16 Principle Stresses Examples
Problem 1:
An element in plane stress is subjected to stresses σ x = 84.8 MPa, σ y = −28.9 MPa, and τ xy = −32.4 MPa, as shown in Fig. a. (1) Determine the principal stresses and show them on a sketch of a properly oriented element; (2) Determine the maximum shear stresses and show them on a sketch of a properly oriented element.
(a) Element in plane stress; (b) principal stresses; and (c) maximum shear stresses
Solution 1:
(1) Calculation of principal stresses. The principal angles θ p that locate the principal planes can be obtained from Eq. (3.19): 2τ xy 2 ( −32.4 MPa ) tan 2θ p = = = −0.5697. σ x − σ y 84.8 MPa − ( −28.9 MPa ) Solving for the angles, we get the following two sets of values: 2θ p = 150.3° and θ p = 75.2°, 2θ p = 330.3° and θ p = 165.2°. The principal stresses may be obtained by substituting the two values of 2θ p into the transformation equation for σ x1 (Eq. (3.8)). Determine preliminary the following quantities: σ x + σ y 84.8 MPa − 28.9 MPa = = 27.9 MPa , A= 2 2 σ x − σ y 84.8 MPa + 28.9 MPa B= = = 56.8 MPa . 2 2 Now we substitute the first value of 2θ p into Eq. (3.8) and obtain σx +σ y σx −σ y cos 2θ + τ xy sin 2θ = σ x1 = + 2 2 = 27.9 MPa + ( 56.8 MPa )( cos150.3° ) − ( 32.4 MPa )( sin150.3° ) = −37.5 MPa . By the similar way, we substitute the second value of 2θ p and obtain σ x1 = 93.4 MPa. In result, the principal stresses and their corresponding principal angles are σ1 = 93.4 MPa and θ p1 = 165.2°
σ 3 = −37.5 MPa and θ p2 = 75.2° .
Keep in mind, that σ 2 = 0 acts in z direction. Note that θ p1 and θ p 2 differ by 90° and that σ1 + σ 3 = σ x + σ y . The principal stresses are shown on a properly oriented element in the Fig. b. Of course, the principal planes are free from shear stresses. The principal stresses may also be calculated directly from Eq. (3.26):
σ1,2(3) =
σx +σ y
= 27.9 MPa ±
2
2
⎛σx −σ y ⎞ 2 ± ⎜ ⎟ + τ xy = 2 ⎝ ⎠
( 56.8 MPa )2 + ( −32.4 MPa )2 ,
σ1,2(3) = 27.9 MPa ± 65.4 MPa .
Therefore,
σ1 = 93.4 MPa , σ 3 = −37.5 MPa , (σ 2 = 0 ) .
(2) Maximum shear stresses. The maximum in-plane shear stresses are given by Eq. (3.38): 2
⎛σ x −σ y ⎞ 2 τ max = ⎜ ⎟ + τ xy = 2 ⎝ ⎠
( 56.8 MPa )2 + ( −32.4 MPa )2 = 65.4 MPa .
The angle θ s1 to the plane having the maximum positive shear stress is calculated from Eq. (3.37): θ s1 = θ p1 − 45° = 165.2° − 45° = 120.2° . It follows that the maximum negative shear stress acts on the plane for which θ s2 = 120.2° − 90° = 30.2° . The normal stresses acting on the planes of maximum shear stresses are calculated from Eq. (3.40): σx +σ y σ aver = = 27.9 MPa . 2 Finally, the maximum shear stresses and associated normal stresses are shown on the stress element of Fig. c. Problem 2:
The plane stress state is described in the figure. (1) Write the principalstress-transformation formulas. (2) Calculate the principal stresses. (3) Calculate the maximum shearing stresses and the associated normal stresses. Sketch the results on properly oriented elements. Solution 2: Solution (1) Principal-stress-transformation formulas. Equations (3.26), (3.30), and (3.8) are written as
( ) , 1/2 σ 3 = A − ( B2 + C 2 ) , 2
σ1 = A + B + C
2 1/2
1 C , 2 B σ x1 = A + B cos 2θ + C sin 2θ ,
θ p = arctan
where A=
(
)
(
)
1 1 σ x + σ y , B = σ x − σ y , C = τ xy . 2 2
(2) Calculation of principal stresses. The principal angles θ p that locate the principal planes can be obtained from Eq. (3.19): tan 2θ p =
2τ xy
σx −σ y
=
2 ( +5 MPa ) = +2.0. 7 MPa − ( +2 MPa )
Solving for the angles, we get the following two sets of values: 2θ p = 63.4° and θ p = 31.7°, 2θ p = 243.4° and θ p = 121.7°. The principal stresses may be obtained by substituting the two values of 2θ p into the transformation equation for σ x1 (Eq. 3.8). Determine preliminary the following quantities: A= B=
σx +σ y 2
σx −σ y 2
=
7 MPa + 2 MPa = 4.5 MPa, 2
=
7 MPa − 2 MPa = 2.5 MPa. 2
Now we substitute the first value of 2θ p into Eq. (3.8) and obtain σx +σ y σx −σ y cos 2θ + τ xy sin 2θ = σ x1 = + 2 2 = A + B cos 2θ + C sin 2θ = = 4.5 MPa + ( 2.5 MPa )( cos63.4° ) + ( 5 MPa )( sin 63.4° ) = +10.09 MPa. By the similar way, we substitute the second value of 2θ p and obtain σ x1 = −1.09 MPa. In result, the principal stresses and their corresponding principal angles are
σ1 = +10.09 MPa and θ p1 = 31.7° σ 3 = −1.09 MPa and θ p2 = 121.7° . Note that θ p1 and θ p 2 differ by 90° and that σ1 + σ 3 = σ x + σ y . The principal stresses are shown on a properly oriented element in Fig. b. Of course, the principal planes are free from shear stresses. The principal stresses may also be calculated directly from the Eq. (3.26):
σ1,2(3) =
σx +σ y 2
2
⎛σx −σ y ⎞ 2 2 2 ± ⎜ ⎟ + τ xy = A ± B + C = 2 ⎝ ⎠
= 4.5 MPa ±
( 2.5 MPa )2 + ( 5.0 MPa )2 .
σ1,2(3) = 4.5 MPa ± 5.59 MPa.
Therefore,
σ1 = 10.09 MPa , σ 3 = −1.09 MPa
(σ 2 = 0 ) .
(3) Maximum shear stresses. The maximum in-plane shear stresses are given by Eq. (3.38): 2
2 ⎛σx −σ y ⎞ ⎛7−2⎞ 2 2 τ max = ⎜ ⎟ + τ xy = ⎜ ⎟ + 5 = 5.59MPa. 2 ⎝ 2 ⎠ ⎝ ⎠
The angle θ s1 to the plane having the maximum positive shear stress is calculated from Eq. (3.37): θ s1 = θ p1 − 45° = 31.7° − 45° = −13.3° . It follows that the maximum negative shear stress acts on the plane for which θ s2 = −13.3° + 90° = 76.7° . The normal stresses acting on the planes of maximum shear stresses are calculated from Eq. (3.40): σx +σ y = 4.5 MPa. σ aver = 2 Finally, the maximum shear stresses and associated normal stresses are shown on the stress element of Fig. c.
Note. The direction of the τ max may also be readily predicted by recalling that they act toward the shear diagonal.
An element in plane stress is subjected to stresses σ x = 84.8 MPa, σ y = −28.9 MPa, and τ xy = −32.4 MPa, as shown in Fig. a. (1) Determine the principal stresses and show them on a sketch of a properly oriented element; (2) Determine the maximum shear stresses and show them on a sketch of a properly oriented element.
(a) Element in plane stress; (b) principal stresses; and (c) maximum shear stresses
Solution 1:
(1) Calculation of principal stresses. The principal angles θ p that locate the principal planes can be obtained from Eq. (3.19): 2τ xy 2 ( −32.4 MPa ) tan 2θ p = = = −0.5697. σ x − σ y 84.8 MPa − ( −28.9 MPa ) Solving for the angles, we get the following two sets of values: 2θ p = 150.3° and θ p = 75.2°, 2θ p = 330.3° and θ p = 165.2°. The principal stresses may be obtained by substituting the two values of 2θ p into the transformation equation for σ x1 (Eq. (3.8)). Determine preliminary the following quantities: σ x + σ y 84.8 MPa − 28.9 MPa = = 27.9 MPa , A= 2 2 σ x − σ y 84.8 MPa + 28.9 MPa B= = = 56.8 MPa . 2 2 Now we substitute the first value of 2θ p into Eq. (3.8) and obtain σx +σ y σx −σ y cos 2θ + τ xy sin 2θ = σ x1 = + 2 2 = 27.9 MPa + ( 56.8 MPa )( cos150.3° ) − ( 32.4 MPa )( sin150.3° ) = −37.5 MPa . By the similar way, we substitute the second value of 2θ p and obtain σ x1 = 93.4 MPa. In result, the principal stresses and their corresponding principal angles are σ1 = 93.4 MPa and θ p1 = 165.2°
σ 3 = −37.5 MPa and θ p2 = 75.2° .
Keep in mind, that σ 2 = 0 acts in z direction. Note that θ p1 and θ p 2 differ by 90° and that σ1 + σ 3 = σ x + σ y . The principal stresses are shown on a properly oriented element in the Fig. b. Of course, the principal planes are free from shear stresses. The principal stresses may also be calculated directly from Eq. (3.26):
σ1,2(3) =
σx +σ y
= 27.9 MPa ±
2
2
⎛σx −σ y ⎞ 2 ± ⎜ ⎟ + τ xy = 2 ⎝ ⎠
( 56.8 MPa )2 + ( −32.4 MPa )2 ,
σ1,2(3) = 27.9 MPa ± 65.4 MPa .
Therefore,
σ1 = 93.4 MPa , σ 3 = −37.5 MPa , (σ 2 = 0 ) .
(2) Maximum shear stresses. The maximum in-plane shear stresses are given by Eq. (3.38): 2
⎛σ x −σ y ⎞ 2 τ max = ⎜ ⎟ + τ xy = 2 ⎝ ⎠
( 56.8 MPa )2 + ( −32.4 MPa )2 = 65.4 MPa .
The angle θ s1 to the plane having the maximum positive shear stress is calculated from Eq. (3.37): θ s1 = θ p1 − 45° = 165.2° − 45° = 120.2° . It follows that the maximum negative shear stress acts on the plane for which θ s2 = 120.2° − 90° = 30.2° . The normal stresses acting on the planes of maximum shear stresses are calculated from Eq. (3.40): σx +σ y σ aver = = 27.9 MPa . 2 Finally, the maximum shear stresses and associated normal stresses are shown on the stress element of Fig. c. Problem 2:
The plane stress state is described in the figure. (1) Write the principalstress-transformation formulas. (2) Calculate the principal stresses. (3) Calculate the maximum shearing stresses and the associated normal stresses. Sketch the results on properly oriented elements. Solution 2: Solution (1) Principal-stress-transformation formulas. Equations (3.26), (3.30), and (3.8) are written as
( ) , 1/2 σ 3 = A − ( B2 + C 2 ) , 2
σ1 = A + B + C
2 1/2
1 C , 2 B σ x1 = A + B cos 2θ + C sin 2θ ,
θ p = arctan
where A=
(
)
(
)
1 1 σ x + σ y , B = σ x − σ y , C = τ xy . 2 2
(2) Calculation of principal stresses. The principal angles θ p that locate the principal planes can be obtained from Eq. (3.19): tan 2θ p =
2τ xy
σx −σ y
=
2 ( +5 MPa ) = +2.0. 7 MPa − ( +2 MPa )
Solving for the angles, we get the following two sets of values: 2θ p = 63.4° and θ p = 31.7°, 2θ p = 243.4° and θ p = 121.7°. The principal stresses may be obtained by substituting the two values of 2θ p into the transformation equation for σ x1 (Eq. 3.8). Determine preliminary the following quantities: A= B=
σx +σ y 2
σx −σ y 2
=
7 MPa + 2 MPa = 4.5 MPa, 2
=
7 MPa − 2 MPa = 2.5 MPa. 2
Now we substitute the first value of 2θ p into Eq. (3.8) and obtain σx +σ y σx −σ y cos 2θ + τ xy sin 2θ = σ x1 = + 2 2 = A + B cos 2θ + C sin 2θ = = 4.5 MPa + ( 2.5 MPa )( cos63.4° ) + ( 5 MPa )( sin 63.4° ) = +10.09 MPa. By the similar way, we substitute the second value of 2θ p and obtain σ x1 = −1.09 MPa. In result, the principal stresses and their corresponding principal angles are
σ1 = +10.09 MPa and θ p1 = 31.7° σ 3 = −1.09 MPa and θ p2 = 121.7° . Note that θ p1 and θ p 2 differ by 90° and that σ1 + σ 3 = σ x + σ y . The principal stresses are shown on a properly oriented element in Fig. b. Of course, the principal planes are free from shear stresses. The principal stresses may also be calculated directly from the Eq. (3.26):
σ1,2(3) =
σx +σ y 2
2
⎛σx −σ y ⎞ 2 2 2 ± ⎜ ⎟ + τ xy = A ± B + C = 2 ⎝ ⎠
= 4.5 MPa ±
( 2.5 MPa )2 + ( 5.0 MPa )2 .
σ1,2(3) = 4.5 MPa ± 5.59 MPa.
Therefore,
σ1 = 10.09 MPa , σ 3 = −1.09 MPa
(σ 2 = 0 ) .
(3) Maximum shear stresses. The maximum in-plane shear stresses are given by Eq. (3.38): 2
2 ⎛σx −σ y ⎞ ⎛7−2⎞ 2 2 τ max = ⎜ ⎟ + τ xy = ⎜ ⎟ + 5 = 5.59MPa. 2 ⎝ 2 ⎠ ⎝ ⎠
The angle θ s1 to the plane having the maximum positive shear stress is calculated from Eq. (3.37): θ s1 = θ p1 − 45° = 31.7° − 45° = −13.3° . It follows that the maximum negative shear stress acts on the plane for which θ s2 = −13.3° + 90° = 76.7° . The normal stresses acting on the planes of maximum shear stresses are calculated from Eq. (3.40): σx +σ y = 4.5 MPa. σ aver = 2 Finally, the maximum shear stresses and associated normal stresses are shown on the stress element of Fig. c.
Note. The direction of the τ max may also be readily predicted by recalling that they act toward the shear diagonal.
