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

Videos

Textbook Question
Book Icon
Chapter 4, Problem 4.5.39P

A plane frame (see figure) consists of column AB and beam BC that carries a triangular distributed load (see figure part a). Support A is fixed, and there is a roller support at C. Beam BC has a shear release just right of joint B.

  1. Find the support reactions at A and C then plot axial-force (N), shear-force (V), and bending-moment (M) diagrams for both members. Label all critical N,K and M values and also the distance to points where any critical ordinates are zero.

  • Repeat part (a) if a parabolic lateral load acting to the right is now added on column AB (figure part b).
  • Chapter 4, Problem 4.5.39P, A plane frame (see figure) consists of column AB and beam BC that carries a triangular distributed

    a.

    Expert Solution
    Check Mark
    To determine

    The support reaction at point A and C and plot the shear, moment, and axial force diagram.

    Answer to Problem 4.5.39P

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  1

    Explanation of Solution

    Given: .

    The given figure.

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  2

    AB column and BC beam forms the plane frame that carries a load that is distributed in the triangular shape. At C there is roller support and support A is fixed. Below the B joint there is a moment release at column AB

    Concept Used: .

      .

      .

    Vertical force equilibrium is given as,

    FV=0.

    Ay+Cy=q0L2.

    Horizontal force equilibrium is given as,

      FH=0Ax=0.

    Calculation: .

    Vertical force equilibrium is given as,

    FV=0.

    Ay+Cyq0L2=0.

    Ay+Cy=q0L2.

    Horizontal force equilibrium is given as,

      FH=0Ax=0.

    At the top of the moment release the moment is given as,

    MB=0q0L2(2L/3)CyL=0q0L2(2L/3)=CyLCy=q0L3.

    In equation (1),Cy is substituted,

    Ay=q0L2q0L3q0L6.

    Below B there is moment release at point A,

    MA=0.

    At x the shear force is given and equated to 0,

      q0L6=(1/2)x(q0/L)xq0L6=q02Lx2x2=q0L6(2L/q0).

      =L23.

      0.57735L.

    At point x bending moment is maximum,

      Mx=q0L6(L/3)q02LL23(L/33).

      =q0L263q0L2183=3q0L2q0L2183.

      =2q0L2183.

      Mx=q0L2930.0641q0L2.

    Axial force critical valuesNmax=q0L6.

    Shear force critical valuesVmax=q0L3.

    Moment critical valuesMmax=0.06415q0L2.

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  3

    Conclusion: .

    Thus, the support reaction at point A and C and plot the shear, moment and axial force diagram.

    b.

    Expert Solution
    Check Mark
    To determine

    For the load that is parabolic, lateral acts to the right added to AB column and part (a) is repeated.

    Answer to Problem 4.5.39P

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  4

    Explanation of Solution

    Given: .

    The given figure:.

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  5

    AB column and BC beam forms the plane frame that carries a load that is distributed in the triangular shape. At C there is roller support and support A is fixed. Below the B joint, there is a moment release at column AB.

    Concept Used: .

      .

    Vertical force equilibrium is given as,

    FV=0.

    Ay+Cy=q0L2.

    Horizontal force equilibrium is given as,

      FH=0Ax=0.

    Calculation: .

    With the equilibrium force being vertical,

      Fy=0Ay+Cy=q0L2(2).

    Moment at point B above moment release,

      MB=0CyLq0L2(2L/3)=0CyL=q0L2(2L/3)Cy=q0L3.

    In (2) substituteCy.

      Ay=q0L2q0L3.

      =3q0L3q0L6=q0L6.

    With the equilibrium force being horizontal,

      FH=0.

    As left of x axis is negative,f(x)0.

      Ax=02L(1y/2L)q0dy.

      =q002L(1y/2L)dy.

      =q0(2L(1y/2L)3/23/2)02L=4q0L3((12L/2L)(10))=4q0L3(01).

      Ax=4q0L3.

    At A moment below the moment release,

      MA=02L(q0y1y/2L)dy+12q0L(2L/3)q0L3(L).

      =I1+q0L23q0L23.

      I1=02L(q0y1y/2L)dyu=y,dv=(1y/2L)1/2du=dy,v=(1y/2L)3/22L3/2.

      =4L3(1y/2L)3/2.

      d(uv)=uvvdu.

      =q0{(y(4L/3)(1y/2L)3/2)02L02L((4L/3)(1y/2L)3/2)dy}=q0(0+(4L/3)((1y/2L)5/25/2(1/2L))02L=q0(4L3(0(4L/5))=q0((4L/3)(4L/5)).

      MA=16q0L215.

    At x shear force is calculated and equated to 0,

      q0L6=(1/2)x(q0/L)xq0L6=q02Lx2.

      x2=q0L6(2L/q0).

      =L23x=0.57735L.

    At x bending moment is calculated

      Mx=q0L6(L/3)q02LL23(L/33).

      =q0L263q0L2183=3q0L2q0L2183=2q0L2183q0L293Mx=0.06415q0L2.

    For plane frame and critical values of N, V, M, the axial, shear and moment is given.

    Mechanics of Materials (MindTap Course List), Chapter 4, Problem 4.5.39P , additional homework tip  6

    Axial force critical valuesNmax=q0L6.

    Shear force critical values in beamVmax=q0L3.

    Shear force critical values in columnVmax=4q0L3.

    Moment critical values in beamMmax=0.06415q0L2.

    Moment critical values in columnMmax=16q0L215.

    Conclusion: .

    Thus, for the load that is parabolic lateral acts to the right added to AB column and part (a) is repeated.

    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!

    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
    Mechanics of Materials Lecture: Beam Design; Author: UWMC Engineering;https://www.youtube.com/watch?v=-wVs5pvQPm4;License: Standard Youtube License