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Welcome to our shaft size calculator, your one-stop site for designing a shaft ideal for your requirements. When designing a shaft, we start with calculating the minimum diameter the shaft must have to withstand various loads. Whether you wish to calculate the diameter of a hollow shaft or a solid shaft, our calculator can assist you.

You would be forgiven for being confused with the shaft design process, for there is much to consider. Fret not — in the article below, we shall discuss the essential fundamentals of calculating shaft design for various loads.

Considerations when calculating shaft design

A shaft is a rotating member that transmits power between two parts through a twisting moment or torque. Machine parts such as gears or pulleys can be mounted onto the shaft to transmit power from or to the shaft.

Most shafts are of circular cross-section. Hence, calculating shaft size or designing a shaft entails the calculation of shaft diameter. While designing a shaft, one has to ensure that the stresses acting on the shaft are within the permissible limits. The stresses one needs to consider for shaft design are:

  1. Shear stress while carrying torque (torsion load).

  2. Bending stress due to members like gears and pulleys, along with the shaft's weight.

  3. Stresses developed due to combined torsional and bending loads.

So we must consider the types of load the shaft will undergo before calculating shaft size. We can base the shaft design on any one of the following:

  1. Shaft transmitting torque only.

  2. Shaft under bending moment only.

  3. Shaft subjected to a combination of twisting and bending moments.

  4. Shaft under fluctuating twisting and bending moments.

  5. Shaft with a desired torsional rigidity.

In the following sections, we shall discuss the shaft design process for each case listed above one by one.

Calculating shaft diameter for torque only

Some shafts are designed for carrying torque alone from one place to another. Any effect from other loads is negligible. Calculation of propeller shaft size could fall into this category.

In such cases, we can use the torsion equation to calculate the shaft size:

TJ=τr(1)\tag{1} \frac{T}{J} = \frac{\tau}{r}


  • TTTwisting moment acting on the shaft;

  • JJPolar moment of inertia about the shaft's rotational axis;

  • τ\tauTorsional shear stress; and

  • rrDistance from the neutral axis to the outermost fiber, equal to the shaft radius for circular cross-section.

r=d2(2)\qquad \tag{2} r = \frac{d}{2}

where dd is the diameter of the shaft.

The polar moment of inertia JJ for a solid circular shaft is given by:

J=π32×d4(3)\tag{3} J = \frac{\pi}{32} \times d^4

🔎 Are you curious why this is the case? Head to our polar moment of inertia calculator.

Substituting (2) and (3) in (1), we get:

T=π16×τd3(4)\tag{4} T = \frac{\pi}{16} \times \tau \cdot d^3

Thus, we can calculate the shaft diameter from the torque it carries and the allowable shear stress.

🙋 Remember that we just calculated the minimum shaft diameter. You can find a standard-sized shaft with a diameter greater than this value.

Let's now calculate the diameter of a hollow shaft. The polar moment of inertia for a hollow shaft is given by:

J=π32[do4di4](5)\tag{5} J = \frac{\pi}{32} \left[ d_o^4 - d_i^4 \right]


  • dod_o — Outer diameter of the hollow shaft; and

  • did_i — Inner diameter of the hollow shaft.

Similarly, rr is given by:

r=do2(6)\tag{6} r = \frac{d_o}{2}

Substituting (5) and (6) in (1), we arrive at:

T=π16×τ[do4di4do](7)\tag{7} T = \frac{\pi}{16} \times \tau \left[ \frac{d_o^4 - d_i^4}{d_o}\right]

Introducing k=di/dok = d_i / d_o, the ratio of inner to the outer diameter, we can simplify this to:

T=π16×τdo3(1k4)(8)\tag{8} T = \frac{\pi}{16} \times \tau \cdot d_o^3 (1-k^4)

So, in addition to torque and the allowable shear stress, we need to know kk, the ratio between inner and outer diameter.

🙋 You can also calculate a shaft's diameter from the power it transmits. This is possible due to the relation between torque and power, given by P=2πNT/60P = 2 \pi NT / 60. To learn more about this torque-power conversion, visit our torque to hp calculator.

Calculating shaft design for bending moment only

Shafts are rarely designed to handle bending moment only but say you are designing a shaft with negligible torsional loads. In such cases, the allowable stress in the shaft is given by the bending equation:

MI=σby(9)\tag{9} \frac{M}{I} = \frac{\sigma_b}{y}


  • MMBending moment acting on the shaft;

  • IIMoment of inertia of the shaft's cross-section;

  • σb\sigma_bBending stress acting on the shaft; and

  • yyDistance from the neutral axis to the outermost fiber.

For a circular cross-section solid shaft, the moment of inertia would be:

I=π64×d4(10)\tag{10} I = \frac{\pi}{64} \times d^4

🔎 Learn more with our moment of inertia calculator.

And yy would be:

y=d2(11)\tag{11} y = \frac{d}{2}

Substituting (10) and (11) in (9), we get the formula to calculate shaft diameter:

M=π32×σbd3(12)\tag{12} M = \frac{\pi}{32} \times \sigma_b \cdot d^3

Now let's calculate the diameter of a hollow shaft. The moment of inertia would be:

I=π64×[do4di4](13)\tag{13} I = \frac{\pi}{64} \times [d_o^4 - d_i^4]

And yy would be:

y=do2(14)\tag{14} y = \frac{d_o}{2}

Substituting (13) and (14) in (9) and using k=di/dok = d_i / d_o, we arrive at:

M=π32×σbdo3(1k4)(15)\tag{15} M = \frac{\pi}{32} \times \sigma_b \cdot d_o^3 (1-k^4)

Calculating shaft design for combined twisting and bending moment

When a combination of twisting and bending moments are acting on a shaft, the combined stresses must be accounted for. Two important theories for calculating these combined stresses are:

  1. Maximum shear stress theory (or Guest's theory) states that the maximum shear stress τmax\tau_{\rm max} in the shaft is given by:
τmax=12σb2+4τ2(16)\qquad \tag{16}\tau_{\rm max} = \frac{1}{2} \sqrt{\sigma_b^2 + 4\tau^2}
  1. Maximum normal stress theory (or Rankine's theory) states that the maximum normal stress σb(max)\sigma_{\rm b(max)} in the shaft is given by:
σb(max)=12(σb+σb2+4τ2)(17)\small \sigma_{\rm b(max)} = \frac{1}{2} \left(\sigma_{\rm b} + \sqrt{\sigma_b^2 + 4\tau^2} \right)(17)

Let's start with a solid shaft. Rewriting equation (4), we get:

τ=16Tπd3(18)\tag{18} \tau = \frac{16 T}{\pi d^3}

Similarly, rewriting equation (12), we get:

σb=32Mπd3(19)\tag{19} \sigma_b = \frac{32 M}{\pi d^3}

Using (18) and (19) in (16), we get:

 ⁣τmax=12(32Mπd3)2+4(16Tπd3)2=16πd3M2+T2\!\small \begin{align*} \tau_{\rm max} &= \frac{1}{2} \sqrt{\left( \frac{32 M}{\pi d^3}\right)^2 + 4\left( \frac{16 T}{\pi d^3}\right)^2}\\[1em] &= \frac{16 }{\pi d^3} \sqrt{M^2 + T^2}\\[1em] \end{align*}
or: M2+T2=π16×τmaxd3\small \begin{align*} \text{or: } \sqrt{M^2 + T^2} &= \frac{\pi}{16} \times \tau_{\rm max} \cdot d^3 \end{align*}

We call the expression M2+T2\sqrt{M^2 + T^2} the equivalent twisting moment TeT_e in the shaft.

Te=M2+T2=π16×τmaxd3(20)\small \tag{20}\begin{split}T_e &= \sqrt{M^2 + T^2}\\[.5em] &= \frac{\pi}{16} \times \tau_{\rm max} \cdot d^3\end{split}

Let's give a similar consideration to the maximum normal stress theory. Using (18) and (19) in (17), we get:

 ⁣σb(max)=12[32Mπd3+(32Mπd3)2+4(16Tπd3)2]=32πd3(12[M+M2+T2])\!\scriptsize \begin{align*} \sigma_{\rm b(max)}\! &= \frac{1}{2}\!\! \left[\frac{32 M}{\pi d^3}\! +\! \sqrt{\!\left( \frac{32 M}{\pi d^3} \right)^2\! + 4\!\left( \frac{16 T}{\pi d^3} \right)^{\!2}\!} \right]\\[1.5em] &= \frac{32}{\pi d^3} \left( \frac{1}{2}\left[ M + \sqrt{M^2 + T^2} \right] \right) \end{align*}

The expression 12[M+M2+T2]\frac{1}{2}\left[ M + \sqrt{M^2 + T^2} \right] is called the equivalent bending moment MeM_e. Rearranging the above equation:

Me=12[M+M2+T2]=π32×σb(max)d3(21)\small \tag{21} \begin{split}M_e &= \frac{1}{2}\left[ M + \sqrt{M^2 + T^2} \right]\\[1em] &= \frac{\pi}{32} \times \sigma_{\rm b (max)} \cdot d^3 \end{split}

🙋 We will get two diameter values — one from equation (20) for TeT_e and one from equation (21) for MeM_e. We must pick whichever value is larger among the two for shaft design so that the shaft can withstand the combined stresses.

For a hollow shaft, the equivalent twisting moment is given by:

Te=M2+T2=π16×τmaxdo3(1k4)(22)\tag{22}\small \begin{split}T_e &= \sqrt{M^2 + T^2}\\ &= \frac{\pi}{16} \times \tau_{\rm max} \cdot d_o^3 (1-k^4)\end{split}

Similarly, the equivalent bending moment is given by:

Me=12[M+M2+T2]=π32×σb(max)do3(1k4)(23)\small \tag{23} \begin{split}M_e &= \frac{1}{2}\left[ M + \sqrt{M^2 + T^2} \right]\\[1em] &= \frac{\pi}{32} \times \sigma_{\rm b (max)} \cdot d_o^3(1-k^4) \end{split}

Again, we will obtain two outer diameter values from equations (22) and (23). We must pick whichever value is larger for our shaft design calculation.