Combined Stresses
Intended Learning Outcomes:
1. Illustrate a variety of combined stresses on stress elements.
2. Analyze a load-carrying member subjected to combined stress to determine the principal stresses, the maximum
normal stress and the maximum shear stress on any given element.
3. Determine the coordinate system in which the principal stresses are aligned.
4. Determine the state of stress on an element in any specified coordinate system.
5. Draw the complete Mohr’s circle as an aid in completing the analyses for the maximum stresses.
6. Design members under combined stresses.
A. Combined Normal Stresses
When the same cross section of a load-carrying member is subjected to both a direct tensile or compressive
stress and a stress due to bending, the resulting normal stress can be computed by the method of superposition.
The formula is
𝑀𝑐 𝐹
𝜎 = ∓ 𝐼 ±𝐴
where tensile stresses are positive and compressive stresses are negative.
An example of a load-carrying member subjected to combined bending and axial tension is shown in
figure. It shows a beam subjected to a load applied downward and to the right through a bracket below the
beam. Resolving the load into horizontal and vertical components shows that its effect can be broken into three
parts:
1. The vertical component tends to place the beam in bending with tension on the top and compression on
the bottom.
2. The horizontal component, because it acts away from the neutral axis of the beam, causes bending with
tension on the bottom and compression on the top.
3. The horizontal component causes direct tensile stress across the entire cross section.
B. Combined Shear Stresses
This applies to eccentric loading of bolted and riveted connections. If the resultant of the external load on a
joint pass through the centroid of the bolt pattern, such a joint is called an axial shear joint. Under these
conditions, all the fasteners in the joint can be assumed to see an equal shear load. If the resultant of the applied
load passes through some point other than the centroid of the bolt group, there will be a net moment on the bolt
pattern. Each of the bolts will help the joint resist this moment. A joint loaded this way is said to be under an
eccentric shear load.
The analysis would go as follows:
1. Determine the centroid of the bolt group.
∑𝐴 𝑥 ∑𝐴 𝑦
𝑥̅ = ∑ 𝐴𝑖 𝑖 𝑦̅ = ∑ 𝐴𝑖 𝑖
𝑖 𝑖
2. Determine the stresses in the bolts: primary stress refers to direct shear stress and the secondary stress
refers to torsional stress induced on the bolt.
𝑉 𝑇𝑟
𝜏1 = 𝑎𝑛𝑑 𝜏2 =
𝐴 𝐽
3. Combine primary shear stress and secondary shear stress using vector addition. The resultant stress is
the combined shear stress of the bolt/rivet.
, C. Combined Shear and Normal Stresses
The general two-dimensional stress element in the figure shows two normal stresses 𝜎𝑥 𝑎𝑛𝑑 𝜎𝑦 , both
positive, and two shear stresses 𝜏𝑥𝑦 𝑎𝑛𝑑 𝜏𝑦𝑥 , positive also. The element is in static equilibrium, and hence
𝜏𝑥𝑦 = 𝜏𝑦𝑥 ,. The stress state depicted by the figure is called plane or biaxial stress.
Figure b shows an element face whose normal makes an angle ∅ to the x axis. It can be shown that the stress
components 𝜎 𝑎𝑛𝑑 𝜏 acting on this face are given by the equations:
𝜎𝑥 +𝜎𝑦 𝜎𝑥 −𝜎𝑦
𝜎= + 𝑐𝑜𝑠2∅ + 𝜏𝑥𝑦 𝑠𝑖𝑛2∅
2 2
𝜎𝑥 −𝜎𝑦
𝜏 = − 2 𝑠𝑖𝑛2∅ + 𝜏𝑥𝑦 𝑐𝑜𝑠2∅
It can be shown that when the angle ∅ is varied in the equation above, the normal stress 𝜎 has two extreme
values. These are called the principal stresses, and they are given by the equation:
1/2
𝜎𝑥 +𝜎𝑦 𝜎𝑥 −𝜎𝑦 2 2
𝜎1 , 𝜎2 = ± [( ) + 𝜏𝑥𝑦 ]
2 2
The corresponding values of 𝜎 are called the principal directions. These directions can be obtained from:
2𝜏𝑥𝑦
2∅ = arctan 𝜎 −𝜎
𝑥 𝑦
The shear stresses are always zero when the element is aligned in the principal directions. However, it also turns
out that the shear stress 𝜏 has two extreme values. These and the angles at which they occur may be found from
1/2
𝜎𝑥 −𝜎𝑦 2 2 𝜎𝑥 −𝜎𝑦
𝜏1 , 𝜏2 = ± [( ) + 𝜏𝑥𝑦 ] and 2∅ = arctan
2 2𝜏𝑥𝑦
The two normal stresses are equal when the element is aligned in the directions given by above equation.
The act of referring stress components to another reference system is called transformation of stress. Such
transformations are easier to visualize, and to solve, using a Mohr's circle diagram. We create a 𝜎𝜏 coordinate
system with normal stresses plotted as abscissa and shear as the ordinates. On the abscissa, tensile (positive)
normal stresses are plotted to the right of the origin O, and compression (negative) normal stresses are plotted to
the left. The sign convention for shear stresses is that clockwise (cw) shear stresses are plotted above the abscissa
and counterclockwise (ccw) shear stresses are plotted below.
The stress state is shown on the diagram in figure above. Points A and C represent𝜏𝑥𝑦 𝑎𝑛𝑑 𝜏𝑦𝑥 , respectively,
and point E is midway between them. Distance AB is 𝜏𝑥𝑦 and distance CD is 𝜏𝑦𝑥 . The circle of radius ED is
Mohr's circle. This circle passes through the principal stresses at F and G and through the extremes of the shear
stresses at H and /. It is important to observe that an extreme of the shear stress may not be the same as the
maximum.
Intended Learning Outcomes:
1. Illustrate a variety of combined stresses on stress elements.
2. Analyze a load-carrying member subjected to combined stress to determine the principal stresses, the maximum
normal stress and the maximum shear stress on any given element.
3. Determine the coordinate system in which the principal stresses are aligned.
4. Determine the state of stress on an element in any specified coordinate system.
5. Draw the complete Mohr’s circle as an aid in completing the analyses for the maximum stresses.
6. Design members under combined stresses.
A. Combined Normal Stresses
When the same cross section of a load-carrying member is subjected to both a direct tensile or compressive
stress and a stress due to bending, the resulting normal stress can be computed by the method of superposition.
The formula is
𝑀𝑐 𝐹
𝜎 = ∓ 𝐼 ±𝐴
where tensile stresses are positive and compressive stresses are negative.
An example of a load-carrying member subjected to combined bending and axial tension is shown in
figure. It shows a beam subjected to a load applied downward and to the right through a bracket below the
beam. Resolving the load into horizontal and vertical components shows that its effect can be broken into three
parts:
1. The vertical component tends to place the beam in bending with tension on the top and compression on
the bottom.
2. The horizontal component, because it acts away from the neutral axis of the beam, causes bending with
tension on the bottom and compression on the top.
3. The horizontal component causes direct tensile stress across the entire cross section.
B. Combined Shear Stresses
This applies to eccentric loading of bolted and riveted connections. If the resultant of the external load on a
joint pass through the centroid of the bolt pattern, such a joint is called an axial shear joint. Under these
conditions, all the fasteners in the joint can be assumed to see an equal shear load. If the resultant of the applied
load passes through some point other than the centroid of the bolt group, there will be a net moment on the bolt
pattern. Each of the bolts will help the joint resist this moment. A joint loaded this way is said to be under an
eccentric shear load.
The analysis would go as follows:
1. Determine the centroid of the bolt group.
∑𝐴 𝑥 ∑𝐴 𝑦
𝑥̅ = ∑ 𝐴𝑖 𝑖 𝑦̅ = ∑ 𝐴𝑖 𝑖
𝑖 𝑖
2. Determine the stresses in the bolts: primary stress refers to direct shear stress and the secondary stress
refers to torsional stress induced on the bolt.
𝑉 𝑇𝑟
𝜏1 = 𝑎𝑛𝑑 𝜏2 =
𝐴 𝐽
3. Combine primary shear stress and secondary shear stress using vector addition. The resultant stress is
the combined shear stress of the bolt/rivet.
, C. Combined Shear and Normal Stresses
The general two-dimensional stress element in the figure shows two normal stresses 𝜎𝑥 𝑎𝑛𝑑 𝜎𝑦 , both
positive, and two shear stresses 𝜏𝑥𝑦 𝑎𝑛𝑑 𝜏𝑦𝑥 , positive also. The element is in static equilibrium, and hence
𝜏𝑥𝑦 = 𝜏𝑦𝑥 ,. The stress state depicted by the figure is called plane or biaxial stress.
Figure b shows an element face whose normal makes an angle ∅ to the x axis. It can be shown that the stress
components 𝜎 𝑎𝑛𝑑 𝜏 acting on this face are given by the equations:
𝜎𝑥 +𝜎𝑦 𝜎𝑥 −𝜎𝑦
𝜎= + 𝑐𝑜𝑠2∅ + 𝜏𝑥𝑦 𝑠𝑖𝑛2∅
2 2
𝜎𝑥 −𝜎𝑦
𝜏 = − 2 𝑠𝑖𝑛2∅ + 𝜏𝑥𝑦 𝑐𝑜𝑠2∅
It can be shown that when the angle ∅ is varied in the equation above, the normal stress 𝜎 has two extreme
values. These are called the principal stresses, and they are given by the equation:
1/2
𝜎𝑥 +𝜎𝑦 𝜎𝑥 −𝜎𝑦 2 2
𝜎1 , 𝜎2 = ± [( ) + 𝜏𝑥𝑦 ]
2 2
The corresponding values of 𝜎 are called the principal directions. These directions can be obtained from:
2𝜏𝑥𝑦
2∅ = arctan 𝜎 −𝜎
𝑥 𝑦
The shear stresses are always zero when the element is aligned in the principal directions. However, it also turns
out that the shear stress 𝜏 has two extreme values. These and the angles at which they occur may be found from
1/2
𝜎𝑥 −𝜎𝑦 2 2 𝜎𝑥 −𝜎𝑦
𝜏1 , 𝜏2 = ± [( ) + 𝜏𝑥𝑦 ] and 2∅ = arctan
2 2𝜏𝑥𝑦
The two normal stresses are equal when the element is aligned in the directions given by above equation.
The act of referring stress components to another reference system is called transformation of stress. Such
transformations are easier to visualize, and to solve, using a Mohr's circle diagram. We create a 𝜎𝜏 coordinate
system with normal stresses plotted as abscissa and shear as the ordinates. On the abscissa, tensile (positive)
normal stresses are plotted to the right of the origin O, and compression (negative) normal stresses are plotted to
the left. The sign convention for shear stresses is that clockwise (cw) shear stresses are plotted above the abscissa
and counterclockwise (ccw) shear stresses are plotted below.
The stress state is shown on the diagram in figure above. Points A and C represent𝜏𝑥𝑦 𝑎𝑛𝑑 𝜏𝑦𝑥 , respectively,
and point E is midway between them. Distance AB is 𝜏𝑥𝑦 and distance CD is 𝜏𝑦𝑥 . The circle of radius ED is
Mohr's circle. This circle passes through the principal stresses at F and G and through the extremes of the shear
stresses at H and /. It is important to observe that an extreme of the shear stress may not be the same as the
maximum.
