𝑋 +∈𝑍 −−−− −(𝑒)
1 − 𝜗2
𝐸
𝜎𝑍 = ∈𝑍 +𝜗 ∈𝑋 +∈𝑌 −−−− −(𝑒)
1 − 𝜗2
For shearing stresses,
𝐸
𝜏 = 𝐺𝛾 = 𝛾 −−−−− −(𝑓)
2 1+𝜗
22 PROBLEMS: POISSON’S RATIO
Problem 223: A rectangular steel block is 3 in. long in x-direction, 2 in. long in y-direction
and 4 in. long in z-direction. The block is subjected to a triaxial loading consisting of three
uniformly distributed forces as follows: 48 kips tension in the x-direction, 60 kips
compression in the y-direction and 54 kips tension in the z-direction. If 𝜗 = 0.3 and E =
29x106 psi, determine the single uniformly distributed load in the x-direction that would
produce the same deformation in the y-direction as the original loading.
Solution:
σX = +48/(2 x 4) = +6.0 ksi
σY = -60/(3 x 4) = -5.0 ksi
σZ = +54/(2 x 3) = +9.0 ksi
Strain in y-direction is,
1 1
∈𝑌 = 𝜎𝑌 − 𝜗 𝜎𝑋 + 𝜎𝑍 = −5.0 − 0.3 6.0 + 9.0 = −327.586𝑥10−6
𝐸 29𝑥103
23 PROBLEMS: POISSON’S RATIO
Problem 223 (Cont..):
Now strain in y-direction due to a load P in x-direction is given by:
ЄY = -  ЄX = -  (σX /E ) = -  (P /AE ) = - 0.3 {P/(2x4x29000)} = -1.293x10-6 P
Now equating strains in both cases,
-1.293x10-6 P = -327.586x10-6
=> P = +253.3 kips
24 STATICALLY INDETERMINATE MEMBERS
• A member is said to be statically indeterminate if it cannot be analyzed using available
static equilibrium equations, i.e. the number of unknowns are greater than the number of
available equilibrium equations.
• In case axially loaded members, the available equilibrium equation is one. Therefore if
the number of unknowns are greater than one, it is said to be statically indeterminate.
• Extra equations are required to analyze such members. These equations depend upon
the deformed shape and are called compatibility conditions.
• Compatibility conditions are the conditions of deformations/strains. Stress-strain relation
(Hook’s law) is used to convert deformations/strains to corresponding forces/stresses.
25 PROBLEM: INDETERMINATE MEMBERS
Problem 233: A steel bar 50 mm in diameter and 2 m long is surrounded by a shell of cast
iron 5 mm thick. Compute the load that will compress the combined bar a total of 0.8 mm in
the length of 2 m. For steel E = 200 GPa, and for cast iron, E = 100 GPa.
Solution:
Diameter of steel rod = 50 mm
Thickness of cast iron shell = 5 mm.
Length of combined bar = 2 m
deformation combined bar = 0.8 mm
Area of Steel Bars, Ast = πd2/4 = πx 502/4 = 1963 mm2
Area of Cast Iron Shell = Aci = π/4 (602-502) = 864 mm2
26 PROBLEM: INDETERMINATE MEMBERS
Problem 233 (Cont..):
Let Pst and Pci are the loads resisted by steel rod and cast iron shell respectively. The total
load resisted by the combined bar is:
P = Pst + Pci
Load required to produce 0.8 mm deformation in steel rod and cast iron shell are
calculated from:
𝛿 st = (PL/AE) st => Pst = (AE/L) 𝛿 = (1963 x 200,000/2000) x 0.8 = 157,040 N
𝛿 ci = (PL/AE) ci => Pci = (AE/L) 𝛿 = (864 x 100,000/2000) x 0.8 = 34,560 N
Now, P = Pst + Pci = 157,040 + 34,560 = 191,600 N = 191.600 kN
27 PROBLEM: INDETERMINATE MEMBERS
Problem 235: A timber column, 8 in. x 8 in. in cross section, is reinforced on each side by steel
plate 8 in wide and t in. thick. Determine the thickness t so that the column will support an axial
load of 300 kips without exceeding a maximum timber stress of 1200 psi or a maximum steel
stress of 20 ksi. The moduli of elasticity are 1.5 x 106 psi for timber and 29 x 106 psi for steel.
Solution:
Total Load supported by column, P = 300 kips
Area of timber column, Atimber = 8 x 8 = 64 in2
Area of steel plates = Ast = 4 x 8 x t = 32 t in2
Allowable stress in timber, σtimber = 1200 psi
Allowable stress in steel, σst = 20 ksi
Etimber = 1.5 x 106 psi
Est = 29 x 106 psi
28 PROBLEM: INDETERMINATE MEMBERS
Problem 235 (Cont..):
The equilibrium condition, P = Ptimber + Pst => Ptimber + Pst = 300 ------- (a)
And the compatibility condition, 𝛿 timber = 𝛿 st => (PL/AE)timber = (PL/AE) st
=> (σ /E)timber = (σ /E) st =>Ptimber/(1.5x106) = Pst/(29x106)
=> σst = 19.33 σtimber
Now limiting stress in timber to σtimber = 1.2 ksi, σst = 19.33 σtimber =19.33 x 1.2 = 23.2 ksi
And Limiting stress in steel to σst = 20 ksi, σtimber = σst / 19.33 = 20/19.333 = 1.034 ksi
In first case the stress in steel is exceeding its limiting value of 20 psi while in the 2nd case the
stress in timber is with in its limit of 1.2 ksi, therefore the 2nd case governs.
Therefore Ptimber = (σ x A)timber = 1.034 x 64 = 66.2 kips Compatibility condition are generated from
From equation (a), Pst = 300 – 66.2 = 233.8 kips
deform shapea and no specific rule
but decide seeing to efrom shape.

Pst = (σ x A)st = 20 x 32t = 640 t kips

=> t = Pst/640 = 233.8/640 = 0.365 in
29 PROBLEM: INDETERMINATE MEMBERS
Problem 236: A rigid block of mass M is supported by three symmetrically spaced rods as
shown in Fig. P-236. Each copper rod has an area of 900 mm2; E = 120 GPa; and the
allowable stress is 70 MPa. The steel rod has an area of 1200 mm2; E = 200 GPa; and the
allowable stress is 140 MPa. Determine the largest mass M which can be supported.
Solution:
The equilibrium condition is:
Pst + 2Pco = Mg ⇒ σA st + 2 σA co = Mg
1200σst + 2x900σco = 9.81M ⇒ M = 122.3σst + 183.5σco −− −(a)
The deformation id both steel and copper rod is same,
Therefore compatibility condition is:
σL σL σst 240 σco 160
δst = δco ⇒ = ⇒ = ⇒ σco = 0.90σst −− −(b)
E st E co 200 120
30 PROBLEM: INDETERMINATE MEMBERS
Problem 236 (Cont..):
Now using the allowable stress of 140 in steel, the stress in copper is: σco = 0.90σst = 0.9 140 =
126 MPa
which is greater than the allowable value of 70 MPa, therefore allowable stress in steel cannot be
used.
Using allowable stress in copper, the stress in steel is:
σ co 70
σst = 0.90 = 0.9 = 77.78 MPa

The stress in steel is less than the allowable stress, therefore it is OK.
Now using equation (a) to calculate the mass M of block.
M = 122.3σst + 183.5σco
M = 122.3 77.78 + 183.5 70
M = 22,357 kg
31 PROBLEM: INDETERMINATE MEMBERS
Problem 239: The rigid platform in Fig. P-239 has negligible mass and rests on two steel bars, each 250.00 mm
long. The center bar is aluminum and 249.90 mm long. Compute the stress in the aluminum bar after the center
load P = 400 kN has been applied. For each steel bar, the area is 1200 mm2 and E = 200 GPa. For the aluminum bar,
the area is 2400 mm2 and E = 70 GPa.
Solution:
Assuming that after application of load P,
the deflection in steel bar is greater than 0.1 mm.
Equilibrium condition is:
2Pst + Pal = 400,000 - - - - - - (a)
The compatibility condition is:
δst = δal + 0.1 => (PL/AE)st = (PL/AE)st + 0.1
=> (Pst x 250)/(1200 x 200,000) = (Pal x 249.9)/(2400 x 70,000) + 0.1
=> 1.0417 Pst – 1.4881 Pal = 100,000 - - - - - - (b)
Solving equations (a) and (b) simultaneously, Pst = 173.035 N, Pal = 53,928 N
Stress in aluminum bar is, σal = 53,928/2400 = 22.47 MPa
32 PROBLEM: INDETERMINATE MEMBERS
Problem 241: As shown in Fig. P-241, three steel wires, each 0.05 in2 in area, are used to lift a load
W = 1500 lb. Their unstressed lengths are 74.98 ft, 74.99 ft, and 75.00 ft. (a) What stress exists in
the longest wire? (b) Determine the stress in the shortest wire if W = 500 lb.
Solution:
L1 = 74.98 ft = 899.76 in.
L2 = 74.99 ft = 899.88 in.
L3 = 75.00 ft = 900 in.
(a) Assume that after application of load, the deformation is all three bars is
such that the final length of shortest bar is greater than or equal to 50 ft.
In such case, all three wires will contribute to support the weight W = 1500 lb
Therefore, the equilibrium condition is:
P1 + P2 + P3 = 1500 - - - - - - (a)
This equation contain three unkown so need of two compatibility condition
33 PROBLEM: INDETERMINATE MEMBERS
Problem 241 (Cont..):
The compatibility conditions are:
δ1 = δ2 + 0.12 => (PL/AE)1 = (PL/AE)2 + 0.12 => P1 – P2 = 0.12 (AE/L)
=> P1 - P2 = 0.12 (0.05 x 29 x 106/900) => P1 - P2 = 193.33 - - - - - (b)
and δ2 = δ3 + 0.12 => (PL/AE)2 = (PL/AE)3 + 0.12 => P2 – P3 = 0.12 (AE/L)
=> P2 – P3 = 0.12 (0.05 x 29 x 106/900) => P2 – P3 = 193.33 - - - - - (c)
Solving equations (a), (b) and (c) simultaneously: (If i got zero or negative it means
that threr is problem in assumption)
P1 = 693.33 lb, P2 = 500.00 lb, P3 = 306.67 lb
Stress in longest bar 3 is: σ3 = 306.67/0.05 = 6133 psi
(b) Let for weight W = 500 lb, the deformation is shortest bar is greater
than 0.12 in and less than 0.24 in. In such case, load in bar 3 will be zero.
Therefore, the equilibrium condition is: P1 + P2 = 500 - - - - - (d)
34 PROBLEM: INDETERMINATE MEMBERS
Problem 241 (Cont..):
The compatibility condition is:
δ1 = δ2 + 0.12 => (PL/AE)1 = (PL/AE)2 + 0.12 => P1 – P2 = 0.12 (AE/L)
=> P1 - P2 = 0.12 (0.05 x 29 x 106/900) => P1 - P2 = 193.33 - - - - - (e)
and Solving equations (d) and (e) simultaneously:
P1 = 346.67 lb, P2 = 153.33 lb
Stress in shortest bar 1 is: σ1 = 346.67/0.05 = 6933 psi
35 PROBLEM: INDETERMINATE MEMBERS
Problem 247: The composite bar in Fig. P-247 is stress-free before the axial loads P1 and P2 are
applied. Assuming that the walls are rigid, calculate the stress in each material if P1 = 150 kN
and P2 = 90 kN.
Solution:
Let R1 and R2 are the support reactions.
The equilibrium condition is:
R1 + R2 = 150 + 90 = 240 - - - - - - (a)
The compatibility condition is that the total
elongation in the bar zero, i.e. sum of the change in length in all three segments is zero.
δal + δst + δbr = 0 => (PL/AE)al + (PL/AE)st + (PL/AE)br = 0 - - - - - (b)
From the free body diagram,
Pal = R1, Pst = R1 + 150, Pbr = R1 + 150 + 90 = R1 + 240
36 PROBLEM: INDETERMINATE MEMBERS
Problem 247 (Cont..):
Putting values in equation (a)
(R1x500/(900x70,000) +(R1+150) x250/(2000x200,000) + (R1+240) x 350/(1200x83,000) = 0
=> 12.076 R1 + 937.13 = 0 => R1 = -77.6 kN
Now force and stress in each segment is:
Pal = -77.6 kN,
Pst = -77.6 + 150 = 72.4 kN
Pbr = -77.6 + 240 = 162.4 kN
σal = -77,600/900 = -86.2 MPa
σst = 72,400/2000 = 36.2 MPa
σal = 162,400/1200 = 135.33 MPa
37 PROBLEM: INDETERMINATE MEMBERS
Problem 252: The light rigid bar ABCD shown in Figure is
pinned at B and connected to two vertical rods. Assuming
that the bar was initially horizontal and the rods stress-free,
determine the stress in each rod after the load P = 20 kips is
applied.
Solution:
Let Pst and Pal are the forces in steel and aluminum rods.
Apply equilibrium condition to the free-body diagram:
ΣMB = 0 => 20 x 4 –Pst x 2 – Pal x 4 = 0
=> Pst +2 Pal = 40 - - - - - (a)
From the deform shape, considering similar triangles:
δal /4 = δst /2 => δal = 2δst
38 PROBLEM: INDETERMINATE MEMBERS
Problem 252 (Cont..):
=> (PL/AE)al = 2(PL/AE)st
=> (Pal x 48) /(0.75x10x106) = 2 (Pstx36)/(0.5 x 29x106)
=> Pal = 0.7758 Pst- - - - (b)
Putting value of Pal from (b) in to equation (a):
Pst +2 (0.7758 x Pst) = 40 => Pst = 15.68 kips
Putting value of Pst in equation (b)
Pal = 0.7758 (15.68) => Pal = 12.16 kips
The corresponding stresses are:
σst = 15.68 / 0.5 = 31.36 ksi
σal = 12.16 / 0.75 = 16.21 ksi
39 THERMAL STRESS
The stress induced in a body due to change in its temperature is known as thermal stress.
When a body is subjected to a change in temperature, it will expand or contract. If a
member of length “L” is subjected to change in temperature ΔT , then the change in length
of the member is:
δT = α L ΔT - - - - - (a)
Where α is the coefficient of linear expansion. The unit of α is oF-1 or oC-1.
If a member is free to expand or contract, no thermal stress will be induced in the member.
However, if the member is restrained to expand or contract, thermal stress will be induced
in the member. Expasnion joint are provided in builing or rails to relaes thermal stresses.Expasion joint is
opening providede.

Consider a fix-ended member of length “L” is subjected to a change in temperature ΔT. It is

desired to calculate thermal stress in the member.
40 THERMAL STRESS
The expression can be developed by first allowing the member to expand and then a force
P is applied such that the deformation caused by the force P is equal to the thermal
expansion.
δT = α L ΔT - - - - - (a)
δP = -PL/AE - - - - - (b)
Here δP = δT
=> -PL/AE = α L ΔT
Poistive delta T means increase
=> σT = -α E ΔT in temprature.Which causes comp
forces so -ve is to show comp forces
This is expression for thermal stress in a member which depends upon the coefficient of
linear expansion, modulus of elasticity and change in temperature.
Increase in temperature cause compressive stress while decrease in temperature results in
tensile stress.
41 PROBLEMS: THERMAL STRESS
Problem 261: A steel rod with a cross-sectional area of 0.25 in2 is stretched between two
fixed points. The tensile load at 70°F is 1200 lb. What will be the stress at 0°F? At what
temperature will the stress be zero? Assume α = 6.5 × 10−6 in/(in·°F) and E = 29 × 106 psi.
Solution:
Cross-sectional area of rod, A = 0.25 in2.
Tensile Load at 70°F, P = 1200 lb
Tensile stress at 70°F, σP = P/A = 1200/0.25 = 4,800 psi
Thermal stress induced in rod due to temperature change from 70°F to 0°F is,
ΔT = 0-70 = -70 °F
σT = -α E ΔT = -(6.5 × 10−6) (29 × 106) (-70) = 13,195 psi (tension)
Net stress in the rod at 0°F is: σ = σP + σT = 4,800 + 13,195 = 17,995 psi
42 PROBLEMS: THERMAL STRESS
Problem 261:
The temperature at which the stress in the rod will be zero can be calculated from equation:
σ = σP + σT = 0 => 4,800 + (-α E ΔT) = 0
=> ΔT = 4,800/(6.5 × 10−6 x 29 × 106) = 25.46 °F
=> (T – 70 °F) = 25.46 °F => T = 25.46 + 70 = 95.46 °F
43 PROBLEMS: THERMAL STRESS
Problem 266: Calculate the increase in stress for each segment of the compound bar shown in
figure if the temperature increases by 100°F. Assume that the supports are unyielding and that
the bar is suitably braced against buckling.
Solution:
For the compound bar, the equilibrium condition is:
R1 – R2 = 0 => R1 = R2.
From the free-body diagram, forces in each segment are:
Pal = -R1 and Pst = -R2 = -R1
As the net total deformation in the bar is zero,
therefore the compatibility condition is:
δal + δst = 0 => (PL/AE + α L ΔT )al + (PL/AE + α L ΔT )st = 0
=> (-R1x10)/(2x10x106) +(12.8x10-6x10x100) + (-R1x15)/(1.5x29x106) +(6.5x10-6x15x100) = 0
=> 0.845 R1 = 22550 => R1 = 26,686 lb.
44 PROBLEMS: THERMAL STRESS
Problem 266 (Cont..):
Now forces in each segment are:
Pal = Pst = -R1 = -26,686 lb
The corresponding stresses are:
σal = P/A = -26,686 /2.0 = -13,343 psi
σst = P/A = -26,686 /1.5 = -17,790 psi
45 PROBLEMS: THERMAL STRESS
Problem 268: The rigid bar ABC in figure is pinned at B and attached to the two vertical
rods. Initially, the bar is horizontal and the vertical rods are stress-free. Determine the stress
in the aluminum rod if the temperature of the steel rod is decreased by 40°C. Neglect the
weight of bar ABC.
Solution:
Apply equilibrium condition to the rigid bar ABC:
ΣMB = 0 => Pal x 1.2 = Pst x 0.6 => Pst = 2 Pal.- - - - (a)
From the deform shape of the system, the compatibility condition is:
δal /0.6 = δst /1.2 => δal = 2 δst.- - - - - (b)
(PL/AE + α L ΔT )al = 2 (PL/AE + α L ΔT )st.
[(Palx1200)/(1200x70,000) + 0] = 2 [(Pstx900)/(300x200,000) +(11.7x10-6x900x(-40))]
Pal/70,000= 3 Pst/100,000+ (-0.8424) => Pal = 2.1 Pst - 58,968 - - - - -(c)
46 PROBLEMS: THERMAL STRESS
Problem 268 (Cont..):
The Putting value of Pas from equation (a) in (c):
Pal = 2.1 (2Pal) - 58,968 => Pal = 18,428N
Putting value of Pal in equation (a):
Pst = 2 (18,428) = 36,856 N
The stress in aluminum and steel bars are:
σal = P/A = 18,428 / 1200 = 15.36 MPa.
σst = P/A = 36,856/300 = 122.85 MPa.
47 PROBLEMS: THERMAL STRESS
Problem 269: As shown in figure, there is a gap between the aluminum bar and the rigid slab
that is supported by two copper bars. At 10°C, Δ = 0.18 mm. Neglecting the mass of the slab,
calculate the stress in each rod when the temperature in the assembly is increased to 95°C. For
each copper bar, A = 500 mm2, E = 120 GPa, and α = 16.8 µm/(m·°C). For the aluminum bar, A =
400 mm2, E = 70 GPa, and α = 23.1 µm/(m·°C).
Solution:
When temperature of the assembly is increased, initially the assembly will move freely until the
aluminum bar touches the rigid block. Let T1 is the temperature at which the aluminum bar
touches the block.
δal = δco + 0.18 => (α L ΔT)al = (α L ΔT)co + 0.18
23.1x10-6 x 750 x (T1-10) = 16.8x10-6 x 750 x (T1-10) + 0.18
4.725 x 10-3T1 = 0.04725 + 0.18 => T1 = 48.1 °C
So at temperature of 48.1 °C, the assembly will be stress free.
If the temperature is increased beyond 48.1 °C to 95 °C, stress will be produced.

Beyond 48.1 C srress will produce becayuse aluminum will try to expnad fastly.But slab is attacd with copper
so cannot mve freely so stresses will produce.
48 PROBLEMS: THERMAL STRESS
Problem 269 (Cont..):
Now beyond 48.1 °C to 95 °C (ΔT = 46.9 °C), stresses will be produced, the compatibility condition is:
δal = δco => (PL/AE + α L ΔT )al = (PL/AE + α L ΔT )co.
=>(Palx750)/(400x70,000)+23.1x10-6x750x46.9=(Pcox750)/(500x120,000)+16.8x10-6x750x 46.9
=> 26.786 x 10-6 Pal – 12.5 x 10-6 Pco = -0.2216
=> Pco = 2.1429 Pal + 17,728 - - - - - (a)
The equilibrium condition is:
2Pco + Pal = 0 => Pal = -2 Pco - - - - - -(b)
(a) => Pco = 2.1429 (-2 Pco) + 17,728 => Pco = 3,353 N
(b) => Pal = -2 (3,353) = -6,706 N
Stresses are: σco = P/A = 3,353/500 = 6.71 MPa. (Tension)
σal = P/A = -6,706/400 = -16.76 MPa. (Compression)