Page 244 - Marks Calculation for Machine Design

P1: Shibu/Sanjay
January 4, 2005
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U.S. Customary 14:35 STRENGTH OF MACHINES SI/Metric
Example 3. For normal stresses (σ) and (2σ) Example 3. For normal stresses (σ) and (2σ)
acting on a biaxial stress element, find the acting on a biaxial stress element, find the
principal stresses (σ 1 ) and (σ 2 ), the maximum principal stresses (σ 1 ) and (σ 2 ), the maximum
and minimum shear stresses (τ max ) and (τ min ), and minimum shear stresses (τ max ) and (τ min ),
and the special angles (φ p ) and (φ s ), using and the special angles (φ p ) and (φ s ), using
the graphical Mohr’s circle process shown in the graphical Mohr’s circle process shown in
Fig. 5.39 through 5.44, where Fig. 5.39 through 5.44, where
σ = 8 kpsi and 2σ = 16 kpsi σ = 56 MPa and 2σ = 112 MPa
solution solution
Step 1. Plot points (8,0) and (16,0) as in Step 1. Plot points (56,0) and (112,0) as in
Fig. 5.39, and locate the center of Mohr’s circle, Fig. 5.39, and locate the center of Mohr’s circle,
which is the average stress, like that shown in which is the average stress, like that shown in
Fig. 5.40. Fig. 5.40.
σ + 2σ (8 + 16) kpsi σ + 2σ (56 + 112) MPa
σ avg = = σ avg = =
2 2 2 2
24 kpsi 168 MPa
= = 12 kpsi = = 84 MPa
2 2
Step 2. Draw Mohr’s circle like that shown Step 2. Draw Mohr’s circle like that shown
in Fig. 5.41 using a radius of (4 kpsi), so that in Fig. 5.41 using a radius of (28 MPa), so that
where the circle crosses the (σ) axis it gives the where the circle crosses the (σ) axis it gives the
principal stresses like that shown in Fig. 5.42. principal stresses like that shown in Fig. 5.42.
σ 1 = σ avg + τ max = (12 + 4) kpsi σ 1 = σ avg + τ max = (84 + 28) MPa
= 16 kpsi = 112 MPa
σ 2 = σ avg − τ max = (12 − 4) kpsi σ 2 = σ avg − τ max = (84 − 28) MPa
= 8 kpsi = 56 MPa
Step3. FromFig.5.43,themaximumandmin- Step3. FromFig.5.43,themaximumandmin-
imum shear stresses are shown 90 to the prin- imum shear stresses are shown 90 to the prin-
◦
◦
cipal stresses, and equal to the following value. cipal stresses, and equal to the following value.
σ 8 kpsi σ 56 MPa
τ max = = = 4 kpsi τ max = = = 28 MPa
2 2 2 2
τ min =−τ max =−4 kpsi τ min =−τ max =−28 MPa
Step 4. As the biaxial stress element is actually Step 4. As the biaxial stress element is actually
the principal stress element, the rotation angle the principal stress element, the rotation angle
(2φ p ), and therefore the angle (φ p ), is zero. (2φ p ), and therefore the angle (φ p ), is zero.
2φ p = 0or φ p = 0 2φ p = 0or φ p = 0
Step 5. Using Fig. 5.44, the rotation angle Step 5. Using Fig. 5.44, the rotation angle
◦
◦
(2φ s )forthemaximumshearstressis90 clock- (2φ s )forthemaximumshearstressis90 clock-
wise, or negative, meaning wise, or negative, meaning
◦
2φ s = 2φ p − 90 = 0 − 90 =−90 ◦ 2φ s = 2φ p − 90 = 0 − 90 =−90 ◦
◦
◦
◦
φ s =−45 ◦ φ s =−45 ◦