Question 10.5: It has been determined that a point in a load-carrying membe...
It has been determined that a point in a load-carrying member is subjected to the following stress condition:
\sigma_{x} = 400 MPa \sigma_{y} = -300 MPa \tau_{xy} = 200 MPa (CW)
Perform the following:
(a) Draw the initial stress element.
(b) Draw the complete Mohr’s circle, labeling critical points.
(c) Draw the complete principal stress element.
(d) Draw the maximum shear stress element.
The 16-step Procedure for Drawing Mohr’s Circle is used here to complete the problem. The numerical results from steps 1 through 12 are summarized here and shown in Figure 10–33:
Step 1. The initial stress element is shown at the upper left of Figure 10–33.
Step 2. Point 1 is plotted at σ_{x} = 400 MPa and τ_{xy} = 200 MPa in quadrant 1.
Step 3. Point 2 is plotted at σ_{y} = −300 MPa and τ_{yx} = −200 MPa in quadrant 3.
Step 4. The line from point 1 to point 2 has been drawn.
Step 5. The line from step 4 crosses the σ-axis at the average applied normal stress, called point O in Figure 10–33, computed from
\sigma_{avg} = \frac{1}{2} (\sigma_{x} + \sigma_{y} ) = \frac{1}{2} [400 +(-300)] = 50 MPa
Step 6. Point O is the center of the circle. The line from point O through point 1 is labeled as the x-axis to correspond with the x-axis on the initial stress element.
Step 7. The values of a, b, and R are found using the triangle formed by the lines from point O to point 1 to σ_{x} = 400 MPa and back to point O. The length of the lower side of the triangle on the horizontal axis is called a, found from
a = \frac{1}{2} (\sigma_{x} + \sigma_{y} ) = \frac{1}{2} [400 – (-300)] = 350 MPa
The vertical side of the triangle, b, is computed from
b = \tau_{xy} = 200 MPa
The radius of the circle, R, is computed from
R = \sqrt{a² + b²} = \sqrt{(350)² + (200)²} = 403 MPa
Step 8. This is the drawing of the circle with point O as the center at σ_{avg} = 50 MPa and a radius of R = 403 MPa.
Step 9. The vertical diameter of the circle has been drawn through point O. The intersection of this line with the circle at the top indicates the value of τ_{max} = 403 MPa, the same as the value of R.
Step 10. The maximum principal stress, σ_{1} , is at the right end of the horizontal diameter of the circle and the minimum principal stress, σ_{2} , is at the left.
Step 11. The values for or σ_{1} and σ_{2} are
\sigma_{1} = O + R = 50 + 403 = 453 MPa
\sigma_{2} = O – R = 50 – 403 = -353 MPa
Step 12. The angle 2ϕ is shown on the circle, using a single-headed arrow, as the angle from the x-axis to the σ_{1} -axis, a CW rotation. The value is computed from
2\phi = \tan^{-1} \left\lgroup \frac{b}{a}\right\rgroup = \tan^{-1} \left\lgroup \frac{200}{350}\right\rgroup = 29.74°
Note that 2ϕ is the angle moving from the x-axis to the σ_{1} -axis and in a CW direction. We will need the value of the actual angle, ϕ, computed here:
\phi = \frac{29.74°}{2} = 14.87°
Step 13. Using the results from steps 11 and 12, the principal stress element is drawn as shown in Figure 10–34(b). The element is rotated 14.87° CW from the original x-axis to the face on which the tensile stress σ_{1} = 453 MPa acts. The compressive stress σ_{2} = -353 MPa acts on the faces perpendicular to the σ_{1} faces.
Step 14. The angle 2ϕ′ is shown in Figure 10–33, using a single-headed arrow, from the x-axis CCW to the vertical diameter that locates τ_{max} at the top of the circle. Its value can be found in either of two ways. First, using Equation (10–16), observe that the numerator is the same as the value of a and the denominator is the same as the value of b from the construction of the circle. Then,
2\phi ' = \tan^{-1} (a/b) = \tan^{-1} (350/200) = 60.26° CCW
Or using the geometry of the circle, we can compute
2ϕ’ = 90° -2ϕ = 90° – 29.74° = 60.26° CCW
Then the angle ϕ′ is one-half of 2ϕ′:
ϕ′ = \frac{60.26°}{2} = 30.13°
Step 15. The maximum shear stress element is drawn in Figure 10–34(c), rotated 30.13° CCW from the original x-axis to the face on which the positive τ_{max} acts. The maximum shear stress of 403 MPa is shown on all four faces with vectors that create the two pairs of opposing couples characteristic of shear stresses on a stress element. Also shown is the tensile stress σ_{avg} = 50 MPa acting on all four faces of the element.
Step 16. Here is the summary of results for Example Problem 10–5, using Mohr’s circle.
Given \sigma_{x} = 400 MPa; \sigma_{y} = -300 MPa; \tau_{xy} = 200 MPa; CW
Results Figures 10-33 and 10-34
\sigma_{1} = 453 MPa tension
\sigma_{2} = -353 MPa compression
ϕ = 14.87° CW from x-axis
\tau_{max} = 403 MPa
\sigma_{avg} = 50 MPa tension
ϕ’ = 30.13° CW from x-axis
Comment The x-axis is in the first quadrant.
