Chapter 2

Simple Strain

2.1 Introduction

In the preceding chapter, the main focus was the strength of the material, more specifically, the
relations between applied loads, area, and stress. Another important consideration in the field of
strength of materials is the change in length of the body that accompany a loading. The same
assumptions are also considered in the analysis of body’s deformation as mentioned in Chapter 1, i. e.,
load must be axial, material is homogeneous, and cross sectional area is uniform or constant.

2-2 Axial Deformation; Stress-Strain Diagram

The strength of material is an important mechanical property that must be considered in the design of
machine parts or structure. However, another criterion that needs consideration in the design is the
stiffness of the body. It is of equal relevance as hardness, ductility, and toughness but these properties
require laboratory tests. The complete description of tests for each property is not to be discussed but
merely one test which is the tension test of the steel. The results will be considered since it helps to
develop basic ideas for the derivation of equations used to measure the strain developed in the
specimen.

In the tension test for steel, the specimen is gripped between the jaws of machine and then
subjected to an axial load. The machine will tend to elongate the steel slowly until the specimen
ruptures. During the test, two specific data are recorded, readings of the magnitude of the applied load
and the elongation of the gage length. These data are carefully read and converted to stress and strain
diagram as shown in Figure 2-1a.

Figure 2-1a Stress-strain diagram

 As seen in Figure 2-1a, the stress-strain diagram is a straight line from the point of origin O to a
point called proportional limit. The straight line denotes that stress is proportional to strain as
postulated by Robert Hooke. The ordinate represents the stress while the abscissa denotes the strain
incurred by the specimen. This plot is a manifestation of Hooke’s law where stress is proportional to
strain. In symbol,

where E is a property of material called the modulus of elasticity. For steel, its value is 29 x 10 6 psi or
200 GPa. It is noted that Hooke’s law is not applicable to the entire diagram but valid only until the
proportional limit. Beyond this point, stress is no longer proportional to strain.

From the stress-strain diagram, the following important points are noticeable: ( 1 ) The elastic
limit, the stress beyond which the material is no longer elastic as the name implies. A material is elastic if
after being loaded, the material returns to its usual shape when the applied load is removed. The
permanent deformation sustained after the removal of the load is called permanent set. ( 2 ) Yield point
is the point where the diagram becomes almost horizontal. Beyond the yield point there is the yielding
of the material without any corresponding increase in load. ( 3 ) Ultimate strength, also called ultimate
stress is the highest ordinate in the stress-strain diagram. And ( 4 ) Rupture strength or rupture stress is
the stress at which failure occurs. The term is sometimes referred to as the breaking strength. For
structural steel, the value of rupture stress is somewhat lower than the ultimate stress because of the
error caused by a phenomenon called necking.

In general, strain is a geometric quantity that measures the deformation of the body. Stress and
strain are two basic concepts of mechanics of materials. The relationship between the stress and strain
defines the mechanical properties of materials which is very relevant in the actual design. As shown in
Figure 2-1b, the body is elongated by deformation δ ( lower case Greek letter delta ) due to applied
force P. Strain is the ratio of the deformation of the body with respect to its original length.

Figure 2-1b

Simple strain is denoted by Greek letter epsilon, ε and

expressed as:

where : ε = strain ( mm/ mm, in/ in )

δ = deformation ( mm or in )
L = original length ( mm or in )
Another variation of Hooke’s law is obtained by substituting the value of stress, P/ A and
replacing the strain by δ/ L. Since stress is proportional to strain, i. e., σ α ε and σ = E ε therefore P/ A =
E ( δ/ L ). The formula to calculate the deformation is now:
where : δ = deformation ( mm, in )
P = applied load( N or lb )
L = length ( mm or in )
A = cross-sectional area (mm2 or in2 )
E = modulus of elasticity ( MPa, psi )

Illustrative Problems:

1. The rigid beam is supported by a hinge at A and two wires attached to points D and E. The load P
causes the beam to be displaced 15 mm downward at C. Determine the strain developed in wires CE
and BD.

Solution: Consider the deformation diagram;

3m 4m
δBD δCE = 15 mm

By ratio and proportion, calculate the value of δBD;

δBD = 3 m ( 15 mm ) / 7 m
δBD = 6.43 mm
therefore;
εBD = 6.43 mm/ 4000 mm εCE = 15 mm/ 4000 mm
εBD = 0.00161 mm/ mm Ans εCE = 0.00375 mm/ mm Ans

2. The aluminum bar shown is loaded as shown in the figure. The modulus of elasticity is 10 x 10 6 psi
and the cross-sectional area is 0.60 in 2. Assuming that the bar is properly braced to prevent any lateral
buckling, determine the total deformation of the bar.

2.5 ft 5 ft 3.5 ft
Solution: Pass a section through each segment and consider to the left of the cutting section;

Applying the equations of equilibrium;

Σ FH = 0 Σ FH = 0
P1 – 6000 lb = 0 –P 2 – 6000 lb + 7000 lb = 0
P1 = 6000 lb ( + ) P2 = 1000 lb (– )
Σ FH = 0
P3 – 6000 lb + 7000 lb – 5000 lb = 0
P3 = 4000 lb ( + )
Total deformation = Σ PL/ AE
δtotal = 6000 lb ( 2.5’ x 12 )/ ( 0.60 in 2 x 10 x 106 psi ) + 4000 lb ( 3.5’ x 12 )/ ( 0.60 in 2 x 10 x 106 psi ) – 1000
lb ( 5’ x 12 )/ ( 0.60 in2 x 10 x 106 psi )
δtotal = 0.048 in Ans

3. The rigid bar is supported by a smooth pin at B and steel rod AC. The diameter of the steel rod is 0.60
in. with E = 29 x 106 psi. Calculate the change in length of steel rod AC.

Solution: Consider the FBD at joint C;

By equations of equilibrium;
Σ FV = 0
PAC( Sin 400 ) – 2000 lb = 0
PAC = 3111.45 lb ( tension )
Calculate the length of rod AC;
Cos 400 = ( 8’ x 12” ) / LAC
LAC = 125.32 in
therefore:
δAC = 3111.45 lb ( 125.32 in ) / ( π x 0.60 2/ 4 ) (29 x 106 psi )
δAC = 0.04755 in Ans

Exercises:
1. A steel bar 8 m long, hanging vertically supports a tensile load of 2500 N. Neglecting the weight of
the bar, calculate the required diameter so that the axial stress will not exceed 130 MPa and the total
elongation is not to exceed 6 mm. Use E = 200 GPa. Answer: D = 4.95 mm

2. A bar having a length of 5 in. and cross-sectional area of 0.8 in 2 is subjected to an axial force of 8000
lb. If the bar stretches 0.003 in., determine the modulus of elasticity of the material. Assume that the
material behave elastically. Answer: E = 16.67 x 103 ksi

3. A force P is applied to the rigid lever arm ABC as shown. The arm rotates counterclockwise about pin
A through an angle of 0.060. Calculate the normal strain developed in wire BD.
Answer: εBD = 0.0014 mm/ mm

4. The control linkage of an airplane consists of a rigid member DBC and a flexible cable AB. A force P is
applied to the end D of the member and causes it to rotate by θ = 0.35 0. The cable is assumed to be
originally unstretched. Determine the strain developed in the cable.
Answer: εAB = 0.00282 mm/ mm
 450 mm

5. The rigid beam rests on two short posts AC and BD as shown. Post AC is made from steel with E =
200 GPa and BD from aluminum with E = 70 GPa. The diameters of posts AC and BD are 25 mm and 40
mm, respectively. Determine the vertical displacement of point F on AB.
100 kN

6. The timber beam has a cross-sectional area of 2000 mm 2 and its modulus of elasticity is 12 GPa. Axial
loads are applied at points B, C, and D as shown. Calculate the total change in length of the beam.
Answer: δtotal = 2.33 mm

1.60 m 3.20 m 3.20 m

7. The stress-strain diagram for polyester resin is shown in the figure. Both strut AB and column CD,
made from this material, support the rigid beam AC. Calculate the angle of tilt of the beam when it is
subjected to load P = 60 kN. The diameter of the strut is 50 mm and the diameter of the column is 80
mm. Answer: θ = 0.3270
8. The compound bar carries the loads of P and 2P as shown. Determine the maximum value of P that
will satisfy the following requirements: allowable normal stress for steel is 35 ksi, allowable normal
stress for aluminum is 18 ksi, and allowance for total deformation is 0.15 inch. Answer: P = 3.60 kips

9. The rigid bar AB is supported by two rods made of the same material. The bar is initially horizontal
before the load P is applied. Neglect the weight of the bar. Calculate the distance “x” so that the bar
remains horizontal. Answer: x = 3 ft

12 ft

10. The rigid bars shown are separated by a roller and hinged at A and D. Bar AC is supported by a steel
rod at B. Calculate the vertical movement of the roller at C.
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2.3 Poisson’s Ratio: Biaxial and Triaxial Deformations

When a bar is elongated by an axial tensile force, there is a contraction in the transverse side as shown
in Figure 2-3a. In 1811, Simeon-Denis Poisson, a French mathematician and engineer, showed that the
ratio of the transverse strain to the longitudinal strain is constant for stresses within the proportional
limit. Accordingly, the ratio is named after him and is denoted by lowercase Greek letter nu:

where : ν = Poisson’s ratio

εlat = lateral strain( mm/ mm, in/ in )
εlong = longitudinal strain (mm/ mm, in/ in)

Figure 2-3a
The negative sign indicates that a positive strain along longitudinal direction causes a negative
strain along the transverse direction. Poisson’s ratio is a dimensionless value that ranges from 0.25 to
0.33 for metals. Poisson’s ratio permits to extend Hooke’s law of uniaxial stress to biaxial stress. If the
body is subjected to loadings along X and Y directions simultaneously, the strain in the X direction due to
tensile stress σx is σx/ E. On the other hand the tensile stress σy will produce transverse contraction in
the X direction of the amount ν σy/ E, so the resultant strain in the X direction will be

and similarly the strain in the Y direction is

When stresses act on a material element in triaxial loading as shown in Figure 2-3b, a further
extension results to the following equations for strains caused by simultaneous action of triaxial
stresses.
where : ν = Poisson’s ratio
εX = strain along X-direction( mm/ mm, in/ in )
εY = strain along Y-direction( mm/ mm, in/ in )
εZ = strain along Z-direction( mm/ mm, in/ in )
σX, σY, σZ = stresses in the X, Y, and Z directions, respectively ( MPa, psi )
E = modulus of elasticity ( GPa, psi )

Figure 2-3b
Another important relation between the constants E, G, and ν for any given material is
expressed as

where : ν = Poisson’s ratio

G = modulus of rigidity ( GPa, psi )
E = modulus of elasticity ( GPa, psi )
which is useful for computation of ν when the values of G and E are given. The value of Poisson’s ratio is
0.25 to 0.30 for steel, 0.20 for concrete, and 0.33 for other types of metal.

Illustrative Problems:

1. A welded steel cylindrical drum made of a 10-mm plate has an internal diameter of 1.20 m. Assume
that Poisson's ratio is 0.30 and E = 200 GPa. Calculate the change in diameter caused by an internal
pressure of 1.75 MPa.
Solution: From the thin-walled pressure vessels equations;
For the longitudinal stress ( stress along Y-direction );
σY = ρD/ 4t
σY = 1.75 N/ mm2 ( 1200 mm ) / 4 ( 10 mm )
σY = 52.5 MPa
For the tangential stress ( stress along X-direction );
σX = ρD/ 2t
σX = 1.75 N/ mm2 ( 1200 mm ) / 2 ( 10 mm )
σX = 105 MPa
therefore:
εX = ( 1/ E ) (σX – ν σY )
εX = ( 1/ 200 x 103 N/ mm2 ) ( 105 MPa – 0.30 x 52.5 MPa )
εX = 4.4625 x 10-3
but εX = ΔD/ D
4.4625 x 10-3 = ΔD/ 1200 mm
ΔD = 0.5355 mm Ans

2. The steel bar shown is subjected to loads P = 90 kN and assumed to behave elastically. The length of
the bar is 1.5 m with ν = 0.30 and E = 200 GPa. Calculate the change in its dimension along X and Y axes.

Solution: Calculate the strain in the Z-direction;

εZ = δZ / LZ
but δZ = PLZ / AE
δZ = 90 x 103 N ( 1500 mm )/ [ ( 50 ) ( 100 ) ] (200 x 10 3 N/ mm2 )
δZ = 0.135 mm
εZ = 0.135 mm/ 1500 mm
εZ = 9 x 10-5 mm/ mm
Solving for strains in the X and Y directions;
εX = εY = εZ ( ν )
εX = εY = 9 x 10-5 mm/ mm ( 0.30 )
εX = εY = 2.7 x 10-5 mm/ mm
thus the changes in the dimensions are:
δX = εXLX δY = εYLY
-5
δX = 2.7 x 10 mm/ mm ( 100 mm ) δY = 2.7 x 10-5 mm/ mm ( 50 mm )
δX = 0.0027 mm Ans δY = 0.00135 mm Ans

Exercises:
1. A 60-mm diameter steel tube with a wall thickness of 3 mm just fits in a rigid hole. Determine the
tangential stress developed if an axial compressive load of 12 kN is applied. Use ν = 0.30 and E = 200
GPa. Answer: σt = 6.37 MPa

2. A 200-mm long bronze tube closed at both ends fits without clearance in a 70-mm hole in a rigid
block. It has a diameter of 70 mm and a wall thickness of 5 mm. The tube then sustained an internal
pressure of 4.5 MPa. Use ν = 0.33 and E = 83 GPa. Compute the tangential stress in the tube.
Answer: σt = 5.20 MPa

3. The aluminum block has the rectangular cross-section and is subjected to an axial compressive force
of 8 kips. The 1.5-in side changed its length to 1.50014 inches. Use E = 10 x 10 3 ksi. Calculate the new
length of the 2-inch side. Answer: Lnew = 2.00019 in

4. A solid cylinder of diameter D is subjected to an axial load P. Show that its change in diameter is δ D =
4Pν / πDE.

5. A rectangular aluminum block is 90 mm long in the X-direction, 70 mm wide in the Y-direction, and 40
mm thick in the Z-direction. The block is subjected to axial loads of 220 kN tensile force in the X
direction, 150 kN compressive force in the Y direction, and 200 kN tensile force in the Z-direction.
Assume that ν = 0.33 and E = 70 GPa. Determine the single load in the X-direction that would produce
the same Z deformation as the original loading.
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2-4 Statically Indeterminate Members

There are problem situations for axially loaded members in which the equations of static equilibrium are
not enough to compute the reaction elements. This condition exists when the number of reactions
exceeded the number of independent equations of equilibrium. Such types of problem are called
statically indeterminate and need additional relations which are dependent upon the elastic
deformations of the members. An example of statically indeterminate member is the propped beam
shown in Figure 2-4. There are four reaction elements that will be computed against three equations of
static equilibrium, thus called statically indeterminate beam.

Figure 2-4
The support reactions for statically indeterminate members are first computed using the
equations of equilibrium, compatibility, and force-displacement requirement for the member. Express
the compatibility conditions in terms of the deformations caused by the applied forces. Lastly, use the
load-displacement relation to establish relationship between unknown displacements and reactions.

Illustrative Problems:

1. The rigid bar shown is supported by a smooth pin at A and two rods, one made of steel ( A = 500 mm 2
and E = 200 GPa ) and the other of bronze ( A = 250 mm 2 and E = 83 GPa ). Determine the normal stress
developed in each bar caused by the 50 kN load.

Solution: Consider the FBD of the bar;

By equation of equilibrium;
Σ MA = 0
– PSt ( 0.60 m ) – PBr ( 1.60 m ) + 50 kN ( 2.40 m )= 0 Equation ( 1 )
Consider now the deformation diagram;

By ratio and proportion;

1.60 δSt = 0.60 δBr
1.60 [ PSt ( 1 m )/ 500 mm2 ( 200 GPa ) ] = 0.60 [ PBr ( 2 m )/ 250 mm2 ( 83 GPa ) ]
PSt = 3.614458PBr Equation ( 2 )
Solve equations ( 1 ) and ( 2 ) simultaneously;
PBr = 31.84 kN
PSt = 115.09 kN
Calculate the stress;
σBr = 31.84 x 103 N / 250 mm2 σSt = 115.09 x 103 N / 500 mm2
σBr = 127.36 MPa Ans σSt = 230.18 MPa Ans

2. Two steel bars support the rigid platform shown in the figure. Each steel bar has a length of 250 mm,
an area of 1100 mm2, and E = 200 GPa. The center bar is made of aluminum with an area of 2200 mm 2,
E = 70 GPa and 249.90-mm long. Neglect the mass of the rigid platform. Compute the stress developed
in the steel and aluminum bars caused by an applied load P = 400 kN.

Solution: Consider the FBD of the rigid platform;

By equation of equilibrium;
Σ FV = 0
2PSt + PAl – 400 kN = 0 Equation ( 1 )
Consider the deformation diagram;

δSt = 0.10 + δAl

PL/ AE)St = 0.10 + PL/ AE)Al
PSt ( 250 mm )/ 1100 mm2 ( 200 x 103 N/ mm2 ) = 0.10 + PAl ( 249.90 mm )/ 2200 mm2 ( 70 x 103 N/ mm2 )
1.136 x 10-6PSt = 0.10 + 1.623 x 10-6PAl Equation ( 2 )
Solve equations ( 1 ) and ( 2 ) simultaneously;
PAl = 58.09 kN
PSt = 170.95 kN
Calculate the stress;
σAl = 58091.29 N / 2200 mm2 σSt = 170954.36 N / 1100 mm2
σAl = 26.41 MPa Ans σSt = 155.41 MPa Ans

Exercises:
1. A steel bar is 50 mm in diameter and 2000 mm long. It is surrounded by a shell of cast iron 5 mm
thick. Calculate the load P that will contract the combined bar a total of 0.90 mm. For modulus of
elasticity use Est = 200 GPa and Ecast = 100 GPa. Answer: P = 215.60 kN

2. A reinforced concrete column 300 mm in diameter supports an axial compressive force of 500 kN.
Determine the required area of reinforcing steel that will satisfy the following conditions: allowable
stress for steel is 120 MPa whereas for concrete, the stress is limited to 6 MPa. Use Est = 200 GPa
and Econ = 14 GPa.
3. A homogeneous bar with a cross-sectional area of 400 mm 2 is attached to fixed supports as shown in
the figure. It is subjected to lateral forces P 1 = 20 kN and P2 = 50 kN. Determine the normal stress
developed in segments AB and BC. Answer: σAB = 80.55 MPa

4. The two vertical rods attached to the rigid bar are identical except for the length. The bar is pinned at
O and initially horizontal when the 6000-lb was applied. Determine the normal force exerted by the two
rods. The weight of the bar is negligible. Answer: PA = 4090.91 lb

6000 lb
5. The composite bar, composed of aluminum and steel bars, is firmly attached to rigid supports. An
axial load P = 60 kips is applied as shown. Calculate the stress in the steel and aluminum bar.

6. The rigid beam is supported by a smooth pin at O and attached to two vertical rods. Assume that the
bar has negligible weight and the rods were initially stress-free. Calculate the force P that can be applied
such that the allowable stresses for steel rod and bronze rod are 140 MPa and 75 MPa, respectively.
Answer: P = 115 kN
7. The rigid slab which weighs 600 kN is supported by three rods of the same cross-section and material.
The lower ends of the rods prior to slab attachment were of the same level. Calculate the tensile force
in each rod. Answer: PA = 239 kN

8. The homogeneous rod of uniform cross-section is attached to rigid supports. The rod is then
subjected to axial load P as indicated in the figure. Prove that the reactions are given by R 1 = Pb/ L and
R2 = Pa/ L.

9. Three steel rods having an area of 300 mm 2, jointly support the 8-kN load. Assume that there was no
slack or stress in the rods before the load was applied. Determine the axial force exerted by each rod.
Use E = 200 GPa.

8 kN
8 kN
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2.5 Thermal Stresses

When the deformation due to change in temperature is allowed to occur in the body, no stress will be
induced in the structure. However, there are some instances where change in dimension due to
temperature is not permitted. In that case an internal stress is created. The internal stress induced in
the body is called thermal stress.

An increase in temperature will cause the body to expand whereas a temperature decrease
produces contraction. This deformation is isotropic or the same in all direction and proportional to the
change in temperature. When the temperature change is uniform throughout the body so is the
thermal strain. Consequently, the change in length of the member is given by

where : δT = deformation due to change in temperature ( mm, in )

α = coefficient of thermal expansion ( m/ m0C, in/ in0F )
L = length of the body ( mm, in )
ΔT = change in temperature ( 0C, 0F )

The forces resulted from change in temperature cannot be computed by equilibrium analysis
alone since the system is statically indeterminate. The same principle will be used as in statically
indeterminate members using the equations of equilibrium, compatibility, and force-displacement
requirement for the member. The only difference is the inclusion of thermal expansion in the analysis of
deformation.

Illustrative Problems:

1. The horizontal steel rod with a cross-sectional area of 1200 mm 2 is stretched between two rigid walls.
If the rod is stress free at 100C , calculate the stress when the temperature has dropped to -20 0C. Use α =
11.7 x 10-6/ 0C and E = 200 GPa.

Solution: Assume that the rod is detach from the wall so that the contraction due to decrease in
temperature occurs freely. To reconnect the rod to the wall, it is stretched to its original length by
subjecting to axial load P.
By compatibility of deformations, the resulting elongation δP must be equated to contraction δT;

δT = δ P
ΔT = 100C – (-200C ) = 300C
αL ΔT = σL / E
σ = α ( ΔT ) ( E )
σ =11.7 x 10-6/ 0C ( 300C ) ( 200 x 103 MPa )
σ = 70.2 MPa Ans

2. Three symmetrically arranged rods support a rigid block weighing 12 kips as shown. The lower ends
of the rods were at the same level prior to block attachment. Determine the axial stress in each rod after
the block is attached and when the temperature is increased to 100 0F. Use the tabulated data:

Solution: Consider the FBD of the block;

Σ FV = 0
2PSt + PBr – 12 kips = 0
2PSt + PBr = 12 kips Equation ( 1 )
Consider the deformation diagram;

From the diagram:

δT)St + δP)St = δT)Br + δP)Br
α L ΔT )St + PL/ AE)St = α L ΔT )St + PL/ AE)St
6.5 x 10-6/ ( in/in0F ) ( 2’ x 12” ) ( 1000F ) + PSt ( 2’ x 12” ) / 0.75 in2 ( 29 x 106 lb/ in2 ) = 10 x 10-6/(in/in 0F ) (
3’ x 12” ) ( 1000F ) + PBr ( 3 x 12” ) / 1.5 in2 ( 12 x 106 lb/ in2 ) Equation ( 2 )
Solve equations ( 1 ) and ( 2 ) simultaneously;
PSt = 8700 lb
PBr = - 5400 lb
Calculate the stress;
σSt = 8700 lb / 0.75 in2
σSt = 11600 psi Ans
σSt = 5400 lb / 1.5 in2
σSt = 3600 psi Ans

Exercises:
1. A steel rod with a cross-sectional area of 140 mm 2 is placed between two fixed supports. The
0 0 3
tensile force at 20 C is 4000 N. What will be the stress at – 15 C? Use E = 200 x 10 MPa and α = 11.7 x
10-6/ 0C. Answer: σ = 110.5 MPa

2. A 3-m long bronze bar with a cross-sectional area of 350 mm 2 is secured between two rigid walls as
shown. At a temperature of 15 0C, the gap Δ = 3 mm. Calculate the temperature at which the
compressive stress developed in the bar is 30 MPa. Assume α = 18 x 10-6/ 0C and E = 80 GPa.
Answer: Tf = 61.40C

3. Three bars each made from different materials are placed between walls when the temperature is
140C. Determine the support reaction when the temperature becomes 19 0C. The material properties
are given in the figure. Answer: R = 3502.30 N

4. The rigid slab is supported by two identical copper rods as shown. The copper rod has an area of 500
mm2 with α = 16.8 μm/ ( mm/ ( m0C ) and E = 120 GPa. There is a gap of 0.18 mm between the slab and the
aluminum rod. The aluminum rod has an area of 400 mm 2 with α = 23.1 μm/ ( mm/ ( m0C ) and E = 70 GPa.
Compute the stress developed in each rod when the temperature is raised to 85 0C. Assume that the
weight of the slab is negligible. Answer: σAl = 16.80 MPa
5. The rigid bar is supported by a smooth pin and two rods as shown. Neglect the weight of the rigid
bar. The assembly is initially stress-free. Determine the stress in each rod if the temperature increases
by 250C after a load W = 100 kN is applied. The steel rod has an area of 320 mm 2 with α = 11.7 μm/ ( mm/
( m0C ) and E = 200 GPa. For the bronze rod, it has an area of 1380 mm 2 with α = 18.9 μm/ ( mm/ ( m0C ) and E
= 83 GPa.

6. Three wires are used to support the 150-lb force. The wires AB and AC are made of steel, and wire
AD is made of copper. Assume that the three wires have constant cross-sectional area A = 0.0123 in 2.
For steel wire, α = 8 x 10-6 in/ (in0F ) and E = 29000 ksi and for copper wire, α = 9.6 x 10 -6 in/ (in0F ) and E
= 17000 ksi. Calculate the axial force exerted by the three wires if the temperature is raised by 80 0F.
Answer: PSt = 10 lb, Pcu = 136 lb