20.3718
MPa
0.010315
rad
—
°
20.3718
MPa
0.010315
rad
—
°
The Torsion Calculator analyzes shafts and members subjected to twisting loads, computing the shear stress and angle of twist using classical torsion theory. Torsion is one of the fundamental loading modes in mechanics of materials alongside axial loading, bending, and shear, and it governs the design of drive shafts, axles, bolts, springs, and countless other mechanical components.
When a torque \(T\) is applied to a circular shaft, the resulting shear stress varies linearly from zero at the center to a maximum at the outer surface. The torsion formula is:
$$\tau = \frac{Tr}{J}$$
where \(\tau\) is the shear stress at radial distance \(r\) from the center, and \(J\) is the polar moment of inertia of the cross-section. For a solid circular shaft of diameter \(d\), \(J = \pi d^4 / 32\). The maximum stress occurs at the outer surface where \(r\) equals the shaft radius.
Equally important is the angle of twist, which measures the rotational deformation along the shaft's length:
$$\theta = \frac{TL}{GJ}$$
where \(L\) is the shaft length and \(G\) is the shear modulus (modulus of rigidity) of the material. For steel, \(G \approx 79{,}000\) MPa; for aluminum, \(G \approx 26{,}000\) MPa; for copper, \(G \approx 44{,}000\) MPa.
These formulas assume the shaft is prismatic (constant cross-section), the material is linearly elastic and isotropic, and the cross-section is circular. Non-circular cross-sections experience warping and require different approaches, such as the membrane analogy or finite element analysis.
In practice, engineers must ensure that the maximum shear stress remains below the allowable shear stress (typically 40–60% of the tensile yield strength), and that the angle of twist stays within acceptable limits, often specified as 0.25° to 1° per meter of shaft length depending on the application. This calculator handles both checks simultaneously, enabling quick verification of shaft adequacy.
Torsion analysis extends beyond simple shafts to helical springs (where wire torsion provides the spring force), bolted connections (where tightening torque induces bolt shear), and structural tubes in buildings and bridges. Mastering the torsion formula is foundational for any engineer dealing with rotating machinery or torque-transmitting components.
The calculator applies the classical torsion formulas for circular shafts:
Shear Stress:
$$\tau = \frac{Tr}{J}$$
where \(T\) is torque (converted to N·mm), \(r\) is radial distance (mm), and \(J\) is the polar moment of inertia (mm⁴). The result is in MPa (N/mm²).
Angle of Twist:
$$\theta = \frac{TL}{GJ}$$
where \(L\) is shaft length (mm) and \(G\) is shear modulus (MPa). The result is in radians, then converted to degrees by multiplying by \(180/\pi\).
For a solid circular shaft: \(J = \pi d^4 / 32\). For a hollow shaft: \(J = \pi (d_o^4 - d_i^4) / 32\). Enter the appropriate \(J\) value for your cross-section.
The shear stress \(\tau\) should be compared with the material's allowable shear stress. For structural steel (yield ≈ 250 MPa), the allowable shear stress is typically 100–150 MPa. The angle of twist should be checked against serviceability limits — excessive twist can cause vibration, misalignment, or gear meshing problems. A common rule of thumb is to limit twist to 0.25°–1.0° per meter of shaft length.
Inputs
Results
A 50 mm diameter solid steel shaft (J = π × 50⁴/32 ≈ 613,592 mm⁴) carries 500 N·m torque over 1 m length. The maximum shear stress is 20.4 MPa (well below the ~100 MPa allowable for steel), and the twist is 0.59° per meter — within typical limits.
Inputs
Results
A hollow aluminum tube (outer dia 60 mm, inner dia 40 mm, J ≈ 1,021,018 mm⁴) subjected to 200 N·m over 2 m. Shear stress is only 5.9 MPa, but the twist is 0.86° — aluminum's lower shear modulus means more deformation than steel for the same geometry.
The torsion formula \(\tau = Tr/J\) gives the shear stress at any radial distance \(r\) in a shaft subjected to torque \(T\), where \(J\) is the polar moment of inertia. This assumes a circular cross-section, elastic material behavior, and that cross-sections remain plane after twisting.
The polar moment of inertia \(J\) measures a cross-section's resistance to torsional deformation. For a solid circle: \(J = \pi d^4 / 32\). For a hollow circle: \(J = \pi(d_o^4 - d_i^4)/32\). Larger \(J\) means less stress and less twist for the same applied torque.
The shear modulus \(G\) is a material property. Common values: steel ≈ 79,000 MPa, aluminum ≈ 26,000 MPa, copper ≈ 44,000 MPa, titanium ≈ 44,000 MPa. It relates to Young's modulus and Poisson's ratio by \(G = E / [2(1+\nu)]\). Check material data sheets for precise values.
Allowable twist depends on the application. General machinery shafts: 0.25°–1.0° per meter. Precision shafts (machine tools): less than 0.25°/m. Structural members: often limited by code provisions. Excessive twist causes vibration, noise, and misalignment in gear trains and couplings.
No. The torsion formula \(\tau = Tr/J\) is only valid for circular (solid or hollow) cross-sections. Non-circular sections (rectangular, L-shapes, open thin-walled) experience warping, and the stress distribution is quite different. For those cases, use St. Venant's torsion theory or finite element analysis.
Hollow shafts are significantly more efficient than solid ones. Material near the center of a solid shaft carries very little stress. By removing that underutilized core, a hollow shaft achieves nearly the same torsional strength at a fraction of the weight. For example, a hollow shaft with inner diameter 60% of the outer diameter retains 87% of the torsional strength with only 64% of the material.
Roboculator Team
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