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

Independent reference results

Open handcalculations

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

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

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

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

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

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

Diamonds results and comparison

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 2024r01.

Article Attachments

.bsf

SLL04 198 KB

Was this article helpful?

Yes No

Need Support?

Can't find the answer you're looking for? Don't worry we're here to help!

CONTACT SUPPORT