What is the shaft size required to transmit 20 kW at 200 rpm?

Assuming the shaft is mild steel, the minimum required shaft diameter is 48.7 mm . To calculate this result, follow these steps: Calculate the torque acting on the shaft from the power and shaft speed , using T = 60⋅P/(2π⋅n) = (60 × 20 × 10 3 )/(2π⋅200) = 955 N-m . Assume the allowable shear stress of mild steel τ = 42 MPa . Substituting the values in the shaft diameter formula for torsion loads, T = π ⋅τ⋅d 3 /16 , we get: 955 = π × 42 × 10 6 × d 3 /16 . Solving this equation, we get d = 0.487 m = 48.7 mm . Read more

What is the difference between a shaft and an axle?

A shaft is a rotating machine member transmitting torque and power from one part to another. On the other hand, an axle is a stationary member designed to support rotating bodies like car wheels by transmitting bending moment. Read more

Why is torsional rigidity important when designing camshafts?

A camshaft in an internal combustion engine is used for timing valve opening and closing. If the angle of twist exceeds an acceptable amount of 0.25° per meter , the timing of these valves will be affected. Torsional rigidity is thus an essential factor to consider when designing camshafts. Read more

What is the maximum permissible shear stress for a transmission shaft?

According to the American Society of Mechanical Engineers (ASME) code for transmission shaft design, the maximum permissible shear stress is: 56 MPa without allowance for key ways. 42 MPa with allowance for keyways. Read more

Shaft Size Calculator

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:

Shear stress while carrying torque (torsion load).
Bending stress due to members like gears and pulleys, along with the shaft's weight.
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:

Shaft transmitting torque only.
Shaft under bending moment only.
Shaft subjected to a combination of twisting and bending moments.
Shaft under fluctuating twisting and bending moments.
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:

\tag{1} \frac{T}{J} = \frac{\tau}{r}

where:

$T$ — Twisting moment acting on the shaft;
$J$ — Polar moment of inertia about the shaft's rotational axis;
$\tau$ — Torsional shear stress; and
$r$ — Distance from the neutral axis to the outermost fiber, equal to the shaft radius for circular cross-section.

\qquad \tag{2} r = \frac{d}{2}

where $d$ is the diameter of the shaft.

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

\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:

\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:

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

where:

$d_o$ — Outer diameter of the hollow shaft; and
$d_i$ — Inner diameter of the hollow shaft.

Similarly, $r$ is given by:

\tag{6} r = \frac{d_o}{2}

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

\tag{7} T = \frac{\pi}{16} \times \tau \left[ \frac{d_o^4 - d_i^4}{d_o}\right]

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

\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 $k$ , 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 $P = 2 \pi NT / 60$ . 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:

\tag{9} \frac{M}{I} = \frac{\sigma_b}{y}

where:

$M$ — Bending moment acting on the shaft;
$I$ — Moment of inertia of the shaft's cross-section;
$\sigma_b$ — Bending stress acting on the shaft; and
$y$ — Distance from the neutral axis to the outermost fiber.

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

\tag{10} I = \frac{\pi}{64} \times d^4

🔎 Learn more with our moment of inertia calculator.

And $y$ would be:

\tag{11} y = \frac{d}{2}

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

\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:

\tag{13} I = \frac{\pi}{64} \times [d_o^4 - d_i^4]

And $y$ would be:

\tag{14} y = \frac{d_o}{2}

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

\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:

Maximum shear stress theory (or Guest's theory) states that the maximum shear stress $\tau_{\rm max}$ in the shaft is given by:

\qquad \tag{16}\tau_{\rm max} = \frac{1}{2} \sqrt{\sigma_b^2 + 4\tau^2}

Maximum normal stress theory (or Rankine's theory) states that the maximum normal stress $\sigma_{\rm b(max)}$ in the shaft is given by:

\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:

\tag{18} \tau = \frac{16 T}{\pi d^3}

Similarly, rewriting equation (12), we get:

\tag{19} \sigma_b = \frac{32 M}{\pi d^3}

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

\!\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*}

\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 $\sqrt{M^2 + T^2}$ the equivalent twisting moment $T_e$ in the shaft.

\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:

\!\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 $\frac{1}{2}\left[ M + \sqrt{M^2 + T^2} \right]$ is called the equivalent bending moment $M_e$ . Rearranging the above equation:

\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 $T_e$ and one from equation (21) for $M_e$ . 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:

\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:

\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.