Chap. 6 Structural Analysis

Chapter Outline

Simple TrussesThe Method of JointsZero-Force MembersThe Method of SectionsSpace TrussesFrames and Machines

6-3

APPLICATIONS

Trusses are commonly used to support a roof

For a given truss geometry and load, how can we determine the forces in the truss members and select their sizes?

A more challenging question is that for a given load, how can we design the trusses geometry to minimize cost?

6-4

APPLICATIONS (continued)

Trusses are also used in a variety of structures like cranes and the frames of aircraft or space stations.

How can we design a light weight structure that will meet load, safety, and cost specifications?

6-5

DEFINING A SIMPLE TRUSS (Section 6.1)

A truss is a structure composed of slender members joined together at their end points.If a truss, along with the imposed load, lies in a single plane (as shown at the top right), then it is called a planar truss.

A simple truss is a planar truss which begins with a a triangular element and can be expanded by adding two members and a joint. For these trusses, the number of members (M) and the number of joints (J) are related by the equation M = 2 J 3.

6-6

ANALYSIS and DESIGN ASSUMPTIONS When designing both the member and the joints of a truss, first it is necessary to determine the forces in each truss member. This is called the force analysis of a truss. When doing this, two assumptions are made:

1. All loads are applied at the joints. The weight of the truss members is often neglected as the weight is usually small as compared to the forces supported by the members.

2. The members are joined together by smooth pins. This assumption is satisfied in most practical cases where the joints are formed by bolting or welding.

6-7

With these two assumptions, the members act as two-force members. They are loaded in either tension or compression. Often compressive members are made thicker to prevent buckling.

6-8

THE METHOD OF JOINTS (Section 6.2)

In this method of solving for the forces in truss members, the equilibrium of a joint (pin) is considered. All forces acting at the joint are shown in a FBD. This includes all external forces (including support reactions) as well as the forces acting in the members. Equations of equilibrium (

FX = 0 and

FY = 0) are used to solve for the unknown forces acting at the joints.

6-9

STEPS FOR ANALYSIS

1. If the support reactions are not given, draw a FBD of the entire truss and determine all the support reactions using the equations of equilibrium.

2. Draw the free-body diagram of a joint with one or two unknowns. Assume that all unknown member forces act in tension (pulling the pin) unless you can determine by inspection that the forces are compression loads.

3. Apply the scalar equations of equilibrium,

FX = 0 and

FY = 0, to determine the unknown(s). If the

answer is positive, then the assumed direction (tension) is correct, otherwise it is in the opposite direction (compression).

4. Repeat steps 2 and 3 at each joint in succession until all the required forces are determined.

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6-11

6-12

ZERO-FORCE MEMBERS (Section 6.3)

If a joint has only two non-colinear members and there is no external load or support reaction at that joint, then those two members are zero-force members.

In this example members DE, CD, AF, and AB are zero force members.

6-13

Zero-force members can be removed (as shown in the figure) when analyzing the truss.

6-14

Zero-Force Members

for : D

for : A

6-15

Zero-Force Members

6-16

Zero-Force Members

6-17

6-18

6-19

p.277, Problem 6-12Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 1200 N, P2 = 500 N.

6-20

6-21

6-22

p.279, Problem 6-27Determine the force in each member of the double scissors truss in terms of the load P and state if the members are in tension or compression.

6-23

6-24

6-25

6-26

6-27

6-28

Long trusses are often used to construct bridges.

The method of joints requires that many joints be analyzed before we can determine the forces in the middle part of the truss.

Is there another method to determine these forces directly?

THE METHOD OF SECTIONS

6-29

THE METHOD OF SECTIONS

In the method of sections, a truss is divided into two parts by taking an imaginary cut (shown here as a-a) through the truss.Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut member will also be either tensile or compressive with the same magnitude. This result is based on the equilibrium principle and Newtons third law.

6-30

STEPS FOR ANALYSIS

1. Decide how you need to cut the truss. This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general).

6-31

3. If required, determine the necessary support reactions by drawing the FBD of the entire truss and applying the EofE.

2. Decide which side of the cut truss will be easier to work with (minimize the number of reactions you have to find).

6-32

PROCEDURE (continued)

4. Draw the FBD of the selected part of the cut truss.

6-33

PROCEDURE (continued)

5. Apply the equations of equilibrium (EofE) to the selected cut section of the truss to solve for the unknown member forces. Please note that in most cases it is possible to write one equation to solve for one unknown directly.

sections 3member

6-34

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6-36

6-37

6-38

p.287, Problem 6-36Determine the force in members BC, CG, and GF of the Warren truss. Indicate if the members are in tension or compression.

6-39

6-40

p.289, Problem 6-52Determine the force in members KJ, NJ, ND, and CD of the K truss. Indicate if the members are in tension or compression. Hint: Use sections aa and bb.

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6-42

6-43

PROBLEM CountersThe diagonal members in the center panels of the truss shown arevery slender and can act only in tension; such members are known as counters. Determine the forces in the counters which are acting under the given loading.

6-44

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6-2-1

Space Trusses

Simple Truss : m = 3j 6, use tetrahedron as the basic element

6-2-2

p.292, Problem 6-57Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by a ball-and-socket at A, B and E. Set F = {-200 i + 400 j} N.

(1, 2, 2)

(6, 3.5, 0)

(0, 3.5, 0)

6-2-3

6-2-4

6-2-5

FRAMES AND MACHINES: DEFINITIONS

Frames and machines are two common types of structures that have at least one multi-force member. (Recall that trusses have nothing but two-force members).Frames are generally stationary and support external loads.Machines contain moving parts and are designed to alter the effect of forces.

6-2-6

6.6 Frames and MachinesExample 6.11For the frame, draw the free-body diagrams of (a) the entire frame including the pulleys and cords, (b) the frame without the pulleys and cords, and (c) each of the pulley.

6-2-7

SolutionPart (a)

Consider the entire frame, interactions at the points where the pulleys and cords are connected to the frame become pairs of internal forces which cancel each other and not shown on the FBD

6.6 Frames and Machines

6-2-8

6.6 Frames and MachinesSolutionPart (b) and (c)

When cords and pulleys are removed, their effect on the frame must be shown

6-2-9

6.6 Frames and MachinesExample 6.12Draw the free-body diagrams of the bucket and the vertical boom of the back hoe. The bucket and its content has a weight W. Neglect the weight of the members.

6-2-10

6.6 Frames and MachinesSolution

Idealized model of the assemblyMembers AB, BC, BE and HI are two force members

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6.6 Frames and Machines

Example 6.14Determine the horizontal and vertical components of the force which the pin C exerts on member CB of the frame.

6-2-12

6.6 Frames and MachinesSolution

Identify member AB as two force memberFBD of the members AB and BC

6-2-13

Solution

NC

CNN

FNC

C

FNF

mFmN

M

y

y

y

x

x

x

AB

AB

C

1000

0200060sin7.1154

;0

577060cos7.1154

;07.1154

0)4(60sin)2(2000

;0

=

=

=+

==

=+=

=

=

o

o

o

6.6 Frames and Machines

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6.6 Frames and MachinesExample 6.15The compound beam is pin connected at B. Determine the reactions at its support. Neglect its weight and thickness.

6-2-15

6.6 Frames and MachinesSolution

FBD of the entire frameDismember the beam into two segments since there are 4 unknowns but 3 equations of equilibrium

6-2-16

SolutionSegment BC

08

;0

0)2()1(8;0

0;0

=+

=+

=+=

==+

yy

y

y

B

x

x

CkNB

F

mCmkNM

BF

6.6 Frames and Machines

6-2-17

SolutionMember AB

kNCkNBBmkNMkNAkNA

BkNA

F

mBmkNM

M

BkNA

F

yyxAyx

yy

y

yA

A

xx

x

4;4;0;.32;12;6

054)10(

;0

0)4()2(54)10(

;0

053)10(

;0

======

=

=+

=

=

=+

=+

6.6 Frames and Machines

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6.6 Frames and MachinesExample 6.17The smooth disk is pinned at D and has a weight of 20N. Neglect the weights of others member, determine the horizontal and vertical components of the reaction at pins B and D

6-2-19

6.6 Frames and MachinesSolution

FBD of the entire frameFBD of the members

6-2-20

SolutionEntire Frame

NANAF

NANAF

NCcmCcmNM

y

yy

x

xx

x

xA

20

020;0

1.1701.17;0

1.170)5.3()3(20;0

=

==+

===+

==+=

6.6 Frames and Machines

6-2-21

SolutionMember AB

NBBNNF

NNcmNcmNM

NBBNF

y

yy

D

DB

x

xx

20

04020;040

0)3()6(20;01.17

01.17;0

=

=+=+

==+=

===+

6.6 Frames and Machines

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SolutionDisk

NDDNN

FD

F

y

y

y

x

x

20

02040

;0

0;0

=

=

=+

==+

6.6 Frames and Machines

6-2-23

p.311, 6-68. Determine the force required to hold the 150-kg crate in equilibrium.

6-2-24

p.314, 6-83. Determine the horizontal and vertical components of reaction that pins A and C exert on the two-member arch.

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p.316, Problem 6-89By squeezing on the hand brake of the bicycle, the rider subjects the brake cable to a tension of 50 lb. If the caliper mechanism is pin-connected to the bicycle frame at B, determine the normal force each brake pad exerts on the rim of the wheel. Is this the force that stops the wheel from turning?

6-2-29

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p.316, Problem 6-96If the wooden block exerts a force of on the toggle clamp, determine the force P applied to the handle.

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p.321, Problem 6-120Determine the couple moment M that must be applied to member DC for equilibrium of the quick-return mechanism. Express the result in terms of the angles and , dimension L, and the applied vertical force P. The block at C is confined to slide within the slot of member AB.

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P. 327, 6-132 Determine the horizontal and vertical components of reaction that the pins A and B exert on the two-member frame.Set F = 0.

6-2-37

Chap6-1-2012(compressed) 1Chapter Outline 3 4 5 6 7 8 9 10 11 12 13Zero-Force Members Zero-Force MembersZero-Force Members 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Chap6-2-2012(compressed)Space Trusses 2 3 4 56.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines6.6 Frames and Machines 23 24p.314, 6-83. Determine the horizontal and vertical components of reaction that pins A and C exert on the two-member arch. 26 27 28 29 30 31 32 33 34 35 36 37