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Solid Mechanic- TORSION


Strength of Materials and Structures (4th Edition)
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TORSION
Consider a bar to be rigidly attached at one end and twisted at the other end by a torque or twisting moment T equivalent to F × d, which is applied perpendicular to the axis of the bar, as shown in the figure. Such a bar is said to be in torsion.

Bar in torsion

TORSIONAL SHEARING STRESS, τ
For a solid or hollow circular shaft subject to a twisting moment T, the torsional shearing stress τ at a distance ρ from the center of the shaft is

$ \tau = \dfrac{T \rho}{J} \, \text{ and } \, \tau_{max} = \dfrac{Tr}{J} $

where J is the polar moment of inertia of the section and r is the outer radius.

For solid cylindrical shaft:
Section of a solid shaft$ J = \dfrac{\pi}{32} D^4 $
$ \tau_{max} = \dfrac{16T}{\pi D^3} $

For hollow cylindrical shaft:
Hollow cylindrical shaft$ J = \dfrac{\pi}{32}(D^4 - d^4) $
$ \tau_{max} = \dfrac{16TD}{\pi(D^4 - d^4)} $

ANGLE OF TWIST
The angle θ through which the bar length L will twist is

$ \theta = \dfrac{TL}{JG} \, \text{in radians} $

where T is the torque in N·mm, L is the length of shaft in mm, G is shear modulus in MPa, J is the polar moment of inertia in mm4, D and d are diameter in mm, and r is the radius in mm.

POWER TRANSMITTED BY THE SHAFT
A shaft rotating with a constant angular velocity ω (in radians per second) is being acted by a twisting moment T. The power transmitted by the shaft is

$ P = T\omega = 2\pi T f $



Problem 304
A steel shaft 3 ft long that has a diameter of 4 in is subjected to a torque of 15 kip·ft. Determine the maximum shearing stress and the angle of twist. Use G = 12 × 106 psi.

Solution 304
$ \tau_{max} = \dfrac{16T}{\pi D^3} = \dfrac{16(15)(1000)(12)}{\pi (4^3)} $
$ \tau_{max} = 14 324 \, \text{psi} $
$ \tau_{max} = 14.3 \, \text{ksi} \, $answer

$ \theta = \dfrac{TL}{JG} = \dfrac{15(3)(1000)(12^2)}{\frac{1}{32} \pi (4^4)(12 \times 10^6)} $
$ \theta = 0.0215 \, \text{rad} $
$ \theta = 1.23^\circ $           answer



Problem 305

What is the minimum diameter of a solid steel shaft that will not twist through more than 3° in a 6-m length when subjected to a torque of 12 kN·m? What maximum shearing stress is developed? Use G = 83 GPa.

 

Solution 305

$ \theta = \dfrac{TL}{JG} $
$ 3^\circ \left( \dfrac{\pi}{180^\circ} \right) = \dfrac{12(6)(1000^3)}{\frac{1}{32}\pi d^4 (83\,000)} $
$ d = 113.98 \, \text{mm} \, $answer

 

$ \tau_{max} = \dfrac{16T}{\pi d^3} = \dfrac{16(12)(1000^2)}{\pi (113.98^3)} $
$ \tau_{max} = 41.27 \, \text{MPa} \, $answer

Solution 306

$ T = \dfrac{P}{2 \pi f} = \dfrac{5000(396\,000)}{2\pi(189)} $
$ T = 1\,667\,337.5 \, \text{lb}\cdot\text{in} $

 

$ \tau_{max} = \dfrac{16T}{\pi d^3} = \dfrac{16(1\,667\,337.5)}{\pi (14^3)} $
$ \tau_{max} = 3094.6 \, \text{psi} \, $answer

 



Problem 309

A steel propeller shaft is to transmit 4.5 MW at 3 Hz without exceeding a shearing stress of 50 MPa or twisting through more than 1° in a length of 26 diameters. Compute the proper diameter if G = 83 GPa.

 

Solution 309

$ T = \dfrac{P}{2\pi f} = \dfrac{4.5(1\,000\,000)}{2\pi(3)} $
$ T = 238\,732.41 \, \text{N}\cdot\text{m} $

 

Based on maximum allowable shearing stress:
$ \tau_{max} = \dfrac{16T}{\pi d^3} $
$ 50 = \dfrac{16(238\,732.41)(1000)}{\pi d^3} $
$ d = 289.71 \, \text{mm} $

 

Based on maximum allowable angle of twist:
$ \theta = \dfrac{TL}{JG} $
$ 1^\circ \left( \dfrac{\pi}{180^\circ} \right) = \dfrac{238\,732.41(26d)(1000)}{\frac{1}{32}\pi d^4 (83\,000)} $
$ d = 352.08 \, \text{mm} $

 

Use the bigger diameter, d = 352 mm answer


Problem 311

An aluminum shaft with a constant diameter of 50 mm is loaded by torques applied to gears attached to it as shown in Fig. P-311. Using G = 28 GPa, determine the relative angle of twist of gear D relative to gear A.

 

Three segments aluminum shaft

 

Problem 311

Free Body Diagram

 

$ \theta = \dfrac{TL}{JG} $
Rotation of D relative to A:
$ \theta_{D/A} = \dfrac{1}{JG} \Sigma TL $
$ \theta_{D/A} = \dfrac{1}{\frac{1}{32}\pi (50^4)(28\,000)} \, [ \, 800(2) - 300(3) + 600(2) \, ] \, (1000^2) $
$ \theta_{D/A} = 0.1106 \, \text{rad} $
$ \theta_{D/A} = 6.34^\circ \, $answer


Problem 312
A flexible shaft consists of a 0.20-in-diameter steel wire encased in a stationary tube that fits closely enough to impose a frictional torque of 0.50 lb·in/in. Determine the maximum length of the shaft if the shearing stress is not to exceed 20 ksi. What will be the angular deformation of one end relative to the other end? G = 12 × 106 psi.

Solution 312
312-wire-encased-in-tube_0.jpg$ \tau_{max} = \dfrac{16T}{\pi d^3} $
$ 20(1000) = \dfrac{16T}{\pi (0.20)^3} $
$ T = 10\pi \, \text{lb}\cdot\text{in} $

$ L = \dfrac{T}{0.50 \, \text{lb}\cdot\text{in/in}} $
$ L = \dfrac{10\pi \, \text{lb}\cdot\text{in}}{0.50 \, \text{lb}\cdot\text{in/in}} $
$ L = 20\pi \, \text{in} = 62.83 \, \text{in} $

$ \theta = \dfrac{TL}{JG} $
If θ = dθ, T = 0.5L and L = dL
$ \displaystyle \int d\theta = \frac{1}{JG} \int_0^{20\pi} (0.5L) \, dL $
$ \theta = \left[ \dfrac{0.5L^2}{2} \right]_0^{2\pi} = \dfrac{1}{JG} \, [ \, 0.25(20\pi)^2 - 0.25(0)^2 \, ] $
$ \theta = \dfrac{100\pi^2}{\frac{1}{32}\pi (0.20^4)(12 \times 10^6)} $
$ \theta = 0.5234 \, \text{rad} = 30^\circ \, $answer


Problem 316

A compound shaft consisting of a steel segment and an aluminum segment is acted upon by two torques as shown in Fig. P-316. Determine the maximum permissible value of T subject to the following conditions: τst ≤ 83 MPa, τal ≤ 55 MPa, and the angle of rotation of the free end is limited to 6°. For steel, G = 83 GPa and for aluminum, G = 28 GPa.

 

Figure P-316

 

Solution 316

 

316-torque-diagram.jpg

 

Based on maximum shearing stress τmax = 16T / πd3:

Steel
$ \tau_{st} = \dfrac{16(3T)}{\pi(50^3)} = 83 $
$ T = 679\,042.16 \, \text{N}\cdot\text{mm} $
$ T = 679.04 \, \text{N}\cdot\text{m} $
Aluminum
$ \tau_{al} = \dfrac{16T}{\pi (40^3)} = 55 $
$ T = 691\,150.38 \, \text{N}\cdot\text{mm} $
$ T = 691.15 \, \text{N}\cdot\text{m} $

 

Based on maximum angle of twist:

$ \theta = \left( \dfrac{TL}{JG} \right)_{st} + \left( \dfrac{TL}{JG} \right)_{al} $
$ 6^\circ \left( \dfrac{\pi}{180^\circ} \right) = \dfrac{3T(900)}{\frac{1}{32}\pi (50^4)(83\,000)} + \dfrac{T(600)}{\frac{1}{32}\pi (40^4)(28\,000)} $
$ T = 757\,316.32 \, \text{N}\cdot\text{mm} $
$ T = 757.32 \, \text{N}\cdot\text{m} $

 

Use T = 679.04 N·m answer

 



Problem 317

A hollow bronze shaft of 3 in. outer diameter and 2 in. inner diameter is slipped over a solid steel shaft 2 in. in diameter and of the same length as the hollow shaft. The two shafts are then fastened rigidly together at their ends. For bronze, G = 6 × 106 psi, and for steel, G = 12 × 106 psi. What torque can be applied to the composite shaft without exceeding a shearing stress of 8000 psi in the bronze or 12 ksi in the steel?

 

Solution 317

Encased shaft

 

$ \theta_{st} = \theta_{br} $
$ \left( \dfrac{TL}{JG} \right)_{st} = \left( \dfrac{TL}{JG} \right)_{br} $
$ \dfrac{T_{st} L}{\frac{1}{32}\pi (2^4)(12 \times 10^6)} = \dfrac{T_{br} L}{\frac{1}{32}\pi (3^4 - 2^4)(6 \times 10^6)} $
$ \dfrac{T_{st}}{192 \times 10^6} = \dfrac{T_{br}}{390 \times 10^6} \, \to \, $Equation (1)

 

Applied Torque = Resisting Torque
$ T = T_{st} + T_{br} \, \to \, $Equation (2)

 

Equation (1) with Tst in terms of Tbr and Equation (2)
$ T = \dfrac{192 \times 10^6}{390 \times 10^6} T_{br} + T_{br} $
$ T_{br} = 0.6701T $

 

Equation (1) with Tbr in terms of Tst and Equation (2)
$ T = T_{st} + \dfrac{390 \times 10^6}{192 \times 10^6} T_{br} $
$ T_{st} = 0.3299T $

 

Based on hollow bronze (Tbr = 0.6701T)
$ \tau_{max} = \left[ \dfrac{16TD}{\pi(D^4 - d^4)} \right]_{br} $
$ 8000 = \dfrac{16(0.6701T)(3)}{\pi (3^4 - 2^4)} $
$ T = 50 789.32 \, \text{lb}\cdot\text{in} $
$ T = 4232.44 \, \text{lb}\cdot\text{ft} $

 

Based on steel core (Tst = 0.3299T):
$ \tau_{max} = \left[ \dfrac{16TD}{\pi D^3} \right]_{st} $
$ 12\,000 = \dfrac{16(0.3299T)}{\pi (2^3)} $
$ T = 57\,137.18 \, \text{lb}\cdot\text{in} $
$ T = 4761.43 \, \text{lb}\cdot\text{ft} $

 

Use T = 4232.44 lb·ft answer


Problem 318

A solid aluminum shaft 2 in. in diameter is subjected to two torques as shown in Fig. P-318. Determine the maximum shearing stress in each segment and the angle of rotation of the free end. Use G = 4 × 106 psi.

 

318-solid-aluminum-shaft.jpg

 

Solution 318

 

318-torque-diagram.jpg

 

$ \tau_{max} = \dfrac{16T}{\pi D^3} $
For 2-ft segment:
$ \tau_{max2} = \dfrac{16(600)(12)}{\pi (2^3)} = 4583.66 \, \text{psi} \, $answer

 

For 3-ft segment:
$ \tau_{max3} = \dfrac{16(800)(12)}{\pi (2^3)} = 6111.55 \, \text{psi} \, $answer

 

$ \theta = \dfrac{TL}{JG} $
$ \theta = \dfrac{1}{JG} \Sigma TL $
$ \theta = \dfrac{1}{\frac{1}{31}\pi (2^4)(4 \times 10^6)} [ \, 600(2) + 800(3) \, ] \, (12^2) $
$ \theta = 0.0825 \, \text{rad} $
$ \theta = 4.73^\circ $answer


Problem 319
The compound shaft shown in Fig. P-319 is attached to rigid supports. For the bronze segment AB, the diameter is 75 mm, τ ≤ 60 MPa, and G = 35 GPa. For the steel segment BC, the diameter is 50 mm, τ ≤ 80 MPa, and G = 83 GPa. If a = 2 m and b = 1.5 m, compute the maximum torque T that can be applied.

Compound Shaft

Solution 319

319-fbd.jpg

$ \Sigma M = 0 $
$ T = T_{br} + T_{st} \, \to \, $Equation (1)
$ \theta_{br} = \theta_{st} $
$ \left( \dfrac{TL}{JG} \right)_{br} = \left( \dfrac{TL}{JG} \right)_{st} $
$ \dfrac{T_{br}(2)(1000)}{\frac{1}{32}\pi (75^4)(35\,000)} = \dfrac{T_{st}(1.5)(1000)}{\frac{1}{32}\pi (50^4)(83\,000)} $
$ T_{br} = 1.6011T_{st} \, \to \, $Equation (2a)
$ T_{st} = 0.6246T_{br} \, \to \, $Equation (2b)

$ \tau_{max} = \dfrac{16T}{\pi D^3} $

Based on τbr ≤ 60 MPa
$ 60 = \dfrac{16T_{br}}{\pi (75^3)} $
$ T_{br} = 4\,970\,097.75 \, \text{N}\cdot\text{mm} $
$ T_{br} = 4.970 \, \text{kN}\cdot\text{m} \, \to \, $Maximum allowable torque for bronze

$ T_{st} = 0.6246(4.970) \, \to \, $From Equation (2b)
$ T_{st} = 3.104 \, \text{kN}\cdot\text{m} $

Based on τbr ≤ 80 MPa
$ 80 = \dfrac{16T_{st}}{\pi (50^3)} $
$ T_{st} = 1\,963\,495.41 \, \text{N}\cdot\text{mm} $
$ T_{st} = 1.963 \, \text{kN}\cdot\text{m} \, \to \, $Maximum allowable torque for steel

$ T_{br} = 1.6011(1.963) \, \to \, $From Equation (2a)
$ T_{br} = 3.142 \, \text{kN}\cdot\text{m} $

Use Tbr = 3.142 kN·m and Tst = 1.963 kN·m
$ T = 3.142 + 1.963 \, \to \, $From Equation (1)
$ T = 5.105 \, \text{kN}\cdot\text{m} \, $answer


Problem 323
A shaft composed of segments AC, CD, and DB is fastened to rigid supports and loaded as shown in Fig. P-323. For bronze, G = 35 GPa; aluminum, G = 28 GPa, and for steel, G = 83 GPa. Determine the maximum shearing stress developed in each segment.

Three-segment shaft

Solution 323
Stress developed in each segment with respect to TA:

Torque diagram

The rotation of B relative to A is zero.
$ \theta_{A/B} = 0 $
$ \left( {\Large \Sigma} \dfrac{TL}{JG} \right)_{A/B} = 0 $
$ \dfrac{T_A (2)(1000^2)}{\frac{1}{32}\pi (25^4)(35\,000)} + \dfrac{(T_A - 300) (2)(1000^2)}{\frac{1}{32}\pi (50^4)(28\,000)} + \dfrac{(T_A - 1000) (2.5)(1000^2)}{\frac{1}{32}\pi (25^4)(83\,000)} = 0 $
$ \dfrac{2T_A}{(25^4)(35)} + \dfrac{2(T_A - 300)}{(50^4)(28)} + \dfrac{2.5(T_A - 1000)}{(25^4)(83)} = 0 $
$ \dfrac{16T_A}{35} + \dfrac{T_A - 300}{28} + \dfrac{20(T_A - 1000)}{83} = 0 $
$ \frac{16}{25}T_A + \frac{1}{28}T_A - \frac{75}{7} + \frac{20}{83}T_A - \frac{20\,000}{83} = 0 $
$ \frac{8527}{11\,620}T_A = 251.678 $
$ T_A = 342.97 \, \text{N}\cdot\text{m} $

$ \Sigma M = 0 $
$ T_A + T_B = 300 + 700 $
$ 342.97 + T_B = 1000 $
$ T_B = 657.03 \, \text{N}\cdot\text{m} $

$ T_{br} = 342.97 \, \text{N}\cdot\text{m} $
$ T_{al} = 342.97 - 300 = 42.97 \, \text{N}\cdot\text{m} $
$ T_{st} = 342.97 - 1000 = -657.03 \, \text{N}\cdot\text{m} = -T_B \, $(ok!)

$ \tau_{max} = \dfrac{16T}{\pi D^3} $
$ \tau_{br} = \dfrac{16(342.97)(1000)}{\pi (25^3)} = 111.79 \, \text{MPa} \, $answer
$ \tau_{al} = \dfrac{16(42.97)(1000)}{\pi (50^3)} = 1.75 \, \text{MPa} \, $answer
$ \tau_{st} = \dfrac{16(657.03)(1000)}{\pi (25^3)} = 214.16 \, \text{MPa} \, $answer


Problem 325
The two steel shaft shown in Fig. P-325, each with one end built into a rigid support have flanges rigidly attached to their free ends. The shafts are to be bolted together at their flanges. However, initially there is a 6° mismatch in the location of the bolt holes as shown in the figure. Determine the maximum shearing stress in each shaft after the shafts are bolted together. Use G = 12 × 106 psi and neglect deformations of the bolts and flanges.

Shafts connected through flanges

Solution 325
$ \theta_{of\,\,6.5'\,\,shaft} + \theta_{of\,\,3.25'\,\,shaft} = 6^\circ $
$ \left( \dfrac{TL}{JG} \right)_{of\,\,6.5'\,\,shaft} + \left( \dfrac{TL}{JG} \right)_{of\,\,3.25'\,\,shaft} = 6^\circ \left( \dfrac{\pi}{180^\circ} \right) $
$ \dfrac{T(6.5)(12^2)}{\frac{1}{32}\pi (2^4)(12 \times 10^6)} + \dfrac{T(3.25)(12^2)}{\frac{1}{32}\pi (1.5^4)(12 \times 10^6)} = \dfrac{\pi}{30} $
$ T = 817.32 \, \text{lb}\cdot\text{ft} $

$ \tau_{max} = \dfrac{16T}{\pi D^3} $
$ \tau_{of\,\,6.5'\,\,shaft} = \dfrac{16(817.32)(12)}{\pi (2^3)} = 6243.86 \, \text{psi} \, $answer
$ \tau_{of\,\,3.25'\,\,shaft} = \dfrac{16(817.32)(12)}{\pi (1.5^3)} = 14\,800.27 \, \text{psi} \, $answer



Chapter 4 - Shear and Moment in Beams
Shear and moment diagrams by shear and moment equationsThe vertical shear at C in the figure shown in previous section (also shown to the right) is taken as
$ V_C = (\Sigma F_v)_L = R_1 - wx $
where R1 = R2 = wL/2
$ V_c = \dfrac{wL}{2} - wx $

 

The moment at C is
$ M_C = (\Sigma M_C) = \dfrac{wL}{2}x - wx \left( \dfrac{x}{2} \right) $
$ M_C = \dfrac{wLx}{2} - \dfrac{wx^2}{2} $
If we differentiate M with respect to x:
$ \dfrac{dM}{dx} = \dfrac{wL}{2} \cdot \dfrac{dx}{dx} - \dfrac{w}{2} \left( 2x \cdot \dfrac{dx}{dx} \right) $
$ \dfrac{dM}{dx} = \dfrac{wL}{2} - wx = \text{shear} $

 

thus,
$ \dfrac{dM}{dx} = V $

 

Thus, the rate of change of the bending moment with respect to x is equal to the shearing force, or the slope of the moment diagram at the given point is the shear at that point.

 

Differentiate V with respect to x gives
$ \dfrac{dV}{dx} = 0 - w $

 

thus,
$ \dfrac{dV}{dx} = \text{Load} $

 

Thus, the rate of change of the shearing force with respect to x is equal to the load or the slope of the shear diagram at a given point equals the load at that point.

 

Properties of Shear and Moment Diagrams

The following are some important properties of shear and moment diagrams:
  1. The area of the shear diagram to the left or to the right of the section is equal to the moment at that section.
  2. The slope of the moment diagram at a given point is the shear at that point.
  3. The slope of the shear diagram at a given point equals the load at that point.
  4. concavity-of-beam.jpgThe maximum moment occurs at the point of zero shears. This is in reference to property number 2, that when the shear (also the slope of the moment diagram) is zero, the tangent drawn to the moment diagram is horizontal.
  5. When the shear diagram is increasing, the moment diagram is concave upward.
  6. When the shear diagram is decreasing, the moment diagram is concave downward.

Sign Convention

The customary sign conventions for shearing force and bending moment are represented by the figures below. A force that tends to bend the beam downward is said to produce a positive bending moment. A force that tends to shear the left portion of the beam upward with respect to the right portion is said to produce a positive shearing force.

 

Positive and negative bending and shear

 

An easier way of determining the sign of the bending moment at any section is that upward forces always cause positive bending moments regardless of whether they act to the left or to the right of the exploratory section.

 

INSTRUCTION:

Without writing shear and moment equations, draw the shear and moment diagrams for the beams specified in the following problems. Give numerical values at all change of loading positions and at all points of zero shear. (Note to instructor: Problems 403 to 420 may also be assigned for solution


Problem 425
Beam loaded as shown in Fig. P-425. See the instruction.

Concentrated loads in simple beam with one end overhangs

Solution 425
$ \Sigma M_A = 0 $
$ 6R_2 = 2(60) + 7(30) $
$ R_2 = 55 \, \text{kN} $

$ \Sigma M_C = 0 $
$ 6R_1 + 1(30) = 4(60) $
$ R_1 = 35 \, \text{kN} $

425-load-shear-moment.jpgTo draw the Shear Diagram:
  1. VA = R1 = 35 kN
  2. VB = VA + Area in load diagram - 60 kN
    VB = 35 + 0 - 60 = -25 kN
  3. VC = VB + area in load diagram + R2
    VC = -25 + 0 + 55 = 30 kN
  4. VD = VC + Area in load diagram - 30 kN
    VD = 30 + 0 - 30 = 0
To draw the Moment Diagram:
  1. MA = 0
  2. MB = MA + Area in shear diagram
    MB = 0 + 35(2) = 70 kN·m
  3. MC = MB + Area in shear diagram
    MC = 70 - 25(4) = -30 kN·m
  4. MD = MC + Area in shear diagram
    MD = -30 + 30(1) = 0



Problem 426
Cantilever beam acted upon by a uniformly distributed load and a couple as shown in Fig. P-426. See the instruction.

Uniform and moment loads in cantilever beam

Solution 426
426-load-shear-moment.jpgTo draw the Shear Diagram
  1. VA = 0
  2. VB = VA + Area in load diagram
    VB = 0 - 5(2)
    VB = -10 kN
  3. VC = VB + Area in load diagram
    VC = -10 + 0
    VC = -10 kN
  4. VD = VC + Area in load diagram
    VD = -10 + 0
    VD = -10 kN
To draw the Moment Diagram
  1. MA = 0
  2. MB = MA + Area in shear diagram
    MB = 0 - ½(2)(10)
    MB = -10 kN·m
  3. MC = MB + Area in shear diagram
    MC = -10 - 10(2)
    MC = -30 kN·m
    MC2 = -30 + M = -30 + 60 = 30 kN·m
  4. MD = MC2 + Area in shear diagram
    MD = 30 - 10(1)
    MD = 20 kN·m
Home » Strength of Materials (Mechanics of Materials) » Chapter 4 - Shear and Moment in Beams » Relationship Between Load, Shear, and Moment

Solution to Problem 427 | Relationship Between Load, Shear, and Moment

A D V E R T I S E M E N T



Problem 427

Beam loaded as shown in Fig. P-427. See the instruction.

 

Concentrated and uniform loads in simple beam

 

Solution 427

$ \Sigma M_C = 0 $
$ 12R_1 = 100(12)(6) + 800(3) $
$ R_1 = 800 \, \text{lb} $

 

$ \Sigma M_A = 0 $
$ 12R_2 = 100(12)(6) + 800(9) $
$ R_2 = 1200 \, \text{lb} $

 

427-load-shear-moment.jpgTo draw the Shear Diagram

  1. VA = R1 = 800 lb
  2. VB = VA + Area in load diagram
    VB = 800 - 100(9)
    VB = -100 lb
    VB2 = -100 - 800 = -900 lb
  3. VC = VB2 + Area in load diagram
    VC = -900 - 100(3)
    VC = -1200 lb
  4. Solving for x:
    x / 800 = (9 - x) / 100
    100x = 7200 - 800x
    x = 8 ft

To draw the Moment Diagram

  1. MA = 0
  2. Mx = MA + Area in shear diagram
    Mx = 0 + ½(8)(800) = 3200 lb·ft;
  3. MB = Mx + Area in shear diagram
    MB = 3200 - ½(1)(100) = 3150 lb·ft
  4. MC = MB + Area in shear diagram
    MC = 3150 - ½(900 + 1200)(3) = 0
  5. The moment curve BC is downward parabola with vertex at A'. A' is the location of zero shear for segment BC.

 



Problem 428

Beam loaded as shown in Fig. P-428. See the instruction.

 

Uniform and moment loads in overhang beam

 

Solution 428

$ \Sigma M_D = 0 $
$ 5R_1 = 50(0.5) + 25 $
$ R_1 = 10 \, \text{kN} $

 

$ \Sigma M_A = 0 $
$ 5R_2 + 25 = 50(4.5) $
$ R_2 = 40 \, \text{kN} $

 

428-load-shear-moment.jpgTo draw the Shear Diagram

  1. VA = R1 = 10 kN
  2. VB = VA + Area in load diagram
    VB = 10 + 0 = 10 kN
  3. VC = VB + Area in load diagram
    VC = 10 + 0 = 10 kN
  4. VD = VC + Area in load diagram
    VD = 10 - 10(3) = -20 kN
    VD2 = -20 + R2 = 20 kN
  5. VE = VD2 + Area in load diagram
    VE = 20 - 10(2) = 0
  6. Solving for x:
    x / 10 = (3 - x) / 20
    20x = 30 - 10x
    x = 1 m

To draw the Moment Diagram

  1. MA = 0
  2. MB = MA + Area in shear diagram
    MB = 0 + 1(10) = 10 kN·m
    MB2 = 10 - 25 = -15 kN·m
  3. MC = MB2 + Area in shear diagram
    MC = -15 + 1(10) = -5 kN·m
  4. Mx = MC + Area in shear diagram
    Mx = -5 + ½(1)(10) = 0
  5. MD = Mx + Area in shear diagram
    MD = 0 - ½(2)(20) = -20 kN·m
  6. ME = MD + Area in shear diagram
    ME = -20 + ½ (2)(20) = 0


Problem 430

Beam loaded as shown in Fig. P-430. See the instruction.

 

Uniform and point loads in overhanging beam

 

Solution 430

$ \Sigma M_D = 0 $
$ 20R_1 = 1000(25) + 400(5)(22.5) + 2000(10) + 200(10)(5) $
$ R_1 = 5000 \, \text{lb} $

 

$ \Sigma M_B = 0 $
$ 20R_2 + 1000(5) + 400(5)(2.5) = 2000(10) + 200(10)(15) $
$ R_2 = 2000 \, \text{lb} $

 

430-load-shear-moment.jpgTo draw the Shear Diagram

  1. VA = -1000 lb
  2. VB = VA + Area in load diagram; VB = -1000 - 400(5) = -3000 lb; VB2 = -3000 + R1 = 2000 lb
  3. VC = VB2 + Area in load diagram; VC = 2000 + 0 = 2000 lb; VC2 = 2000 - 2000 = 0
  4. VD = VC2 + Area in load diagram; VD = 0 + 200(10) = 2000 lb

To draw the Moment Diagram

  1. MA = 0
  2. MB = MA + Area in shear diagram
    MB = 0 - ½ (1000 + 3000)(5)
    MB = -10000 lb·ft
  3. MC = MB + Area in shear diagram
    MC = -10000 + 2000(10) = 10000 lb·ft
  4. MD = MC + Area in shear diagram
    MD = 10000 - ½ (10)(2000) = 0
  5. For segment BC, the location of zero moment can be accomplished by symmetry and that is 5 ft from B.
  6. The moment curve AB is a downward parabola with vertex at A'. A' is the location of zero shear for segment AB at point outside the beam.


Problem 431

Beam loaded as shown in Fig. P-431. See the instruction.

 

Uniform and point loads in overhanging beam

 

$ \Sigma M_D = 0 $
$ 7R_1 + 40(3) = 5(50) + 10(10)(2) + 20(4)(2) $
$ R_1 = 70 \, \text{kN} $

 

$ \Sigma M_A = 0 $
$ 7R_2 = 50(2) + 10(10)(5) + 20(4)(5) + 40(10) $
$ R_2 = 200 \, \text{lb} $

 

431-load-shear-moment.jpgTo draw the Shear Diagram

  1. VA = R1 = 70 kN
  2. VB = VA + Area in load diagram
    VB = 70 - 10(2) = 50 kN
    VB2 = 50 - 50 = 0
  3. VC = VB2 + Area in load diagram
    VC = 0 - 10(1) = -10 kN
  4. VD = VC + Area in load diagram
    VD = -10 - 30(4) = -130 kN
    VD2 = -130 + R2
    VD2 = -130 + 200 = 70 kN
  5. VE = VD2 + Area in load diagram; VE = 70 - 10(3) = 40 kN
    VE2 = 40 - 40 = 0

To draw the Moment Diagram

  1. MA = 0
  2. MB = MA + Area in shear diagram
    MB = 0 + ½ (70 + 50)(2) = 120 kN·m
  3. MC = MB + Area in shear diagram
    MC = 120 - ½ (1)(10) = 115 kN·m
  4. MD = MC + Area in shear diagram
    MD = 115 - ½ (10 + 130)(4)
    MD = -165 kN·m
  5. ME = MD + Area in shear diagram
    ME = -165 + ½ (70 + 40)(3) = 0
  6. Moment curves AB, CD and DE are downward parabolas with vertices at A', B' and C', respectively. A', B' and C' are corresponding zero shear points of segments AB, CD and DE.
  7. Solving for point of zero moment:
    a / 10 = (a + 4) / 130
    130a = 10a + 40
    a = 1/3 m

    y / (x + a) = 130 / (4 + a)
    y = 130(x + 1/3) / (4 + 1/3)
    y = 30x + 10

    MC = 115 kN·m
    Mzero = MC + Area in shear
    0 = 115 - ½ (10 + y)x
    (10 + y)x = 230
    (10 + 30x + 10)x = 230
    30x2 + 20x - 230 = 0
    3x2 + 2x - 23 = 0
    x = 2.46 m

    Zero moment is at 2.46 m from C


Problem 436

A distributed load is supported by two distributed reactions as shown in Fig. P-436. See the instruction.

 

Beam with two distributed reactions

 

Solution 436

$ \Sigma M_{midpoint\,\,of\,\,CD} = 0 $
$ 4w_1 (11) = 440(8)(5) $
$ w_1 = 400 \, \text{lb/ft} $
$ \Sigma M_{midpoint\,\,of\,\,AB} = 0 $
$ 2w_2 (11) = 440(8)(6) $
$ w_2 = 960 \, \text{lb/ft} $

 

436-load-shear-moment.jpgTo draw the Shear Diagram

  1. VA = 0
  2. VB = VA + Area in load diagram
    VB = 0 + 400(4) = 1600 lb
  3. VC = VB + Area in load diagram
    VC = 1600 - 440(8) = -1920 lb
  4. VD = VC + Area in load diagram
    VD = -1920 + 960(2) = 0
  5. Location of zero shear:
    x / 1600 = (8 - x) / 1920
    x = 40/11 ft = 3.636 ft from B

To draw the Moment Diagram

  1. MA = 0
  2. MB = MA + Area in shear diagram
    MB = 0 + ½ (1600)(4) = 3200 lb·ft
  3. Mx = MB + Area in shear diagram
    Mx = 3200 + ½ (1600)(40/11)
    Mx = 6109.1 lb·ft
  4. MC = Mx + Area in shear diagram
    MC = 6109.1 - ½ (8 - 40/11)(1920)
    MC = 1920 lb·ft
  5. MD = MC + Area in shear diagram
    MD = 1920 - ½ (1920)(2) = 0

 Problem 439

A beam supported on three reactions as shown in Fig. P-439 consists of two segments joined by frictionless hinge at which the bending moment is zero. See the instruction.

 

Continuous beam with internal hinge

 

Solution 439

439-fbd-to-the-left-of-hinge.jpg$ \Sigma M_H = 0 $
$ 8R_1 = 4000(4) $
$ R_1 = 2000 \, \text{lb} $
$ \Sigma M_A = 0 $
$ 8V_H = 4000(4) $
$ V_H = 2000 \, \text{lb} $

 

439-fbd-to-the-right-of-hinge.jpg$ \Sigma M_D = 0 $
$ 10R_2 = 2000(14) + 400(10)(5) $
$ R_2 = 4800 \, \text{lb} $
$ \Sigma M_H = 0 $
$ 14R_3 + 4(4800) = 400(10)(9) $
$ R_3 = 1200 \, \text{lb} $

 

439-load-shear-moment.jpgTo draw the Shear Diagram

  1. VA = 0
  2. VB = 2000 lb
    VB2 = 2000 - 4000 = -2000 lb
  3. VH = -2000 lb
  4. VC = -2000 lb
    VC = -2000 + 4800 = 2800 lb
  5. VD = 2800 - 400(10) = -1200 lb
  6. Location of zero shear:
    x / 2800 = (10 - x) / 1200
    1200x = 28000 - 2800x
    x = 7 ft

To draw the Moment Diagram

  1. MA = 0
  2. MB = 2000(4) = 8000 lb·ft
  3. MH = 8000 - 4000(2) = 0
  4. MC = -400(2)
    MC = -8000 lb·ft
  5. Mx = -800 + ½ (2800)(7)
    Mx = 1800 lb·ft
  6. MD = 1800 - ½(1200)(3)
    MD = 0
  7. Zero M is 4 ft from R2

 




Problem 440
A frame ABCD, with rigid corners at B and C, supports the concentrated load as shown in Fig. P-440. (Draw shear and moment diagrams for each of the three parts of the frame.) See the instruction.

Cantilever frame

Solution 440
440-load-shear-moment.jpg

 

 

 

 

Problem 442

Beam carrying the uniformly varying load shown in Fig. P-442. See the instruction.

 

Triangular load in simple beam

 

Solution 442

442-fbd.jpg$ \Sigma M_{R2} = 0 $
$ LR_1 = \frac{1}{3}L \,(\frac{1}{2}Lw_o) $
$ R_1 = \frac{1}{6}Lw_o $

 

$ \Sigma M_{R1} = 0 $
$ LR_2 = \frac{2}{3}L \, (\frac{1}{2}Lw_o) $
$ R_2 = \frac{1}{3}Lw_o $

442-load-shear-moment.jpg 

To draw the Shear Diagram

  1. VA = R1 = 1/6 Lwo
  2. VB = VA + Area in load diagram
    VB = 1/6 Lwo - 1/2 Lwo
    VB = -1/3 Lwo
  3. Location of zero shear C:
    By squared property of parabola:
    x2 / (1/6 Lwo) = L2 / (1/6 Lwo + 1/3 Lwo)
    6x2 = 2L2
    x = L / √3
  4. The shear in AB is a parabola with vertex at A, the starting point of uniformly varying load. The load in AB is 0 at A to downward wo or -wo at B, thus the slope of shear diagram is decreasing. For decreasing slope, the parabola is open downward.

To draw the Moment Diagram

  1. MA = 0
  2. MC = MA + Area in shear diagram
    MC = 0 + 2/3 (L/√3)(1/6 Lwo)
    MC = 0.06415L2wo = Mmax
  3. MB = MC + Area in shear diagram
    MB = MC - A1 (see figure for solving A1)

    442-solving-a1.jpg

    For A1:
    A1 = 1/3 L(1/6 Lwo + 1/3 Lwo) - 1/3 (L/√3)(1/6 Lwo) - 1/6 Lwo (L - L/√3)
    A1 = 0.16667L2wo - 0.03208L2wo - 0.07044L2wo
    A1 = 0.06415L2wo

    MB = 0.06415L2wo - 0.06415L2wo = 0

The shear diagram is second degree curve, thus the moment diagram is a thir

 

 

  1. d degree curve. The maximum moment (highest point) occurred at C, the location of zero shear. The value of shears in AC is positive then the moment in AC is increasing; at CB the shear is negative, then the moment in CB is decreasing.









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