A truss is an assemblage of straight members connected at ...tycnw01.vtc.edu.hk/con4331/3 -...

Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 3 – Plane Truss

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Plane Truss

A truss is an assemblage of straight members connected at their ends by flexible connections to form a rigid configuration. The members are usually formed into triangular patterns to produce an efficient, lightweight and load bearing structure. Although joints are typically formed by welding or bolting truss bars to gusset plates, in most structural analysis, the members are commonly assumed to be connected at the joints by frictionless pins. Since no moment can be transferred through a frictionless pin joint, truss members are assumed to carry only axial force – either tension or compression. All the members of a truss and the applied loads lie in a single plane, the truss is called a plane truss. The upper and lower members of a truss, which can be either horizontal or sloping, are called top and bottom chords. The chords are connected by a vertical and diagonal members.


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Basic truss element The simplest internally stable plane truss can be formed by connecting three members at their ends by hinges to form a triangle. This triangular truss is called the basic truss element. Note that this triangular truss is internally stable in the sense that it is a rigid body that will not change its shape under loads. In contrast, a rectangular truss formed by connecting four members at their ends by hinges is internally unstable because it will change its shape and collapse when subject to a general system of coplanar forces.

Simple truss The basic truss element ABC can be enlarged by attaching two new members, BD and CD, to two of the existing joints B and C and by connecting them to form a new joint D. The truss can be further enlarged by repeating the same procedure. Truss constructed by this procedure is called simple trusses.


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Assumptions for analysis of trusses Assumptions: 1. All members are connected only at their ends by frictionless hinges in

plane trusses. 2. All loads and support reactions are applied only at the joints. 3. The centroidal axis of each member coincides with the line connecting

the centers of the adjacent joints.

Method for analysis of plane trusses The member forces in a statically determinate truss can be found by making use of the equations of equilibrium. The process is to consider different free-body diagrams of parts of the structures. The two common methods are:

1. Method of joint 2. Method of section

In practice, it is often convenient to use a combination of the two methods. The key is to choose the most convenient free-body diagram.


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Method of joint If a truss is in equilibrium, then each of its joints must also be in equilibrium. Hence, the method of joints consists of satisfying the equilibrium conditions ΣFx = 0 and ΣFy = 0 for the forces exerted on the pin at each joint of the truss. Steps for a truss analysis: 1. In this method, a free-body of each joint is considered, one joint at a time. 2. Two independent equilibrium equations are available for each joint. 3. You should work each time with at a joint with only TWO unknown

member forces. 4. Once the unknown forces at one joint are determined, they become

known forces for other joints.


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Example Determine the magnitudes and the types of forces in all members of the following truss.

Solution Step 1, In order to simplify the analysis, we just only consider part of the truss. Take joint B as free body,

Step 2, If a truss is in equilibrium, then each of its joints must also be in equilibrium. ΣFx = 0

FBC cos 450 + 500 = 0 FBC = -707.1 kN (Negative value means that we assume the wrong

sense of the member force BC. Member BC should be in compression.)

ΣFy = 0

FBC sin 450 + FBA = 0 -707.1 sin 450 + FBA = 0


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FBA = 500 kN (Positive value means that we assume the right sense of

the member force BA. Member BA should be in tension.)

Step 3 Repeat steps 1 to 2 for joint C.

ΣFx = 0 FAC + (-707.1) cos 450 = 0 FAC = 500 kN (Positive value means that we assume the right sense of

the member force AC. Member AC should be in tension.)


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Example Determine the magnitudes and the types of forces in all members of the following truss.


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Solution Take joint C as free body,

ΣFy = 0

FCD * sin 26.60 = -115

FCD = -257 kN (Negative value means that we assume the wrong sense of the member force CD. Member CD should be in compression.)

ΣFx = 0

FBC + FCD * cos 26.60 = 0 FBC + (-257) * cos 26.60 = 0

FBC = 230 kN (Positive value means that we assume the right sense of

the member force BC. Member BC should be in tension.)


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Take joint B as free body,

ΣFy = 0

FBD = 120 kN (Tension)

ΣFx = 0

FAB = 230 kN (Tension)

Take joint D as free body,


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ΣFx = 0

FDE * cos 26.60 + FAD * cos 450 = -257 * cos 26.60 FDE + 0.8 * FAD = -257

ΣFy = 0

FDE * sin 26.60 = FAD * sin 450 + 120 +(-257) * sin 26.60 FDE = 1.58 * FAD +11

By solving the above equations,

FAD = -112.6 kN (Compression)

FDE = -167 kN (Compression) Take joint E as free body,

ΣFy = 0

FAE +(-167) * sin 26.60 = 0 FAE = 75 kN (Tension)


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Important notes for using method of joints. 1. It is necessary to choose a joint having two unknowns. 2. Realize that once the force in a member is found from the analysis of a

joint at one of its ends, the result can be used to analyze the forces acting on the joint at its other end. Strict adherence to the principle of action, equal but opposite force reaction, must, of course, is observed.

3. Remember that a member in compression ‘pushes’ on the joint and a member in tension ‘pulls’ on the joint.

Method of section This method is particularly useful when only certain bar forces are required. It consists of passing an imaginary section through the truss, thus cutting it into two parts. Provided the entire truss is in equilibrium, each of the two parts must also be in equilibrium; and as a result, the three equations of equilibrium may be applied to either one of these two parts to determine the member forces at the ‘cut section’. Steps for a truss analysis: 1. Make a decision as to how to ‘cut’ or section the truss through the

members where forces are to be determined. 2. Determine the support reactions. 3. Draw the free body diagram of that part of the sectioned truss, which has

the least number of forces on it. 4. By inspection, attempt to show the unknown member forces acting in the

correct sense of direction. 5. All three equations of equilibrium are available and THREE unknown

bar forces can be determined.


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Example Determine the magnitudes and the types of force in members DE, EK and JK of the following truss.

Solution Take moment at point A, VH *(4*600) = 5*(600*tan 400) + 8*(2*600) VH = 5.05 kN ΣFy = 0, 8 = VA + VH VA =2.95 kN ΣFx = 0, HA = 5 kN


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Take EGHJ as free body,

Take moment at point E, FJK *(600*tan 400) = 5.05*600 FJK = 6.02 kN (Tension) Take moment at point K, FDE *(600*tan 400) + 5.05*600*2 = 0 FDE = -12 kN (Compression) ΣFy = 0, FEK * sin 400 = 5.05 FEK = 7.85 kN (Tension)


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Example Determine the forces in members BG and BC for the following truss under the given loading condition.


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Solution Take ABH as free body,

From similar triangles,

r = 24 m Take moment at point O, FBG * cos 36.90 * 24 + FBG * sin 36.90 * 8 + (50+200) * 24

-18*287.5 = 0

FBG = -34.4 kN (Compression) ΣFy = 0, 50 + 200 - 287.5 - 34.4 * cos 36.90 = FBC sin 18.40 FBC = -206 kN (Compression)

6108+

=rr


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Applying the Method of Sections In applying the method of sections, two decisions are made:

(1) Choosing the free body, (2) Choosing the points for taking moments about.


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In choosing the free-body, remember that the “cut” does not need to be a straight line. In choosing a point for taking moment, remember that it can be any point in the plane. It does not have to be a joint or a support. Zero Force Members Truss analysis using the method of joints is greatly simplified if one is able to first determine those members that support no loading. These zero-force members may be necessary for the stability of the truss during construction and to provide support if the applied loading is changed. The zero-force members of a truss can generally be determined by inspection of the joints. Case 1 If no external load is applied to a joint that consists of two bars, the

force in both bars must be zero.

ΣFx = 0 requires F1 = 0 ΣFx’ = 0 requires F2 = 0

Case 2 If no external load acts at a joint composed of three bars – two of which are collinear, the force in the bar that is not collinear is zero.


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Example - Zero-Force Members



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Example Using the method of joints, indicate all the members of the truss shown that have zero forces.

Solution: Joint D, fig b ΣFy = 0, FDC sinθ = 0, FDC = 0 ΣFx = 0, FDE + 0 = 0, FDE = 0 Joint E, fig c ΣFx = 0 FEF = 0 (Note that FEC = P and an analysis of joint C would yield a force in member CF.) Joint H, fig d ΣFy = 0 FHB = 0 Joint G, fig e The roller support at G exerts only an x component of force on the joint. Hence ΣFy = 0 FGA = 0


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Determining Direction of A Member Forces in Trusses by Inspection By visualizing the way a truss deflects under given loading, it is often possible to determine quickly whether the force in a member is tension or compression. This provides a useful check for the results of an analysis. Consider the illustrated example below:

1. The top members are in compression. 2. The bottom members are in tension. 3. The diagonals are in tension. 4. The verticals are in compression.

The three zero-force members should also be identified easily.


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(a) Basic truss assemblies. (b) The senses of the forces in

the diagonals can be determined by first imagining them to be removed and then ascertaining their role in preventing the probable type of truss deformation that would occur. Thus, a diagonal placed between B and F in truss A would have to be in tension because its role is to prevent B and F from drawing apart in the manner indicated.

(c) Final force distribution in

trusses: C, compression; T, tension. (d) A "cable" or "arch"

analogy can also be used to determine the senses of the forces in different members. In the truss to the left, member FBD is imagined to be a "cable" and is obviously in tension. Other members serve roles related to maintaining the equilibrium of this basic "cable" configuration.

Forces in truss members: the senses of the forces in some simple truss configurations can be determined through intuitive approaches. More complex trusses require quantitative approaches.


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Problems Q1 to Q4 Determine the forces in all the members, indicating whether they are tensile or compression.


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Solution to Q1 to Q4


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Problems Q5 to Q7 Solutions


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Problems Q8 to Q11 Solutions


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