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TLL 01: Beam under temperature load

Description

Material Modulus of elasticity E=210 000 N/mm^2
Thermal dilation \alpha =0.000 012/ ^{\circ}C
Geometry Cross-section IPE 500
Boundary conditions Case 1: cantilever
Case 2: fixed beam
Loads Load 1: Global temperature change \Delta T = 40 ^{\circ}C
Load 2: Linear gradient in local y’ direction \Delta T = 40 ^{\circ}C
Load 3: Linear gradient in local z’ direction \Delta T = 40 ^{\circ}C
Mesh No. of divisions 8

Independent reference results

Open handcalculations

Load 1: global temperature change \Delta T = 40 ^{\circ}C

  • Case 1: cantilever beam
    Since the beam can lengthen/ shorten, the temperature load will extend the beam

    \[ \delta_x=\alpha \cdot \Delta T\cdot L=0.000 012 / ^{\circ}C\cdot 40^{\circ}C\cdot 5m=2.4mm\]

  • Case 2: fixed beam
    Since the beam cannot lengthen/ shorten, the temperature load will result in axial force.
    The increase in temperature will want to lengthen the beam. To counteract this deformation, a compressive force must be created. So the normal force in Diamonds will have a negative sign.

    \[N=\alpha \cdot \Delta T\cdot A\cdot E=0.000 012 / ^{\circ}C\cdot 40^{\circ}C\cdot 11553mm^2\cdot 210000N/mm^2=1164.46kN\]

Load 2: linear gradient in local y’ direction \Delta T = 40 ^{\circ}C

  • Case 1: cantilever beam
    Since the beam can lengthen/ shorten, the gradient will cause an elongation of the upper fiber and a shortening of the non-fiber. The combination of both results in a downward deformation.

        \[ \varphi_z=\frac{\alpha \cdot \Delta T\cdot L}{h}=\frac{0.000 012 / ^{\circ}C\cdot 40^{\circ}C\cdot 11553mm^2\cdot 5m}{500mm}=0.2750^{\circ}\]

        \[ \delta _y=\frac{\alpha \cdot \Delta T\cdot L^2}{2 \cdot h}=\frac{0.000 012 / ^{\circ}C\cdot 40^{\circ}C\cdot (5m)^2}{2 \cdot 500mm}=12.0000mm\]

  • Case 2: Fixed beam
    Since the beam cannot lengthen/ shorten, the gradient will cause an elongation of the upper fiber and a shortening of the non-fiber. To counteract this deformation, a compressive force must be created in the upper fiber and a tensile force in the lower. So the bending moment in Diamonds will be drawn on the lower side of the beam.

        \[M_{y'}=\frac{\alpha \cdot \Delta T\cdot E\cdot I_y}{h}=\frac{0.000 012 / ^{\circ}C\cdot 40^{\circ}C \cdot 4.82\cdot 10^8 mm^4}{500mm}=97.17kNm\]

Load 3: linear gradient in local z’ direction \Delta T = 40 ^{\circ}C

  • Case 1: cantilever beam
    Since the beam can lengthen/ shorten, the gradient will cause an elongation of the upper fiber and a shortening of the non-fiber. The combination of both results in a downward deformation.

        \[ \varphi_z=\frac{\alpha \cdot \Delta T\cdot L}{b}=\frac{0.000 012 / ^{\circ}C\cdot 40^{\circ}C\cdot 11553mm^2\cdot 5m}{200mm}=0.6875^{\circ}\]

        \[ \delta _y=\frac{\alpha \cdot \Delta T\cdot L^2}{2 \cdot b}=\frac{0.000 012 / ^{\circ}C\cdot 40^{\circ}C\cdot (5m)^2}{2 \cdot 200mm}=30.0000mm\]

  • Case 2: Fixed beam
    Since the beam cannot lengthen/ shorten, the gradient will cause an elongation of the upper fiber and a shortening of the non-fiber. To counteract this deformation, a compressive force must be created in the upper fiber and a tensile force in the lower. So the bending moment in Diamonds will be drawn on the lower side of the beam.

        \[M_{y'}=\frac{\alpha \cdot \Delta T\cdot E\cdot I_y}{b}=\frac{0.000 012 / ^{\circ}C\cdot 40^{\circ}C \cdot 2.14\cdot 10^7 mm^4}{200mm}=10.79kNm\]

Diamonds results and comparison

Load 1: global temperature change \Delta T = 40 ^{\circ}C


Independent reference Diamonds Difference
Case 1 Horizontal deformation \delta_x 2.4mm 2.4mm 0,00%
Case 2 Axial force N -1164.46kN -1164.46kN 0,00%

Load 2: linear gradient in local y’ direction \Delta T = 40 ^{\circ}C

Independent reference Diamonds Difference
Case 1 Angular rotation \varphi_z 0.2750° 0.2750° 0,00%
Horizontal deformation \delta_y 12.0000mm 12.0000mm 0,00%
Case 2 Bending moment M_y 97,17kNm 97,17kNm 0,00%

Load 3: linear gradient in local z’ direction \Delta T = 40 ^{\circ}C

Independent reference Diamonds Difference
Case 1 Angular rotation \varphi_y 0.6875° 0.6875° 0,00%
Horizontal deformation \delta_z 30.0000mm 30.0000mm 0,00%
Case 2 Bending moment M_z 10.79kNm 10.79kNm 0,00%

References

  • Hibbeler, R. (2006). Sterkteleer, 2/e. Pearson Education.
  • Weaver, W., & Gere, J. M. (1990). Matrix analysis of framed structures Table B-2. In Springer eBooks.
  • Tested in Diamonds 2025.

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