Internal Forces in Frames Due to Temperature Difference

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;
α = coefficient of thermal expansion of the material
h = depth of the member
∆t = change in temperature
t_av = average temperature for the member
∫M ̅ds = Area of the bending moment diagram for the member
∫N ̅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 Mohr’s integral or Vereschagin’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 (t₁) is 50°C while the temperature outside (t₂) is 21°C. Draw the bending moment, shearing force, and axial force diagram for the temperature difference action on the frame (E = 2.17 × 10⁷ KN/m², α = 11 × 10^-6 /°C).

SOLUTION
Geometric properties and temperature data;
Moment of inertia of column I_C = (bh³)/12 = (0.3 × 0.3³)/12 = 6.75 × 10^-4 m⁴
Moment of inertia of beam I_B = (bh³)/12 = (0.3 × 0.6³)/12 = 5.4 × 10^-4 m⁴
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 × 10⁷) × (6.75 × 10^-4) = 14647.5 KNm²

For the frame above;
∆t = t₂ – t₁ = 50 – 21 = 29°C; t_av = (50 + 21)/2 = 35.5°C
For member AB, ∆t/h = 29/0.3 = 96.667; For member BC, ∆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) – 3n – s ————————— (2)
So that R_D = (3 × 2) + 5 – (3 × 3) – 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; X₁ = 1.0, X₂ = 0
The simple static analysis of the structure loaded with case 1 loading is as shown below in terms of the internal stresses diagram.

 Analysis of case 2; X₁ = 0, X₂ = 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;

δ₁₁X₁ + δ₁₂X₂ + Δ_1t = 0
δ₂₁X₁ + δ₂₂X₂ + Δ_2t = 0

Computation of the influence coefficients
Influence coefficients are based on Mohr’s integral such that δ_i = 1/EI ∫Mmds. When we wish to handle this by using the graphical method (making use bending moment diagrams), we directly employ Verecshagin’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 Mohr’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.

δ₁₁ (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.

δ₁₁ = (5 × 5 × 4) + (1/3 × 5 × 5 × 5 × 0.125) = 105.208

δ₂₁ = δ₁₂ (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 Maxwell’s theorem and Betti’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.

δ₂₁ = (1/2 × 5 × 4 × 4) = 40

δ₂₂ (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.

δ₂₂ = (1/3 × 4 × 4 × 4) = 21.333

Influence coefficients due to temperature difference

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

Δ_1t/EI_C = α∆t/h ∫M ̅ds + αt_av ∫N ̅ds
Considering the first term of the equation for members AB and BC = [(11 × 10^-6 × 96.667 × 5 × 4) + (11 × 10^-6 × 48.333 × (1/2) × 5 × 4)] = 0.02791
Considering the second term of the equation for member AB = (11 × 10^-6 × 35.5 × 1 × 4) = 0.001562
Hence, Δ_1t = α∆t/h ∫M ̅ds + αt_av ∫N ̅ds = 0.02791 + 0.001562 = 0.029473EI_C
Δ_1t = 0.029473 × 14647.5 = 431.691

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

Δ_2t/EI_C = α∆t/h ∫M ̅ds + αt_av ∫N ̅ds
Considering the first term of the equation for members AB = (11 × 10^-6 × 96.667 × (1/2) × 4 × 4) = 0.0085067
Considering the second term of the equation for member BC = (11 × 10^-6 × 35.5 × -1 × 5) = -0.0019525
Hence, Δ_2t = α∆t/h ∫M ̅ds + αt_av ∫N ̅ds = 0.0085067 – 0.0019525 = 0.0065542EI_C
Δ2_t = 0.0065542 × 14647.5 = 96.0026

The appropriate canonical equation now becomes;

105.208X₁ + 40X₂ = -431.691
40X₁ + 21.333X₂ = -96.0026

On solving the equations simultaneously; X₁ = -8.332 KN; X₂ = 11.122 KN

The final value of the internal stresses is given by the equations below;
M_def = M₁X₁ + M₂X₂
Q_def = Q₁X₁ + Q₂X₂
N_def = N₁X₁ + N₂X₂

Final Moment (M_def)
M_A = (-8.332 × 5) + (11.122 ×4) = 2.828 KNm
M_B = (-8.332 × 5) = -41.660 KNm
M_C = hinged support = 0

Final Shear Force (Q_def)
Q_A – Q_B^B = (11.122 × -1) = -11.122 KN
Q_B^R – Q_C^L = (-8.332 × 1) = -8.332 KN

Final Axial Force (N_def)
N_A – N_B^B = (-8.332 × 1) = -8.332 KN
N_B^R – N_C^L = (11.122 × -1) = -11.122 KN

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