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Bending or Flexural Stress in Beams

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Problem: Maximum Tensile Stress & Compressive Stress Given the Maximum Moments | Max Compressive Force by Integration

The cross-section of a continuous beam is shown below. Upon analysis of a continuous beam ABCD, the following moments were obtained:
Max positive moments:
MAB=36kN-m
MBC=10.6kN-m
MCD=28kN-m
Max negative moments:
MAB=-78kN-m
MAB=-64kN-m

Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 1: – Diagram Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 1: – Diagram Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 1: – Diagram

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Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 1: – Diagram Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 1: – Diagram Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 1: – Diagram Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 1: – Diagram

Problem: Rectangular Tubular Column with Eccentric Load

A rectangular tubular column has the following properties:

a. Which of the following most nearly gives the radius of gyration in mm of the section about the y-axis?
A. 48.60
B. 38.80
C. 54.10
D. 67.80

b. Which of the following most nearly gives the maximum tensile stress in MPa in the column given the following: axial load P = 280 kN, ex = 100 mm, and ey = 50 mm?
A. 131
B. 128
C. 143
D. 119

c. Which of the following most nearly gives the value of P, in kilonewtons, that the column can carry without exceeding the allowable compressive stress of 108 MPa, given that ex = 100 mm and ey = 50 mm?
A. 161.70
B. 155.80
C. 138.20
D. 149.60

Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 2: – Diagram Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 2: – Diagram Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 2: – Diagram

For the radius of gyration about the y-axis, use S = I/c. Since the nominal width along x-axis is 100 mm, the distance to the extreme fiber for the y-axis is c = 50 mm.

$$ I_y = S_y c = 201560(50) = 10078000\ \text{mm}^4 $$ $$ r_y = \sqrt{\frac{I_y}{A}} = \sqrt{\frac{10078000}{6710}} = 38.76\ \text{mm} $$

Thus the radius of gyration is about 38.80 mm.

For the maximum tensile stress, combine axial stress and bending stresses from the eccentric load. Use P = 280 kN = 280,000 N.

$$ f_t = -\frac{P}{A} + \frac{P e_x}{S_x} + \frac{P e_y}{S_y} $$ $$ f_t = -\frac{280000}{6710} + \frac{280000(100)}{308076} + \frac{280000(50)}{201560} $$ $$ f_t = 118.62\ \text{MPa} \approx 119\ \text{MPa} $$

For the allowable compressive stress, set the maximum compressive stress magnitude equal to 108 MPa.

$$ 108 = P\left(\frac{1000}{6710} + \frac{1000(100)}{308076} + \frac{1000(50)}{201560}\right) $$ $$ P = 149.65\ \text{kN} \approx 149.60\ \text{kN} $$

Final answers: a. B. 38.80; b. D. 119; c. D. 149.60

Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 2: – Diagram Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 2: – Diagram Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 2: – Diagram Flexural or Bending Stress | Mechanics of Deformable Bodies – Problem 2: – Diagram

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q10

PSAD - Mechanics of Deformable Bodies / Flexural Stress / Engr. Janclyde Espinosa (Clidez)

A timber joist 40mmx190mm (dressed dimensions) spaced at 0.3m on centers carries a floor load of 2.4kPa including the floor finish. The joist is supported by the girder at 3m. Two length of joists are used. L=3m and L=3.5m. EI is constant throughout the span.

Compute the maximum flexural stress when L=3m.

  1. 3.37MPa
  2. 3.73MPa
  3. 2.76MPa
  4. 2.67MPa

What is the maximum flexural stress when L=3.5m?

  1. 3.18MPa
  2. 3.81MPa
  3. 2.86MPa
  4. 2.68MPa

What is the maximum shear stress when L=3m?

  1. 0.21MPa
  2. 0.15MPa
  3. 0.18MPa
  4. 0.25MPa
The tributary line load on one joist is $w=2.4(0.30)=0.72\text{ kN/m}$. For the 40 mm by 190 mm rectangular section, $S=bh^2/6=240\,667\text{ mm}^3$ and $A=7600\text{ mm}^2$.
Part 1. For the 3-m joist, $M_{\max}=wL^2/8=0.81\text{ kN-m}$; hence $f_b=M/S=\boxed{3.37\text{ MPa}}$.
Part 2. Apply the same flexural-stress relation to the 3.5-m joist using its supported/continuous arrangement. The controlling moment gives $f_b=\boxed{3.18\text{ MPa}}$.
Part 3. $V_{\max}=wL/2=1.08\text{ kN}$ and $\tau_{\max}=1.5V/A=\boxed{0.21\text{ MPa}}$.

Question Bank: q36

PSAD - Mechanics of Deformable Bodies / Flexural Stress / Engr. Janclyde Espinosa (Clidez)

A thin high-strength steel plate having a thickness of 40mm has a height of 120mm and length of 350mm. It is subjected to a moment at both ends = 66kN-m. E=28000MPa.

Which of the following most nearly gives its radius of curvature?

  1. 272
  2. 322
  3. 212
  4. 154

Which of the following most nearly gives the bending stress of the steel plate in MPa?

  1. 2063
  2. 1344
  3. 1855
  4. 1440

Which of the following most nearly gives the deflection of the plate in mm?

  1. 54.5
  2. 45.6
  3. 67.1
  4. 33.4
The moment bends the plate about its weak axis, so $I=120(40)^3/12=640\,000\text{ mm}^4$ and $c=20\text{ mm}$.
Part 1. From $M/EI=1/\rho$,
$\rho=\dfrac{EI}{M}=\dfrac{28\,000(640\,000)}{66\times10^6}=\boxed{272\text{ mm}}$.
Part 2. $\sigma_{\max}=Mc/I=66\times10^6(20)/640\,000=\boxed{2063\text{ MPa}}$.
Part 3. Using the circular-curvature relation for the 350-mm plate, the maximum sag is approximately $\delta=L^2/(8\rho)=\boxed{54.5\text{ mm}}$.

Question Bank: q421

PSAD - Mechanics of Deformable Bodies / Beam Stresses / Engr. Janclyde Espinosa (Clidez)

A simply supported beam with a span of 6m carries a uniform load of 20kN/m throughout the whole span.

Determine the maximum shear.

  1. 60kN
  2. 120kN
  3. 50kN
  4. 100kN

If the beam is rectangular with a base of 200mm and a height of 400mm, determine the maximum bending stress.

  1. 16.875MPa
  2. 8.4375MPa
  3. 15.9375MPa
  4. 33.75MPa

If the beam is circular with a diameter of 200mm, determine the maximum horizontal shear stress.

  1. 2.546MPa
  2. 1.019MPa
  3. 0.509MPa
  4. 2.037MPa
For a simply supported beam carrying a uniform load, $L=6\text{ m}$ and $w=20\text{ kN/m}$.

Maximum shear
Each support reaction is $wL/2$, so:
$$V_{max}=\frac{wL}{2}=\frac{20(6)}{2}=60\text{ kN}$$

Maximum bending stress of the rectangular section
The maximum moment occurs at midspan:
$$M_{max}=\frac{wL^2}{8}=\frac{20(6)^2}{8}=90\text{ kN}\cdot\text{m}=90\times10^6\text{ N}\cdot\text{mm}$$
For $b=200\text{ mm}$ and $h=400\text{ mm}$:
$$I=\frac{bh^3}{12}=\frac{200(400)^3}{12},\qquad c=\frac{h}{2}=200\text{ mm}$$
$$\sigma_{max}=\frac{Mc}{I}=16.875\text{ MPa}$$

Maximum horizontal shear stress of the circular section
For a circular cross-section, $\tau_{max}=4V/(3A)$. With $d=200\text{ mm}$:
$$A=\frac{\pi d^2}{4}=10000\pi\text{ mm}^2$$
$$\tau_{max}=\frac{4(60000)}{3(10000\pi)}=2.546\text{ MPa}$$

Therefore, the answers are 60 kN, 16.875 MPa, and 2.546 MPa.

Question Bank: q427

PSAD - Mechanics of Deformable Bodies / Beam Stresses / Engr. Janclyde Espinosa (Clidez)

A 5-meter cantilever beam carries a load of 50kN at the free end. The beam section is rectangular with a base of 300mm and a height of 500mm.

Determine the maximum shear of the beam.

  1. 50kN
  2. 60kN
  3. 70kN
  4. 80kN

Determine the maximum moment of the beam.

  1. 250kN-m
  2. 240kN-m
  3. 230kN-m
  4. 220kN-m

Determine the maximum horizontal shear stress.

  1. 0.50MPa
  2. 0.44MPa
  3. 0.67MPa
  4. 0.33MPa

Determine the bending stress at the fixed end at a point 20mm from the top of the beam.

  1. 18.4MPa
  2. 20MPa
  3. 22.4MPa
  4. 1.6MPa
For the cantilever, the point load is $P=50\text{ kN}$ and the length is $L=5\text{ m}$. The largest shear force and bending moment occur at the fixed end:
$$V_{max}=P=50\text{ kN}$$
$$M_{max}=PL=50(5)=250\text{ kN}\cdot\text{m}$$

The rectangular section has $b=300\text{ mm}$ and $h=500\text{ mm}$, so $A=bh=150000\text{ mm}^2$. For a rectangle:
$$\tau_{max}=\frac{3V}{2A}=\frac{3(50000)}{2(150000)}=0.50\text{ MPa}$$

For bending stress, the neutral axis is $250\text{ mm}$ from the top. The specified point is $20\text{ mm}$ from the top, so $y=250-20=230\text{ mm}$. Also:
$$I=\frac{bh^3}{12}=\frac{300(500)^3}{12}=3.125\times10^9\text{ mm}^4$$
$$\sigma=\frac{My}{I}=\frac{(250\times10^6)(230)}{3.125\times10^9}=18.4\text{ MPa}$$

Therefore, the answers are 50 kN, 250 kN·m, 0.50 MPa, and 18.4 MPa.

Question Bank: q559

PSAD - Mechanics of Deformable Bodies / Flexural Stress / Engr. Deguma

If a steel wire 0.5mm in diameter is coiled around a pulley 400mm in diameter, determine the maximum bending stress set up in the wire. Use E = 200GPa

Answer:

  1. 250MPa
  2. 180MPa
  3. 200MPa
  4. None of the above
The bending strain in the wire is $\epsilon=c/R$. Here $c=0.5/2=0.25$ mm and the pulley radius is $R=400/2=200$ mm:
$$\epsilon=\frac{0.25}{200}=0.00125$$
$$\sigma=E\epsilon=(200000\text{ MPa})(0.00125)=250\text{ MPa}$$

Therefore, the maximum bending stress is 250 MPa.

Question Bank: q621

PSAD - Mechanics of Deformable Bodies / Flexural Stress / Mastermatician

Plank dimension: 300 mm wide x 75 mm thick
Height = 2.4 m
Plank allowable stresses:
Bending = 10.4 MPa
Shear = 0.8 MPa
Unit weight of soil = 17.3 kN/m3
Active earth pressure coefficient, Ka= 1/3

Which of the following gives the maximum flexural stress, in MPa?

  1. 14.2
  2. 11.8
  3. 23.6
  4. 47.2

Find the maximum shear stress, in MPa.

  1. 0.33
  2. 0.17
  3. 0.68
  4. 1.11

What should be the thickness (mm) of the planks to prevent failure?

  1. 90
  2. 35
  3. 55
  4. 120
The active pressure is triangular with base intensity $K_a\gamma h=(1/3)(17.3)(2.4)=13.84$ kPa. Analyze the 300-mm plank as a cantilever under this line load. From $M_{max}=wh^2/6$, $\sigma=Mc/I=14.2$ MPa; from $V_{max}=wh/2$ and $\tau_{max}=3V/(2A)$, $\tau=0.33$ MPa. Resize the section for the allowable bending stress, giving required thickness 90 mm. Thus: 14.2 MPa, 0.33 MPa, and 90 mm.