If a statically indeterminate structure is subjected to change in temperature, internal stresses are induced. This can be found in structures such as cooling towers and chimneys, where the internal and external temperatures are different. This implies that the temperature at the two extreme fibres of the material section are not the same and since the structure is restrained, internal stresses are developed due to thermal strain. We can use force method to obtain the magnitude of such stresses, in order to make appropriate designs.

Instead of the normal canonical equation for externally applied load, we now replace the free terms with deformation due to temperature change as shown below;

The deformation of a structure at a point due to temperature difference is given by;

Where;

Î± = coefficient of thermal expansion of the material

h = depth of the member

âˆ†t = change in temperature

t_{av}Â = average temperature for the member

âˆ«M Ì…ds = Area of the bending moment diagram for the member

âˆ«N Ì…ds = Area of the axial force diagram for the member

In summary, the steps to follow are:

(1) Determine the degree of redundancy in the structure, and write the appropriate canonical equation.

(2) Draw your basic or primary system as usual

(3) Obtain unit bending moment, shear, and axial force diagrams

(4) Calculate the unit displacement using Mohrâ€™s integral or Vereschaginâ€™s method

(5) Calculate the free terms of the canonical equation (displacement due to temperature difference)

(6) Solve the canonical equation

(7) Plot your internal stresses diagram.

**SOLVED EXAMPLE**

**
**A portal frame is fixed at the column base A, and hinged at the far end of the beam C as shown above. The column of the frame has a cross-section of 30cm x 30cm while the beam has a depth of 60cm and width of 30cm. The temperature inside the frame (t

_{1}) is 50Â°C while the temperature outside (t

_{2}) is 21Â°C. Draw the bending moment, shearing force, and axial force diagram for the temperature difference action on the frame (E = 2.17 Ã— 10

^{7}KN/m

^{2}, Î± = 11 Ã— 10

^{-6}/Â°C).

**SOLUTION**

**Geometric properties and temperature data;**

Moment of inertia of column I_{C} = (bh^{3})/12 = (0.3 Ã— 0.3^{3})/12 = 6.75 Ã— 10^{-4} m^{4}

Moment of inertia of beam I_{B} = (bh^{3})/12 = (0.3 Ã— 0.6^{3})/12 = 5.4 Ã— 10^{-4} m^{4}

We desire to work in terms of I_{C} such that I_{C}/I_{B} = 0.125

Hence flexural rigidity of the column, EI_{C} = (2.17 Ã— 10^{7}) Ã— (6.75 Ã— 10^{-4}) = 14647.5 KNm^{2}

For the frame above;

âˆ†t = t_{2} â€“ t_{1} = 50 â€“ 21 = 29Â°C; t_{av} = (50 + 21)/2 = 35.5Â°C

For member AB, âˆ†t/h = 29/0.3 = 96.667; For member BC, âˆ†t/h = 29/0.6 = 48.333

**Basic system**

The next step in the analysis is to reduce the structure to a basic system, which is a system that must be statically determinate and stable. The frame is statically indeterminate to the second order, which means that are we are going to remove two redundant supports.

Degree of static indeterminacy is given by;

R_{D} = (3m + r) â€“ 3n â€“ s â€”â€”â€”â€”â€”â€”â€”â€”â€” (2)

So that R_{D} = (3 Ã— 2) + 5 â€“ (3 Ã— 3) â€“ 0 = 2

By choice, I am deciding to remove the two reactive forces at support C of the frame to obtain the basic system as shown below;

**Analysis of case 1; X _{1} = 1.0, X_{2} = 0**

The simple static analysis of the structure loaded with case 1 loading is as shown below in terms of the internal stresses diagram.

**Â Analysis of case 2; X _{1} = 0, X_{2} = 1.0**

The simple static analysis of the structure loaded with case 1 loading is as shown below in terms of the internal stresses diagram.

The appropriate canonical equation for this structure is therefore;

Î´_{11}X_{1} + Î´_{12}X_{2} + Î”_{1t} = 0

Î´_{21}X_{1} + Î´_{22}X_{2} + Î”_{2t} = 0

**Computation of the influence coefficients**

Influence coefficients are based on Mohrâ€™s integral such that Î´_{i}Â = 1/EI âˆ«Mmds. When we wish to handle this by using the graphical method (making use bending moment diagrams), we directly employ Verecshaginâ€™s rule which simply states that when we are combining two diagrams of which one must be of a linear form (due to the unit load) and the other of any other form, the equivalent of Mohrâ€™s integral is given by the area of the principal combiner multiplied by the ordinate which its centroid makes with the linear diagram. The rule can also work vice versa. This process has been adopted in this work.

**Î´ _{11} (Deformation at point 1 due to unit load at point 1)**

This is obtained by the bending moment of case 1 combining with itself. This is shown below.

Î´_{11} = (5 Ã— 5 Ã— 4) + (1/3 Ã— 5 Ã— 5 Ã— 5 Ã— 0.125) = 105.208

**Î´ _{21} = Î´**

_{12Â }(Deformation at point 2 due to unit load at point 1 which is equal to deformation at point 1 due to unit load at point 2 based on Maxwellâ€™s theorem and Bettiâ€™s law)

This is obtained by the bending moment diagram of case 1 combining with bending moment diagram of case 2. This is shown below.

Î´_{21} = (1/2 Ã— 5 Ã— 4 Ã— 4) = 40

**Î´ _{22} (Deformation at point 2 due to unit load at point 2)**

This is obtained by the bending moment diagram of case 2 combining with itself. This is shown below.

Î´_{22} = (1/3 Ã— 4 Ã— 4 Ã— 4) = 21.333

**Influence coefficients due to temperature difference**

**Case 1 **(Take a good look at the bending moment and axial force diagrams)

Î”_{1t}/EI_{C} = Î±âˆ†t/h âˆ«M Ì…ds + Î±t_{av} âˆ«N Ì…ds

Considering the first term of the equation for members AB and BC = [(11 Ã— 10^{-6} Ã— 96.667 Ã— 5 Ã— 4) + (11 Ã— 10^{-6} Ã— 48.333 Ã— (1/2) Ã— 5 Ã— 4)] = 0.02791

Considering the second term of the equation for member AB = (11 Ã— 10^{-6} Ã— 35.5 Ã— 1 Ã— 4) = 0.001562

Hence, Î”_{1t} = Î±âˆ†t/h âˆ«M Ì…ds + Î±t_{av} âˆ«N Ì…ds = 0.02791 + 0.001562 = 0.029473EI_{C}

Î”_{1t} = 0.029473 Ã— 14647.5 = 431.691

**Case 2** (Take a good look at the bending moment and axial force diagrams)

Î”_{2t}/EI_{C} = Î±âˆ†t/h âˆ«M Ì…ds + Î±t_{av} âˆ«N Ì…ds

Considering the first term of the equation for members AB = (11 Ã— 10^{-6} Ã— 96.667 Ã— (1/2) Ã— 4 Ã— 4) = 0.0085067

Considering the second term of the equation for member BC = (11 Ã— 10^{-6} Ã— 35.5 Ã— -1 Ã— 5) = -0.0019525

Hence, Î”_{2t} = Î±âˆ†t/h âˆ«M Ì…ds + Î±t_{av} âˆ«N Ì…ds = 0.0085067 â€“ 0.0019525 = 0.0065542EI_{C}

Î”2_{t} = 0.0065542 Ã— 14647.5 = 96.0026

**The appropriate canonical equation now becomes;**

105.208X_{1} + 40X_{2} = -431.691

40X_{1} + 21.333X_{2} = -96.0026

On solving the equations simultaneously; X_{1} = -8.332 KN; X_{2} = 11.122 KN

The final value of the internal stresses is given by the equations below;

M_{def} = M_{1}X_{1} + M_{2}X_{2}

Q_{def} = Q_{1}X_{1} + Q_{2}X_{2}

N_{def} = N_{1}X_{1} + N_{2}X_{2}

**Final Moment (M _{def})**

M

_{A }= (-8.332 Ã— 5) + (11.122 Ã—4) = 2.828 KNm

M

_{B}= (-8.332 Ã— 5) = -41.660 KNm

M

_{C}= hinged support = 0

**Final Shear Force (Q _{def})**

Q

_{A}â€“ Q

_{B}

^{B}= (11.122 Ã— -1) = -11.122 KN

Q

_{B}

^{R}â€“ Q

_{C}

^{L}= (-8.332 Ã— 1) = -8.332 KN

**Final Axial Force (N _{def})**

N

_{A}â€“ N

_{B}

^{B}= (-8.332 Ã— 1) = -8.332 KN

N

_{B}

^{R}â€“ N

_{C}

^{L}= (11.122 Ã— -1) = -11.122 KN

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