TLL 01: Beam under temperature load

Description

Material	Modulus of elasticity	$E=210 000 N/mm^2$
Material	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 temperatur change $\Delta T = 40 ^{\circ}C$
		Load 2: Lineair gradient in local y’ direction $\Delta T = 40 ^{\circ}C$
		Load 3: Lineair gradient in local z’ direction $\Delta T = 40 ^{\circ}C$
Mesh	No. of divisions	8

Handcalculation

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.

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

		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%

Handcalculation

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.

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

		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%

Handcalculation

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.

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

		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%

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. https://doi.org/10.1007/978-1-4684-7487-9

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