18 Mohrs Circle Examples
Problem 1:
At a point on the surface of a cylinder, loaded by internal pressure, the material is subjected to biaxial stresses σ x = 90 MPa and σ y = 20MPa, as shown on the stress element of figure (a). Using Mohr's circle, determine the stresses acting on an element inclined at an angle θ = 30° . (Consider only the in-plane stresses, and show the results on a sketch of a properly oriented element).
(a)
(c) (a) Element in plane stress; (b) the corresponding Mohr’s circle; (c) stresses acting on a n element oriented at an angle θ = 30° (Note: All stresses on the circle have units of MPa)
(b)
Solution 1:
(1) Construction of Mohr’s circle. Let us set up the axes for the normal and shear stresses, with σ x1 positive to the right and τ x y1 1 positive
downward, as shown in figure (b). Then we place the center C of the circle on the σ x1 axis at the point where the stress equals the average normal stress: σ x + σ y 90 MPa + 20 MPa σ aver = = = 55 MPa . 2 2 Point A, representing the stresses on the x face of the element (θ = 0 ) , has coordinates σ x1 = 90 MPa , τ x1 y1 = 0 . Similarly, the coordinates of point B, representing the stresses on the y face (θ = 90° ) , are σ x1 = 20 MPa , τ x1 y1 = 0 . Now we draw the circle through points A and B with center at C and radius R equal to 2
2 ⎛σ x −σ y ⎞ ⎛ 90 MPa − 20 MPa ⎞ 2 R= ⎜ ⎟ + τ xy = ⎜ ⎟ + 0 = 35 MPa . 2 2 ⎝ ⎠ ⎝ ⎠ (2) Stresses on an element inclined at θ = 30D . The stresses acting on a plane oriented at an angle θ = 30° are given by the coordinates of point D, which is at an angle 2θ = 60° from point A (see figure (b)). By inspection of the circle, we see that the coordinates of point D are σ x1 = σ aver + R cos60° = 55 MPa + ( 35 MPa )( cos60° ) = 72.5 MPa ,
τ x1 y1 = − R cos60° = − ( 35 MPa )( cos60° ) = −30.3 MPa .
In a similar manner, we can find the stresses represented by point D′ , which corresponds to an angle θ = 120° (or 2θ = 240° ): σ x1 = σ aver − R cos 60° = 55 MPa − ( 35 MPa )( cos 60° ) = 37.5 MPa ,
τ x1 y1 = R cos60° = ( 35 MPa )( cos60° ) = 30.3 MPa .
These results are shown in figure (c) on a sketch of an element oriented at an angle θ = 30° , with all stresses shown in their true directions. Note. The sum of the normal stresses on the inclined element is equal to σ x + σ y or 110 MPa.
Problem 2:
An element in plane stress at the surface of a structure is subjected to stresses σ x = 100 MPa, σ y = 35 MPa, and τ xy = 30 MPa, as shown in figure (a). Using Mohr's circle, determine the following quantities: (1) the stresses acting on an element inclined at an angle θ = 40° , (2) the principal stresses, and (3) the maximum shear stresses. Consider only the in-plane stresses, and show all results on sketches of properly oriented elements.
Solution 2:
(1) Construction of Mohr’s circle. Let us set up the axes for Mohr's circle, with σ x1 positive to the right and τ x y1 1 positive downward (see figure (b)). The center C of the circle is located on the σ x1 axis at the point where σ x1 equals the average normal stress:
σ aver =
σx +σ y 2
=
100 MPa + 35 MPa = 67.5 MPa . 2
Point A, representing the stresses on the x face of the element (θ = 0 ) , has coordinates σ x1 = 100 MPa , τ x1 y1 = 30 MPa . Similarly, the coordinates of point B, representing the stresses on the y face (θ = 90 ) , are σ x1 = 35 MPa , τ x1 y1 = −30 MPa . The circle is now drawn through points A and B with center at C. The radius of the circle is 2
2 ⎛σ x −σ y ⎞ ⎛ 100 MPa − 35 MPa ⎞ 2 2 R= ⎜ ⎟ + τ xy = ⎜ ⎟ + ( 30 MPa ) = 44.2 MPa. 2 2 ⎝ ⎠ ⎝ ⎠
(a) Element in plane stress; (b) the corresponding Mohr’s circle; (c) stress acting on an element oriented at
θ = 40D ; (d) principal stresses; (e) maximum shear stresses
(2) The stresses acting on a plane oriented at an angle θ = 40° . They are given by the coordinates of point D, which is at an angle 2θ = 80° from point A (see figure (b)). To calculate these coordinates, we need to know the angle between line CD and the σ x1 axis (that is, angle DCP1 ), which in turn requires that we know the angle between line CA and the σ x1 axis (angle ACP1 ). These angles are found from the geometry of the circle, as follows: 30 MPa tan ACP1 = = 0.857, ACP1 = 40.6°, 35 MPa DCP1 = 80° − ACP1 = 80° − 40.6° = 39.4°. Knowing these angles, we can determine the coordinates of point D directly from the figure: σ x1 = 67.5 MPa + ( 44.2 MPa )( cos39.4° ) = 101.65 MPa,
τ x1 y1 = − ( 44.2 MPa )( sin 39.4° ) = −28.06 MPa .
In an analogous manner, we can find the stresses represented by point D′ , which corresponds to a plane inclined at an angle θ = 130° (or 2θ = 260° ): σ x1 = 67.5 MPa − ( 44.2 MPa )( cos39.4° ) = +33.35 MPa ,
τ x1 y1 = ( 44.2 MPa )( sin 39.4° ) = 28.06 MPa .
These stresses are shown in figure (c) on a sketch of an element oriented at an angle θ = 40° (all stresses are shown in their true directions). Note. The sum of the normal stresses is equal to σ x + σ y or 135 MPa. (3) Principal stresses. The principal stresses are represented by points P1 and P2 on Mohr's circle (see figure (b)). The algebraically larger principal stress (point P1 ) is σ1 = 67.5 MPa + 44.2 MPa = 111.7 MPa ,
as seen by inspection of the circle. The angle 2θ p1 to point P1 from point A is the angle ACP1 on the circle, that is, ACP1 = 2θ p1 = 40.6° , θ p1 = 20.3° .
Thus, the plane of the algebraically larger principal stress is oriented at an angle θ P 1 = 20.3°, as shown in figure (d). The algebraically smaller principal stress (represented by point P2 ) is obtained from the circle in a similar manner:
σ 2 = 67.5 MPa − 44.2 MPa = 23.3 MPa . The angle 2θ P2 to point P2 on the circle is 40.6° + 180° = 220.6° ; thus, the second principal plane is defined by the angle θ P2 = 110.3° . The principal stresses and principal planes are shown in the figure (d). Note. The sum of the normal stresses is equal to 135 MPa. (4) Maximum shear stresses. The maximum shear stresses are represented by points S1 and S2 on Mohr's circle; therefore, the maximum in-plane shear stress (equal to the radius of the circle) is τ max = 44.2 MPa . The angle ACS1 from point A to point S1 is 90° − 40.6° = 49.4° , and therefore the angle 2θ s1 , for point S1 is
2θ s1 = −49.4° . This angle is negative because it is measured clockwise on the circle. The corresponding angle θ s1 to the plane of the maximum positive shear stress is onehalf that value, or θ s1 = −24.7° , as shown in Figs. (b) and (e). The maximum negative shear stress (point S2 on the circle) has the same numerical value as the maximum positive stress ( 44.2 MPa). The normal stresses acting on the planes of maximum shear stress are equal to σ aver , which is the abscissa of the center C of the circle (67.5 MPa). These stresses are also shown in figure (e). Note. The planes of maximum shear stresses are oriented at 45° to the principal planes.
Problem 3:
At a point on the surface of a shaft the stresses are σ x = −50 MPa, σ y = 10 MPa, and τ xy = −40 MPa, as shown in figure (a). Using Mohr's circle, determine the following quantities: (1) the stresses acting on an element inclined at an angle θ = 45° , (2) the principal stresses, and (3) the maximum shear stresses. Solution 3:
(1) Construction of Mohr’s circle. The axes for the normal and shear stresses in the Mohr’s circle are shown in figure (b), with σ x1 positive to the right and τ x1 y1 positive downward. The center C of the circle is located on the
σ x1 axis at the point where the stress equals the average normal stress: σx +σ y
−50 MPa + 10 MPa = −20 MPa . 2 2 Point A, representing the stresses on the x face of the element (θ = 0 ) , has
σ aver =
coordinates
=
σ x1 = −50 MPa , τ x1 y1 = −40 MPa .
Similarly, the coordinates of point B, representing the stresses on the y face (θ = 90° ) , are
σ x1 = 10 MPa , τ x1 y1 = 40 MPa . The circle is now drawn through points A and B with center at C and radius R equal to: 2
2 ⎛σ x −σ y ⎞ ⎛ −50 MPa − 10 MPa ⎞ 2 2 R= ⎜ ⎟ + τ xy = ⎜ ⎟ + ( −40 MPa ) = 50 MPa . 2 2 ⎝ ⎠ ⎝ ⎠
(2) Stresses on an element inclined at θ = 45° . These stresses are given by the coordinates of point D, which is at an angle 2θ = 90° from point A (figure (b)). To evaluate these coordinates, we need to know the angle between line CD and the negative σ x1 axis (that is, angle DCP2 ), which in turn requires that we
know the angle between line CA and the negative σ x1 axis (angle ACP2 ). These angles are found from the geometry of the circle as follows: 40 MPa 4 tan ACP2 = = , ACP2 = 53.13° , 30 MPa 3 DCP2 = 90° − ACP2 = 90° − 53.13° = 36.87° .
(a) Element in plane stress; (b) the corresponding Mohr’s circle; (c) stresses acting on an element oriented at θ=45°; (d) principal stresses, and (e) maximum shear stresses. (Note: All stresses on the circle have units of MPa)
Knowing these angles, we can obtain the coordinates of point D directly from the figure: σ x1 = −20 MPa − ( 50 MPa )( cos36.87° ) = −60 MPa ,
τ x1 y1 = ( 50 MPa )( sin 36.87° ) = 30 MPa .
In an analogous manner, we can find the stresses represented by point D′ , which corresponds to a plane inclined at an angle θ = 135° (or 2θ = 270° ): σ x1 = −20 MPa + ( 50 MPa )( cos36.87° ) = 20 MPa ,
τ x1 y1 = ( −50 MPa )( sin 36.87° ) = −30 MPa .
These stresses are shown in Fig. c on a sketch of an element oriented at an angle θ = 45° (all stresses are shown in their true directions).
Note. The sum of the normal stresses is equal to σ x + σ y or – 40 MPa. (3) Principal stresses. They are represented by points P1 and P2 on Mohr's circle. The algebraically larger principal stress (represented by point P1 ) is σ1 = −20 MPa + 50 MPa = 30 MPa , as seen by inspection of the circle. The angle 2θ p1 to point P1 from point A is the
angle ACP1 measured counterclockwise on the circle, that is, ACP1 = 2θ p1 = 53.13° + 180° = 233.13° , θ p1 = 116.6° . Thus, the plane of the algebraically larger principal stress is oriented at an angle θ p1 = 116.6° .
The algebraically smaller principal stress (point P2 ) is obtained from the circle in a similar manner: σ 3 = −20 MPa − 50 MPa = −70 MPa . The angle 2θ p2 to point P2 on the circle is 53.13°. The second principal plane is defined by the angle 2θ p2 = 26.6° . The principal stresses and principal planes are shown in Fig. (d). Note. The sum of the normal stresses is equal to σ x + σ y or – 40 MPa.
(4) Maximum shear stresses. The maximum positive and negative shear stresses are represented by points S2 and S2 on Mohr's circle (figure (b)). Their magnitudes, equal to the radius of the circle, are τ max = 50 MPa . The angle ACS1 from point A to point S1 is 90° + 53.13° = 143.13° , and therefore the angle 2θ s1 for point S1 is 2θ s1 = 143.13° .
The corresponding angle θ s1 to the plane of the maximum positive shear stress is one-half that value, or θ s1 = 71.6° , as shown in figure (e). The maximum negative shear stress (point S2 on the circle) has the same numerical value as the positive stress (50 MPa). The normal stresses acting on the planes of maximum shear stress are equal to σ aver , which is the coordinate of the center C of the circle ( −20 MPa). These stresses are also shown in figure (e). Note. The planes of maximum shear stress are oriented at 45° to the principal planes.
At a point on the surface of a cylinder, loaded by internal pressure, the material is subjected to biaxial stresses σ x = 90 MPa and σ y = 20MPa, as shown on the stress element of figure (a). Using Mohr's circle, determine the stresses acting on an element inclined at an angle θ = 30° . (Consider only the in-plane stresses, and show the results on a sketch of a properly oriented element).
(a)
(c) (a) Element in plane stress; (b) the corresponding Mohr’s circle; (c) stresses acting on a n element oriented at an angle θ = 30° (Note: All stresses on the circle have units of MPa)
(b)
Solution 1:
(1) Construction of Mohr’s circle. Let us set up the axes for the normal and shear stresses, with σ x1 positive to the right and τ x y1 1 positive
downward, as shown in figure (b). Then we place the center C of the circle on the σ x1 axis at the point where the stress equals the average normal stress: σ x + σ y 90 MPa + 20 MPa σ aver = = = 55 MPa . 2 2 Point A, representing the stresses on the x face of the element (θ = 0 ) , has coordinates σ x1 = 90 MPa , τ x1 y1 = 0 . Similarly, the coordinates of point B, representing the stresses on the y face (θ = 90° ) , are σ x1 = 20 MPa , τ x1 y1 = 0 . Now we draw the circle through points A and B with center at C and radius R equal to 2
2 ⎛σ x −σ y ⎞ ⎛ 90 MPa − 20 MPa ⎞ 2 R= ⎜ ⎟ + τ xy = ⎜ ⎟ + 0 = 35 MPa . 2 2 ⎝ ⎠ ⎝ ⎠ (2) Stresses on an element inclined at θ = 30D . The stresses acting on a plane oriented at an angle θ = 30° are given by the coordinates of point D, which is at an angle 2θ = 60° from point A (see figure (b)). By inspection of the circle, we see that the coordinates of point D are σ x1 = σ aver + R cos60° = 55 MPa + ( 35 MPa )( cos60° ) = 72.5 MPa ,
τ x1 y1 = − R cos60° = − ( 35 MPa )( cos60° ) = −30.3 MPa .
In a similar manner, we can find the stresses represented by point D′ , which corresponds to an angle θ = 120° (or 2θ = 240° ): σ x1 = σ aver − R cos 60° = 55 MPa − ( 35 MPa )( cos 60° ) = 37.5 MPa ,
τ x1 y1 = R cos60° = ( 35 MPa )( cos60° ) = 30.3 MPa .
These results are shown in figure (c) on a sketch of an element oriented at an angle θ = 30° , with all stresses shown in their true directions. Note. The sum of the normal stresses on the inclined element is equal to σ x + σ y or 110 MPa.
Problem 2:
An element in plane stress at the surface of a structure is subjected to stresses σ x = 100 MPa, σ y = 35 MPa, and τ xy = 30 MPa, as shown in figure (a). Using Mohr's circle, determine the following quantities: (1) the stresses acting on an element inclined at an angle θ = 40° , (2) the principal stresses, and (3) the maximum shear stresses. Consider only the in-plane stresses, and show all results on sketches of properly oriented elements.
Solution 2:
(1) Construction of Mohr’s circle. Let us set up the axes for Mohr's circle, with σ x1 positive to the right and τ x y1 1 positive downward (see figure (b)). The center C of the circle is located on the σ x1 axis at the point where σ x1 equals the average normal stress:
σ aver =
σx +σ y 2
=
100 MPa + 35 MPa = 67.5 MPa . 2
Point A, representing the stresses on the x face of the element (θ = 0 ) , has coordinates σ x1 = 100 MPa , τ x1 y1 = 30 MPa . Similarly, the coordinates of point B, representing the stresses on the y face (θ = 90 ) , are σ x1 = 35 MPa , τ x1 y1 = −30 MPa . The circle is now drawn through points A and B with center at C. The radius of the circle is 2
2 ⎛σ x −σ y ⎞ ⎛ 100 MPa − 35 MPa ⎞ 2 2 R= ⎜ ⎟ + τ xy = ⎜ ⎟ + ( 30 MPa ) = 44.2 MPa. 2 2 ⎝ ⎠ ⎝ ⎠
(a) Element in plane stress; (b) the corresponding Mohr’s circle; (c) stress acting on an element oriented at
θ = 40D ; (d) principal stresses; (e) maximum shear stresses
(2) The stresses acting on a plane oriented at an angle θ = 40° . They are given by the coordinates of point D, which is at an angle 2θ = 80° from point A (see figure (b)). To calculate these coordinates, we need to know the angle between line CD and the σ x1 axis (that is, angle DCP1 ), which in turn requires that we know the angle between line CA and the σ x1 axis (angle ACP1 ). These angles are found from the geometry of the circle, as follows: 30 MPa tan ACP1 = = 0.857, ACP1 = 40.6°, 35 MPa DCP1 = 80° − ACP1 = 80° − 40.6° = 39.4°. Knowing these angles, we can determine the coordinates of point D directly from the figure: σ x1 = 67.5 MPa + ( 44.2 MPa )( cos39.4° ) = 101.65 MPa,
τ x1 y1 = − ( 44.2 MPa )( sin 39.4° ) = −28.06 MPa .
In an analogous manner, we can find the stresses represented by point D′ , which corresponds to a plane inclined at an angle θ = 130° (or 2θ = 260° ): σ x1 = 67.5 MPa − ( 44.2 MPa )( cos39.4° ) = +33.35 MPa ,
τ x1 y1 = ( 44.2 MPa )( sin 39.4° ) = 28.06 MPa .
These stresses are shown in figure (c) on a sketch of an element oriented at an angle θ = 40° (all stresses are shown in their true directions). Note. The sum of the normal stresses is equal to σ x + σ y or 135 MPa. (3) Principal stresses. The principal stresses are represented by points P1 and P2 on Mohr's circle (see figure (b)). The algebraically larger principal stress (point P1 ) is σ1 = 67.5 MPa + 44.2 MPa = 111.7 MPa ,
as seen by inspection of the circle. The angle 2θ p1 to point P1 from point A is the angle ACP1 on the circle, that is, ACP1 = 2θ p1 = 40.6° , θ p1 = 20.3° .
Thus, the plane of the algebraically larger principal stress is oriented at an angle θ P 1 = 20.3°, as shown in figure (d). The algebraically smaller principal stress (represented by point P2 ) is obtained from the circle in a similar manner:
σ 2 = 67.5 MPa − 44.2 MPa = 23.3 MPa . The angle 2θ P2 to point P2 on the circle is 40.6° + 180° = 220.6° ; thus, the second principal plane is defined by the angle θ P2 = 110.3° . The principal stresses and principal planes are shown in the figure (d). Note. The sum of the normal stresses is equal to 135 MPa. (4) Maximum shear stresses. The maximum shear stresses are represented by points S1 and S2 on Mohr's circle; therefore, the maximum in-plane shear stress (equal to the radius of the circle) is τ max = 44.2 MPa . The angle ACS1 from point A to point S1 is 90° − 40.6° = 49.4° , and therefore the angle 2θ s1 , for point S1 is
2θ s1 = −49.4° . This angle is negative because it is measured clockwise on the circle. The corresponding angle θ s1 to the plane of the maximum positive shear stress is onehalf that value, or θ s1 = −24.7° , as shown in Figs. (b) and (e). The maximum negative shear stress (point S2 on the circle) has the same numerical value as the maximum positive stress ( 44.2 MPa). The normal stresses acting on the planes of maximum shear stress are equal to σ aver , which is the abscissa of the center C of the circle (67.5 MPa). These stresses are also shown in figure (e). Note. The planes of maximum shear stresses are oriented at 45° to the principal planes.
Problem 3:
At a point on the surface of a shaft the stresses are σ x = −50 MPa, σ y = 10 MPa, and τ xy = −40 MPa, as shown in figure (a). Using Mohr's circle, determine the following quantities: (1) the stresses acting on an element inclined at an angle θ = 45° , (2) the principal stresses, and (3) the maximum shear stresses. Solution 3:
(1) Construction of Mohr’s circle. The axes for the normal and shear stresses in the Mohr’s circle are shown in figure (b), with σ x1 positive to the right and τ x1 y1 positive downward. The center C of the circle is located on the
σ x1 axis at the point where the stress equals the average normal stress: σx +σ y
−50 MPa + 10 MPa = −20 MPa . 2 2 Point A, representing the stresses on the x face of the element (θ = 0 ) , has
σ aver =
coordinates
=
σ x1 = −50 MPa , τ x1 y1 = −40 MPa .
Similarly, the coordinates of point B, representing the stresses on the y face (θ = 90° ) , are
σ x1 = 10 MPa , τ x1 y1 = 40 MPa . The circle is now drawn through points A and B with center at C and radius R equal to: 2
2 ⎛σ x −σ y ⎞ ⎛ −50 MPa − 10 MPa ⎞ 2 2 R= ⎜ ⎟ + τ xy = ⎜ ⎟ + ( −40 MPa ) = 50 MPa . 2 2 ⎝ ⎠ ⎝ ⎠
(2) Stresses on an element inclined at θ = 45° . These stresses are given by the coordinates of point D, which is at an angle 2θ = 90° from point A (figure (b)). To evaluate these coordinates, we need to know the angle between line CD and the negative σ x1 axis (that is, angle DCP2 ), which in turn requires that we
know the angle between line CA and the negative σ x1 axis (angle ACP2 ). These angles are found from the geometry of the circle as follows: 40 MPa 4 tan ACP2 = = , ACP2 = 53.13° , 30 MPa 3 DCP2 = 90° − ACP2 = 90° − 53.13° = 36.87° .
(a) Element in plane stress; (b) the corresponding Mohr’s circle; (c) stresses acting on an element oriented at θ=45°; (d) principal stresses, and (e) maximum shear stresses. (Note: All stresses on the circle have units of MPa)
Knowing these angles, we can obtain the coordinates of point D directly from the figure: σ x1 = −20 MPa − ( 50 MPa )( cos36.87° ) = −60 MPa ,
τ x1 y1 = ( 50 MPa )( sin 36.87° ) = 30 MPa .
In an analogous manner, we can find the stresses represented by point D′ , which corresponds to a plane inclined at an angle θ = 135° (or 2θ = 270° ): σ x1 = −20 MPa + ( 50 MPa )( cos36.87° ) = 20 MPa ,
τ x1 y1 = ( −50 MPa )( sin 36.87° ) = −30 MPa .
These stresses are shown in Fig. c on a sketch of an element oriented at an angle θ = 45° (all stresses are shown in their true directions).
Note. The sum of the normal stresses is equal to σ x + σ y or – 40 MPa. (3) Principal stresses. They are represented by points P1 and P2 on Mohr's circle. The algebraically larger principal stress (represented by point P1 ) is σ1 = −20 MPa + 50 MPa = 30 MPa , as seen by inspection of the circle. The angle 2θ p1 to point P1 from point A is the
angle ACP1 measured counterclockwise on the circle, that is, ACP1 = 2θ p1 = 53.13° + 180° = 233.13° , θ p1 = 116.6° . Thus, the plane of the algebraically larger principal stress is oriented at an angle θ p1 = 116.6° .
The algebraically smaller principal stress (point P2 ) is obtained from the circle in a similar manner: σ 3 = −20 MPa − 50 MPa = −70 MPa . The angle 2θ p2 to point P2 on the circle is 53.13°. The second principal plane is defined by the angle 2θ p2 = 26.6° . The principal stresses and principal planes are shown in Fig. (d). Note. The sum of the normal stresses is equal to σ x + σ y or – 40 MPa.
(4) Maximum shear stresses. The maximum positive and negative shear stresses are represented by points S2 and S2 on Mohr's circle (figure (b)). Their magnitudes, equal to the radius of the circle, are τ max = 50 MPa . The angle ACS1 from point A to point S1 is 90° + 53.13° = 143.13° , and therefore the angle 2θ s1 for point S1 is 2θ s1 = 143.13° .
The corresponding angle θ s1 to the plane of the maximum positive shear stress is one-half that value, or θ s1 = 71.6° , as shown in figure (e). The maximum negative shear stress (point S2 on the circle) has the same numerical value as the positive stress (50 MPa). The normal stresses acting on the planes of maximum shear stress are equal to σ aver , which is the coordinate of the center C of the circle ( −20 MPa). These stresses are also shown in figure (e). Note. The planes of maximum shear stress are oriented at 45° to the principal planes.
