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Question 1
ప్రశ్న 1
For a soil element in which the major and minor principal compressive stresses are 300 kPa and 100 kPa, respectively, what are the centre and radius of the corresponding Mohr circle?
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Explanation:
The centre of a Mohr circle is the average of the major and minor principal stresses. Thus C = (300 + 100)/2 = 200 kPa. Its radius is half their difference. Therefore R = (300 - 100)/2 = 100 kPa. The principal stresses are represented by the rightmost and leftmost points of the circle on the normal-stress axis.
వివరణ:
Question 2
ప్రశ్న 2
A Mohr circle is centred at 180 kPa on the compressive normal-stress axis and has a radius of 70 kPa. What are the major and minor principal stresses represented by the circle?
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Explanation:
The principal stresses occur at the intersections of the Mohr circle with the normal-stress axis. The major principal stress is the centre plus the radius. It equals 180 + 70 = 250 kPa. The minor principal stress is the centre minus the radius, which is 180 - 70 = 110 kPa. Both values use the stated convention in which compressive normal stress is positive.
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Question 3
ప్రశ్న 3
For a two-dimensional soil stress tensor with compressive normal stresses sigma_x = 180 kPa and sigma_y = 100 kPa and symmetric shear components tau_xy = tau_yx = 30 kPa, what are the major and minor in-plane principal stresses?
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Explanation:
The mean normal stress is C = (180 + 100)/2 = 140 kPa. The circle radius is R = sqrt[((180 - 100)/2)^2 + 30^2]. This becomes R = sqrt(40^2 + 30^2) = 50 kPa. The principal stresses are C + R and C - R. Hence the major and minor principal stresses are 190 kPa and 90 kPa.
వివరణ:
Question 4
ప్రశ్న 4
If the major and minor principal stresses at a point are 260 kPa and 100 kPa, what are the maximum in-plane shear-stress magnitude and the normal stress acting on a plane carrying that maximum shear?
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Explanation:
The maximum in-plane shear stress is the radius of the Mohr circle. The radius equals (260 - 100)/2 = 80 kPa. Maximum shear occurs at the top or bottom point of the circle. The normal stress at either of these points equals the centre, (260 + 100)/2 = 180 kPa. Thus the required pair is 80 kPa shear stress and 180 kPa normal stress.
వివరణ:
Question 5
ప్రశ్న 5
Two physical planes through a soil element differ in orientation by 35 degrees. What is the magnitude of the angle between their corresponding radii on the Mohr circle, using the double-angle construction?
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Explanation:
The angular separation measured around the Mohr circle is twice the change in orientation between the corresponding physical planes. The stated physical orientation difference is 35 degrees. Doubling this angle gives 70 degrees on the Mohr circle. The direction of rotation on the diagram depends on the shear-stress sign convention being used. The magnitude of the doubled angle remains 70 degrees.
వివరణ:
Question 6
ప్రశ్న 6
The principal compressive stresses in a soil element are sigma_1 = 280 kPa and sigma_3 = 120 kPa. A plane is inclined at 30 degrees to the major principal plane, the plane on which sigma_1 acts normally. What are the normal stress and shear-stress magnitude on the inclined plane? Take cos 60 degrees = 0.5 and sin 60 degrees = 0.8660.
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Explanation:
The centre and radius of the principal-stress circle are C = 200 kPa and R = 80 kPa. Because the specified physical plane makes 30 degrees with the major principal plane, the double angle is 60 degrees. The normal stress is C + R cos 60 degrees = 200 + 80(0.5) = 240 kPa. The shear-stress magnitude is R sin 60 degrees = 80(0.8660) = 69.28 kPa. The rounded pair is 240 kPa and 69.3 kPa.
వివరణ:
Question 7
ప్రశ్న 7
At a point with principal compressive stresses sigma_1 = 350 kPa and sigma_3 = 150 kPa, what are the normal stress and shear-stress magnitude on a plane inclined at 45 degrees to the major principal plane?
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Explanation:
The circle centre is C = (350 + 150)/2 = 250 kPa. The radius is R = (350 - 150)/2 = 100 kPa. A physical plane at 45 degrees to a principal plane corresponds to 90 degrees around the Mohr circle. At this location the normal stress is C = 250 kPa and the shear-stress magnitude is R = 100 kPa. Therefore the required stress pair is 250 kPa and 100 kPa.
వివరణ:
Question 8
ప్రశ్న 8
The principal compressive stresses are 300 kPa and 100 kPa. For a plane inclined at 15 degrees to the major principal plane, calculate the normal stress and shear-stress magnitude. Take cos 30 degrees = 0.8660 and sin 30 degrees = 0.5.
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Explanation:
The centre of the Mohr circle is (300 + 100)/2 = 200 kPa. Its radius is (300 - 100)/2 = 100 kPa. A plane rotation of 15 degrees gives a double angle of 30 degrees. The normal stress is 200 + 100(0.8660) = 286.6 kPa. The shear-stress magnitude is 100(0.5) = 50.0 kPa.
వివరణ:
Question 9
ప్రశ్న 9
In a soil element with sigma_1 = 300 kPa and sigma_3 = 100 kPa, a plane has shear-stress magnitude 50 kPa and normal stress greater than the circle centre. If its inclination to the major principal plane lies strictly between 0 and 45 degrees, what is that inclination?
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Explanation:
The Mohr-circle radius is (300 - 100)/2 = 100 kPa. For a plane inclined by theta to the major principal plane, the shear-stress magnitude is R sin(2 theta) in the stated interval. Hence 50 = 100 sin(2 theta), giving sin(2 theta) = 0.5. The interval 0 < theta < 45 degrees and the stated normal-stress condition select 2 theta = 30 degrees. Therefore theta = 15 degrees.
వివరణ:
Question 10
ప్రశ్న 10
The in-plane compressive-positive stress tensor is [[200, 50], [50, 100]] kPa in x-y coordinates. A new plane has its unit normal directed 45 degrees counterclockwise from the positive x-axis. What are the normal stress and magnitude of shear stress on that plane?
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Explanation:
The average normal stress is (200 + 100)/2 = 150 kPa. The normal stress on a plane whose normal rotates counterclockwise through theta is C + [(sigma_x - sigma_y)/2] cos(2 theta) + tau_xy sin(2 theta). At theta = 45 degrees, the cosine term vanishes and the sine term equals one. The transformed normal stress is therefore 150 + 50 = 200 kPa. The transformed shear component is -50 kPa under the specified tensor convention, giving a shear-stress magnitude of 50 kPa.
వివరణ:
Question 11
ప్రశ్న 11
A soil stress tensor has sigma_x = 160 kPa, sigma_y = 80 kPa and symmetric shear component tau_xy = +40 kPa, all using one consistent compressive-positive tensor convention. What is the counterclockwise angle from the positive x-axis to the unit normal of the major in-plane principal plane?
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Explanation:
The principal direction satisfies tan(2 theta_p) = 2 tau_xy/(sigma_x - sigma_y). Substitution gives tan(2 theta_p) = 80/80 = 1. Because sigma_x exceeds sigma_y and tau_xy is positive, the major principal normal lies in the first quadrant at the smaller positive angle. Thus 2 theta_p = 45 degrees. The required counterclockwise angle is theta_p = 22.5 degrees.
వివరణ:
Question 12
ప్రశ్న 12
Which statement correctly relates the principal planes of a two-dimensional stress state to their representations on a Mohr circle when the two in-plane principal stresses are distinct?
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Explanation:
Distinct in-plane principal stresses act normally on two mutually perpendicular principal planes. The shear stress on each principal plane is zero. On the Mohr circle, these stress states appear at the two intercepts with the normal-stress axis. The two intercepts lie at opposite ends of a diameter. Their physical angular separation of 90 degrees corresponds to a Mohr-circle angular separation of 180 degrees.
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Question 13
ప్రశ్న 13
A saturated soil has total principal compressive stresses sigma_1 = 300 kPa and sigma_3 = 140 kPa, and uniform pore-water pressure u = 40 kPa. What are the centre and radius of its effective-stress Mohr circle?
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Explanation:
The effective principal stresses are obtained by subtracting pore pressure from the corresponding total principal stresses. They are sigma_1' = 300 - 40 = 260 kPa and sigma_3' = 140 - 40 = 100 kPa. The effective circle centre is (260 + 100)/2 = 180 kPa. Its radius is (260 - 100)/2 = 80 kPa. The uniform pore pressure shifts the circle horizontally without changing its radius.
వివరణ:
Question 14
ప్రశ్న 14
In a saturated soil element, uniform positive pore-water pressure increases by 30 kPa while the total stress tensor remains unchanged. How does its effective-stress Mohr circle move on a plot with compressive normal stress positive to the right?
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Explanation:
Effective normal stresses equal total normal stresses minus pore-water pressure. A uniform 30 kPa rise in pore pressure reduces both effective principal stresses by 30 kPa. The average of the principal effective stresses therefore decreases by 30 kPa. Their difference, and hence the Mohr-circle radius, remains unchanged. Thus the effective-stress circle translates 30 kPa leftward on the compressive-positive normal-stress axis.
వివరణ:
Question 15
ప్రశ్న 15
For a straight effective-stress Mohr-Coulomb failure envelope plotted with positive normal stress on the horizontal axis and positive shear stress on the vertical axis, what do the shear-axis intercept and slope represent?
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Explanation:
The effective-stress Mohr-Coulomb equation is tau_f = c' + sigma_n' tan(phi'). At zero effective normal stress, the line intersects the shear-stress axis at c'. The increase in shear strength per unit increase in effective normal stress is tan(phi'). Therefore the line slope is tan(phi'), not the friction angle expressed directly as a number of degrees. The envelope characterizes limiting strength, whereas a Mohr circle characterizes a particular stress state.
వివరణ:
Question 16
ప్రశ్న 16
At triaxial compression failure, a cohesionless soil has effective major and minor principal stresses of 300 kPa and 100 kPa. If its linear Mohr-Coulomb failure envelope passes through the origin, what is its effective friction angle?
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Explanation:
For the stated principal stresses, the failure-circle centre is C = 200 kPa and the radius is R = 100 kPa. For a cohesionless Mohr-Coulomb envelope tangent to the circle, R = C sin(phi'). It follows that sin(phi') = 100/200 = 0.50. The corresponding effective friction angle is 30 degrees. This value also satisfies the cohesionless principal stress ratio of three for triaxial compression.
వివరణ:
Question 17
ప్రశ్న 17
A cohesionless soil has phi' = 30 degrees, and its upper effective-stress Mohr circle at failure has centre 200 kPa and radius 100 kPa. What are the effective normal stress and positive shear stress at the point where the circle touches the failure envelope? Use sin 30 degrees = 0.5 and cos 30 degrees = 0.8660.
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Explanation:
At tangency, the radius joining the circle centre to the failure point is perpendicular to the failure envelope. The horizontal coordinate of the upper tangency point is sigma_n' = C - R sin(phi'). Therefore sigma_n' = 200 - 100(0.5) = 150 kPa. Its positive shear coordinate is tau = R cos(phi') = 100(0.8660) = 86.6 kPa. This point also satisfies the cohesionless equation tau = sigma_n' tan(30 degrees) within rounding.
వివరణ:
Question 18
ప్రశ్న 18
For triaxial compression at Mohr-Coulomb failure, a soil has effective cohesion c' = 20 kPa, effective friction angle phi' = 30 degrees and minor effective principal stress sigma_3' = 100 kPa. Using the principal-stress failure relation and sqrt(3) = 1.732, what is sigma_1'?
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Explanation:
The triaxial compression failure relation is sigma_1' = N_phi sigma_3' + 2 c' sqrt(N_phi). For phi' = 30 degrees, N_phi = (1 + sin(phi'))/(1 - sin(phi')) = 3. The frictional part is 3(100) = 300 kPa. The cohesion part is 2(20)sqrt(3) = 40(1.732) = 69.28 kPa. Therefore sigma_1' = 369.28 kPa, or approximately 369.3 kPa.
వివరణ:
Question 19
ప్రశ్న 19
A saturated clay is represented in undrained total-stress analysis by phi_u = 0 degrees and c_u = 55 kPa. If the minor total principal stress in triaxial compression is 120 kPa at failure, what are the major principal stress and radius of its total-stress Mohr circle?
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Explanation:
For the ideal phi_u = 0 model, the undrained shear strength equals the total-stress Mohr-circle radius at failure. Hence R = c_u = 55 kPa. The principal-stress difference is 2R = 110 kPa. Therefore sigma_1 = sigma_3 + 110 = 120 + 110 = 230 kPa. The corresponding failure envelope is horizontal in the total-stress Mohr diagram.
వివరణ:
Question 20
ప్రశ్న 20
Two effective-stress Mohr circles at failure for the same soil have principal stress pairs (sigma_1', sigma_3') of (300, 100) kPa and (500, 200) kPa. Assuming both circles are tangent to one straight Mohr-Coulomb envelope, determine approximately phi' and c'. Use sin^-1(1/3) = 19.47 degrees and cos(19.47 degrees) = 0.9428.
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Explanation:
The first failure circle has centre C1 = 200 kPa and radius R1 = 100 kPa. The second has centre C2 = 350 kPa and radius R2 = 150 kPa. The tangency relation R = C sin(phi') + c' cos(phi') gives sin(phi') = (150 - 100)/(350 - 200) = 1/3. Therefore phi' = 19.47 degrees. From the first circle, c' = [100 - 200/3]/0.9428 = approximately 35.4 kPa.
వివరణ:
Question 21
ప్రశ్న 21
A cohesionless soil has phi' = 30 degrees, and an effective-stress Mohr circle has centre C = 160 kPa and radius R = 70 kPa. How should this circle be interpreted relative to the ideal linear Mohr-Coulomb failure envelope through the origin?
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Explanation:
For a cohesionless material, the envelope passes through the origin and is inclined by phi' to the normal-stress axis. The perpendicular distance from the circle centre to this line is C sin(phi'). For C = 160 kPa and phi' = 30 degrees, this limiting radius equals 160(0.5) = 80 kPa. The actual circle radius is only 70 kPa, which is less than that limiting distance. Therefore the circle lies inside the ideal strength envelope and does not reach the shear failure condition.
వివరణ:
Question 22
ప్రశ్న 22
An effective-stress Mohr circle has fixed centre C = 200 kPa. For a soil with c' = 20 kPa and phi' = 30 degrees, what circle radius is required for tangency to the upper Mohr-Coulomb failure envelope? Use sin 30 degrees = 0.5 and cos 30 degrees = 0.8660.
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Explanation:
The upper failure envelope has equation tau = c' + sigma_n' tan(phi'). At tangency, the perpendicular distance from the circle centre to the failure line equals the circle radius. The relevant relation is R = C sin(phi') + c' cos(phi'). Substitution gives R = 200(0.5) + 20(0.8660) = 117.32 kPa. The required limiting radius is approximately 117.3 kPa.
వివరణ:
Question 23
ప్రశ్న 23
Under the ideal Mohr-Coulomb triaxial compression failure construction, a soil has friction angle phi' = 30 degrees. What acute angle does the predicted failure plane make with the major principal plane, defined as the plane carrying sigma_1' normally?
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Explanation:
The inclination of the Mohr-Coulomb failure plane is conventionally measured relative to the major principal plane. Under triaxial compression, that inclination is 45 degrees plus half the friction angle. The specified friction angle is 30 degrees. Hence the angle is 45 + 15 = 60 degrees. This plane inclination is distinct from the angle between the failure plane and the major principal stress direction.
వివరణ:
Question 24
ప్రశ్న 24
Which procedure correctly describes use of the pole, or origin-of-planes, method on Mohr's circle when the physical orientation of a reference plane and its stress point on the circle are known?
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Explanation:
The pole method connects physical plane orientations with their corresponding stress points on Mohr's circle. From a known stress point, a line is drawn parallel to the physical plane on which those stresses act. Its other intersection with the circle defines the pole, with the coincident point used in a tangent case. A line drawn from the pole parallel to any target plane locates the corresponding stress point on the circle. This construction avoids guessing angles from an unavailable drawing and must use a consistent shear-stress plotting convention.
వివరణ:
Question 25
ప్రశ్న 25
At a point in saturated soil, the total major and minor principal stresses are 330 kPa and 130 kPa, and uniform pore-water pressure is 30 kPa. If c' = 0 and phi' = 30 degrees, what is the relationship between the effective-stress Mohr circle and the Mohr-Coulomb envelope?
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Explanation:
The effective principal stresses are 330 - 30 = 300 kPa and 130 - 30 = 100 kPa. The effective Mohr circle therefore has centre C = (300 + 100)/2 = 200 kPa. Its radius is R = (300 - 100)/2 = 100 kPa. For a cohesionless soil with phi' = 30 degrees, the tangent radius at that centre is C sin(phi') = 200(0.5) = 100 kPa. Equality of the actual and limiting radii means the effective-stress circle is exactly tangent to the failure envelope.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Centre = 200 kPa; radius = 100 kPa. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. 250 kPa and 110 kPa. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. 190 kPa and 90 kPa. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 80 kPa and 180 kPa. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 70 degrees. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. 240 kPa and 69.3 kPa. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: C. 250 kPa and 100 kPa. సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: B. 286.6 kPa and 50.0 kPa. సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: D. 15 degrees. సమాధానం: D.
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Question 10 ప్రశ్న 10Answer: B. 200 kPa and 50 kPa. సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: C. 22.5 degrees. సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: A. The principal planes are perpendicular physically, and their stress points are opposite ends of a diameter on the normal-stress axis. సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: D. Centre = 180 kPa; radius = 80 kPa. సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: A. It shifts 30 kPa to the left without a change in radius. సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: B. Intercept = c'; slope = tan(phi'). సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: D. 30.0 degrees. సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: C. 150.0 kPa and 86.6 kPa. సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: B. 369.3 kPa. సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: A. sigma_1 = 230 kPa; R = 55 kPa. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: D. phi' = 19.47 degrees; c' = 35.4 kPa. సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: B. It lies strictly within the strength envelope because the limiting tangent radius at this centre is 80 kPa. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. 117.3 kPa. సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: A. 60 degrees. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. Draw through the known stress point a line parallel to its physical plane to locate the pole at the other circle intersection, then draw from the pole a line parallel to the target plane to locate its stress point. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: D. The circle is exactly tangent to the envelope, representing the ideal Mohr-Coulomb failure condition. సమాధానం: D.