SLL 04: torsion on a stepped cantilever

Description

Material		Modulus of elasticity	$G=80 000 N/mm^2$
Geometry	Cross-section	For bar AB	⌀ 80mm
		For bar BC	⌀ 60mm
		For bar CD	⌀ 40mm
	Boundary conditions	Point A	Fixed support
Loads		In point B	$M_x=3 kNm$
		In point C	$M_x=2 kNm$
		In point D	$M_x=0.8 kNm$
Mesh		No. of divisions	8

Results

Handcalculation

$I_{t.AB}=\frac{\pi \cdot D^4}{32}=\frac{\pi \cdot (80mm)^4}{32}=4.021 \cdot 10^6mm^4$

$I_{t.BC}=\frac{\pi \cdot D^4}{32}=\frac{\pi \cdot (60mm)^4}{32}=1.272 \cdot 10^6mm^4$

$I_{t.CD}=\frac{\pi \cdot D^4}{32}=\frac{\pi \cdot (40mm)^4}{32}=0.251 \cdot 10^6mm^4$

$\varphi_{x.B}=\frac{L \cdot (M_{x.B}+M_{x.C}+M_{x.D})}{I_{t.AB} \cdot G}=\frac{0.5m \cdot (5.8kNm)}{4.021 \cdot 10^6mm^4 \cdot 80000N/mm^2}=0.000009rad$

$\varphi_{x.C}=\varphi_{x.B}+\frac{L \cdot (M_{x.C}+M_{x.D})}{I_{t.BC} \cdot G}=0.000009rad+\frac{0.5m \cdot (3.8kNm)}{1.272 \cdot 10^6mm^4 \cdot 80000N/mm^2}=0.000023rad$

$\varphi_{x.D}=\varphi_{x.B}+\varphi_{x.C}+\frac{L \cdot M_{x.D}}{I_{t.CD} \cdot G}=0.000009rad+0.000023rad+\frac{0.5m \cdot (3.8kNm)}{0.251 \cdot 10^6mm^4 \cdot 80000N/mm^2}=0.000043rad$

Angle of twist $\varphi_x$ in Diamonds

Point	Which result	Independent reference	Diamonds	Difference
B	Angle of twist $\varphi_x$	0.000009rad	0.000009rad	0%
C	Angle of twist $\varphi_x$	0.000023rad	0.000023rad	0%
D	Angle of twist $\varphi_x$	0.000043rad	0.000043rad	0%

References

Hibbeler, R. (2006). Sterkteleer, 2/e. Pearson Education.

Gere J. M., and Timoshenko, S. P., Mechanics of Materials, 2nd Edition, PWS Engineering, Page 171, Problem 3.3 – 1

Tested in Diamonds 2023r01.

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