Quantitative Problems Problem 4-3 Figure 4-45 shows the state of stress at a point within a soil deposit. Using both an analytical approach and the pole method, determine: (a) the direction of the principal planes, (b) the magnitude of the principal stresses, and (c) the normal and shear stress in a plane at 45 o with the horizontal. Figure 4-45 State of stress at a point for Problem 4-3. SOLUTION: Pole method (a) The directions of the principal planes: we can obtain principal planes π and π 3 by rotating π and π 33 clockwise by about 6 degrees (Refer to the Mohr circle below). (b) The magnitude of the principal stresses: the maximum principal stress is.0 MPa, and the minimum principal stress is approximately 0.09 MPa. (c) The dotted line is the plane at 45 with the horizontal, and the intersection with Mohr circle is approximately (0., 0.4). So that the normal stress is 0. MPa and the shear stress is 0.4 MPa.
0.6 Pole 0.4 0. τ (Mpa) 0 0 0. 0.4 0.6 0.8. -0. -0.4-0.6 (Mpa) S-Figure 4- Analytical approach (a) The directions of the principal planes: = MPa, 33 = 0.MPa, 3 = 0.MPa (angle decreasing) ( 0.) 3 tan p = = = 0. 33 0. p = 6.6 Rotate π and π 33 planes clockwise by 6.6 Therefore, π : -6.6 and π 3 : 83.74 from the π plane.
(b) The magnitude of the principal stresses: = ( + ) + ( ) + 4 33 33 3 = ( + 0.) + ( 0. ) + ( 0.) 4 =.0MPa = ( + ) ( ) + 4 3 33 33 3 = ( + 0.) ( 0. ) + ( 0.) 4 = 0.089MPa (c) The rotation angle is 45 +30 =75 with π plane (counterclockwise) End = ( + 33 ) + ( 33 ) cos +3 sin = 0.55 + 0.45cos 75 0.sin 75 = 0.0MPa = 0.38MPa ( ) ( ) τ = ( 33 ) sin 3 cos = 0.45sin ( 75 ) 0.cos( 75 ) Problem 4-4 For the state of stress defined in Fig. 4-46: (a) draw the Mohr circle; (b) find the pole, identify it in the Mohr circle, and write its coordinates below; (c) find the principal planes and stresses (both graphically and analytically); (d) find the stresses (
and τ) on the planes making angles of ±45 o with the horizontal, indicate which is which, and solve both graphically and analytically. Figure 4-46 State of stress for Problem 4-4. SOLUTION: (a) Coordinates of given stress states: (0.4, -0.) and (, 0.) Center of the Mohr circle: (0.7, 0) (b) Coordinate of the pole(refer to the S-Figure 4-): (0.936, 0.0) (c) Find the principal planes and stress: From Mohr circle (Refer to the S-Figure 4-) maximum principal stress,.0 MPa, minimum principal stress, 0.38 MPa. From analytical solution = MPa, 33 = 0.4MPa, 3 = 0.MPa (angle decreasing) ( 0.) 3 tan p = = = 0.333 33 0.4 p = 9. for π 3 : 80.78, for π : - 9. from π
= ( + ) + ( ) + 4 33 33 3 = + + + 4 ( 0.4) ( 0.4) ( 0.) =.06MPa = ( + ) ( ) + 4 3 33 33 3 = + + 4 ( 0.4) ( 0.4) ( 0.) = 0.384MPa 3 (d) Find the normal and shear stress on the planes at angle of 45 and -45 with horizontal: From Mohr circle (Refer to the S-Figure 4-) The plane at angle of 45 with horizontal has the normal stress, 0.5 MPa, and shear stress, -0.4 MPa. The plane at angle of -45 with horizontal has the normal stress, 0.9 MPa, and shear stress, 0.4 MPa. From analytical solution +45 with horizontal: 75 clockwise (-75 ) from π = ( + 33 ) + ( 33 ) cos + 3 sin = 0.7 + 0.3cos ( 75 ) + 0.sin ( 75 ) = 0.490MPa [ ] { [ ]}
τ = ( 33 ) sin 3 cos = 0.3sin ( 75 ) 0.cos ( 75 ) = 0.37MPa [ ] { [ ]} -45 with horizontal: 65 clockwise (-65 ) from π [ ] { [ ]} = 0.7+ 0.3cos ( 65 ) + 0.sin ( 65 ) = 0.90MPa [ ] { [ ]} τ = 0.3sin ( 65 ) 0.cos ( 65 ) = 0.37MPa 0.4 Pole 0. τ (MPa) 0 0 0. 0.4 0.6 0.8. -0. -0.4 (MPa) S-Figure 4- End
Problem 4-5 Repeat Problem 4-4 for the state of stress of Fig. 4-47. Figure 4-47 State of stress for Problem 4-5. SOLUTION: (a) Coordinates of given stress states: (00, 0) and (500, 0) Center of the Mohr circle: (350, 0) (b) Coordinate of the pole (Refer to the S-Figure 4-3): (45, 30) (c) Find the principal planes and stress: The stress state of the element is the principal stress state so that the planes of element are the principal planes. π is at the angle 60 (clockwise) and π 3 is at the angle 30 (counter-clockwise) with horizontal. (d) Find the normal and shear stress on the planes at angle of 45 and -45 with horizontal: From Mohr circle (Refer to the S-Figure 4-3) The plane at angle of 45 with horizontal has the normal stress, 0 kpa, and shear stress, -75 kpa. The plane at angle of -45 with horizontal has the normal stress, 480 kpa, and shear stress, 75 kpa. From analytical solution
= 500kPa, 33 = 00kPa, and 3 = 0 +45 with horizontal: 75 clockwise (-75 ) from π = ( + 33 ) + ( 33 ) cos +3 sin = 350 + 50cos ( 75 ) = 0kPa [ ] [ ] τ = ( 33 ) sin 3 cos = 50sin ( 75 ) = 75kPa -45 with horizontal: 65 clockwise (-65 ) from π [ ] = 350 + 50cos ( 65 ) = 480kPa [ ] τ = 50sin ( 65 ) = 75MPa
00 Pole 00 τ (kpa) 0 0 00 00 300 400 500 600-00 -00 (kpa) S-Figure 4-3 End Problem 4-6 Consider the stress state of Problem 4-3. Redo the problem for shear stresses of 0, 0., and 0.3 MPa. What effect do you observe on (a) the magnitude and (b) the direction of the principal stresses as the applied shear stresses increase? SOLUTION: The Mohr circles are shown in S-Figure 4-4, S-Figure 4-5 and S-Figure 4-6. We observe that: () as the applied shear stress increases, the maximum principal stress increases and the minimum principal stress decreases in magnitude. () as the applied shear stress increases, the principal planes rotate. The major
principal plane rotates away from the π plane in the clockwise direction. The minor principal plane comes closer to π plane. 0.6 0.4 Pole 0. τ (Mpa) 0 0 0. 0.4 0.6 0.8. -0. -0.4-0.6 (Mpa) S-Figure 4-4
0.6 Pole 0.4 0. τ (Mpa) 0-0. 0 0. 0.4 0.6 0.8. -0.4-0.6 (Mpa) S-Figure 4-5 0.6 0.4 Pole 0. τ (Mpa) 0-0. 0 0. 0.4 0.6 0.8. -0.4-0.6 (Mpa) S-Figure 4-6
End Problem 4-7 * The unit block abcd is acted upon by loadings as shown in Fig. 4-48. (a) Draw the Mohr circle of stress. (b) Locate the points on the Mohr circle indicating the stresses on faces ad and dc. (c) Find the pole. (d) Find the directions of principal planes and the magnitudes of the major and minor principal stresses. (e) Draw a smaller element inside abcd to represent the principal planes and stresses acting on them. (f) Solve the problem analytically and compare your results. Figure 4-48 Element corresponding to Problem 4-7. SOLUTION: For the given load on a unit block, we can calculate the, 33, 3. 00 cos30 = = 86.6kN / m 00 cos 30 33 = = 86.6kN / m 00 sin 30 3 = = 50.0kN / m
33 30 00kN/m 3 a b 30 d c 3 00kN/m 33 S-Figure 4-7
(a) and (b) τ (kn/m ) 60 50 Plane dc (86.6, 50) 40 0 (36.6, 0) C (86.6, 0) (36.6, 0) 0 40 60 80 0 4 (kn/m ) -0-40 -50-60 Plane ad (86.6, -50) S-Figure 4-8
If we draw a line parallel to the plane ad from the point on the mohr circle representing the stresses acting on the plane ad, the drawn line intersects the circle at point P. Then from the definition of pole of a mohr circle, P is the pole. Similarly, we can also find the Pole from drawing a horizontal line from the point on Mohr cicle, which to (c) represents the stress state on a horizontal plane dc. τ (kn/m ) 60 50 Pole P 40 0 (36.6, 0) C (86.6, 0) (36.6, 0) 0 40 60 80 0 4 (kn/m ) -0-40 -50-60 Stressses on plane ad S-Figure 4-9
Since we know the Pole, the directions of principal planes can be determined by drawing straight lines from the Pole to the points where principal stresses act. τ (kn/m ) 60 50 Pole P 40 36.6kN/m 36.6kN/m 0 (36.6, 45 C (86.6, 0) 45 (36.6, 0) 0 40 60 80 0 4 (kn/m ) -0-40 -50-60 Stresses on plane ad S-Figure 4-0 From the figure, to (d) The major principal stress is equal to 36.6kN/m to (d) The minor principal stress is equal to 36.6kN/m The major and minor principal planes are at angles 45 and -45 from to (d) horizontal.
(e) 30 00kN/m 30 a b 36.6kN/m 36.6kN/m 45 36.6kN/m 36.6kN/m d c 00kN/m S-Figure 4- (f). Analytical Solution 00 cos30 = = 86.6kN / 00 cos30 33 = = 86.6kN / m = 00 sin 30 3 = = 50.0kN / m ; Positive (+) m 3 tan p = = ; This means p is equal to 90. 33
Since is equal to 33, we expect the absolute value of direction of major principal stress, p is equal to the absolute value of the direction of minor principal stress, p3. Therefore, p = p3 = 45, p = 45 and p3 = -45 from the plane where acts. We can calculate the major and minor principal stresses as + 4 86.6 + 86.6 33 = + ( 33) + 3 = + 50 = 36.6kN / m alternatively, + 4 86.6 + 86.6 33 3 = ( 33 ) + 3 = 50 = 36.6kN / + 33 33 = + cos + 3 sin 86.6 + 86.6 86.6 86.6 cos90 o 50 sin 90 o = + + = 36.6kN / m 86.6 + 86.6 86.6 86.6 3 cos( 90 o ) 50 sin( 90 o = + + ) = 36.6kN / m The results are in agreement with the solution from the graphical method. End m Problem 4-8 For the soil element of Example 4-5, consider the major principal strain increment to be vertical and (a) find the pole and (b) find the directions of the zero- extension lines (slip lines) through the element. SOLUTION: Referring to S-Figure 4-, the dilatancy angle is given by: