Chapter 6: Torsion

6.2.1 Sign Convention and Representation of Torque

In this text counterclockwise torques are considered positive and clockwise torques negative. However, the apparent direction of a torque depends on the perspective from which the rotation is observed. In Figure 6.5, the applied torque would appear clockwise if viewed from the left end of the bar toward the right and counterclockwise if viewed from the right end of the bar toward the left. As long as you are consistent, which perspective is used isn't usually important. However, a typical convention (and the one followed in this text) is to look from the free end of the shaft (if there is one) toward the fixed end, or from the positive end of the longitudinal toward the negative.

To visually determine whether a torque appears clockwise or counterclockwise can sometimes be difficult, so one helpful tool that can be used is the right-hand rule: Curl the fingers of your right hand in the direction of the rotation represented by the torque arrow. If in doing this your thumb points in the direction of the positive longitudinal axis (the x-axis in Figure 6.6), a positive torque is indicated, and if it points away, a negative torque.

For bodies that are sectioned to find internal torques, consider the direction of the torsional moment from the perspective of looking from the cut edge towards the other end of the cut section. Imagine your right hand to be at the cut edge and curl your fingers in the direction of the torque. If your thumb points away from the cut edge, the internal torque is positive, and if your thumb points toward the cut edge, the internal torque is negative. This is illustrated in Figure 6.7.

Finally, since circular arrows can be difficult to draw in a way that clearly conveys direction, double-headed line arrows are frequently used to represent torque. Note that conventions can vary between texts and other sources. This text uses the following convention:

1. On a whole body FBD used to determine external reactions, arrows facing in the direction of the positive longitudinal axis indicate positive torques (counterclockwise) and those facing the opposite direction negative torques.

2. When applying equilibrium, the positive or negative sign from step 1 is used.

3. For FBDs of cut sections used to determine internal torques, positive torques are indicated by arrows pointed away from the section's cut edge and negative torques by arrows pointed toward the cut edge. However, when summing moments, the regular convention of adding right arrows and subtracting left arrows is used.

The direction of the internal torque is unimportant in calculating shear stress (the sign on the stress requires concepts discussed in Chapter 12) but is important when calculating angle of twist, as evident in Section 6.2.

Example 6.1 demonstrates how to determine internal torques in a multisection body as well as how to apply the described sign conventions.

Example 6.1

A multisection steel bar is supported by a fixed support at A and is subjected to torque T_B = 70 N·mm and T_E = 30 N·mm.

Determine the internal torque at all points in the shaft.

Solution

With torques applied only at points B and E (there is also a reaction moment from the wall at A), section AB and section BE are sections of constant internal torque. The reaction torque at A could be obtained by applying equilibrium to the whole shaft, but this example shows how to avoid that.

Section AB:

To determine the internal torque in section AB, we can cut the shaft anywhere between points A and B and draw an FBD of the section to the left of the cut or the section to the right of the cut.

Note that in the following illustration of the cut and the resulting FBDs, the internal torque T_AB is drawn in opposite directions on the two sections. This must be the case if the bar is in equilibrium.

T_AB is drawn as a positive internal torque on both the left and the right section diagrams since in both instances it points away from the cut. We assume that T_AB looks counterclockwise if we view the left section from the cut toward A or the right section from the cut toward E.

For the applied loads, since T_A is drawn in the direction opposite of T_AB on the section to the left of the cut, it is assumed to appear clockwise from the cut toward A.

On the right side diagram, T_B is drawn in the direction opposite of T_AB since it appears clockwise from the cut toward E. Applied torque T_E is drawn in the same direction as T_AB since it appears counterclockwise from the cut toward E.

In writing the equilibrium equation to find T_AB, we use either the left section FBD or the right section FBD and obtain the same result. To avoid needing to find T_A, we will choose to use the right section FBD.

Recall that torques are moments around the longitudinal axis, which is the x-axis in this case. Applying the x-axis moment equilibrium equation to the right section gives

$$\begin{array}{lll}{{\displaystyle \sum }}^{}{M}_x\hfill & =\hfill & -T_{AB}+T_B-T_E=0\hfill \\ {{\displaystyle \sum }}^{}{M}_x\hfill & =\hfill & -T_{AB}+70N\cdot mm-30N\cdot mm=0\hfill \\ T_{AB}\hfill & =\hfill & 40N\cdot mm.\hfill \end{array}$$

Note that in performing the summation, right directed arrows were added and vice versa. The opposite convention can also be used with the same end result.

The positive result for T_AB in this case confirms that the assumed direction for T_AB is the correct one, so it is a counterclockwise internal torque.

Section BE:

Similarly, the internal torque in all parts of the shaft between B and E can be determined by cutting the shaft at a nonspecific point between B and E and applying equilibrium to either the section of shaft to the left of the cut or the section to the right of the cut, as shown in the next two FBDs.

Once again the internal torque (T_BE in this case) is drawn in the direction away from the cut on both sections, so it is assumed to be counterclockwise on both sections.

Applying the moment equilibrium equation on the right section results in

$$\begin{array}{lll}{{\displaystyle \sum }}^{}{M}_x\hfill & =\hfill & -T_{BE}-T_E=0\hfill \\ {{\displaystyle \sum }}^{}{M}_x\hfill & =\hfill & -T_{BE}-30N\cdot mm=0\hfill \\ T_{BE}\hfill & =\hfill & -30N\cdot mm(or30N\cdot mmclockwise).\hfill \end{array}$$

The negative result indicates that T_BE goes in the opposite direction than what was assumed on the FBD, so it is clockwise.

Answer:

Maximum shear stress in the bar is 19.1 kPa and occurs in section CD.

6.4.1 Units

If values for T, L, J, and G are used in Equation 6.2 such that the units are all consistent with each other, the overall unit works out to be unitless, as is expected for an angle calculation. Since the calculation is for an angle, the actual unit is in radians. To obtain an answer in degrees, convert the result of the equation by multiplying it by $\frac{{180}^{\circ }}{\pi }.$

Because the angle of twist often works out to be very small, you may want to express the value in terms of μrad, which is 10⁻⁶ radians. To convert from rad to μrad, multiply the number of radians by 10⁶.

6.4.2 Notation and Relative Angle of Twist

When finding the angle of twist, we encounter the angle at any given point along the length of the shaft relative to some reference point. Often the reference point is a fixed point (e.g., at a fixed support), so the relative angle is an absolute angle. In this case, the symbol ϕ is subscripted with the point along the shaft where the angle was calculated. For example, given the shaft assembly in Figure 6.13, the notation ϕ_Β denotes the angle of twist at point B relative to the fixed point A.

When finding the relative angle of twist between two nonfixed points on the same shaft section (same T, J, and G), we notate the angle of one point relative to another on the same section by using two subscripts specifying the two points. For example, the angle of twist of D relative to C in Figure 6.13 is notated as ${\varphi }_{\text{CD}}$. In this case we perform the calculation for the angle of twist at D as if point C is fixed.

$${\varphi }_{CD}=\frac{T_{CD}{L}_{CD}}{J_{CD}{G}_{CD}}$$

To find the relative angle of twist between two nonfixed points (let's say B and D in Figure 6.13) on different shaft sections (i.e., one or more sections change between them due to changing T, J, or G), sum the angles of twist of each section between them. In this case, the notation would be ϕ_D/B, which is the angle of twist of D relative to B. If point C is where the section change occurs, the calculation would be

$${\varphi }_{D/B}={\varphi }_{BC}+{\varphi }_{CD}.$$

These concepts are illustrated in Example 6.3, which continues with the same bar assembly and loading used in Example 6.1 and Example 6.2.

Example 6.3

The multisection steel bar of Example 6.1 and Example 6.2 is repeated here with the same loading (T_B = 70 N·mm and T_E= 30 N·mm). The diameters are given as d_AC = 50 mm, d_CD = 20 mm, and d_DE = 35 mm, and section lengths as L_AB = 300 mm, L_BC = 200 mm, L_CD = 400 mm, and L_DE = 700 mm. All sections have a shear modulus of G = 40 GPa.

Determine the angle of twist of points C, D, and E, all relative to fixed point A.

A negative result indicates a clockwise twist, and a positive result indicates a counterclockwise twist.

Solution

As was the case when calculating shear stress in Example 6.2, the changes in torque and geometry result in four distinct sections for which we need to calculate angle of twist: AB, BC, CD, and DE. Example 6.1 showed that the internal torque in section AB is counterclockwise and equal to 40 N·mm. It also showed that the internal torque in all sections between B and E is clockwise and equal to 30 N·mm. The directions in this case are from the positive end of the x-axis to the negative (i.e., point E to point A). This means the angle of twist in section AB is positive while the angles of twist in BC, CD, and DE are negative.

Since point A is a fixed point, ${\varphi }_{\mathrm{B}\text{/}\mathrm{A}}={\varphi }_{\mathrm{B}}={\varphi }_{\text{AB}}.$

$${\varphi }_B={\varphi }_{AB}=\frac{T_{AB}{L}_{AB}}{J_{AB}{G}_{AB}}=\frac{0.04N\cdot m\ast \left(0.300m\right)}{\frac{\pi }{32}{(0.05m)}^4\left(40\times {10}^9\frac{N}{m^2}\right)}=4.889\times {10}^{-7}rad$$

To find the angle of twist of point C relative to A, ϕ_C/A, add the angle of twist of section BC to that of section AB.

$$\begin{array}{lll}{\varphi }_{C/A}\hfill & =\hfill & {\varphi }_C={\varphi }_{AB}+{\varphi }_{BC}=4.889\times {10}^{-7}rad+\frac{T_{BC}{L}_{BC}}{J_{BC}{G}_{BC}}\hfill \\ \hfill & =\hfill & 4.889\times {10}^{-7}rad+\frac{-0.03N\cdot m\ast (0.200m)}{\frac{\pi }{32}{(0.05m)}^4(40\times {10}^9\frac{N}{m^2})}\hfill \\ \hfill & =\hfill & 2.445\times {10}^{-7}rad\hfill \end{array}$$

Note that T_BC was substituted as a negative value because it is clockwise.

To find the angle of twist of point D relative to A, ${\varphi }_{\text{D/A}}$, the angle of twist of section CD is added to that of AB and BC.

$$\begin{array}{lll}{\varphi }_{D/A}\hfill & =\hfill & {\varphi }_D={\varphi }_{AB}+{\varphi }_{BC}+{\varphi }_{CD}=2.445\times {10}^{-7}rad+\frac{T_{CD}{L}_{CD}}{J_{CD}{G}_{CD}}\hfill \\ \hfill & =\hfill & 2.445\times {10}^{-7}rad+\frac{-0.03N\cdot m\ast (0.400m)}{\frac{\pi }{32}{(0.02m)}^4(40\times {10}^9\frac{N}{m^2})}\hfill \\ \hfill & =\hfill & -1.885\times {10}^{-5}rad\hfill \end{array}$$

Finally, to find the angle of twist of E relative to A, ${\varphi }_{\text{E/A}}$, we add the angle of twist of section DE to that of AB, BC, and CD.

$$\begin{array}{lll}{\varphi }_{E/A}\hfill & =\hfill & {\varphi }_E={\varphi }_{AB}+{\varphi }_{BC}+{\varphi }_{CD}+{\varphi }_{DE}=-1.885\times {10}^{-5}rad+\frac{T_{DE}{L}_{DE}}{J_{DE}{G}_{DE}}\hfill \\ \hfill & =\hfill & -1.885\times {10}^{-5}rad+\frac{-0.03N\cdot m\ast (0.700m)}{\frac{\pi }{32}{(0.035m)}^4(40\times {10}^9\frac{N}{m^2})}\hfill \\ \hfill & =\hfill & -2.242\times {10}^{-5}rad\hfill \end{array}$$

Step-by-Step: Angle of Twist

1. Use equilibrium to determine reaction torques at any supports (if necessary).

2. Cut a section through the body at the point of interest and draw an FBD of everything either to the left or to the right of the cut.

3. Using the FBD from the previous step, write the moment equilibrium equation around the longitudinal axis to determine the internal torque in the chosen section.

4. Determine the segment's angle of twist using $\varphi =\frac{\text{TL}}{\text{JG}}.$

5. The total angle of twist between two points can be determined by summing the angle of twist for each segment between those points. Be careful with signs, as some segments may rotate clockwise, while others rotate counterclockwise.

6.5.1 Method of Superposition

For the case of redundant supports, an alternative is to use the method of superposition. The method of superposition is based on the idea that the effects of individual loads on the deformation of a body can be calculated individually and then added together to obtain the total effect. To apply this method do the following:

1. Draw the FBD of the whole body and write out the relevant equilibrium equations.

2. Choose one of the supports to consider redundant, remove the corresponding reaction torque, and find the total angle of twist from the remaining loads.

3. Put the reaction torque from the redundant support back, remove the applied loads, and then find the total angle of twist due to the replaced reaction torque in terms of the corresponding variable.

4. Sum the angle of twist from step 2 and step 4 and set the total equal to zero (or a given angle as appropriate).

5. Solve the equation resulting from step 4 for the reaction at the redundant support and then use the whole body equilibrium to solve for the reaction at the other support(s).

Example 6.5 demonstrates the use of superposition to solve part a of Example 6.4.

Example 6.4

A shaft assembly consists of a 12 in. long steel section AB (d = 10 in., G = 11,600 ksi) bonded rigidly to a 9 in. long solid brass section BC (d = 7 in., G = 5,800 ksi). The assembly is to be bolted to walls A and C.

If a torque T_B of magnitude 2,000 kip·in. is applied at B, determine the shear stress in the steel section and the brass section after the ends are bolted for each of the following end conditions:

1. Both ends A and C are tightly connected immediately at installation.

2. A slight misalignment exists at end A that necessitates turning that end by 0.5° to lock in.

Solution

To calculate stresses, we need to know the internal torque in each section. To find the internal torque, we need to know at least one of the reaction torques exerted by the walls at A and C. The steps below can be used to find the reaction torques.

• Step 1: Draw the FBD of the whole body and write the relevant equilibrium equation(s).

  

  In this case the only helpful equilibrium equation is $\sum {\mathrm{M}}_{\mathrm{x}}=0.$

  Note that the assumed directions for T_A and T_C (the reaction torques exerted by the bolts) would usually be completely assumed, with the signs of the answers indicating if the assumed directions are correct. For case (a), this is still true.

  For case (b) we must be more careful. The direction of the 0.5° turn is not specified, but it must be in the direction of T_B, since that is the only applied load. The direction of T_A, which is the reaction torque exerted by the bolt at A, should be in the direction opposite to the way the shaft turns since the bolt is there to fix the assembly in place.

  The assumed direction of T_C can be in either direction for both case (a) and case (b).

  

  The positive x-axis goes from right to left in the whole body FBD. As long as the choice of signs applied to torques for any whole body equation is based on the same assumption, it does not matter which direction is chosen to be the positive x direction. In this case, the signs on the torques will be based on the way the torques are perceived when looking in the direction from A towards C.

  Since the direction of T_B is counterclockwise from that perspective, it is drawn in the positive x direction. T_A is drawn in the negative x direction. T_C is arbitrarily assumed to be the same direction as T_A. Write the moment equilibrium equation.

  $$\begin{array}{l}\sum M_x=-T_A+T_B-T_C=0\\ T_C+T_A=T_B\end{array}$$

  Because there are two unknowns (T_A and T_C) but only one equilibrium equation, the problem is statically indeterminate.

• Step 2: Apply kinematic constraint.

  Applying the kinematic constraint will provide the second equation needed to solve for T_A and T_C. For case (a), both ends are fixed at the time of installation so the total angle of twist should be zero. Thus

  $${\varphi }_A=\sum \frac{TL}{JG}={\varphi }_{AB}+{\varphi }_{BC}=0.$$

  For case (b), the misalignment means end A must be turned by 0.5° before it gets locked in, so the magnitude of the total angle of twist is also 0.5°. In step 1, it was discussed that the twist will be in the same direction as T_B, which is counterclockwise looking from A toward C.

  So for case (b), the kinematic equation is

  $${\varphi }_A=\sum \frac{TL}{JG}={\varphi }_{AB}+{\varphi }_{BC}={0.5}^{\circ }\ast \frac{\pi }{{180}^{\circ }}.$$

• Step 3: Write the compatibility equation.

  Write the compatibility equation by substituting the appropriate torque-twist equations for the angles of twist for the angles of twist in the kinematic constraint equation.

  To implement this step, establish the internal torque (and sign) for each section by cutting sections and applying equilibrium.

  The FBD for a section made by cutting through section AB is as shown.

  

  The equilibrium equation for this section is

  $$\begin{array}{c}{{\displaystyle \sum }}^{}{M}_x=T_A+T_{AB}=0\\ T_{AB}=-T_A.\end{array}$$

  Thus, the angle of twist in section A_B is

  $${\varphi }_{AB}=-\frac{T_{AB}{L}_{AB}}{J_{AB}{G}_{AB}}=\frac{-T_A\ast (12in.)}{\frac{\pi }{32}{(10\ast in.)}^4\ast (11,600ksi)}.$$

  The FBD for a section made by cutting through section BC is as shown.

  

  The equilibrium equation for this section is

  $$\begin{array}{c}{{\displaystyle \sum }}^{}{M}_x=-T_{BC}+T_C=0\\ T_{BC}=T_C.\end{array}$$

  The angle of twist in section BC is

  $${\varphi }_{BC}=+\frac{T_{BC}{L}_{BC}}{J_{BC}{G}_{BC}}=\frac{T_C\ast (9in.)}{\frac{\pi }{32}{(7in.)}^4\ast (5,800ksi)}.$$

  Substituting the expressions for ϕ_AB and ϕ_BC into the kinematic constraint equation for case (a) gives the compatibility equation

  $$\frac{-T_A\ast (12in.)}{\frac{\pi }{32}{(10in.)}^4\ast (11,600ksi)}+\frac{T_C\ast (9in.)}{\frac{\pi }{32}{(7in.)}^4\ast (5,800ksi)}=0.$$

  Case (b) gives the compatibility equation

  $$\frac{-T_A\ast (12in.)}{\frac{\pi }{32}{(10in.)}^4\ast (11,600ksi)}+\frac{T_C\ast (9in.)}{\frac{\pi }{32}{(7in.)}^4\ast (5,800ksi)}={0.5}^{\circ }\ast \frac{\pi }{{180}^{\circ }}.$$

• Step 4: Solve the equilibrium and compatibility equations simultaneously for T_A and T_C.

  • Case (a):

    $$\begin{array}{l}{T}_A=T_C\left[\frac{9in.\ast \frac{\pi }{32}{(10in.)}^4\ast (11,600ksi)}{12in.\ast \frac{\pi }{32}{(7in.)}^4\ast (5,800ksi)}\right]\\ T_A=T_C[6.247]\\ \text{Substitute}\text{into}{T}_A+T_C=2,000kip\cdot in.\\ 6.247T_C+T_C=2,000kip\cdot in.\\ \\ T_C=276kip\cdot in.\\ T_A=1,724kip\cdot in.\end{array}$$

  • Case (b):

    $$\begin{array}{l}{T}_A=\left(\frac{T_C\ast (9in.)}{\frac{\pi }{32}{(7in.)}^4\ast (5,800ksi)}-{0.5}^{\circ }\ast \frac{\pi }{{180}^{\circ }}\right)\ast \left(\frac{\frac{\pi }{32}{(10in.)}^4\ast 11,600ksi}{12in.}\right)\\ T_A=T_C\left[6.247\right]-8,282kip\cdot in.\\ \text{Substitute}\text{into}{T}_A+T_C=2,000kip\cdot in.\\ 6.247T_C+T_C-8,282kip\cdot in.=2,000kip\cdot in.\\ T_C=1,419kip\cdot in.\\ T_A=581kip\cdot in.\end{array}$$

    The results are positive for both cases, so the assumed directions are the actual directions.

• Step 5: Calculate stresses.

  Now that the reaction torques are known and it was established in step 3 that T_AB = T_A and T_BC = T_C in magnitude, the shear stresses in the two sections can be calculated using Equation 6.1.

  • Case (a):

    $$\begin{array}{l}{\tau }_{AB}=\frac{T_{AB}\ast c_{AB}}{J_{AB}}=\frac{1,724kip\cdot in.\ast 5in.}{\frac{\pi }{32}{(10in.)}^4}=8.78ksi\\ {\tau }_{BC}=\frac{T_{BC}\ast c_{BC}}{J_{BC}}=\frac{276kip\cdot in.\ast 3.5in.}{\frac{\pi }{32}{(7in.)}^4}=4.10ksi.\end{array}$$

  • Case (b):

    $$\begin{array}{l}{\tau }_{AB}=\frac{T_{AB}\ast c_{AB}}{J_{AB}}=\frac{581kip\cdot in.\ast 5in.}{\frac{\pi }{32}{(10in.)}^4}=2.96ksi\\ {\tau }_{BC}=\frac{T_{BC}\ast c_{BC}}{J_{BC}}=\frac{1,419kip\cdot in.\ast 3.5in.}{\frac{\pi }{32}{(7in.)}^4}=21.1ksi.\end{array}$$

Example 6.5

Solve part (a) of Example 6.4 using the method of superposition. The problem is repeated here for convenience.

A shaft assembly consisting of a 12 in. long steel section AB (d = 10 in., G = 11,600 ksi) is bonded rigidly to a 9 in. long solid brass section BC (d = 7 in., G = 5,800 ksi). The assembly is bolted to walls A and C.

If both ends A and C are tightly connected to the walls and a torque T_B of magnitude 2,000 kip·in. is applied at B, determine the shear stress in the steel section and the brass section.

Solution

The first step to solving (global equilibrium) remains the same. The method of superposition is applied in step 2 and step 3.

• Step 1: Draw the FBD of the whole body and write out the relevant equilibrium equations.

  Draw the whole body FBD and write out the relevant equilibrium equations. Shown on the FBD are the assumed directions for T_A and T_C as well as the direction used for the positive x direction. Recall that the direction of the torque should be determined by looking from the positive x-axis toward the negative (A to C).

  

  Applying equilibrium in this case reduces to the sum of the moments about the x-axis as the only nontrivial equation.

  $$\begin{array}{c}\sum M_x=-T_A+T_B-T_C=0\\ T_C+T_A=T_B\end{array}$$

  Because there are two unknowns (T_A and T_C) but only one equilibrium equation, the problem is statically indeterminate.

• Step 2: Choose one of the supports to consider redundant, remove the corresponding reaction torque, and then find the total angle of twist from the remaining loads.

  

  Let's consider the support at C to be redundant. Note that the support at A could alternatively be considered redundant with the same results.

  With no support at C, a section cut within BC shows that the internal torque T_BC must be 0 for equilibrium.

  A section cut through AB shows that the internal torque T_AB = T_B with the direction of T_AB opposite of T_B. Viewed from the cut towards C, T_B is counterclockwise, so T_AB is clockwise. Therefore, the angle of twist is negative and equal to

  $${\varphi }_{C/A}=\sum \frac{TL}{JG}=\frac{T_{AB}{L}_{AB}}{J_{AB}{G}_{AB}}=\frac{-(2,000kip\cdot in.)\ast (12in.)}{\frac{\pi }{32}{(10in.)}^4\ast (11,600ksi)}=-2.107\times {10}^{-3}rad.$$

• Step 3: Put the reaction torque from the redundant support back, remove the applied loads, and then find the total angle of twist due to the replaced support torque in terms of the corresponding variable.

  

  The section FBDs in this case reveals that both T_AB and T_BC are equal in magnitude and opposite in direction from T_C. Since both are pointed away from the cut, the direction is taken to be counterclockwise, or positive. So the angle of twist with this loading is

  $$\begin{array}{lll}{\varphi }_{C/A}\hfill & =\hfill & {{\displaystyle \sum }}^{}\frac{TL}{JG}=\frac{T_{AB}{L}_{AB}}{J_{AB}{G}_{AB}}+\frac{T_{BC}{L}_{BC}}{J_{BC}{G}_{BC}}\hfill \\ \hfill & =\hfill & \frac{T_C\ast (12in.)}{\frac{\pi }{32}{(10in.)}^4\ast (11,600ksi)}+\frac{T_C\ast (9in.)}{\frac{\pi }{32}{(7in.)}^4\ast (5,800ksi)}\hfill \\ \hfill & =\hfill & 7.637\times {10}^{-6}{T}_C.\hfill \end{array}$$

• Step 4: Set the sum of the angles of twist equal to 0 (or other fixed angle when applicable) and solve for the unknown reaction torque.

  The total angle of twist should be equal to zero when the ends are tightly bolted to the walls without extra twisting.

  $$2.107\times {10}^{-3}rad+7.637\times {10}^{-6}Tc=0$$

  Solving for T_C gives T_C = 276 kip·in which is the same result found in Example 6.4.

• Step 5: Use the whole body equilibrium equation from step 1 to solve for the other reaction torque.

  Substituting the above result for T_C into the whole body equilibrium equation from step 1 gives T_A = 1,724 kip in as was found in Example 6.4. Once again the corresponding shear stresses are

  $$\begin{array}{l}{\tau }_{AB}=\frac{T_{AB}\ast c_{AB}}{J_{AB}}=\frac{1,724kip\cdot in.\ast 5in.}{\frac{\pi }{32}{(10in.)}^4}=8.78ksi\\ {\tau }_{BC}=\frac{T_{BC}\ast c_{BC}}{J_{BC}}=\frac{276kip\cdot in.\ast 3.5in.}{\frac{\pi }{32}{(7in.)}^4}=4.10ksi.\end{array}$$

Example 6.6

The vertical shaft assembly is fixed to the bottom surface and consists of a solid brass (G = 40 GPa) circular shaft completely bonded in section BC to a steel (G = 80 GPa) reinforcing tube. The diameter of the brass is 25 mm along the whole length, and the outer diameter of the steel is 50 mm. The assembly is subjected to torques T_B = 800 N·m and T_A = 300 N·m in the directions shown.

Determine the magnitude of the shear stress in the brass in sections AB and BC.

Solution

First we'll find the shear stress in the brass in section BC. To begin, we determine how much of the internal torque within that section is supported by the brass versus how much is supported by the steel.

• Step 1: Draw the FBD of a section cut through BC to examine the internal torque in BC.

  Since the reaction at C is not strictly necessary to solve this problem, the whole body FBD is not needed here. Start by cutting a section through BC and drawing the FBD. Note that instead of representing the internal reaction at the cut as T_BC, it is split into the components that make up T_BC: T_Br and T_St.

  

  With only one nontrivial equilibrium equation and two unknowns, the problem is statically indeterminate.

  $$\sum M_x=T_A-T_B-T_{Br}-T_{St}=0$$

• Step 2: Apply the kinematic constraint to obtain a second equation.

  In this case, the materials are fully bonded in section BC, so the materials are constrained to twist as a unit in that section. In BC, the kinematic equation is

  $${\varphi }_{Br}={\varphi }_{st}.$$

• Step 3: Substitute the torque-twist equation into the kinematic equation to get the compatibility equation.

  Substituting Equation 6.2 into the kinematic constraint equation gives us a second equation that relates T_Br and T_St.

  $$\frac{T_{Br}{L}_{Br}}{J_{Br}{G}_{Br}}=\frac{T_{St}{L}_{St}}{J_{St}{G}_{St}}$$

  The lengths are equal and therefore cancel each other. Rearrange this equation to get an expression for T_St in terms of T_Br.

  $$\begin{array}{c}{T}_{St}=\frac{T_{Br}{J}_{St}{G}_{St}}{J_{Br}{G}_{Br}}\hfill \\ =T_{Br}\left[\frac{\frac{\pi }{32}\left({(0.05m)}^4-{(0.025m)}^4\right)\left(80\times {10}^9\frac{N}{m^2}\right)}{\frac{\pi }{32}{(0.025m)}^4\left(40\times {10}^9\frac{N}{m^2}\right)}\right]\hfill \\ =30.0T_{Br}\hfill \end{array}$$

• Step 4: Use the equilibrium equation and the simplified compatibility relationship to solve for T_Br.

  Substite T_st = 30 T_Br into the equilibrium equation.

  $$\begin{array}{l}300N\cdot m-800N\cdot m-T_{Br}-30T_{Br}=0\\ T_{Br}=-16.1N\cdot m\end{array}$$

• Step 5: Apply the torque-shear stress equation to find the shear stress in the brass for sections AB and BC.

  Note that because only the magnitude of the stress is requested, we substitute torque values in as positive values.

  Applying Equation 6.1 in section BC yields

  $${\tau }_{Br\left(BC\right)}=\frac{T_{Br}\ast c_{Br}}{J_{Br}}=\frac{16.1N\cdot m\ast (0.0125m)}{\frac{\pi }{32}{(0.025m)}^4}=5.26MPa.$$

  In section AB, which consists only of brass, all the internal torque is carried by the brass. Making a cut within that section and applying equilibrium shows that the magnitude of the internal torque T_AB = T_A = 300 N·m.

  

  Applying the shear stress equation yields

  $${\tau }_{Br\left(AB\right)}=\frac{T_{Br}\ast c_{Br}}{J_{Br}}=\frac{300N\cdot m\ast (0.0125m)}{\frac{\pi }{32}{(0.025m)}^4}=97.8MPa.$$

Step-by-Step: Statically Indeterminate Structures

1. Draw the whole body FBD and apply whichever equilibrium equations are relevant given the loading and problem geometry. For this chapter's pure torsion problems, only the moment equilibrium about the longitudinal axis is useful. All other equilibrium equations are identically 0.

2. Write the equation of kinematic constraint or the equation that expresses the restriction on the deformation. These equations would generally be

   • $\sum \varphi =0$ if a shaft assembly is tightly supported by fixed supports on the ends.

   • $\sum \varphi =\text{angle}$ if shaft assembly is supported by fixed supports on both ends but one is loose, misaligned, and so on such that the end to twist has some degree of freedom.

   • ${\varphi }_1={\varphi }_2$, where 1 and 2 represent two different shafts bonded together co-axially.

3. Write the compatibility equation by substituting the appropriate torque-twist relationship ($\varphi =\frac{\text{TL}}{\text{JG}}$) into the kinematic constraint equation.

4. You should now have the same number of equations as unknowns. Solve the equations.

6.6.1 Derivation of Power-Torque Relationship

Recall from other subjects of study that power is defined to be the time rate of change of work. The definition of work is force times distance. So power can be calculated as

$$P=\frac{W}{t}=\frac{F\times d}{t}.$$

Within the context of rotating shafts, the force is applied to the shaft's outer edge resulting in torque, and the distance is the arc length of a point on the shaft's edge that travels around the circumference in a given amount of time, t. Figure 6.14 illustrates this point.

If the arc length over which point B travels is ${\text{BB}}^{\prime }=\mathrm{r}\theta \text{then}\mathrm{W}=\mathrm{F}(\text{rθ})$. Since torque is equal to (F)(r), we can see that W = Tθ. Substituting this expression for work into the power equation above, we then have $\mathrm{P}=\frac{\mathrm{T}\theta }{\mathrm{t}}.\text{The}\frac{\theta }{\mathrm{t}}$ term is angular velocity, which is denoted here with the Greek letter ω. So the power equation can finally be expressed as

$$\boxed{P=T\omega },$$(6.3)

where

• P = Power [Watts, in.-lb/sec, horsepower]

• T = Torque [N·m, lb·in.]

• $\omega$ = Angular velocity [rad/s, rad/sec]

6.6.3 Torque Transfer Between Connected Gears (Gear Ratio)

Consider two connected gears as shown in Figure 6.15. The interaction of the teeth results in a force applied to each gear from one gear to the other. For equilibrium, this force must be equal in magnitude but opposite in direction. The force results in a torque that can be expressed as T = Fr.

Given that the force on each gear is equal in magnitude to the force on the other, we can write

$$F=\frac{T_A}{r_A}=\frac{T_B}{r_B},$$

which then allows us to express the torque exerted by one gear in terms of the torque exerted by the other as well as the ratio of radii.

$$\boxed{T_A=\frac{r_A}{r_B}{T}_B}$$(6.4)

The ratio of radii is known as the gear ratio. It can be expressed as the ratio of gear diameters or the number of teeth on each gear.

While the forces that result in torque are opposite in direction for each gear of the connected pair, the torques themselves are in the same direction. In Figure 6.15, the rightwards force gear A exerts on gear B results in a clockwise torque on gear B. Likewise, the leftward force gear B exerts on gear A results in a clockwise torque on gear A.

6.6.4 Angular Velocity Ratio of Connected Shafts

When two shafts are connected by a pair of gears, conservation of energy dictates that the power transferred into one shaft equals the power transferring out of the other (power in = power out, assuming losses are negligible).

For two connected gears A and B, if P_A = T_Aω_A and P_B = T_Bω_B and P_A = P_B, then ${\omega }_{\mathrm{A}}=\frac{{\mathrm{T}}_{\mathrm{B}}}{{\mathrm{T}}_{\mathrm{A}}}{\omega }_{\mathrm{B}}$.

Substituting in Equation 6.4 gives us

$$\boxed{{\omega }_A=\frac{r_B}{r_A}{\omega }_B}.$$(6.5)

Note that the gear ratio in this case is inverse of the one used to relate torques.

6.6.5 Angle of Twist

When two connected gears, each attached to a separate shaft, rotate they travel the same distance for a given amount of shaft rotation, or angle of twist. The distance of travel is equal to the arc length r⋅ϕ. Given that the gears exert forces in opposite directions (as illustrated in Figure 6.15), the direction of travel of each gear will also be opposite of the other, as depicted in Figure 6.16. For connected gears A and B, attached to shafts A and B respectively this can be expressed as

$$r_A{\varphi }_A=-r_B{\varphi }_B.$$

where ${\varphi }_{\mathrm{A}}$ is the angle of twist of shaft A and ${\varphi }_{\mathrm{A}}$ is the angle of twist of shaft B. Rearranging this relationship gives

$$\boxed{{\varphi }_A=-\frac{r_B}{r_A}{\varphi }_B}.$$(6.6)

As with the angular velocities, the gear ratio is the inverse of that used to relate torques.

Example 6.7 presents a problem of power transmission with connected gears that utilizes the above concepts.

Example 6.7

Power of 60 kW is transmitted at 33 Hz to gear D of the assembly. Shaft CD is allowed to turn freely within its bearings in the direction shown at D. End A of shaft AB is attached rigidly to the wall.

Determine the angle of twist of point D relative to A.

Assume gear C has 35 teeth (N_C = 25) around its circumference and gear B has 50 teeth (N_B= 50). In addition, shaft AB (d = 40 mm) is 450 mm long and shaft CD (d = 30 mm) is 700 mm long. Both shafts are made of steel (G = 80 GPa).

Solution

In general, think of the angle of twist at D as equal to whatever the angle of twist of point C is plus the angle of twist of shaft CD.

$${\varphi }_{D/A}={\varphi }_C+{\varphi }_{CD}$$

With the connection of gear C to gear B, we have

$${\varphi }_C=-\frac{r_B}{r_A}{\varphi }_B=-\frac{N_B}{N_A}{\varphi }_B,$$

so

$${\varphi }_{D/A}=-\frac{N_B}{N_A}{\varphi }_B+{\varphi }_{CD}.$$

With N_B and N_A given, we need to find ${\varphi }_{\mathrm{B}}$ and ${\varphi }_{\text{CD}}$.

1. Given the power information, calculate the torque being applied at D. When considering units, recall that a Watt is an N m/s and a Hz is a revolution per second.

   $$T_D=\frac{P_D}{\omega }=\frac{60,000W}{33Hz\left(\frac{2\pi \text{rad}}{Hz}\right)}=289.4N\cdot m$$

2. Given the torque at D, calculate the torque at C by applying equilibrium to shaft CD.

   

   $$\begin{array}{c}{{\displaystyle \sum }}^{}{M}_x=T_D-{\mathrm{T}}_C=0\\ {\mathrm{T}}_C=T_D\end{array}$$

Looking from D toward C, note that the given direction of T_D is counterclockwise, so T_C = 289.4 N·m (clockwise).

3. Use the gear ratio to relate the torque exerted on gear C to the torque exerted on gear B. Recall from section 6.6.3 that T_B will be the same direction as T_C.

   $$T_B=\frac{N_B}{N_C}{T}_c=\frac{50}{35}(289.4N\cdot m)=413.4N\cdot m\left(\text{clockwise}\right)$$

4. Calculate ϕ_B = ϕ_AB (since point A is fixed).

We can see from applying equilibrium to the FBD of a section cut through shaft AB that the internal torque T_AB = T_B (clockwise). This means that T_AB results in a clockwise (negative) angle of twist ϕ_AB.

$${\varphi }_B=\frac{T_{AB}{L}_{AB}}{J_{AB}{G}_{AB}}=-\frac{413.4N\cdot m\ast (0.450m)}{\frac{\pi }{32}{(0.04m)}^4\left(80\times {10}^9\frac{N}{m^2}\right)}=-0.009252rad$$

5. Determine and use the internal torque T_CD to determine ϕ_CD.

   Cutting a section through shaft CD and drawing the FBD shows that T_CD = T_D (counterclockwise), so

   

   $${\varphi }_{CD}=\frac{T_{CD}{L}_{CD}}{J_{CD}{G}_{CD}}=\frac{289.4N\cdot m\ast (0.700m)}{\frac{\pi }{32}{(0.03m)}^4\left(80\times {10}^9\frac{N}{m^2}\right)}=0.03184rad.$$

6. Calculate ${\varphi }_{\text{D/A}}=-\frac{{\mathrm{N}}_{\mathrm{B}}}{{\mathrm{N}}_{\mathrm{A}}}{\varphi }_{\mathrm{B}}+{\varphi }_{\text{CD}}.$

   $${\varphi }_{D/A}=-\frac{50}{35}(-0.009252rad)+0.03184rad=0.04506rad$$

Step-by-Step: Power Transmission

1. Use P = Tw to find power, torque, or angular velocity depending on the specific problem.

2. Find internal torques by cutting sections and applying equilibrium.

3. If the problem involves connected gears (A and B), and the torque or angle of twist of one shaft is already known, determine the torque and/or angle of twist for the other shaft by using ${\mathrm{T}}_{\mathrm{A}}=\frac{{\mathrm{r}}_{\mathrm{A}}}{{\mathrm{r}}_{\mathrm{B}}}{\mathrm{T}}_{\mathrm{B}}\text{and}{\varphi }_A=-\frac{{\mathrm{r}}_{\mathrm{B}}}{{\mathrm{r}}_{\mathrm{A}}}{\varphi }_{\mathrm{B}}$.

4. Calculate the total angle twist for assemblies with multiple shafts by summing angles of twist for each shaft individually, accounting for the effect of connected gears using Equation 6.7, and then sum the results together.

Summary

Key Takeaways

Torque is a moment applied around the longitudinal axis of a body. The corresponding twisting deformation is quantified by the angle of twist. In a circular cross-section, the cross-section remains plane after deformation.

Torsional loading results in shear stress. Both stress and strain vary from 0 at the center of a circular cross-section to a maximum value on the outer edge of the cross-section.

Internal torques can be determined by cutting a section through the member, drawing an FBD of the cut section, and applying equilibrium equations.

Internal torques are considered positive if they act counterclockwise when viewed from the positive end of a section toward the negative. For cut sections, the direction away from the cut edge of a body is considered the positive direction.

Sign for shear stress has not been discussed in this text yet (this will come later), but the sign of the angle of twist is important. Many problems involve multiple segments experiencing different stresses and deformations. The stress and angle of twist may be determined for each segment independently, but we need to sum deformations across all the segments to determine the total angle of twist. Therefore the correct signs must be maintained.

One common application where torques are present is power transmission. We usually know the amount of power and the angular velocity but need to determine the torque. Problems involving power transmission often involve shafts with connected gears. These gears alter the torque and angle of twist according to the ratio of their radii. If we know the torque and angle of twist at one gear, we can determine the torque and angle of twist of the connected gear and attached shaft.

Key Equations

Torsional stress:

$$\tau =\frac{T\rho }{J}$$

$${\tau }_{max}=\frac{Tc}{J}$$

Angle of twist:

$$\varphi =\sum \frac{TL}{JG}$$

Relationship between power and torque:

$$P=T\omega$$

Transfer of torque, angular velocity, and angle of twist between connected gears:

$$T_A=\frac{r_A}{r_B}{T}_B$$

$${\omega }_A=\frac{r_B}{r_A}{\omega }_B$$

$${\varphi }_A=-\frac{r_B}{r_A}{\varphi }_B$$