Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 4, Problem 4.5.14P

The beam AB shown in the figure supports a uniform load of intensity 3000 N/m acting over half the length of the beam. The beam rests on a foundation that produces a uniformly distributed load over the entire length.

  1. Draw the shear-force and bending-moment diagrams for this beam.

Repeat part (a) for the distributed load variation shown in Fig. b.

  Chapter 4, Problem 4.5.14P, The beam AB shown in the figure supports a uniform load of intensity 3000 N/m acting over half the , example  1

  Chapter 4, Problem 4.5.14P, The beam AB shown in the figure supports a uniform load of intensity 3000 N/m acting over half the , example  2

(a)

Expert Solution
Check Mark
To determine

The shear force and bending moment diagram for the given beam.

Answer to Problem 4.5.14P

Maximum shear force Vmax= y13 1200Ν

Maximum bending moment Mmax= 960Νm

Explanation of Solution

Given information: The given beam and parameters are shown in the figure below:

  Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.14P , additional homework tip  1

For calculating the maximum shear force (V) and bending moment (M) of the given figure, we need to find the amount of force acting upwards on the entire span length.

  fq=0

  q(1.6+2×0.8)3000×1.6=03.2q=4800q=48003.2q=1500Ν/m

Shear Force Diagram:

To find the shear force of the given figure, we divide the above figure in a number of sections.

  1. Firstly taking a section from length 0 to 0.8 m.
  2.   V1=qxV1=1500x

    The value of shear force when x= 0 at point A is

      V1=qxV1=1500×0V1=0

    The value of shear force when x= 0.8 at point A is

      V1=qxV1=1500×0.8V1=1200Ν

  3. Again taking a section from length 0.8 m to mid-span.
  4.   V2=1500x3000(x0.8)V2=1500x3000x+2400V2=24001500x

    The value of shear force when x= 0.8 is,

      V2=24001500×0.8V2=24001200V2=1200Ν

    The value of shear force at mid- span when x= 1.6 is,

      V2=24001500×1.6V2=24002400V2=0Ν

  5. For the next half of the beam, the shear values can be obtained from the concept of symmetry and the obtained values is shown below in the given shear force diagram.

Bending Moment Diagram:

To find the bending moments of the given figure, we divide the above figure in a number of sections.

  1. Firstly taking a section from length 0 to 0.8 m.
  2.   M1=qx(x2)M1=1500x(x2)M1=750x2

    As the equation is of second order of degree, the curve obtained is a parabola.

    The value of Moment when x= 0 at point A is

      M1=750x2M1=750×0M1=0

    The value of moment when x= 0.8 at point A is

      M1=750x2M1=750×(0.8)2M1=480Νm

  3. Again taking a section from length 0.8 m to mid-span.
  4.   M2=1500x(x2)3000(x0.8)(x0.82)M2=750x21500(x0.8)2

    The value of bending moment when x= 0.8 is,

      M2=750×(0.8)21500(0.80.8)2M2=480Νm

    The value of bending moment at mid- span when x= 1.6 is,

      M2=750×(1.6)21500(1.60.8)2M2=19201500×(0.8)2M2=1920960M2=960Νm

  5. For the next half of the beam, the bending moment values can be obtained from the concept of symmetry and the obtained values is shown below in the given bending moment diagram.

On the basis of above calculation the shear force and bending moment diagram for the given beam is as follows:

  Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.14P , additional homework tip  2

(b)

Expert Solution
Check Mark
To determine

The shear force and bending moment diagram for the given beam.

Answer to Problem 4.5.14P

Maximum shear force Vmax= 4.5ΚΝ

Maximum bending moment Mmax= 12.67ΚΝ

Explanation of Solution

Given information: The given beam and parameters are shown in the figure below:

  Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.14P , additional homework tip  3

For calculating the maximum shear force (V) and bending moment (M) of the given figure, we need to find the amount of force acting upwards on the entire span length.

  fq=0

  q(1.6+2×0.8)12×1.6×3000=03.2q=2400q=24003.2q=750Ν/m

Shear Force Diagram:

To find the shear force of the given figure we divide the above figure in a number of sections.

  1. Firstly taking a section from length 0 to 0.8 m.
  2.   V1=qxV1=750x

    The value of shear force when x= 0 at point A is

      V1=qxV1=750×0V1=0

    The value of shear force when x= 0.8 at point A is

      V1=qxV1=750×0.8V1=600Ν

  3. Again taking a section from length 0.8 m to mid-span.
  4. For obtaining the required shear force value, consider a section X-X for the given uniformly varying load.

    From the triangle similarity,

      qxx0.8=q01.6qx=30001.6(x0.8)qx=1875(x0.8)

      V2=qx12×(x0.8)×1875(x0.8)V2=qx12×(x0.8)2×1875

    The value of shear force when x= 0.8 is,

      V2=750×0.812×(0.80.8)2×1875V2=600

    The value of shear force at mid- span when x= 2.4 is,

      V2=750×2.412×(2.40.8)2×1875V2=18002400V2=600Ν

  5. For the next half of the beam, the shear values can be obtained from the concept of symmetry and the obtained values is shown below in the given shear force diagram.

Bending Moment Diagram:

To find the bending moments of the given figure we divide the above figure in a number of sections.

  1. Firstly taking a section from length 0 to 0.8 m.
  2.   M1=qx(x2)M1=750x(x2)M1=375x2

    As the equation is of second order of degree, the curve obtained is a parabola.

    The value of Moment when x= 0 at point A is

      M1=375x2M1=375×0M1=0

    The value of moment when x= 0.8 at point A is

      M1=375x2M1=375×(0.8)2M1=240Νm

  3. For obtaining the required shear force value, consider a section X-X for the given uniformly varying load. From the triangle similarity,
  4.   qxx0.8=q01.6qx=30001.6(x0.8)qx=1875(x0.8)

      M2=750x(x2)12(x0.8)×1875(x0.8)(x0.83)M2=375x218756(x0.8)3

    The value of bending moment when x= 0.8 is,

      M2=375(0.8)218756(0.80.8)3M2=240Νm

    The value of bending moment at mid- span when x= 1.6 is,

      M2=375(1.6)218756(1.60.8)3M2=960250M2=710Νm

  5. For the next half of the beam, the bending moment values can be obtained from the concept of symmetry and the obtained values is shown below in the given bending moment diagram.

On the basis of above calculation the shear force and bending moment diagram for the given beam is as follows:

  Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.14P , additional homework tip  4

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
3. Figure 4 shows a simply supported timber beam AB with 4.5 m long carrying the uniformly distributed load of 50 kN/m at span AC. An additional moment 20 kNm is applied at support B. The cross section of the timber beam is 75 mm x 250 mm. (a) Determine the equations of shear force and bending-moment using cut section method. (b) Draw the shear force and bending-moment diagrams for the beam AB. (C) Determine the maximum normal stress due to bending and maximum shearing stress of the beams. 50 kN/m 3.0 m Figure 4 1.5 m 20 kNm 75 mm 250 mm

Chapter 4 Solutions

Mechanics of Materials (MindTap Course List)

Ch. 4 - A beam ABCD with a vertical arm CE is supported as...Ch. 4 - A simply supported beam AB supports a trapezoid...Ch. 4 - Beam ABCD represents a reinforced-concrete...Ch. 4 - Find shear (V) and moment (M) at x = 3L/4 for the...Ch. 4 - Find expressions for shear force V and moment M at...Ch. 4 - Find expressions for shear force V and moment Mat...Ch. 4 - Find expressions for shear force V and moment Mat...Ch. 4 - Find expressions for shear force V and moment M at...Ch. 4 - Find expressions for shear force V and moment M at...Ch. 4 - Find expressions for shear force V and moment M at...Ch. 4 - A cable with force P is attached to a frame at A...Ch. 4 - Find expressions for shear force V and moment M at...Ch. 4 - A cable with force P is attached to a frame at D...Ch. 4 - Frame ABCD carries two concentrated loads (2P at T...Ch. 4 - Frame ABC has a moment release just left of joint...Ch. 4 - The simply supported beam ABCD is loaded by a...Ch. 4 - The centrifuge shown in the figure rotates in a...Ch. 4 - Draw the shear-Force and bending-moment diagrams...Ch. 4 - A simple beam AB is subjected to a counter...Ch. 4 - Draw the shear-force and bending-moment diagrams...Ch. 4 - The cantilever beam AB shown in the figure is...Ch. 4 - Cantilever beam AB carries an upward uniform load...Ch. 4 - The simple beam AB shown in the figure is...Ch. 4 - A simple beam AB subjected to couples M1and 3M2...Ch. 4 - A simply supported beam ABC is loaded by a...Ch. 4 - A simply supported beam ABC is loaded at the end...Ch. 4 - A beam ABC is simply supported at A and B and has...Ch. 4 - Beam ABCD is simply supported at B and C and has...Ch. 4 - Draw the shear-force and bending-moment diagrams...Ch. 4 - The simple beam AB supports a triangular load of...Ch. 4 - The beam AB shown in the figure supports a uniform...Ch. 4 - A cantilever beam AB supports a couple and a...Ch. 4 - The cantilever beam A B shown in the figure is...Ch. 4 - Beam ABC has simple supports at .A and B. an...Ch. 4 - Beam ABC with an overhang at one end supports a...Ch. 4 - Consider the two beams shown in the figures. Which...Ch. 4 - The three beams in the figure have the same...Ch. 4 - The beam ABC shown in the figure is simply...Ch. 4 - A simple beam AB is loaded by two segments of...Ch. 4 - Two beams (see figure) are loaded the same and...Ch. 4 - The beam A BCD shown in the figure has overhangs...Ch. 4 - A beam ABCD with a vertical arm CE is supported as...Ch. 4 - Beams ABC and CD are supported at A,C, and D and...Ch. 4 - The simple beam ACE shown in the figure is...Ch. 4 - A beam with simple supports is subjected to a...Ch. 4 - A beam of length L is designed to support a...Ch. 4 - The compound beam ABCDE shown in the figure...Ch. 4 - Draw the shear-force and bending-moment diagrams...Ch. 4 - The shear-force diagram for a simple beam is shown...Ch. 4 - The shear-force diagram for a beam is shown in the...Ch. 4 - A compound beam (see figure) has an internal...Ch. 4 - A compound beam (see figure) has an shear release...Ch. 4 - A simple beam AB supports two connected wheel...Ch. 4 - The inclined beam represents a ladder with the...Ch. 4 - Beam ABC is supported by a tie rod CD as shown....Ch. 4 - A plane frame (see figure) consists of column AB...Ch. 4 - The plane frame shown in the figure is part of an...
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Understanding Shear Force and Bending Moment Diagrams; Author: The Efficient Engineer;https://www.youtube.com/watch?v=C-FEVzI8oe8;License: Standard YouTube License, CC-BY
Bending Stress; Author: moodlemech;https://www.youtube.com/watch?v=9QIqewkE6xM;License: Standard Youtube License