Fixed-Fixed Torsion Shaft (Interior Torque)

stressmass-cost

Take a solid circular shaft, build it rigidly into a wall at both ends, and apply a torque $T$ somewhere along its length. A shaft fixed at only one end is statically determinate: the single wall reaction must equal the applied torque, and you are done. Anchor the second end and the problem changes character. Now there are two reaction torques — $T_A$ at the left wall, $T_B$ at the right — but torsional equilibrium gives only one equation, $T_A + T_B = T$. The shaft is statically indeterminate to the first degree, and equilibrium alone cannot say how the applied torque splits between the two walls.

You meet this arrangement wherever a rotating member is restrained at both ends and driven in the middle: a splined shaft keyed to fixed hubs at each end, a torsion-bar suspension anchored to the frame at both ends and loaded by the arm, a fixed-fixed structural member picking up an applied twisting couple.

The missing equation is compatibility

The second equation comes not from force balance but from geometry. The shaft is one continuous piece, so the rotation at the load point must be a single number. Reckoned from the left, that point has twisted by $T_A a / GJ$ — the left segment of length $a$ carries the reaction $T_A$. Reckoned from the right, it has twisted by $T_B b / GJ$, with $b$ the right-segment length. They describe the same rotation, so they must be equal:

$\frac{T_A\,a}{GJ} = \frac{T_B\,b}{GJ} \quad\Longrightarrow\quad T_A\,a = T_B\,b$

Combine that with equilibrium (and $L = a + b$) and the split is fixed:

$T_A = T\,\frac{b}{L}, \qquad T_B = T\,\frac{a}{L}$

The build does not hand you these formulas to trust. It certifies that the two relations (one equilibrium, one compatibility) form a system that is linear in the two unknowns, solves that $2\times2$ system exactly at build time, and checks the solution back through both relations — the same total verification an ordinary closed form gets. This is the solveLinear capability; the propped cantilever is its reference beam case, and this shaft is its smallest — a genuinely coupled two-unknown solve. What the machine proves and what still rests on a book is spelled out on the verification page.

The larger reaction lands on the shorter segment

Read the result again: $T_A = T\,b/L$. The left wall’s share is set by the length of the right segment. Slide the load toward the left wall — small $a$, large $b$ — and $T_A$ grows toward the full torque $T$ while $T_B$ fades. The near wall does the heavy lifting; the short, stiff segment between the load and the near wall is harder to twist, so it carries more torque. Because each segment carries its own reaction, each has its own surface shear stress $\tau = Tr/J$, and the shorter segment is where the higher stress — and first yield — appears. That is why the widget reports $\tau_1$ and $\tau_2$ separately, with a safety factor for each.

The reaction torques do not care what the shaft is made of

Look at the compatibility step in the derivation below. Both twists carry the same torsional rigidity $GJ$, and because the shaft is prismatic — uniform shear modulus and uniform radius — $GJ$ divides out completely. The line $T_A a = T_B b$ has no $G$ and no $r$ in it. That cancellation is machine-checked, and it is the lesson: the reaction torques are material-blind, and radius-blind too. Switch the material in the widget from A36 steel to 6061 aluminium and watch $T_A$, $T_B$, $\tau_1$, and $\tau_2$ not move at all, even as:

• Shear modulus $G$ changes the twist $\varphi = T_A a / GJ$ — aluminium is about a third as stiff in shear ($G \approx 26$ vs $79$ GPa), so the same load twists it nearly three times as far.
• Yield strength $\sigma_y$ changes the safety factors $\mathrm{SF} = (\sigma_y/2)/\tau$.
• Density $\rho$ changes the mass.

A redundant restraint redistributes torque by geometry, not by material — the “sharing” between the two walls is set purely by the segment lengths. (Make the shaft non-prismatic — a stepped shaft, two different diameters spliced at the load point — and $GJ$ no longer cancels; the split would then follow the stiffness distribution. That is a later story.)

What governs, and the sign convention

Throughout: $x$ runs from the left wall $A$ to the right wall $B$, the applied torque $T$ acts at $x = a$, the segment lengths are $a$ and $b = L - a$, and $T_A$, $T_B$ are the reaction-torque magnitudes at the two walls (their senses oppose the applied torque). The shear stresses $\tau_1 = T_A r/J$ and $\tau_2 = T_B r/J$ grow linearly from zero at the axis to their peak at the surface, exactly as in the simple torsion shaft; the shorter segment carries the larger of the two. Shear yield is taken at $\sigma_y/2$ by the maximum-shear-stress criterion, the same convention the torsion-shaft page uses.

Try it

Governing relations

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

How it fails

The widget’s two safety factors guard one thing each — first shear yield at the surface of each segment, where $\tau = Tr/J$ peaks. A real doubly-restrained shaft has more exits, and the most interesting ones are specific to it being statically indeterminate:

• A support that rotates — the assumption that can quietly move every number. The whole solution rests on one premise: both walls are perfectly rigid, so the end rotations are exactly zero. Let one wall give a little — a hub that slips on its key, a foundation that yields in torsion by an angle $\Delta$ — and the compatibility equation changes from $T_A a/GJ = T_B b/GJ$ to one carrying $\Delta$. The torque split redistributes, and the “material-blind” result stops being material-blind: now $GJ$ does appear, multiplied by the give. In a determinate shaft a support that rotates just goes along for the ride; in an indeterminate one, support movement is a load case in its own right.

• Yielding in the shorter segment first, then redistribution. Because the shorter segment carries the larger reaction torque, its surface yields first. Past that point the shaft does not fail outright — the yielded segment sheds stiffness and the split shifts toward the other wall, a redistribution the elastic formulas here cannot see. The $\tau > \sigma_y/2$ warning marks where they stop telling the truth.

• Stress concentration at the walls and the load point. The torsion formula assumes a smooth prismatic shaft. Real fixed ends are splines, keyways, or shrink-fit hubs, and the applied torque enters through a gear or coupling — every one of these is a stress raiser that local shear can exceed the nominal $Tr/J$ by a large factor. Fatigue cracks in torsion shafts almost always start at such a feature, not in the smooth bore.

• Buckling of a thin-walled shaft. If the section is a thin tube rather than a solid bar, torsional shear can buckle the wall into a diagonal wrinkle pattern well before the material yields — a stability failure this solid-shaft page does not model.

• Fatigue. A shaft that reverses or pulses its torque cycles the surface shear at the wall of the shorter segment and at every stress raiser. Fatigue strength in shear sits far below static shear yield, and nothing on this page addresses cyclic life.

• Brittle materials fail differently. The $\sigma_y/2$ shear-yield criterion assumes a ductile metal. A brittle shaft (gray cast iron, a ceramic) fails on the maximum tensile stress, which in pure torsion acts on a 45° helix — which is why a twisted piece of chalk breaks along a slanted spiral, not straight across. Rate such a material against its tensile strength, not the shear-yield factor shown here.

Sources

• Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §3.8 (statically indeterminate torsional members: a bar fixed at both ends with an intermediate torque, solved by combining equilibrium with the compatibility of the angle of twist); §3.3–3.4 (torsion formula τ = Tr/J, angle of twist TL/GJ).
• Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — §5-4 (maximum-shear-stress theory, shear yield S_sy = 0.5 S_y).