Torsion - Structure Bench

🎓 New 3D Beams and Frames Learning Pathway - A Guide to using Structure Bench

Torsion - Structure Bench

Things to try

Press Load the model. The Deformed shape and Reactions are on by default, and the UC's tip has visibly rolled over while the box's tip has simply dropped. Scroll to the Displacements table and read the two tip nodes. Node o2 has deflected 2.779 mm and twisted 334.620 mrad about X, and c2 has deflected 3.423 mm and twisted 1.285 mrad. The box twists 260 times less and deflects more.
Switch on Torsion. A flat band of 3.0 kNm runs the length of both members, from the fixed end to the tip, because every section between the torque and the wall carries all of it. Hover along either member, and the probe card now carries a T entry that reads 3.0 kNm wherever you are.
Switch on Bending moment, with its Mz plane. The diagram is a straight line from 20.0 kNm at the fixed end to zero at the tip, on the tension side (the top), and it's identical on both members, because the moment comes from statics and doesn't depend on the section. Now look at the Reactions table. Each fixed end carries Rz 10.0 kN, Mx 3.0 kNm and My 20.0 kNm. The Mx is the torque arriving back at the wall. The My is the bending moment, about global Y here because the member runs along X. Remember from step 4 that the member's Mz is about its local strong axis, while the reaction's My is in global axes.
Look at the Query stations table at 1 m along each member. The rx value is 167.310 mrad on the UC and 0.643 mrad on the box, half the tip values in each case, and the deflections are 0.868 mm and 1.070 mm. A constant torque gives a twist that grows in a straight line from the wall.
Switch on Member stress and select the τ (T) chip, the torsional shear on its own, then hover over a flange of the UC and over a wall of the box. We won't quote a value here, because the member stress view samples the section on its own grid, but the comparison is what matters. The same torque produces a shear stress in the box that is a small fraction of the UC's.
If you're a member, press Copy to my projects: your copy opens in the editor. Press Loads to open the loads table and, under Nodal loads, set Mx to zero on both tips, then press Solve. The torsion diagram is empty and the Displacements table's rx reads zero on both members. The effect of the bracket has disappeared from the analysis, which is exactly what you get from a load typed "at the tip" of a member. Now put the 3 kNm back, then move it. Set Mx to zero again, and under In-span point loads add a row on member open at x = 1 m, axes local, with Mx = 3 kNm, and solve. The torsion diagram now reads 3.0 kNm from the wall to midspan and zero beyond it, and the tip twist halves to 167.3 mrad, because the outer metre carries no torque, so it rotates with the midspan and adds no twist of its own.
In the Members table, tick all four of the UC's releases boxes (i·My, i·Mz, j·My and j·Mz) and press Solve. The solve fails with Analysis failed, and the message beneath it reads nodal moment p-o is applied at o2, which has no rotational stiffness to carry it (truss-only node, hinge, or released member ends). A member released about both bending axes at both ends can't resist either the 10 kN at its tip or the moment applied there, so Structure Bench stops with an error that names the node, rather than report a meaningless number. Untick the boxes before you move on.

Two cantilevers, one torque

	UC 203 x 203 x 46 (open)	SHS 200 x 200 x 8 (closed)	By hand
Torque along the member	3.0 kNm	3.0 kNm	$T = Pe = 10 \times 0.3 = 3$ kNm
Twist at the tip, rx	334.620 mrad	1.285 mrad	$TL/GJ$ = 334.6 and 1.285 mrad
Twist at midspan	167.310 mrad	0.643 mrad	half the tip twist
Moment at the fixed end	-20.0 kNm	-20.0 kNm	$PL = 10 \times 2 = 20$ kNm
Deflection at the tip	-2.779 mm	-3.423 mm	$PL^3/3EI$ = 2.779 and 3.423 mm
Slope at the tip, ry	2.084 mrad	2.567 mrad	$PL^2/2EI$ = 2.084 and 2.567 mrad
Reactions at the fixed end	Rz 10.0 kN, Mx 3.0 kNm, My 20.0 kNm	the same	statics

Downward deflections and the hogging moment at the wall are negative, as the tables print them. The torque and the twist are printed with the sign of the global X axis.

Hand check: the twist of each section

The torque. The 10 kN hangs 0.3 m to one side of the member's axis, so the torque is,

T = Pe = 10 \times 0.3 = 3\ \text{kNm}

and because nothing else twists the member between the tip and the wall, every section carries this 3 kNm. The reaction Mx of 3.0 kNm at each fixed end is the same torque arriving at the support.

The shear modulus. Steel's modulus is 210 GPa and its Poisson's ratio 0.3, so,

G = \frac{E}{2(1+\nu)} = \frac{210}{2 \times 1.3} = 80.77\ \text{GPa}

The UC. The catalogue gives $J = 22.2$ cm $^4$ , which is $22.2 \times 10^{-8}$ m. Over the 2 m from the wall to the tip, the twist is,

\theta = \frac{TL}{GJ} = \frac{3{,}000 \times 2}{80.77 \times 10^9 \times 22.2 \times 10^{-8}} = \frac{6{,}000}{17{,}931} = 0.3346\ \text{rad}

or 334.6 mrad, against the solver's 334.620. In degrees, that's $0.3346 \times 57.3 = 19.2°$ . At midspan, $L$ is 1 m, so the twist is half, 167.3 mrad, which is what the query station reads.

The box. The catalogue gives $J = 5{,}780$ cm $^4$ for the SHS 200 x 200 x 8, so,

\theta = \frac{6{,}000}{80.77 \times 10^9 \times 5{,}780 \times 10^{-8}} = \frac{6{,}000}{4{,}668{,}500} = 0.001285\ \text{rad}

or 1.285 mrad, the solver's value exactly. The ratio of the two twists is $334.620 / 1.285 = 260.4$ , and the ratio of the two torsion constants is $5{,}780 / 22.2 = 260.4$ .

Where the box's torsion constant comes from. It's worth seeing the thin-walled formula produce that 5,780. The middle line of an 8 mm wall on a 200 mm box is a square of side $200 - 8 = 192$ mm, so it encloses $A_0 = 192^2 = 36{,}864$ mm and has a perimeter of mm. With a constant thickness, , and,

J = \frac{4 A_0^2}{\oint (ds/t)} = \frac{4 \times 36{,}864^2 \times 8}{768} = 56.6 \times 10^6\ \text{mm}^4 = 5{,}660\ \text{cm}^4

within 2% of the catalogue's 5,780, which accounts for the rounded corners more carefully. The $A_0^2$ term is why the box is so stiff. The shear flow round a loop 192 mm square acts on a lever arm of about 192 mm, whereas in each of the UC's plates the lever arm is only about the plate's own thickness.

The bending. The 10 kN at the tip of a 2 m cantilever gives $M = PL = 20$ kNm at the wall, the same on both members. The tip deflection is $PL^3/3EI$ . For the UC, with $I_z = 4{,}570$ cm,

\delta = \frac{PL^3}{3EI} = \frac{10{,}000 \times 2^3}{3 \times 210 \times 10^9 \times 4{,}570 \times 10^{-8}} = 2.779\ \text{mm}

and for the box, with $I = 3{,}710$ cm $^4$ , the same arithmetic gives 3.423 mm. The slope at the tip is $PL^2/2EI$ , which gives rad for the UC, the 2.084 mrad in the ry column, and 2.567 mrad for the box.