All of the members and structures that we have considered so far were assumed to remain at the same temperature while they were being loaded .we are now going to consider various situations involving changes in temperature.

Let us first consider a homogeneous rod AB of uniform cross section .Which rests freely on a smooth horizontal surface. If the temperature of the rod is raised by ∆T. We observe that the rod elongates by an amount δT which is proportional to both the temperature change ∆T and the length L of the rod. We have

δT=α (∆T) L……………… (2.21)

Let us first consider a homogeneous rod AB of uniform cross section .Which rests freely on a smooth horizontal surface. If the temperature of the rod is raised by ∆T. We observe that the rod elongates by an amount δT which is proportional to both the temperature change ∆T and the length L of the rod. We have

δT=α (∆T) L……………… (2.21)

Where α is a constant characteristic of the material. Called the coefficient of thermal expansion. Since δT and L are both expressed in units of length. α represents a quantity per degree C. or per degree F. depending whether the temperature change is expressed in degrees Celsius or in degrees Fahrenheit.

With the deformation δT must be associated a strain ЄT = δT/L. Recalling eq.(2.21) we conclude that :

ЄT=α ∆T……………………………….(2.22)

The strain ЄT is referred to as thermal strain, since it’s caused by the change of temperature of the rod. In the case we are considering here there is no stress associated with the strain ЄT.

Let us now assume that the same rod AB of length L is placed between two fixed supports at a distance L from each other (fig.2.35 a). Again, there is neither stress nor strain in the initial condition. IF we raise the temperature by ∆T .the rod cannot elongate because of the restraints imposed on its ends: the elongation δT of the rod is thus zero. Since the rod is homogeneous and of uniform cross section. The strain ЄT at any point is ЄT= δT/L and thus. Also zero. However, the supports will exert equal and opposite forces P and P’ on the rod after the temperature has been raised. Too keep it from elongating (Fig.2.53b).It thus follows that a state of stress (with no corresponding strain) is created in the rod.

As we prepare to determine the stress σ created by the temperature change ∆T. we observe that the problem we have to solve is statically indeterminate. Therefore, We should first compute the magnitude P of the reactions at the supports from the condition that the elongation of the rod is zero. Using the superposition method described in sec.2.9. We detach the rod from its support B(Fig.2.36b).According to formula (2.21).The corresponding elongation is

δT=α (∆T) L

With the deformation δT must be associated a strain ЄT = δT/L. Recalling eq.(2.21) we conclude that :

ЄT=α ∆T……………………………….(2.22)

The strain ЄT is referred to as thermal strain, since it’s caused by the change of temperature of the rod. In the case we are considering here there is no stress associated with the strain ЄT.

Let us now assume that the same rod AB of length L is placed between two fixed supports at a distance L from each other (fig.2.35 a). Again, there is neither stress nor strain in the initial condition. IF we raise the temperature by ∆T .the rod cannot elongate because of the restraints imposed on its ends: the elongation δT of the rod is thus zero. Since the rod is homogeneous and of uniform cross section. The strain ЄT at any point is ЄT= δT/L and thus. Also zero. However, the supports will exert equal and opposite forces P and P’ on the rod after the temperature has been raised. Too keep it from elongating (Fig.2.53b).It thus follows that a state of stress (with no corresponding strain) is created in the rod.

As we prepare to determine the stress σ created by the temperature change ∆T. we observe that the problem we have to solve is statically indeterminate. Therefore, We should first compute the magnitude P of the reactions at the supports from the condition that the elongation of the rod is zero. Using the superposition method described in sec.2.9. We detach the rod from its support B(Fig.2.36b).According to formula (2.21).The corresponding elongation is

δT=α (∆T) L

Applying now to end B the Force P representing the redundant reaction. And recalling formula (2.7).we obtain a second deformation (Fig.2.36c)

δP=PL/AE

Expressing the total deformation δ must be zero, we have

δ=δT + δP=αL(∆T) +PL/AE = 0

from which we conclude that

P= - AEα(∆T)

and that the stress in the rod due to the temperature change ∆T is ,

σ=P/A = - Eα(∆T)……………………..(2.23)

It should be kept in mind that the result we have obtained here and our earlier remark regarding the absence of any strain in the rod apply only in the case of a homogeneous rod of uniform cross section. Any other problems involving a restrained structure undergoing a change in temperature must be analyzed on its own merits.However, the same general approach can be used; ie, we can consider separately the deformation due to the temperature change and the deformation due to the redundant reaction and superpose the solutions obtained.

δP=PL/AE

Expressing the total deformation δ must be zero, we have

δ=δT + δP=αL(∆T) +PL/AE = 0

from which we conclude that

P= - AEα(∆T)

and that the stress in the rod due to the temperature change ∆T is ,

σ=P/A = - Eα(∆T)……………………..(2.23)

It should be kept in mind that the result we have obtained here and our earlier remark regarding the absence of any strain in the rod apply only in the case of a homogeneous rod of uniform cross section. Any other problems involving a restrained structure undergoing a change in temperature must be analyzed on its own merits.However, the same general approach can be used; ie, we can consider separately the deformation due to the temperature change and the deformation due to the redundant reaction and superpose the solutions obtained.

**Example:**

*Determine the values of the stress in portion AC and CB of the steel bar shown (Fig. 2.37) when the temperature of the bar is -45°C knowing that a close fit exists at both of the rigid supports when the temperature is +24°C. Use the values E=200GPa and α = 11.7* 10-6/°C for steel.

Sol.

we first determine the reactions at the supports. Since the problem is statically in determinate. We detach the bar from its support at B and let it undergo the temperature change

∆T=(-45°C)-(24°C)= -69°C

The corresponding deformation (fig.2.38b) is

δT=α(∆T)L=(11.7*10-6/°C)(-69°C)(600mm)

=-0.484mm

Applying now the unknown force at the end B RB(fig. 2.38c). We use Eq. (2.8) to express the corresponding deformation δR. Substituting

L1=L2=300mm

A1=380 mm2 A2=750mm2

P1=P2=RB E=200GPa

into Eq.(2.8) we write

δR=P1L1/A1E + P2L2/A2E=RB/200GPa (300mm/380mm2 + 300mm/750mm2)=5.95*10-6 mm (N) RB

Expressing that the total deformation of the bar must be zero as a result of the imposed constraints. we write

δ=δT + δR = 0

=-0.484mm + (5.95 * 10-6 mm/(N))RB=0

from which we obtain

RB=81.34 * 103 N = 81.34 KN

The reaction at A is equal and opposite.

Noting that the forces in the two portions of the bar are P1=P2=81.34KN. We obtain the following values of the stress in portions Ac and CB of the bar:

σ1=P1/A1=81.34 /380 = 214.1 M Pa

σ2=P2/A2=81.34/750 = 108.5 M Pa

We cannot emphasize the fact that, while the total deformation of the bar must be zero, the deformation of the portions AC and CB are not zero. A solution of the problem based on the assumption that these deformations are zero would therefore be wrong. Neither can the values of the strain in AC or CB be assumed equal to zero. To amplify this point, let us determine the strain ЄAC in portion AC of the bar. The strain ЄAC can be divided into two component parts; one is the thermal strain ЄT produced in the unrestrained bar by the temperature change ∆T (fig. 2.38b). From Eq. (2.22) we write

ЄT = α (∆T) = 11.7 * 10-6* -69=-807.3*10-6

The other component ЄAC is associated with the stress σ1 due to the force RB applied to the bar (Fig. 2.38c). From Hook’s law, We express the component of the strain as

σ1/E = 214.1 M Pa/200 G Pa = 1070.5 * 10-6

Adding the two components of the strain in AC, we obtain

ЄAC=ЄT + σ1/E = -807.3*10-6 + 1070.5 * 10-6

=263.2 * 10-6

A similar computation yields the strain in portion CB of the bar:

ЄCB=ЄT + σ2/E = -264.8 * 10-6

The deformation δAC and δCB of the two portions of the bar are expressed respectively as

δAC=ЄAC(AC) = (263.2 * 10-6 * 300)

=0.079 mm

δAB=ЄCB(CB)=(-264.8 * 10-6* 300 )

=-0.079mm

We thus check that, while the sum δ = δAC + δCB of the two deformations is zero, neither of the deformations is zero.

L1=L2=300mm

A1=380 mm2 A2=750mm2

P1=P2=RB E=200GPa

into Eq.(2.8) we write

δR=P1L1/A1E + P2L2/A2E=RB/200GPa (300mm/380mm2 + 300mm/750mm2)=5.95*10-6 mm (N) RB

Expressing that the total deformation of the bar must be zero as a result of the imposed constraints. we write

δ=δT + δR = 0

=-0.484mm + (5.95 * 10-6 mm/(N))RB=0

from which we obtain

RB=81.34 * 103 N = 81.34 KN

The reaction at A is equal and opposite.

Noting that the forces in the two portions of the bar are P1=P2=81.34KN. We obtain the following values of the stress in portions Ac and CB of the bar:

σ1=P1/A1=81.34 /380 = 214.1 M Pa

σ2=P2/A2=81.34/750 = 108.5 M Pa

We cannot emphasize the fact that, while the total deformation of the bar must be zero, the deformation of the portions AC and CB are not zero. A solution of the problem based on the assumption that these deformations are zero would therefore be wrong. Neither can the values of the strain in AC or CB be assumed equal to zero. To amplify this point, let us determine the strain ЄAC in portion AC of the bar. The strain ЄAC can be divided into two component parts; one is the thermal strain ЄT produced in the unrestrained bar by the temperature change ∆T (fig. 2.38b). From Eq. (2.22) we write

ЄT = α (∆T) = 11.7 * 10-6* -69=-807.3*10-6

The other component ЄAC is associated with the stress σ1 due to the force RB applied to the bar (Fig. 2.38c). From Hook’s law, We express the component of the strain as

σ1/E = 214.1 M Pa/200 G Pa = 1070.5 * 10-6

Adding the two components of the strain in AC, we obtain

ЄAC=ЄT + σ1/E = -807.3*10-6 + 1070.5 * 10-6

=263.2 * 10-6

A similar computation yields the strain in portion CB of the bar:

ЄCB=ЄT + σ2/E = -264.8 * 10-6

The deformation δAC and δCB of the two portions of the bar are expressed respectively as

δAC=ЄAC(AC) = (263.2 * 10-6 * 300)

=0.079 mm

δAB=ЄCB(CB)=(-264.8 * 10-6* 300 )

=-0.079mm

We thus check that, while the sum δ = δAC + δCB of the two deformations is zero, neither of the deformations is zero.