Problem 1.8. Mohr’s circle Draw Mohr’s circle for the state of stress defined by
= 80 MPa,
= -20 MPa and
= 40 MPa. Using this circle, (1) calculate the stress on axes rotated 60 degrees counterclock- wise from the reference axes, and (2) determine the principal stresses and the corresponding directions. Do these results agree with the results in section 1.3.3? Problem 1.9. Mohr’s circle for the state of pure shear Draw Mohr’s circle for the state of pure shear defined in section 1.3.5. Show how eq. (1.51) can be readily derived from Mohr’s circle. Problem 1.10. Mohr’s circle for the hydrostatic state of stress Draw Mohr’s circle for the state of hydrostatic stress defined in section 1.3.5. Show how eq. (1.50) can be readily derived from Mohr’s circle. Problem 1.11. Stresses in a pressure vessel A cylindrical pressure vessel of radius R and thickness t is subjected to an internal pressure
. At any point in the cylindrical portion of vessel wall, two stress components are acting: the hoop stress,
and the axial stress,
. The radial stress, acting in the direction perpendicular to the wall, is very small,
. The pressure vessel features a weld line at a 45 degree angle with respect to the axis of the cylinder, as shown in fig. 1.19. (1) Find the direct stress acting in the direction perpendicular to the weld line. (2) Find the shear stress acting along the weld line.
The Correct Answer and Explanation is:
Let’s break down the problems one by one, beginning with Problem 1.8 and its associated parts.
Problem 1.8: Mohr’s Circle for the State of Stress
Given:
- σx=80 MPa\sigma_x = 80 \, \text{MPa}σx=80MPa
- σy=−20 MPa\sigma_y = -20 \, \text{MPa}σy=−20MPa
- τxy=40 MPa\tau_{xy} = 40 \, \text{MPa}τxy=40MPa
To construct Mohr’s Circle, we will follow these steps:
Step 1: Draw the axes for Mohr’s Circle
The horizontal axis represents normal stress, σ\sigmaσ, and the vertical axis represents shear stress, τ\tauτ.
- The center of Mohr’s circle is located at: σavg=σx+σy2=80+(−20)2=30 MPa\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{80 + (-20)}{2} = 30 \, \text{MPa}σavg=2σx+σy=280+(−20)=30MPa
- The radius of the circle is: R=(σx−σy2)2+τxy2=(80−(−20)2)2+402=502+402=64 MPaR = \sqrt{\left( \frac{\sigma_x – \sigma_y}{2} \right)^2 + \tau_{xy}^2} = \sqrt{\left( \frac{80 – (-20)}{2} \right)^2 + 40^2} = \sqrt{50^2 + 40^2} = 64 \, \text{MPa}R=(2σx−σy)2+τxy2=(280−(−20))2+402=502+402=64MPa
Step 2: Mark the points on the circle
- (σx,τxy)=(80,40)(\sigma_x, \tau_{xy}) = (80, 40)(σx,τxy)=(80,40)
- (σy,−τxy)=(−20,−40)(\sigma_y, -\tau_{xy}) = (-20, -40)(σy,−τxy)=(−20,−40)
These points lie on the Mohr’s circle. You can sketch the circle by connecting these two points with the center at σavg=30\sigma_{\text{avg}} = 30σavg=30.
Step 3: Calculate the stress on rotated axes
For part (1), we are asked to find the stresses on axes rotated by 60 degrees counterclockwise. Mohr’s Circle allows us to find these stresses using the following relationships:
- The angle on Mohr’s circle corresponding to a physical rotation by θ\thetaθ is 2θ2\theta2θ. So for θ=60∘\theta = 60^\circθ=60∘, the corresponding angle is 2×60∘=120∘2 \times 60^\circ = 120^\circ2×60∘=120∘.
- Using the circle:
- Normal stress on the rotated axis σθ=σavg+Rcos(2θ)\sigma_{\theta} = \sigma_{\text{avg}} + R \cos(2\theta)σθ=σavg+Rcos(2θ)
- Shear stress on the rotated axis τθ=Rsin(2θ)\tau_{\theta} = R \sin(2\theta)τθ=Rsin(2θ)
Substitute values:
- σθ=30+64cos(120∘)=30+64×(−0.5)=30−32=−2 MPa\sigma_{\theta} = 30 + 64 \cos(120^\circ) = 30 + 64 \times (-0.5) = 30 – 32 = -2 \, \text{MPa}σθ=30+64cos(120∘)=30+64×(−0.5)=30−32=−2MPa
- τθ=64sin(120∘)=64×0.866=55.4 MPa\tau_{\theta} = 64 \sin(120^\circ) = 64 \times 0.866 = 55.4 \, \text{MPa}τθ=64sin(120∘)=64×0.866=55.4MPa
Step 4: Principal Stresses and Directions
For part (2), the principal stresses are the maximum and minimum normal stresses, which occur at the two points where the shear stress is zero. The angle to the principal stresses is found by looking at the points where the circle intersects the normal stress axis.
- The principal stresses are given by: σ1=σavg+R=30+64=94 MPa\sigma_1 = \sigma_{\text{avg}} + R = 30 + 64 = 94 \, \text{MPa}σ1=σavg+R=30+64=94MPa σ2=σavg−R=30−64=−34 MPa\sigma_2 = \sigma_{\text{avg}} – R = 30 – 64 = -34 \, \text{MPa}σ2=σavg−R=30−64=−34MPa
- The direction of the principal stresses corresponds to the angle where the shear stress is zero, or where the Mohr’s circle intersects the horizontal axis.
These results agree with the results in Section 1.3.3 if the problem conditions match the examples in that section.
Problem 1.9: Mohr’s Circle for Pure Shear
Pure shear corresponds to a situation where there is no normal stress, but only shear stress. From Section 1.3.5, if σx=σy=0\sigma_x = \sigma_y = 0σx=σy=0 and τxy=τ\tau_{xy} = \tauτxy=τ, the Mohr’s circle is a circle centered at the origin, with a radius equal to the shear stress τ\tauτ.
For pure shear, Mohr’s circle shows that:
- The principal stresses will be equal to ±τ\pm \tau±τ.
- The corresponding angles are 45 degrees from the original axes, as the shear stresses are maximum along the diagonal.
Eq. (1.51) describes the relation between the shear stress and normal stress in pure shear, which can be derived from the geometry of Mohr’s circle by recognizing that the shear stress in Mohr’s circle corresponds to the maximum shear stress.
Problem 1.10: Mohr’s Circle for Hydrostatic Stress
Hydrostatic stress is when all three principal stresses are equal (σ1=σ2=σ3\sigma_1 = \sigma_2 = \sigma_3σ1=σ2=σ3). In Mohr’s circle, this corresponds to a point on the horizontal axis at σ=σ1\sigma = \sigma_1σ=σ1, with no shear stress component. Eq. (1.50) expresses this condition as the average normal stress.
Problem 1.11: Stresses in a Pressure Vessel
Given the problem setup with a cylindrical pressure vessel subjected to internal pressure, the radial stress is negligible, and the hoop stress (σh\sigma_hσh) and axial stress (σa\sigma_aσa) are the primary stresses. When considering a weld line at 45 degrees:
- Direct Stress Perpendicular to the Weld Line:
The direct stress acting along the direction perpendicular to the weld line is the combination of the hoop and axial stresses. This is calculated using the stress transformation equations, based on the angle of the weld. - Shear Stress Along the Weld Line:
The shear stress acting along the weld line is obtained by considering the shear components in the direction of the weld. This can also be calculated using Mohr’s Circle, by transforming the stresses at the given angle.
The specific equations for these stresses depend on the geometry and load conditions. However, using the principles of stress transformation and Mohr’s Circle, you can derive the values for the stress components along the weld.
This is a concise explanation for the given problems and Mohr’s Circle applications.
