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Question 1

Combined Stresses and Mohr Circle

Combined Stresses Visualized

A beam can be subjected to multiple stresses at once—axial normal stress, bending normal stress, torsional shear, and transverse (horizontal) shear. In the linear-elastic range, evaluate each at the same point and add them (superposition) for the net state.

$$ \sigma = \frac{P}{A} $$

$$ f_b = \frac{M c}{I} $$

$$ \tau_{\text{torque}} = \frac{T r}{J} $$

$$ \tau_{\text{shear}} = \frac{V Q}{I b} $$

Answer

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Combined Stresses & Mohr Circle | Mechanics of Deformable Bodies – Problem 1: – Diagram

Answer

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Combined Stresses & Mohr Circle | Mechanics of Deformable Bodies – Problem 2: – Diagram

Answer

1. Nature of the Problem

The load is inclined on the face of the cross-section. Therefore, it has two transverse components:

$$ P_z = P\cos\alpha $$ $$ P_y = P\sin\alpha $$

These two components cause bending about two centroidal axes of the rectangular section. The horizontal component bends the beam about the vertical centroidal axis, while the vertical component bends the beam about the horizontal centroidal axis.

2. Section Properties

For a rectangular cross-section, the centroidal moments of inertia are:

$$ I_y = \frac{hb^3}{12} $$ $$ I_z = \frac{bh^3}{12} $$

Substituting b = 80 mm and h = 140 mm:

$$ I_y = \frac{140(80)^3}{12} = 5.973 \times 10^6 \ \text{mm}^4 $$ $$ I_z = \frac{80(140)^3}{12} = 18.293 \times 10^6 \ \text{mm}^4 $$

3. Maximum Tensile Stress

The maximum tensile stress occurs at the corner where the bending stresses from the two moment components add.

The bending stress due to a rectangular section can also be written using the section modulus form:

$$ f_b = \frac{6M}{bd^2} $$

For the given inclined force, the two moment components are produced by $P\cos30^\circ$ and $P\sin30^\circ$. Since the cantilever length is 2.0 m = 2000 mm, the combined maximum tensile stress is:

$$ f_{bt} = \frac{6(P\cos30^\circ)(2000)}{140(80)^2} + \frac{6(P\sin30^\circ)(2000)}{80(140)^2} $$ $$ f_{bt} = \frac{6(575\cos30^\circ)(2000)}{140(80)^2} + \frac{6(575\sin30^\circ)(2000)}{80(140)^2} $$ $$ f_{bt} = 8.86942 \ \text{MPa} $$

Therefore:

$$ \boxed{f_{bt} = 8.87 \ \text{MPa}} $$

4. Meaning of the Neutral Axis

The neutral axis is the line on the cross-section where the bending stress is zero. On one side of this line, the section is in tension. On the other side, the section is in compression.

Since the load is inclined, the beam bends about both centroidal axes. As a result, the neutral axis is not simply horizontal or vertical. It becomes an inclined line passing through the centroid of the cross-section.

The phrase angle of the neutral axis from the horizontal means the angle measured from the horizontal centroidal axis of the cross-section to the neutral axis.

5. Why the Slope is y/z

The neutral axis is drawn on the cross-section. In this cross-section, the horizontal coordinate is z, and the vertical coordinate is y.

For any line drawn on a coordinate plane, the slope is:

$$ \text{slope} = \frac{\text{vertical change}}{\text{horizontal change}} $$

Since the vertical coordinate is y and the horizontal coordinate is z, the slope of the neutral axis is:

$$ m = \frac{y}{z} $$

Also, the slope of a line is related to its angle from the horizontal by:

$$ m = \tan\theta $$

Therefore, for the neutral axis:

$$ \tan\theta = \frac{y}{z} $$

6. Neutral Axis Equation

The general flexural stress equation for biaxial bending is:

$$ f_b = \frac{M_z}{I_z}y + \frac{M_y}{I_y}z $$

Along the neutral axis, the bending stress is zero:

$$ 0 = \frac{M_z}{I_z}y + \frac{M_y}{I_y}z $$

Rearranging:

$$ \frac{M_z}{I_z}y = -\frac{M_y}{I_y}z $$ $$ \frac{y}{z} = -\frac{M_y I_z}{M_z I_y} $$

The negative sign only indicates the direction of the slope. For the angle magnitude of the neutral axis from the horizontal:

$$ \tan\theta = \left|\frac{y}{z}\right| = \frac{M_y I_z}{M_z I_y} $$

7. Substitution for the Neutral Axis Angle

The moment components are proportional to the force components:

$$ M_y = (P\cos30^\circ)L $$ $$ M_z = (P\sin30^\circ)L $$

Therefore:

$$ \tan\theta = \frac{ \dfrac{(P\cos30^\circ)(2000)}{\dfrac{140(80)^3}{12}} }{ \dfrac{(P\sin30^\circ)(2000)}{\dfrac{80(140)^3}{12}} } $$

The common terms P and 2000 cancel, so:

$$ \tan\theta = \frac{ \dfrac{\cos30^\circ}{\dfrac{140(80)^3}{12}} }{ \dfrac{\sin30^\circ}{\dfrac{80(140)^3}{12}} } $$ $$ \theta = \tan^{-1}\left(\frac{49\sqrt{3}}{16}\right) $$ $$ \theta = 79.32376^\circ $$

Therefore:

$$ \boxed{\theta = 79.3^\circ} $$

8. Final Answers

$$ \boxed{f_{bt,\max} = 8.87 \ \text{MPa}} $$ $$ \boxed{\theta = 79.3^\circ \text{ from the horizontal}} $$

Board Exam Reminder

Do not confuse the load angle with the neutral axis angle. The load is inclined at 30°, but the neutral axis is inclined at 79.3° from the horizontal. They are different because the rectangular section has unequal moments of inertia about its two centroidal axes.

8. Corner Stresses and Minimum Compressive Stress

The maximum tensile stress occurs at the corner where the two bending stress components add as tension. The maximum compressive stress occurs at the opposite corner where the two bending stress components add as compression.

However, the other two corners are mixed-stress corners. At these corners, one bending component gives tension while the other gives compression. Therefore, the stresses are obtained by subtracting the two component stresses.

$$ f_1 = \frac{6(P\cos30^\circ)(2000)}{140(80)^2} $$ $$ f_1 = \frac{6(575\cos30^\circ)(2000)}{140(80)^2} = 6.669 \ \text{MPa} $$ $$ f_2 = \frac{6(P\sin30^\circ)(2000)}{80(140)^2} $$ $$ f_2 = \frac{6(575\sin30^\circ)(2000)}{80(140)^2} = 2.200 \ \text{MPa} $$

Thus, the four corner stresses are:

$$ f_A = +f_1 + f_2 = 6.669 + 2.200 = +8.869 \ \text{MPa} $$ $$ f_B = -f_1 - f_2 = -6.669 - 2.200 = -8.869 \ \text{MPa} $$ $$ f_C = +f_1 - f_2 = 6.669 - 2.200 = +4.469 \ \text{MPa} $$ $$ f_D = -f_1 + f_2 = -6.669 + 2.200 = -4.469 \ \text{MPa} $$

Therefore, the maximum tensile stress is:

$$ \boxed{f_{t,\max} = 8.87 \ \text{MPa}} $$

The maximum compressive stress is:

$$ \boxed{f_{c,\max} = -8.87 \ \text{MPa}} $$

If the question asks for the minimum compressive stress among the corner stresses, use the smaller negative value:

$$ \boxed{f_{c,\min} = -4.47 \ \text{MPa}} $$

In magnitude form:

$$ \boxed{f_{c,\min} = 4.47 \ \text{MPa compression}} $$

If the question instead asks for the minimum tensile stress among the corner stresses, use the smaller positive value:

$$ \boxed{f_{t,\min} = 4.47 \ \text{MPa}} $$

Note that if the entire cross-section is considered, the stress becomes zero along the neutral axis. Therefore, the theoretical minimum compression in the section is zero at the neutral axis. But among the corner stresses, the minimum compressive stress is 4.47 MPa compression.

Combined Stresses & Mohr Circle | Mechanics of Deformable Bodies – Problem 3: – Diagram

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Combined Stresses & Mohr Circle | Mechanics of Deformable Bodies – Problem 4: – Diagram

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Combined Stresses & Mohr Circle | Mechanics of Deformable Bodies – Problem 5: – Diagram

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Combined Stresses & Mohr Circle | Mechanics of Deformable Bodies – Problem 6: – Diagram

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Combined Stresses & Mohr Circle | Mechanics of Deformable Bodies – Problem 7: – Diagram

Answer

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Combined Stresses & Mohr Circle | Mechanics of Deformable Bodies – Problem 8: – Diagram

Answer

Part 1.

From Mohr's circle, the left intercept is 30 MPa and the radius is 25 MPa. Therefore the center is:
$C=30+25=55\text{ MPa}$
The right intercept, which represents $\sigma_x$, is:
$\sigma_x=C+R=55+25=80\text{ MPa}$
$\boxed{\sigma_x=80\text{ MPa}}$

Part 2.

The left intercept of Mohr's circle represents the other normal stress:
$\sigma_y=C-R=55-25=30\text{ MPa}$
$\boxed{\sigma_y=30\text{ MPa}}$

Part 3.

The maximum shearing stress is equal to the radius of Mohr's circle:
$\tau_{max}=R$
$\boxed{\tau_{max}=25\text{ MPa}}$

Question 2

Problem: Cast-Iron with Axial and Bending Stresses

A cast iron is shown below. The allowable stresses are 30MPa in tension and 70MPa in compression.
a. Determine the safe value of P (in kN) based on the allowable tensile stress.
b. Determine the safe value of P (in kN) based on the allowable compressive stress.

Question 3

Problem: Identification of Stresses Given the Mohr Circle

In the Mohr's circle for general state of plane stress as shown below, identify the stresses described below:

a. It refers to the shear stress acting on the y-face of the element.
A. BH
B. CG
C. FD
D. CD

b. It refers to the tensile normal stress acting on the x-face of the element.
A. OA
B. OF
C. OE
D. OH

c. It refers to the average of the normal stress.
A. OC
B. FC
C. OF
D. CH

Question 4

Problem: Rectangular Wood Section with Inclined Load on the Plane of the Cross-Section

A wood cantilever beam of rectangular cross-section and length L = 2.0m supports an inclined load P = 575N at its free end. Determine the maximum tensile stress and the angle of the neutral axis from the horizontal. Given: b × h = 80mm x 140mm; α=30°. Determine also the minimum compressive stress at the corners.

Question 5

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Question 6

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Question 7

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Question 8

Problem:

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Question 9

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Question 10

Question Bank: q134

PSAD - Mechanics of Deformable Bodies / Mohr Circle / Engr. Janclyde Espinosa (Clidez)

In the Mohr's circle for general state of plane stress shown, identify the stresses described below:

It refers to the shear stress acting on the y-face of the element.

FD
BH
CG
CD

It refers to the tensile normal stress acting on the x-face of the element.

OH
OA
OE
OF

It refers to the average of the normal stress.

OC
OF
FC
CH

Solution pending in psadquestions/q134.json.

Question 11

Question Bank: q547

PSAD - Mechanics of Deformable Bodies / Combined Stresses / Engr. Deguma

A shaft 62.5 mm in diameter is subjected to an axial tension of 178 kN together with a twisting moment of 3400 N×m.

Determine the maximum tensile stress of the shaft in MPa.

105.60
112.60
102.60
108.60

Determine the maximum compressive stress of the shaft in MPa.

47.62
62.72
74.72
56.72

Determine the maximum shear stress in the shaft, in MPa.

76.60
67.60
59.60
72.60

Solution pending in psadquestions/q547.json.

Question 12

Question Bank: q564

PSAD - Mechanics of Deformable Bodies / Mohr's Circle/Combined Stresses / Engr. Deguma

The figure shows an element in bi-axial stress and the corresponding Mohr’s circle. Points A and B represent the stresses on the x face and on the y face of the element, respectively.

Determine the normal axial stress, σ_x in MPa.

80
50
25
30

Determine the normal axial stress, σ_y in MPa.

30
80
25
50

Determine the maximum shearing stress in MPa.

25
80
50
30

Question 13

Question Bank: q604

PSAD - Mechanics of Deformable Bodies / Combined Stresses / Mastermatician

A 50-mm diameter shaft is transmitting a torque of T = 2.4 kN-m and a tensile force P = 125 kN.

Determine the maximum tensile stress of the shaft, in MPa.

135
64
98
71

Determine the maximum compressive stress of the shaft, in MPa.

71
135
64
98

Compute the maximum shear stress of the shaft, in MPa.

102.8
67.5
35.5
97.8

Solution pending in psadquestions/q604.json.

Question 14

Question Bank: q607

PSAD - Mechanics of Deformable Bodies / Combined Stresses / Mastermatician

A hollow circular pole 3m high is fixed at the base. It is 6mm thick and its outside diameter is 300mm. The pole is subjected to a torque and a lateral force at the free end.
Torque, T = 25kN-m
Lateral Force, P = 3kN
Shear Modulus of Elasticity = 78GPa
Allowable Shear Stress = 60MPa

What is the maximum shear stress (MPa) at the outside surface of the pole due to the torque, T?

31.3
42.1
4.7
19.5

What is the angle of twist (degrees) due to the torque?

0.46
0.04
0.54
0.15

Find the maximum flexural stress (MPa) at the base of the pole due to the lateral force alone.

22.5
3.4
8.0
14.1

Solution pending in psadquestions/q607.json.

Question 15

Question Bank: q608

PSAD - Mechanics of Deformable Bodies / Combined Stresses / Mastermatician

A hollow pole weighing 150N/m is 3m high, 6mm thick and with outside diameter of 300mm. The pole is fixed at the base and is subjected to a compressive force of 3 kN acting at an eccentricity of 100mm from its centroid and a lateral force of 0.45 kN applied at the top. Calculate the maximum tensile stress at the base of the pole.

3.51
2.51
5.12
1.35

Solution pending in psadquestions/q608.json.

Question 16

Question Bank: q610

PSAD - Mechanics of Deformable Bodies / Combined Stresses / Mastermatician

The cast iron bracket is shown below.

Determine the largest load P that can be supported by the platform if the allowable stress in tension is 30 MPa.

32.9kN
43.4kN
34.3kN
41.3kN

Solution pending in psadquestions/q610.json.

Combined Stresses and Mohr Circle

Combined Stresses Visualized

Problem: Cast-Iron with Axial and Bending Stresses

Problem: Identification of Stresses Given the Mohr Circle

Problem: Rectangular Wood Section with Inclined Load on the Plane of the Cross-Section

1. Nature of the Problem

2. Section Properties

3. Maximum Tensile Stress

4. Meaning of the Neutral Axis

5. Why the Slope is y/z

6. Neutral Axis Equation

7. Substitution for the Neutral Axis Angle

8. Final Answers

Board Exam Reminder

8. Corner Stresses and Minimum Compressive Stress

Problem:

Problem:

Problem:

Problem:

Problem:

Exam Generator Problems

Question Bank: q134

Question Bank: q547

Question Bank: q564

Question Bank: q604

Question Bank: q607

Question Bank: q608

Question Bank: q610