14 Combined Loads
Introduction
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In this chapter, we will use of all of the previously discussed concepts to determine the total normal stress and the total shear stress for a body subjected to various types of loading simultaneously. Axial forces (?sec-axial-loads), bending moments (Chapter 9), and pressure (Chapter 13) may be applied in various combinations to give total normal stress. Transverse loads (Chapter 10) in multiple directions and torsion (Chapter 6) can combine to give total shear stress.
As part of understanding which loads lead to which types of stresses and in which directions, it will be important to have a good understanding of coordinate axes. In particular, we will want to be able to distinguish between the longitudinal axis and the cross-sectional axes. This can change from body to body as different coordinate axes are used, but it can also change within an individual body based on changes in geometry. In Section 14.1, we will review how to distinguish between the coordinate axes as well as define generic terms such that we can express equations in a more generalalized manner.
In Section 14.2, the various sources of normal stress are reviewed and expanded on. They will be re-framed in terms of general axes so as to clarify which other stresses they may combine with. In Section 14.2.1 we will revisit bending stress to express the unsymmetric bending stress equation in terms of general cross-sectional axes. In Section 14.2.2, eccentric loading, which is a specific instance of combined normal stress in which normal forces result in bending, will be considered.
In Section 14.3, we will focus on combining the various sources of shear stress to determine the total shear stress. To expand on the concepts discussed and used in Chapter 6 and Chapter 10, we will perform calculations based on both possible transverse loading directions as well as torsion. As in Section 14.2, this will be done with a general set of coordinate axes.
In Section 14.4, we will further expand the above cases to situations in which all combinations of loading are possible and the general stress state is determined.
Note that the determination of the internal reactions for the general problems presented in this chapter will require the use of 3D equilibrium equations, particularly for determining bending moments and torsion. You may find it helpful to review Section 1.3 to understand these calculations.
14.1 Coordinate Axes and Directions
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To generalize the discussion for any set of coordinate axes, we will use generic terms for the various directions, as shown in Figure 14.1. The vertical axis of the cross-section will be referred to as the V axis, the horizontal axis of the cross-section will be referred to as the H axis, and the longitudinal (axial) direction will be referred to as the L axis.
The longitudinal direction, synonymous with axial direction, is the direction perpendicular to a cross-section which is cut at a point of interest on a body. The cross-sectional axes are the axes one would draw on the cross-section once the cut is made.
These directions could change from point to point depending on the location of the point and the geometry of the body. For example, consider the structure in Figure 14.2 with the x, y, z axes oriented at the origin as shown. If we want to examine point P on the structure, we make a vertical cut at point P. The direction normal to the cut surface is the x-direction, so the x axis is the longitudinal axis. When you view the cut surface looking down the x-axis, you see the y and z axes on the cross-section, so those are the cross-sectional axes.
On the other hand, for point Q, we must make a horizontal cut to examine that point. The normal direction then is the vertical, or yaxis. and the cross-sectional axes are the x and z axes.
In the following sections, we will express equations in terms of these generic directions so that they will apply to any orientation of coordinate axes.
14.2 Combined Normal Stress
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In previous chapters, individual sources of longitudinal normal stress were considered: axial loads (@sec-axial-loading), bending loads (Chapter 9), and pressure loads (Chapter 13). With all three of these forms of loading present, we would calculate the total longitudinal stress as:
\[ \sigma_L =\frac{\sum F_L}{A}+\sigma_{bending}+\frac{p r}{2 t} \]
where:
𝜎L = total normal stress in longitudinal direction {Pa, psi]
FL = force in the longitudinal direction {N, lb]
A = cross-sectional area [m2, in2]
𝜎bending = bending stress (discussed in more detail below) [Pa, psi]
p = internal pressure [Pa, psi]
r = inner radius of pipe or pressure vessel [m, in]
t = wall thickness of pipe or pressure vessel [m, in]
If there is pressure loading, in addition to the contribution to the longitudinal normal stress indicated above, there will also be normal stress in the cross-sectional directions equal to the hoop stress.
\[\sigma_V=\sigma_H=\sigma_{hoop}=\frac{pr} {t}\]
Of course, for problems with no pressure, the pressure terms would be excluded.
Of the three longitudinal normal stresses, bending stresses tend to be the most complex to understand in terms of calculation inputs and signs, especially with varying coordinate axes directions. Section 14.2.1 re-visits the concept of bending stress in order to discuss how to generalize the equations presented in Chapter 9 to any orientation of body with any arrangement of coordinate axes.
14.2.1 Generalized Bending Stress Equation
In Section 9.1 , an equation for bending stress in the x direction based on bending around the cross-sectional axes y and z was derived. That equation is:
\[ \sigma_x=-\frac{M_z y}{I_z}+\frac{M_y z}{I_y} \]
This form of the equation, with the indicated sign in front of each term, is specific to the cross-sectional axes oriented with the z axis positive left and the y axis positive upwards. The negative in front of the Mz term and the positive in front of the My term are part of the derived equation and when the correct signs for Mz, My, y, and z are inserted, the correct sign of the stress is obtained (positive for tensile and negative for compressive). In Chapter 9, no other cross-sectional axes were considered so that focus on the general concept of bending stress could be more easily maintained.
When the cross-sectional axes are not y and z, we apply the same general form of the above bending stress equation. However, for a different set of axes, not only do we need to change the bending moment and moment of inertia axes in the equation, we need to reconsider the signs. In any given bending load situation, the bending moments are the moments around the centroidal axes of the cross-section (V and H) and the resulting normal stress is in the longitudinal direction (L).
We can then re-phrase the bending stress equation as:
\[ \boxed{\sigma_{(bending)}= \pm\left|\frac{M_V\ d_H}{I_V}\right| \pm\left|\frac{M_H\ d_V}{I_H}\right|} \tag{14.1}\]
𝜎(bending) = Bending stress in the longitudinal direction [Pa, psi]
MV = Bending moment around the vertical axis [N⸱m, lb⸱in.]
MH = Bending moment around the horizontal axis [N⸱m, lb⸱in.]
dH = Horizontal distance between the point of interest on the cross-section and the V axis [m, in.]
dV = Vertical distance between the point of interest on the cross-section and the H axis [m, in.]
IV = the moment of inertia about the vertical axis [m4, in.4]
IH = Area moment of inertia about the horizontal axis [m4, in.4]
The absolute values are used for the bending stress terms with the ⨦ in front to indicate that we will calculate the magnitude of each bending stress term and then assess the signs of the terms independently. The assessment of the signs will be done by visualizing the bending direction, or by using visualization tools which will be described below. Recall that compressive stresses are considered negative and tensile stresses are positive.
Figure 14.3 illustrates the distances dH and dV for the case where the stress at the example point P on the cross-section is being calculated.
It will also be helpful to note that the bending stress at a point that lies directly on a centroidal axis will be zero for bending around that axis. This is evident from the fact that dH =0 for a point on the V axis and dV =0 for a point on the H axis.
14.2.2 Assessing Bending Stress Signs For Generalized Bending Stress Equation
The sign on each of the two bending stress terms in Equation 14.1 needs to be determined by inspection of the bending direction. In Section 9.1, we saw that a bending moment around the horizontal axis causes the normal stress to vary along the vertical axis of the cross-section. In Section 9.3, we saw that a bending moment around the vertical axis causes normal stress to vary across the horizontal axis of the cross section. Which side is in tension versus compression in each case depends on the direction (clockwise versus counterclockwise) of the bending moment as well as the orientation of the body.
If we can visualize the expected bending due to loading, we can assess if a point will be the “inside” of the bend (compression) or on the “outside” of the bend (tension). For example, in Figure 14.4, the rightwards horizontal force results in bending around the V axis of the cross-section. The direction of the bending is such that all points to the right of the V axis on the cross-section are on the inside of the bend (compression) and all points to the left of the V axis on the cross-section are on the outside of the bend (tension).
In Figure 14.5, the force in the V direction results in a bending moment around the H axis. The direction of the bending is such that all points above the H axis on the cross-section are on the inside of the bend (compression) and all points below the H axis on the cross-section are on the outside of the bend (tension).
If visualization is difficult, one can use the following tool (refer to Figure 14.6):
- Draw the cross-section as viewed from the end towards the origin. For example, in the case of the columns in Figure 14.4 and Figure 14.5, we would draw the top-down view of the cross-section.
- Draw the bending moment arrows around the V and H axes on the sketch of the cross-section. Exaggerate the arrows so they go all the way across.
- Make sure to draw the bending moment arrows such that they go over (or in front of) the bending axis (as opposed to under or behind)
- The side the arrowhead points to is the compressive side and the side the tail of the arrow is on is the tensile side.
The use of this tool will be demonstrated for the bending stress calculation in Example 14.1.
Example 14.1
The internal bending moments on a 32 mm diameter solid steel rod are found to be Mx = 376 kN mm (clockwise) and Mz = 204 kN mm (counterclockwise) as shown. Determine the bending stress at points P and Q.
1) Normally, we would draw the FBD of the cut section to find the internal bending moments. In this case, for simplicity, the internal moments are already given.
2) Write the general bending stress equation and eliminate unnecessary terms.
\[ \sigma_L= \pm\left|\frac{M_H\ dV}{I_H}\right| \pm\left|\frac{M_V\ dH}{I_V}\right| \]
Looking at the cross-section, we can see that the cross-sectional vertical axis is the z-axis and the cross-sectional horizontal axis is the x axis. The longitudinal axis is the y axis:
\[ \sigma_y= \pm\left|\frac{M_z\ dx}{I_z}\right| \pm\left|\frac{M_x\ dz}{I_x}\right| \]
Evaluating point P:
Because P is on the centroidal z-axis, dx = 0, so the general bending stress equation reduces to:
\[ \sigma_y= \pm\left|\frac{M_x\ dz}{I_x}\right| \]
3) Assess the sign:
With the clockwise bending moment around the x-axis, the beam can be visualized as bending “backwards” such that point P will be on the “inside” of the bend, or in compression. An approximation of the resulting shape is sketched below as it would appear from the side view (looking down the x-axis).
Alternatively, draw the cross-section with the exaggerated moment arrow clockwise around the x axis. Since the arrowhead would be pointed towards P, the point is in compression.
4) Calculate:
|Mx| = given = (376 kN mm)(1000 N/kN) = 376000 N mm
dz = radius = 16 mm
Ix =\(\frac{\pi}{64}d^4=\frac{\pi}{64}(32mm)^4=51.47\times10^3mm^4\)
\[ \begin{aligned} \sigma_{y_P}&=-\frac{376000{~N}\cdot{mm}\ (16mm)}{51.47 \times 10^{3}{~mm}^4}\\[10pt] &=-117{~N/mm^2}\\[10pt] &=-117\ MPa \end{aligned} \]
Evaluating Point Q:
Because point Q is on the centroidal x axis, dz = 0, so the general bending stress equation reduces to:
\[ \sigma_{y_Q}=\pm\left|\frac{M_z\ dx}{I_z}\right| \]
3) Assess the sign:
With the counterclockwise bending moment around the z-axis, the beam can be visulized as bending into somewhat of a backwards “C” shape as viewed from the front (down the z-axis). Point Q would be on the outside of the bend, or in tension.
Alternatively, draw the cross-section with the exaggerated moment arrow counterclockwise around the z-axis. Since Q is at the tail-side, the point is in tension.
4) Calculate:
|Mz| = given = (204 kN mm)(1000 N/kN) = 204000 N mm
dx = radius = 16 mm
For circular cross-section, Iz = Ix = 51.47 x 103 mm4
\[ \begin{aligned} \sigma_{y_Q}&=\frac{204000{~N}\cdot{mm}*(16{~mm})}{51.47 \times 10^{3}{~mm}^4}\\[10pt] &=63.4\ N/mm^2\\[10pt] &=63.4{~MPa} \end{aligned} \]
σyP = 116.9 MPa (compression)
σyQ = 63.4 MPa (tension)
14.2.3 Eccentric Loading
When the line of action of an axial load does not pass through the centroid of a cross section, it will result in both a normal reaction force and a bending moment. For example, consider the traffic light and corresponding free body diagram shown in Figure 14.7.
The vertical support pole is subjected to normal forces from the weights of the traffic lights (the weight of the horizontal poles, cameras, etc., are assumed to be negligible for this example). There are also bending moments around both cross-sectional axes (x and z) that result from the fact that the weights are not directly centered on the the support pole.
The total normal stress in the longitudinal direction, σy, at any given point on the pole would be given by:
\[ \sigma_L =\frac{\sum F_L}{A}+\sigma_{bending} \]
In terms of the general axes discussed in Section 14.1, and substituting in Equation 14.1 for the bending stress, we have:
\[ \boxed{\sigma_L= \pm\left|\frac{F_L}{A}\right| \pm\left|\frac{M_V d_H}{I_V}\right| \pm\left|\frac{M_H d_V}{I_H}\right|} \tag{14.2}\]
𝜎𝐿 = Bending stress in the longitudinal direction [Pa, psi]
FL = Internal normal force in the longitudinal direction [N, lb]
A = Cross-sectional area [m2, in.2]
MV = Bending moment around the vertical axis [N⸱m, lb⸱in.]
MH = Bending moment around the horizontal axis [N⸱m, lb⸱in.]
dH = Horizontal distance between the point on the cross-section for which the stress is being calculated and the V axis [m, in.]
dV = Vertical distance between the point on the cross-section for which the stress is being calculated and the H axis [m, in.]
IV = the moment of inertia about the vertical axis [m4, in.4]
IH = Area moment of inertia about the horizontal axis [m4, in.4]
The sign for the first term is positive if the total force is tensile (directed away from the cross-section) and negative if the total force is compressive (pointed towards the cross-section).
Example 14.2 and Example 14.3 demonstrate the calculation of normal stress in eccentric loading problems.
Example 14.4 demonstrates a loading that includes pressure as part of the loading combination and drawing the resulting stress element.
Example 14.2
The structure is fixed to the ground and subjected to the 75 lb and 50 lb vertical forces shown. Determine the total normal stress at points P and Q on the cross-section located at A-A. Assume the 40 inch horizontal section at the top remains rigid.
1) Draw the FBD of the structure cut at A-A and write the equilibrium equations to find the internal reactions.
Below is the FBD for the section of the structure above the cut. Using the section above the cut allows us to avoid needing to calculate the reactions at the fixed support.
In this case, reaction forces Rx and Rz are 0 by inspection (we can see that there are no forces applied in those directions).
Similarly, reaction moments MRx and MRy are 0 by inspection: All the forces only act in the y direction, so MRy = 0. All the forces pass through the x-axis, so MRx = 0.
Calculating Ry and MRz:
\[\begin{aligned}\sum F_y=R_y-75{~lb}-50{~lb}=0 \quad\rightarrow\quad R_y=125{~lb}\end{aligned}\] \[\begin{aligned}\sum\ M_z &=\sum\pm(Fy*x)+\sum \pm\ (Fx*y)+M_{Rz}=0\\[10pt] &=(75{~lb}*15{~in.})-(50{~lb}*25{~in.})+M_{Rz}=0 \quad\rightarrow\quad M_ {Rz}=125{~lb}\cdot{in.}\end{aligned}\]
Remember when calculating reaction moments to calculate moments about the centroidal axes of the cross-section no matter where on the cross-section the point of interest is.
2) Write the generalized normal stress equation and eliminate unnecessary terms:
\[ \sigma_L= \pm\left|\frac{F_L}{A}\right| \pm\left|\frac{M_V d_H}{I_V}\right| \pm\left|\frac{M_H d_V}{I_H}\right| \]
In this case, the longitudinal axis (L) is the y-axis. The vertical axis of the cross-section is the z-axis and the horizontal axis is the x-axis. With MRx = 0, we have:
\[ \sigma_y= \pm\left|\frac{R_y}{A}\right| \pm\left|\frac{M_{Rz}\ d_x}{I_z}\right| \]
3) Assess signs:
According the equilibrium equations applied in step 2:
Ry points towards the cut section so it is compressive (negative). In addition, the force will affect all points on the cross-section in the same way, so the Ry/A term will be negative for both points P and Q.
MRz is counterclockwise for the section of the body above the cut at A-A, but this means that it is clockwise for the section below the cut (equal/opposite reactions).
It tends to be easier to correctly visualize bending direction if we consider the direction of the bending moment for the section of the body attached to the fixed support. In this case, that bending moment is clockwise, so we visualize the tower as also bending clockwise.
We also use that direction of bending moment for drawing the moment on the cross-section when using the visualization tool discussed and used previously.
Based on the above, we can see that point P will be in tension from the bending stress and point Q will be compression.
4) Calculate the stress:
In addition to Ry and MRz, we need:
A = (6in.)(8in.) = 48 in.2
dx for point P = x-distance between P and the z-axis = 2 in
dx for point Q = x-distance between Q and the z-axis = 4 in
\(I_z = \frac{1}{12}bh^3=\frac{1}{12}(6{~in.})(8{~in.})^3 = 256{~in.^4}\)
Evaluating point P:
\[ \sigma_{yP}=-\left|\frac{125{~lb}}{48{~in.^2}}\right|+\left|\frac{(125{~lb}\cdot{in.})(2{~in.})}{256{~in.^4}}\right| \\ \sigma_{yP}=-1.63{~psi} \]
Evaluating point Q:
\[ \sigma_{yQ}=-\left|\frac{125{~lb}}{48{~in.^2}}\right|-\left|\frac{(125{~lb}\cdot{in.})(4{~in.})}{256{~in.^4}}\right| \\ \sigma_{yQ}=-4.56{~psi} \]
The negative answers indicates that the stress at both points is ovearll compressive.
Answers: σyP = -1.63 psi, σyQ = -4.56 psi
Example 14.3
A post is fixed to the ground and subjected to a 180 kN force pulling up at point P, shown on the cross-section.
Determine the normal stress at point A.
1) Draw the FBD of the structure cut at an arbitrary y-location and write the equilibrium equations to find the internal reactions.
For the given loading, the internal reactions will be the same for any y location, so we can make a cut anywhere to determine the bending moments and normal force reaction.
FBD above the cut:
The reaction forces Rz and Rx are 0 by inspection and the reaction moment MRy is also zero by inspection since the only applied force acts in the y direction.
\(\begin{aligned}\sum F_y=180 kN - R_y=0\quad\rightarrow\quad R_y=180 kN\end{aligned}\)
\(\begin{aligned}\sum M_x&=\sum\pm(Fy*z)+\sum\pm(Fz*y)+M_{Rx}\\[10pt]&=(180{~kN}*50{~mm})+M_{Rx}=0 \quad\rightarrow\quad M_{Rx}=-9000{~kN}\cdot{mm} \end{aligned}\)
\(\begin{aligned}\sum M_z&=\sum\pm(F_x*y)+\sum\pm(F_y*x)+M_{Rz}=0\\[10pt]&=(180kN*40mm)+M_{Rz}=0\quad\rightarrow\quad M_{Rz}=-7200\ kN\cdot mm\end{aligned}\)
2) Write the generalized normal stress equation:
Applying the generalized normal stress equation (with the pressure term omitted):
\[ \sigma_{L}= \pm\left|\frac{F_L}{A}\right| \pm\left|\frac{M_V d_H}{I_V}\right| \pm\left|\frac{M_H d_V}{I_H}\right| \]
In this case, the longitudinal axis of the post is the y-axis, the vertical axis of the cross-section is the z axis, and the horizontal axis of the cross-section is the x-axis. So we have:
\[ \begin{aligned} &\sigma_{yA}=\pm\left|\frac{R_y}{A}\right|\pm\left|\frac{M_z d_x}{I_z}\right|\pm\left|\frac{M_x d_z}{I_x}\right| \\ \end{aligned} \]
3) Assess the signs:
According to the equilibrium analysis, Ry points away from the cross-section, so that force results in tensile stress (positive).
The bending moments were both found to be negative(clockwise) for the section of the body above the cut. As was discussed in Example 14.2, to easier visualize the bending, we will use the direction of the bending moments on the section of the body below the cut (where the fixed support is). These reaction moments will be opposite in direction to what we found in step 1.
Below the cut, we will have MRx = 9000 kN mm (counterclockwise) which will put point A in tension.
We will also have MRz = 7200 kN mm (counterclockwise) which will put point A in compression.
\[ \sigma_{yA}=\left|\frac{180,000{~N}}{(0.2{~m})(0.15{~m})}\right|- \left|\frac{7200{~N}\cdot{m}*0.085{~m}}{\frac{1}{12}(0.15{~m})(0.2 {~m})^3}\right|+\left| \frac{9000{~N}\cdot{m}*0.075{~m})}{\frac{1}{12}(0.2{~m})(0.15{~m})^3}\right| \\[10pt]\sigma_{y A}=11.9 MPa \]
The stress works out to be positive and therefore tensile.
Answer: σyA =11.9 MPa
Example 14.4
For the pressurized pipe loaded as shown, determine the total normal stress at point K. Noting that there would be no shear stress at point K for this loading, represent the general stress state of point H on a stress element. The pipe has an outside diameter of 200 mm and a wall thickness of 10 mm. The load Pz = 18 kN and the internal pressure in the pipe is 1500 kPa.
1) Draw the FBD of the structure cut at point K and write the equilibrium equations to find the internal reactions.
FBD above the cut:
There are no forces in the x or y directions, so Rx= Ry =0.
The 18kN force is parallel to the z-axis and goes through the x-axis so MRz =MRx = 0.
MRy is not zero; however, point K is on the y-axis, so this point will not be affected by MRy (the x distance from the y-axis to K on the cross-section is 0).
Thus, we do not need to calculate any moment reactions.
\(\sum F_y=R_y-18{~kN}=0 \quad\rightarrow\quad R_y=18{~kN}\)
2) Write the general normal stress equation and eliminate unnecessary terms. Note that since we have pressure loading, there will be normal stress in the cross-sectional directions (hoop stress) as well as in the longitudinal direction:
\[ \sigma_L= \pm\left|\frac{F_L}{A}\right| \pm\left|\frac{M_V d_H}{I_V}\right| \pm\left|\frac{M_H d_V}{I_H}\right| + \frac{p r}{2 t} \]
\[ \sigma_{V}=\sigma_{H}=\sigma_{hoop}=\frac{pr}{t} \]
In this problem, the longitudinal direction is the z direction and the cross-sectional directions are x and y. Eliminating both bending stress terms:
\[ \begin{aligned} \sigma_{zK} = \pm\left|\frac{R_z}{A}\right| + \frac{p r}{2 t} \\[10pt] \sigma_{x} =\sigma_y=\frac{p r}{t} \end{aligned} \]
3) Assess signs:
According to the equilibrium analysis, Rz points towards the cut section, so the first term in the stress equation will be negative. Internal pressure stress will always be positive.
4) Calculate:
\[ \begin{aligned} \sigma_{zK}&=-\left|\frac{R_z}{A}\right|+\frac{pr}{2t} \\[10pt] &=-\frac{18,000{~N}}{\pi\left(0.1^2-0.09^2\right){~m^2}}+\frac{(1,500,000{~Pa})(0.1{~m})}{2(0.01{~m})} \\[10pt] &=4.48{~MPa} \end{aligned} \]
\[ \sigma_x=\frac{pr}{t}=\frac{(1,500,000{~Pa})(0.1{~m})}{0.01{~m}}=15{~MPa} \]
5) Sketch the stress element, noting that point K is in the xz-plane.
14.3 General Shear Stress
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Shear stress due to torsion was discussed in Chapter 6 while shear stress due to transverse loading was discussed in Chapter 10. In this section, the additional considerations that go into determining the total shear stress when both forms of loading are present will be discussed.
14.3.1 Shear Stress due to Torsion
As discussed in Section 6.1, shear stress due to torsion in a circular shaft can be calculated as (Equation 6.1):
\[ \tau_{(torsional)}=\frac{T \rho}{J} \]
where:
\(T=Torque=\sum M_L\) (N m, lb in.)
𝜌 = radial distance to point on cross-section (m, in.)
J = polar moment of inertia for circular cross-section (m4, in.4)
For general problems, one may need to sum moments to calculate T, so it is important to remember that torque is the moment about the longitudinal axis of a body, or that component of the moment that causes the body to twist. Since this form of shear stress will be combined with transverse shear stress to get the total, understanding the sign of the shear stress will be important.
As discussed in Section 12.1, the sign of shear stress is based on the orientation of the shear stress arrows on the stress element.To determine in which direction to draw a shear stress arrow on a stress element for a particular point, one must cut the body at one edge of the stress element that represents the point and visualize the direction in which the shear load would act across the cut.
For example, in Figure 14.7, the stress element is shown to be in the L-H plane. When the cylinder is cut at the top of the stress element, we examine the bottom section of the cylinder and note that the reaction torque at the cut surface goes in the opposite direction to that which is applied at the bottom surface. This results in a counterclockwise reaction torque at the cut, which means that the top section tends to want to twist (and exert a force) from left to right across the top edge of the element.
If the longitudinal axis is positive upwards, the top face is a positive face. In addition, if the horizontal axis is positive to the right, the rightwards shear force arrow is in the positive direction. Together this means that the shear stress can be concluded to be positive since we would have a positive shear force on a positive face.
Similarly, when the cut is made at the bottom edge of the element, we can examine the top section of the cylinder for which equilibrium dictates that there is a clockwise reaction moment at the cut edge. This means that the bottom section tends to want twist from right to left across the bottom element edge. If the L axis is positive upwards so that the top face is a positive face, and the H axis is positive to the right, the leftwards shear force would be a negative force acting on a negative face, so the shear stress would still be concluded to be positive.
If the H axis were to be oriented to be positive to the left instead, then this same shear stress would be negative. The same can be said if the H axis is positive to the right but the L axis is positive downwards.
14.3.2 Shear Stress due to Transverse Shear
In Chapter 10, the equation \(\tau=\frac{V Q}{I t}\) (Equation 10.2) was evaluated to find the shear stress due to transverse shear force V. In that chapter, only one direction of shear force was considered for each problem and the sign of the shear stress was not considered. For general loading, we may have two directions of transverse shear force corresponding to the two cross-sectional axes directions, as shown on the illustration of the bar in Figure 14.9, so we will need to apply the transverse shear stress equation for both directions and also understand the signs so that we can combine them with each other and with the torsional shear stress.
When applying the transverse shear stress equations for specific directions of shear force, it is helpful to keep the following in mind:
- The cut one makes to determine the first moment of area Q will be perpendicular to the direction of shear force direction (illustrated on the cross-sections in Figure 14.9).
- The area moment of inertia (I) will be the moment of inertia around the axis perpendicular to the shear force direction.
- The thickness (t), as before, refers to the thickness of the solid part of the cross section that one cuts through to calculate Q (illustrated on the cross-sections in Figure 14.9).
Using the same reference axes as with the general normal stress equation (H for horizontal centroidal axis and V for vertical centroidal axis), the transverse shear stress equation can be written as follows:
\[ \boxed{\tau_{(transverse)}= \pm\left|\frac{V_H Q_V}{I_V t_V}\right| \pm\left|\frac{V_V Q_H}{I_H t_H}\right|} \tag{14.3}\]
𝜏(transverse) = Transverse shear stress at point of interest [Pa, psi]
VH = Internal horizontal shear force [N, lb]
VV = Internal vertical shear force [N, lb]
QV = First moment of area about the vertical axis passing through the point of interest [m3, in.3]
QH = First moment of area about the horizontal axis passing through the point of interest [m3, in.3]
IV = Area moment of inertia about the vertical centroidal axis [m4, in.4]
IH = Area moment of inertia about the horizontal centroidal axis [m4, in.4]
tV = Vertical thickness of the cross-section [m, in.]
tH = Horizontal thickness of the cross-section [m, in.]
When assessing the contributions to shear stress present for any given point, a simplifying observation that can be made is:
1) If a point is on the top or bottom edge of a non-circular cross-section, or is the top or bottom point of a circular cross-section, it will experience no transverse shear stress from a vertical shear force.
2) If a point is on the left or right edge of a non-circular cross-section, or is the left most or right most point of a circular cross-section, it will feel no transverse shear stress from a horizontall shear force.
For example, in Figure 14.9, point P will experience no shear stress from VV and point S will feel no shear stress from VH. Figure 14.10 further illustrates the statements above.
The reason is that, in the case of a vertical shear force, the top and bottom edges of the cross-section are free edges since bending results only around the H axis. Similarly, in the case of a horizontal shear force, the left and right edges of the cross-section are free surfaces since bending results only around the V axis. For a circular cross-sections, the idea is similar, but only applies to the points at the top/bottom and left/right positions along the centroidal axes on the perimeter.
The signs on the shear stress are determined similarly to the manner described in Section 14.2.1 for torsion. The direction of the internal shear force is determined by equilibrium and the resulting shear stress would be drawn in that same direction on the cut surface of the element. For example, in Figure 14.9, point S is on the V-L plane. ?fig-14.11 shows the element drawn from the side view, down the H axis. The vertical shear force acts downwards on the right side of the stress element. If the axes are oriented such that the longitudinal axis is positive to the right, the right side of the stress element is the +L face. If the vertical axis of the cross-section is positive upwards, than VV in the shown direction results in a negative shear stress.
Similarly, point P is on the H-L plane. In ?fig-14.11, the stress element is drawn from the top-down view (viewing down the V axis). The horizontal shear force acts upwards on the right side of the stress element. If the right side is the +L face, and top is the +H face, then the shear stress based on the shown loading is positive and vice versa.
14.3.3 Combined Shear Stress
Now we can consider the case of bodies having both torsion and transverse shear stresses. Given a general loading, one may need to calculate the torque and also identify which of the forces constitute transverse shear forces. It will be helpful to remember that the transverse shear forces are the forces in the directions of the cross-sectional axes and torsion is the moment about the longitudinal axis.
The total shear stress acts in the plane of the point of interest. It can be calculated as:
\[ \boxed{\tau= \pm\left|\frac{T \rho}{J}\right| \pm\left|\frac{V_H Q_V}{I_V t_V}\right| \pm\left|\frac{V_V Q_H}{I_H t_H}\right|} \tag{14.4}\]
𝜏 = Shear stress at point of interest [Pa, psi]
T = Internal torque [N·m, lb·in.]
𝜌 = Radial distance from centroid to point of interest [m, in.]
J = Polar moment of inertia [m4, in.4]
VH = Internal horizontal shear force [N, lb]
VV = Internal vertical shear force [N, lb]
QV = First moment of area about the vertical axis passing through the point of interest [m3, in.3]
QH = First moment of area about the horizontal axis passing through the point of interest [m3, in.3]
IV = Area moment of inertia about the vertical centroidal axis [m4, in.4]
IH = Area moment of inertia about the horizontal centroidal axis [m4, in.4]
tV = Vertical thickness of the cross-section [m, in.]
tH = Horizontal thickness of the cross-section [m, in.]
Example 14.5 demonstrates the calculation of total shear stress when torsion and transverse shear are both acting.
Example 14.6 demonstrates the calculation of transverse shear stress with multiple directions of shear force.
Example 14.5
A solid steel bar with an outside diameter of 6.70 in. is fixed to the wall. A 900 lb force acts vertically and a 1500 lb force acts in the positive x direction as shown. Determine the maximum shear stress at points J and K on the outside of the pipe.
1.) Cut the body at the cross-section where K and J are located and draw the FBD of the cut section.
We will examine the section to the right of the cut to avoid needing to find the reactions at the wall.
The cross-sectional axes are the x and z axes, so these are the bending moment and transverse shear directions. The y axis is perpendicular to the cross-section, so it is the longitudinal axis.
Before proceeding with equilibrium calculations, we note that we are only asked to find shear stress. Therefore, we do not need to know the bending moments MRx and MRz and we do not need to know the normal force Ry.
The shear forces are the forces in the cross-sectional directions (Rx and Rz). We find them by applying the force equilibrium equations:
\(\begin{aligned}& \sum F_x=R_x+1500 lb=0\quad\rightarrow\quad R_x=V_x=-1500 lb \\& \sum F_z=R_z+900{~lb}=0 \quad\rightarrow\quad R_z=V_z=-900{~lb}\end{aligned}\)
The torque is the moment about the longitudinal axis (MRy). We find it by applying the moment equilibrium equation for the y axis. Note that in the FBD, the direction of MRY is assumed to act counterclockwise around the y axis when viewing from the positive end to the negative (right to left).
\(\sum M_y=M_{Ry}+(900{~lb})(18{~in.})=0 \quad\rightarrow\quad M_{Ry}=T=-16,200{~lb\ in} \\\)The negative result means that the torque acting at the cut side of the FBD we drew is clockwise around the y axis when viewing from the positive side towards the negative (right to left).
2) Write the general shear stress equation and eliminate unnecessary terms:
\[ \tau= \pm\left|\frac{T \rho}{J}\right| \pm\left|\frac{V_H Q_V}{I_V t_V}\right| \pm\left|\frac{V_V Q_H}{I_H t_H}\right| \]
The horizontal axis of the cross-section is the x-axis and the vertical axis is the z axis.
Point K is on the yz plane. Since point K is on the x axis, it will not experience shear stress from Vx.
Point J is on the xy plane. Since point J is on z axis, it will not experience shear stress from Vz.
\[ \tau_{Kyz}= \pm\left|\frac{T \rho}{J}\right| \pm\left|\frac{V_z Q_x}{I_x t_x}\right| \]
\[\tau_{Jxy}= \pm\left|\frac{T \rho}{J}\right| \pm\left|\frac{V_x Q_z}{I_z t_z}\right|\]
3) Assess the signs.
For the torsional stress term:
If we consider the stress element for each point to be just inside of the cut section we used to find the internal reactions, as shown in the figure below, the left side of each element will be the side that the calculated internal shear loads act on.
Based on the direction of torque we found, we can see that it acts in the upwards direction across the left face of both stress elements. For point K, this results in a negative contribution to the shear stress but for point J, it will be positive.
For the shear forces:
Rz was found to act in the negative z direction so it will act downwards on the left face of the stress element for point K. This shear stress will be positive.
Rx was found to act in the negative x direction so it will act upwards on the left face of the stress element for point J. This shear stress will also be positive.
Now we have:
\[\tau_{Kyz}= -\frac{T \rho}{J}+\frac{V_z Q_x}{I_x t_x} \]
\[\tau_{Jxy}= \frac{T \rho}{J} +\frac{V_x Q_z}{I_z t_z} \]
4) Calculate:
For a circular cross-section:
\(Q_x=Q_z=\frac{1}{12}d^3=\frac{1}{12}(6.70\ in)^3=25.06\ in^3\)
\(I_x=I_z=\frac{\pi}{64}d^4=\frac{\pi}{64}(6.7\ in)^4=98.92\)
\(J=\frac{\pi}{32}d^4=\frac{\pi}{32}(6.7\ in)^4=197.83\ in^4\)
So for point K:
\[\begin{aligned}\tau_{Kyz}&= -\frac{T \rho}{J}+\frac{V_z Q_x}{I_x t_x}\\[10pt] &=\frac{(-16200\ lb\ in)(\frac{6.70\ in}{2})}{197.83\ in^4}+\frac{(900\ lb)(25.06\ in^3)}{(98.92\ in^4)(6.70\ in)}\\[10pt] &=-274.3\ psi+34.04\ psi\\[10pt] &=-240.3\ psi \end{aligned} \]
For point J, the torsional shear stress will be the same magnitdue, so we have:
\[ \begin{aligned}\tau_{Jxy}&=274.3\ psi+\frac{(1500\ lb)(25.06\ in^3)}{(98.92\ in^4)(6.7\ in)}\\ &=331.0\ psi \end{aligned} \]
Answers: τKyz = -240.27 psi, τJxy = 331.02 psi
Example 14.6
Two forces are applied to a rectangular post as shown. Determine the shear stress at points P and S.
1) Draw the FBD of the structure cut at at the vertical location where P and S are located and apply equilibrium to find the internal reactions.
FBD of the section above the cut:
The cross-sectional axes are the x and y axes, so these are the bending moment and transverse shear directions. The z axis is perpendicular to the cross-section, so it is the longitudinal axis.
As with the previous example, we note that we are only being asked for shear stress, so the bending moments MRx and MRy are not necessary to find for this problem. (However, this problem is re-visited in Example 14.7 and those moments are found in that example).
Finding the internal shear forces using force equilibrium equations:
\(\sum F_x=R_x-94\ kN=0\quad\rightarrow\quad R_x=V_x=94 kN\)
\(\sum F_y=R_y+78\ kN=0\quad\rightarrow\quad R_y=V_y=-78\ kN\)
Since both of the applied forces go through the centroidal z axis, neither will create a moment around that axis, so:
MRy = T = 0
2) Write the general shear stress equation and eliminate unnecessary terms. Given that there is no torsion in this problem, the general shear stress equation can be written:
\[ \tau= \pm\left|\frac{V_H Q_V}{I_V t_V}\right| \pm\left|\frac{V_V Q_H}{I_H t_H}\right| \]
Where the horizontal axis of the cross-section is the y–axis and the vertical axis of the cross section is the x-axis.
As discussed in Section 14.3.2, since point P is on the bottom edge of the cross-section, it will not experience shear stress from Vx.
Point S is on the right edge of the cross-section, so it will not experience shear stress from Vy.
This leaves us with:
\[\tau_{Pyz}=\pm \left|\frac{V_y\ Qx}{I_x\ t_x}\right| \]
\[ \tau_{Sxz}=\pm \left|\frac{V_x\ Qy}{I_y\ t_y}\right| \]
3) Assess the signs.
Since we used the top section of the cut to calculate internal reactions, we consider the stress elements to be at the bottom of the section, just inside the cut surface. The shear forces then act on the bottom face of the stress elements.
Vy was found to act in the negative y direction, so it gets drawn directed to the left at the bottom of the P stress element. This results in a positive shear stress for point P.
\[\tau_{Pyz}=\frac{V_y\ Qx}{I_x\ t_x} \]
Vx was found to act in the positive x direction. so it gets drawn directed to the left at the bottom of the S stress element. This results in a negative shear stress for point S.
\[\tau_{Sxz}=-\frac{V_x\ Qy}{I_y\ t_y} \]
4) Calculate.
With:
\(V_H = V_y = 78{~kN}\)
\(V_V = V_x = 94{~kN}\)
\(I_H=I_y=\frac{1}{12}(160{~mm})(120{~mm})^3=23.04 \times 10^6{~mm}^4=23.04 \times 10^{-6}{~m}^4\)
\(I_V=I_x=\frac{1}{12}(160{~mm})^3(120{~mm})=40.96 \times10^6{~mm}^4=40.96 \times 10^{-6}{~m}^4\)
For point P we need Qx and tx. Cutting vertically through point P on the cross-section:
\[ \begin{aligned}&Q_x=A'y'=\frac{1}{2}(160{~mm})(120{~mm})*\left(\frac{1}{4}\right)(160{~mm})=384000{~mm}^3=384 \times 10^{-6}{~m}^3\\[10pt] &t_x=120\ mm = .12\ m \end{aligned} \]
For point S we need Qy and ty. Cutting horizontally through point S on the cross-section:
\(\begin{aligned}&Q y=A'x'=(160{~mm})(30{~mm})*(60{~mm}-15{~mm})=216000{~mm}^3=216 \times 10^{-6}{~m}^3\\[10pt]&ty=160\ mm=.16\ m\end{aligned}\)
So
\[ \tau_{Pyz}=\frac{V_y Q_x}{I_x t_x}=\frac{(78000{~N})(384 \times 10^{-6}{~m}^3)}{(40.96 \times 10^{-6}{~m}^4)(0.12{~m})}=6.09{~MPa} \]
\[ \tau_{Sxz}=-\frac{V_x Q_y}{I_y t_y}=-\frac{(94000{~N})(216 \times 10^{-6}{~m}^3)}{(23.04 \times 10^{-6}{~m}^4)(0.16{~m})}=-5.51{~MPa} \]
Answers:
τPyz = 6.09 MPa
τSxz~ = -5.51 MPa
14.4 General Combined Stress
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Once the total normal stress and total shear stress at a specified point is calculated, the general stress state can be represented on a stress element and used to determine principal stresses, principal planes, and max/min shear stresses.
In Example 14.7, the problem from Example 14.6 is extended to find the principal stresses and max/min shear stresses.
Example 14.7
Revisit Example 14.6 to find the principal stresses and max/min shear stresses for point P on the post:
In Example 14.6 the total shear stress for point P was found to be positive 10.83MPa.
To find the principal stresses, we must:
1) finish finding the general stress state by also finding the total normal stress and
2) use the principal stress equations or Mohr’s Circle.
To complete the first task:
1) Again, use the FBD of the section of the post above the cut made at the vertical location of P and S.
We know from Example 14.6 that Rx = 94 kN and Ry = - 78 kN.
By inspection, we can see that Rz = 0.
Point P is on the vertical axis of the cross-section (x axis), so we know that it will not experience any bending stress from MRx .
As noted in the previous example, MRy = 0 by inspection, but it would not contribute to the normal stress in any case since that would be a torsional moment.
Thus, the only bending moment reaction we need to calculate is MRy . Using the moment equilibrium equation for the y-axis:
\(\begin{aligned}\sum M_y&=(-94{~kN})(150{~mm})+M_{Ry}=0\\ &=-14,100{~kN\ mm}+M_{Ry}=0\quad\rightarrow\quad M_{Ry}=14,100{~kN\ mm}\end{aligned}\)
The generalized normal stress equation (with no pressures) is:
\[ \sigma_{L}= \pm\left|\frac{F_L}{A}\right| \pm\left|\frac{M_V d_H}{I_V}\right| \pm\left|\frac{M_H d_V}{I_H}\right| \]
With no force in the longitudinal direction and point P located on the x axis (see discussion above), we can reduce the equation to:
\[ \sigma_z=\pm \left |\frac{M_Ry\ x}{I_y}\right| \]
To assess the sign, we note that we determined the reaction moment at the cut surface of the upper section of post to be positive, or counterclockwise. That means the reaction moment at the cut surface for the lower section of post (beneath the cut at P and S) will be negative, or clockwise. As mentioned in Example 14.2 and Example 14.3, we want to use the bending moment on the lower section to visualize the bending since it is attached to the fixed support.
With a clockwise moment around the y-axis for the lower section of post, we visualize the curve of the bend to go “backwards” around the y-axis. From the side view (looking down the y-axis), the post would look like it curves around to the right, placing point P on the outside of the bend (tension).
Alternatively, drawing the cross-section from the top-down view with the clockwise moment arrow drawn around the y-axis shows that point P is on the ithe tail of the cross-section, confirming tension at this point.
So \(\sigma_z=\frac{M_Ry\ x}{I_y}\)
Recalling \(I_y = 23.04 \times 10^{-6}{~m^4}\) from Example 14.6, the normal stress at P is then:
\[ \begin{aligned} &\sigma_{Pz}=\frac{M_y x}{I_y}\\ &\sigma_{Pz}=\frac{14,100{~N}\cdot{m}(0.06{~m})}{23.04 \times 10^{-6}{~m^4}}\\ &\sigma_{Pz}=36.77{~MPa} \end{aligned} \]
Now that the total shear stress and total normal stress are known, the stress state can be represented on a stress element.
Applying the simplified version of the principal stress equations and the max/min shear equation (in terms of the general xy plane), we get:
\[ \begin{aligned} & \sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{x y}{ }^2} \\ & \tau_{\max , \min }= \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{x y}{ }^2} \end{aligned} \]
In this case, since the point P is in the y-z plane instead of the x-y plane, we can replace the x subscripts with y and the y subscripts with z, so the equations become:
\[ \begin{aligned} & \sigma_{1,2}=\frac{36.77{~MPa}+0}{2} \pm \sqrt{(\frac{36.77{~MPa}-0}{2})^2+(6.094{~MPa})^2}\\ &\sigma_{1,2}=37.75{~MPa}, -.984{~MPa} \\ & \tau_{\max , \min }=\sqrt{(\frac{36.77{~MPa}-0}{2})^2+(6.094{~MPa})^2}\\ &\tau_{max,min}=19.37{~MPa},-19.37{~MPa} \end{aligned} \]
σ1 = 37.75 MPa
σ2 = -.984 MPa
τmin = -19.37 MPa
τmax = 19.37 MPa
14.5 Summary
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References
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Figures
All figures in this chapter were created by Kindred Grey in 2025 and released under a CC BY-NC-SA license.