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Engineering Mechanics: STATICS

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Title: Engineering Mechanics: STATICS


1
Engineering Mechanics STATICS
  • Anthony Bedford and Wallace Fowler
  • SI Edition

Teaching Slides Chapter 6 Structures in
Equilibrium
2
Chapter Outline
  • Trusses
  • The Method of Joints
  • The Method of Sections
  • Space Trusses
  • Frames Machines
  • Computational Mechanics

3
6.1 Trusses
  • Truss structures such as the beams supporting a
    house
  • Starting with very simple examples
  • 3 bars pinned together at their ends to form a
    triangle with supports as shown ? structure that
    will support a load F

4
6.1 Trusses
  • Construct more elaborate structures by adding
    more triangles
  • The bars are the members of these structures
    the places where the bars are pinned together are
    called joints
  • Even though these examples are quite simple, they
    begin to resemble the structures used to support
    bridges roofs of houses

5
6.1 Trusses
  • If these structures are supported loaded at
    their joints we neglect the weight of the bars,
    each bar is a 2-force member
  • We call such a structure a truss

6
6.1 Trusses
  • Drawing the free-body diagram of a truss
  • Force T axial force in the member
  • When the forces are directed away from each
    other, the member is in tension
  • When the forces are directed toward each other,
    the member is in compression
  • Because it is a 2-force member, the forces at the
    ends, which are the sums of the forces exerted on
    the member at its joints, must be equal in
    magnitude, opposite in direction directed along
    the line between the joints

7
6.1 Trusses
  • If we cut the member by a plane draw the
    free-body diagram of the part of the member on 1
    side of the plane
  • We represent the system of internal forces
    moments exerted by the part not included in the
    free-body diagram by a force F acting at the
    point P where the plane intersects the axis of
    the member a couple M

8
6.1 Trusses
  • The sum of the moments about P must equal zero,
    so M 0
  • Therefore, we have a 2-force member, which means
    that F must equal in magnitude opposite in
    direction to the force T acting at the joint
  • The internal force is a tension or compression
    equal to the tension or compression exerted at
    the joint
  • Notice the similarity to a rope or cable, in
    which the internal force is a tension equal to
    the tension applied at the ends

9
6.1 Trusses
  • Although many actual structures, including roof
    trusses bridge trusses, consist of bars
    connected at the ends, very few have pinned
    joints
  • E.g. a joint of a bridge truss
  • The ends of the members are welded at the joint
    are not free to rotate
  • Such a joint can exert couples on the members

10
6.1 Trusses
  • However, these structures are designed to
    function as trusses
  • They support loads primarily by subjecting their
    members to axial forces
  • They can usually be modeled as trusses, treating
    the joints as pinned connections under the
    assumption that couples they exert on the members
    are small in comparison to axial forces

11
6.2 The Method of Joints
  • The method of joints involves
  • Drawing free-body diagrams of the joints of a
    truss 1 by 1
  • Using equilibrium equations to determine the
    axial forces in the members

12
6.2 The Method of Joints
  • Before beginning, it is usually necessary to draw
    a free-body diagram of the entire truss (i.e.
    treat the truss as a single object) determine
    the reactions at its supports
  • E.g. consider the Warren truss which has members
    2 m in length support loads at B D

13
6.2 The Method of Joints
  • From the equilibrium equations
  • S Fx Ax 0
  • S Fy Ay E ? 400 N ? 800 N 0
  • S Mpoint A ?(1 m)(400 N) ? (3 m)(800 N) (4
    m)E 0
  • We obtain the reactions
  • Ax 0, Ay 500 N E 700 N
  • The next step is to choose a joint draw its
    free-body diagram
  • Isolate joint A by cutting members AB AC

14
6.2 The Method of Joints
15
6.2 The Method of Joints
  • The terms TAB TAC are the axial forces in
    members AB AC, respectively
  • Although the directions of the arrows
    representing the unknown axial forces can be
    chosen arbitrarily, notice that we have chosen
    them so that a member is in tension if we obtain
    a positive value for the axial force
  • Consistently choosing the directions in this way
    helps avoid errors

16
6.2 The Method of Joints
  • The equilibrium equations for joint A are
  • S Fx TAC TAB cos 60 0
  • S Fy TAB sin 60 500 N 0
  • Solving these equations, we obtain the axial
    force TAB ?577 N TAC 289 N
  • Member AB is in compression member AC is in
    tension

17
6.2 The Method of Joints
  • Next, obtain a free-body diagram of joint B by
    cutting members AB, BC BD
  • From the equilibrium equations for joint B
  • S Fx TBD TBC cos 60 577 cos 60 0
  • S Fy ?400 N 577 sin 60 ? TBC sin 60 0
  • We obtain TBC 115 N TBD ?346 N
  • Member BC is in tension member BD is in
    compression

18
6.2 The Method of Joints
  • By continuing to draw free-body diagrams of the
    joints, we can determine the axial forces of all
    the members
  • In 2 dimensions, you can obtain only 2
    independent equilibrium equations from the
    free-body diagram of a joint
  • Summing the moments about a point does not result
    in an additional independent equation because the
    forces are concurrent

19
6.2 The Method of Joints
  • Therefore when applying the method of joints, you
    should choose joints to analyze that are
    subjected to no more than 2 unknown forces
  • In our example, we analyzed joint A first because
    it was subjected to the known reaction exerted by
    the pin support 2 unknown forces, the axial
    forces TAB TAC
  • We could then analyze joint B because it was
    subjected to 2 known forces 2 unknown forces,
    TBC TBD
  • If we had attempted to analyze joint B first,
    there would have been 3 unknown forces

20
6.2 The Method of Joints
  • When determining the axial forces in the members
    of a truss, it will be simpler if you are
    familiar with 3 particular types of joints
  • 1.Truss joints with 2 collinear members no
    load the sum of the forces must equal zero, T1
    T2. The axial forces are equal.

21
6.2 The Method of Joints
  • 2.Truss joints with 2 noncollinear members no
    load because the sum of the forces in the x
    direction must equal zero, T2 0. therefore T1
    must also equal zero. The axial forces are zero.

22
6.2 The Method of Joints
  • 3.Truss joints with 3 members, 2 of which are
    collinear no load because the sum of the
    forces in the x direction must equal zero, T3
    0. The sum of the forces in the y direction must
    equal zero, so T1 T2. The axial forces in the
    collinear members are equal the axial force in
    the 3rd member is zero.

23
Example 6.1 Applying the Method of Joints
  • Determine the axial forces in the members of
    the truss in Fig. 6.12.
  • Strategy
  • 1st, draw a free-body diagram of the entire
    truss, treating it as a single object determine
    the reactions at the supports. Then apply the
    method of joints, simplifying the task by
    identifying any special joints

24
Example 6.1 Applying the Method of Joints
  • Solution
  • Determine the Reactions at the Supports
  • Draw the free-body diagram of the entire truss

25
Example 6.1 Applying the Method of Joints
  • Solution
  • From the equilibrium equations
  • S Fx Ax B 0
  • S Fy Ay ? 2 kN 0
  • S Mpoint B ?(6 m) Ax ? (10 m)(2 kN) 0
  • We obtain the reactions Ax ?3.33 kN, Ay 2 kN
  • B 3.33 kN.

26
Example 6.1 Applying the Method of Joints
  • Solution
  • Identify Special Joints
  • Because joint C has 3 members, 2 of which are
  • collinear no load, the axial force in member BC
    is
  • zero, TBC 0 the axial forces in the collinear
  • members AC CD are equal, TAC TCD.
  • Draw Free-Body Diagrams of the Joints
  • We know the reaction exerted on joint A by the
  • support joint A is subjected to only 2 unknown
  • forces, the axial forces in members AB AC.

27
Example 6.1 Applying the Method of Joints
  • Solution

28
Example 6.1 Applying the Method of Joints
  • Solution
  • The angle ? arctan (5/3) 59.0
  • The equilibrium equations for joint A are
  • S Fx TAC sin ? ? 3.33 kN 0
  • S Fy 2 kN ? TAB ? TAC cos ? 0
  • Solving these equations, we obtain TAB 0
  • TAC 3.89 kN.

29
Example 6.1 Applying the Method of Joints
  • Solution
  • Now draw the free-body diagram of joint B

30
Example 6.1 Applying the Method of Joints
  • Solution
  • From the equilibrium equation
  • S Fx TBD 3.33 kN 0
  • We obtain TBD ?3.33 kN. The negative sign
  • indicates that member BD is in compression.
  • The axial forces in the members are
  • AB 0
  • AC 3.89 kN in tension (T)
  • BC 0
  • BD 3.33 kN in compression (C)
  • CD 3.89 kN in tension (T)

31
Example 6.1 Applying the Method of Joints
  • Critical Thinking
  • Observe how our solution was simplified by
    recognizing that joint C is the type of special
    joint with 3 members, 2 of which are collinear
    no load
  • This allowed us to determine the axial forces in
    all members of the truss by analyzing only 2
    joints

32
Example 6.2 Determining the Largest Force a Truss
Will Support
  • Each member of the truss in Fig. 6.13 will
    safely support a tensile force of 10 kN a
    compressive force of 2 kN. What is the largest
    downward load F that the truss will safely
    support?

33
Example 6.2 Determining the Largest Force a Truss
Will Support
  • Strategy
  • This truss is identical to the one we analyzed
    in Example 6.1. By applying the method of joints
    in the same way, the axial forces in the members
    can be determined in terms of the load F. The
    smallest value of F that will cause a tensile
    force of 10 kN or a compressive force of 2 kN in
    any of the members is the largest value of F that
    the truss will support.

34
Example 6.2 Determining the Largest Force a Truss
Will Support
  • Solution
  • By using the method of joints in the same way as
  • in Example 6.1, we obtain the axial forces
  • AB 0
  • AC 1.94F (T)
  • BC 0
  • BD 1.67F (C)
  • CD 1.94F (T)

35
Example 6.2 Determining the Largest Force a Truss
Will Support
  • Solution
  • For a given load F, the largest tensile force
    is 1.94F (in members AC CD) the largest
    compressive force is 1.67F (in member BD).
  • The largest safe tensile force would occur
    when 1.94F 10 kN or when F 5.14 kN.
  • The largest safe compressive force would occur
    when 1.67F 2 kN or when F 1.20 kN.
  • Therefore, the largest load F that the truss
    will safely support is 1.20 kN.

36
Example 6.2 Determining the Largest Force a Truss
Will Support
  • Critical Thinking
  • This example demonstrates why engineers analyze
    structures
  • By doing so, they can determine the loads that an
    existing structure will support or design a
    structure to support given loads
  • In this example, the tensile compressive loads
    the members of the truss will support are given

37
Example 6.2 Determining the Largest Force a Truss
Will Support
  • Critical Thinking
  • Information of that kind must be obtained by
    applying the methods of mechanics of materials to
    the individual members
  • Then statics can be used, as we have done in this
    examples, to determine the axial loads in the
    members in terms of the external loads on the
    structure
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