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4.1 · Identify truss assumptions, members, joints, and reactions

Learn to identify truss assumptions, members, joints, and reactions through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Truss Analysis

ENGG 130 Engineering Mechanics: Statics — Study topic 4.1

A truss is an arrangement of straight members joined at their ends. Before analysing one, identify what is included in the system, what counts as a member or joint, and how the supports can exert forces on the structure. These choices define the statics model; they are not merely labels on a drawing. This lesson focuses on recognizing that model and identifying its reactions. The examples use simple equilibrium calculations to make reaction directions and support idealizations concrete.

What you will learn

  • Describe the ideal truss assumptions used in introductory planar statics.
  • Identify members and joints and distinguish internal connections from external supports.
  • Represent pin, roller, and cable reactions with appropriate directions and components.
  • Use whole-truss equilibrium to check support reactions in a simple planar example.

1. The ideal planar truss model

A truss is a framework of slender, straight members connected at joints. A planar truss has its members and applied loads in one plane. In the introductory ideal model, each member is connected at its ends by frictionless pins, and externally applied loads and support reactions act at joints. Member self-weight is often neglected in this model, or represented by equivalent joint loads when that is appropriate to the stated problem.
The central idealization is that a member is a two-force member: the only forces acting on it are at its two ends. For the member to be in equilibrium, those forces must be equal in magnitude, opposite in direction, and along the member's axis. Thus an ideal truss member carries an axial force, either tension (pulling away from its ends) or compression (pushing toward its ends). This statement is a model assumption, not a claim that real connections or members are perfectly ideal.
A joint is the connection point where member ends meet or where a load or support is attached. A member is the straight segment between two joints. A joint label identifies a location; a member label identifies a segment, often by naming its end joints. For example, the segment joining joints A and B may be named member AB. Do not count a crossing of drawn lines as a joint unless the diagram indicates that the members connect there.
∑Fx=0,∑Fy=0,∑M=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M=0
  • Loads in the ideal model act at joints.
  • Each ideal member has two end forces along its own axis.
  • A crossing is not automatically a joint; connectivity must be specified.

2. Identify support reactions

First define the system. For external reactions, isolate the complete truss as one body. Replace each support with the force components it can exert on that body, and show all applied external loads. Choose horizontal and vertical axes, with positive directions to the right and upward. Take counterclockwise moments as positive unless another convention is stated.
In a two-dimensional model, a pin support prevents translation in both directions and is represented by two reaction components, usually AxA_x and AyA_y. Their directions can be assumed for calculation; a negative answer means the actual direction is opposite. A roller on a horizontal surface supplies one reaction normal to that surface, so its reaction is vertical. A cable pulls along its own length and away from the body it supports. It cannot push.
For the isolated whole truss, the planar equilibrium equations are the sums of horizontal forces, vertical forces, and moments equal to zero. These equations provide checks on the external reactions. They do not, by themselves, tell you every member force in a general truss; this topic's focus is recognizing the model and its external reactions.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • A pin generally has two in-plane reaction components.
  • A roller has one reaction normal to its supporting surface.
  • A cable reaction acts along the cable and pulls.
  • Use the whole truss as the body when identifying external reactions.

3. A practical identification routine

Begin by tracing the connectivity: mark each joint once, then identify each straight member between connected joints. Record where external loads are applied and whether the drawing places them at joints. Next, name the supports and replace them with reaction components consistent with their idealized types. A free-body diagram should show the isolated system, loads, reaction directions, dimensions needed for moments, and the chosen axes.
For a whole-truss diagram, do not draw guessed member forces as external forces. Forces that members exert on other members are internal to the complete truss and cancel in pairs in its overall force balance. If the system is instead one isolated joint or member, the forces from the connected parts become external to that smaller system. Keep the system boundary clear.
A useful final check is to substitute the calculated reactions into both force equations and one moment equation. A correct set of reactions must satisfy all three. If the reactions do not balance, revisit the support model, force directions, units, and moment arms.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • Choose the system before drawing forces.
  • Distinguish external reactions from internal member forces.
  • Check horizontal force, vertical force, and moment balance.

Worked example

Count the members and joints

A planar triangular framework has joints A, B, and C. Its straight members connect A to B, B to C, and C to A. Identify the joints and members, and state whether the crossing rule is relevant.
  1. List the joints
    The named connection points are A, B, and C, so there are three joints.
    j=3j=3
  2. List the members
    Each member is one straight segment between a connected pair of joints. The specified connections give AB, BC, and CA, for three members.
    m=3m=3
  3. Interpret the connections
    There are no crossing lines in this description. In a more complex drawing, a crossing would count as a joint only if the members are shown or specified to connect there.
Answer: The framework has joints A, B, and C and members AB, BC, and CA. It has three members. A crossing is not part of this example.
Check: The member list contains each specified connected pair once: AB, BC, and CA.

Worked example

Model pin and roller reactions

A horizontal truss is supported by a pin at A on its left end and a roller at B on a horizontal surface at its right end. Identify the external reaction components before any loads are specified.
  1. Define and isolate the system
    The system is the complete truss. Its support forces are external to that system. With no applied loads specified, this example identifies reaction types rather than calculating their magnitudes.
  2. Replace the pin
    The pin can exert horizontal and vertical force components in the plane. The diagram assumes both components point in positive coordinate directions; actual directions depend on the loading.
    RA=Axi+Ayj\mathbf{R}_A=A_x\mathbf{i}+A_y\mathbf{j}
  3. Replace the roller
    The roller rests on a horizontal surface, so its reaction is normal to that surface and vertical. The reaction shown is upward.
    RB=Byj\mathbf{R}_B=B_y\mathbf{j}
Answer: The pin at A has components AxA_x and AyA_y. The roller at B has one vertical component, ByB_y. No numerical reaction values can be found without applied loading information.
Check: The diagram shows two components at the pin and one normal reaction at the horizontal roller.

Worked example

Calculate reactions for a simple truss system

A whole planar truss is supported by a pin at A and a roller at B, 4.0 m to its right. A 6.0 kN downward load acts at a joint 1.5 m to the right of A. Find the support reactions and verify force and moment balance.
  1. Set axes and signs
    Take right and up as positive, and counterclockwise moments about A as positive. The pin supplies AxA_x and AyA_y; the roller supplies ByB_y. The whole truss is the system, so its internal member forces are not included on this free-body diagram.
    +x right,+y up,+M counterclockwise+x\text{ right},\quad +y\text{ up},\quad +M\text{ counterclockwise}
  2. Use moment equilibrium
    Take moments about A. The reactions at A have zero moment arm about A. The downward load produces a clockwise moment, and the upward reaction at B produces a counterclockwise moment.
    By(4.0 m)−(6.0 kN)(1.5 m)=0B_y(4.0\,\mathrm{m})-(6.0\,\mathrm{kN})(1.5\,\mathrm{m})=0
  3. Solve the vertical reaction
    Solving the moment equation gives the upward reaction at B. The units reduce to force because moment is force multiplied by distance.
    By=2.25 kNB_y=2.25\,\mathrm{kN}
  4. Balance vertical forces
    The vertical forces must sum to zero. Therefore the pin's vertical reaction supplies the remaining upward force.
    Ay+By−6.0 kN=0,Ay=3.75 kNA_y+B_y-6.0\,\mathrm{kN}=0,\quad A_y=3.75\,\mathrm{kN}
  5. Balance horizontal forces
    There are no horizontal applied loads, and the roller has no horizontal reaction in this model. Horizontal equilibrium therefore gives a zero horizontal pin reaction.
    Ax=0A_x=0
  6. Verify equilibrium
    The upward reactions add to the downward load. Taking moments about A also gives equal and opposite contributions from the roller and applied load.
    ∑Fy=3.75+2.25−6.0=0 kN,∑MA=(2.25)(4.0)−(6.0)(1.5)=0 kN⋅m\sum F_y=3.75+2.25-6.0=0\,\mathrm{kN},\quad \sum M_A=(2.25)(4.0)-(6.0)(1.5)=0\,\mathrm{kN\cdot m}
Answer: Ax=0 kNA_x=0\,\mathrm{kN}, Ay=3.75 kNA_y=3.75\,\mathrm{kN} upward, and By=2.25 kNB_y=2.25\,\mathrm{kN} upward.
Check: Horizontal force balance is zero. Vertical reactions total 6.0 kN upward, and the moment contributions about A are both 9.0 kN·m in opposite directions.

Common mistakes and how to avoid them

Treating every crossing in a sketch as a joint.
Correction: Count a crossing only when the drawing or description indicates that the members connect there.
Drawing only one reaction at a pin or two at a roller on a horizontal surface.
Correction: Use two in-plane force components for a pin and one surface-normal force for a roller.
Putting internal member forces on the free-body diagram of the entire truss.
Correction: For the whole-truss system, show external loads and support reactions. Member forces become external only when isolating a smaller part.
Assuming that an ideal member carries a force in any direction.
Correction: In the ideal two-force-member model, its end forces act along the member's axis.
Keeping a negative calculated reaction pointing in the assumed direction.
Correction: A negative result means the actual direction is opposite to the direction assumed in the diagram.

Lesson summary

  • An ideal planar truss is made of straight members joined at their ends; loads are applied at joints in the basic model.
  • Identify each joint and each connected member from the stated connectivity, not from line crossings alone.
  • Replace supports with reactions suited to their idealized type: two components for a pin, a normal force for a roller, and a pulling force along a cable.
  • For the isolated whole truss, check horizontal force, vertical force, and moment equilibrium.

Check your understanding

Question 1

A roller rests on a horizontal surface. Which reaction model is appropriate in a planar statics diagram?
  1. One vertical force normal to the surface
  2. A horizontal and a vertical force component
  3. A force along any chosen direction
  4. A reaction couple only
Show answer and explanation
One vertical force normal to the surface
A roller's ideal reaction is normal to its supporting surface. For a horizontal surface, that direction is vertical.

Question 2

Two member lines cross in a truss sketch, but no connection is shown or specified. What should you do?
  1. Count the crossing as a joint automatically
  2. Treat the crossing as a joint only if connectivity is indicated
  3. Count two new joints at the crossing
  4. Replace the crossing with a support
Show answer and explanation
Treat the crossing as a joint only if connectivity is indicated
A joint represents an actual connection. A visual crossing alone does not establish that the members connect.

Question 3

In the ideal truss model, the two end forces on an isolated member act along which direction?
  1. Perpendicular to the member
  2. Along the member's axis
  3. Along the global vertical axis in every case
  4. In unrelated directions
Show answer and explanation
Along the member's axis
An ideal member with only two end forces in equilibrium has forces along its own axis.

Key terms

Truss
A framework of straight members connected at joints, modelled in this topic as planar.
Member
A straight segment joining two connected joints.
Joint
A connection point where member ends meet, or where a load or support is attached.
Reaction
A force exerted by a support on the structure.
Two-force member
An ideal member acted on only at its two ends; its end forces are equal, opposite, and along its axis.
Pin support
An ideal support represented in planar statics by horizontal and vertical force components.
Roller support
An ideal support represented by one reaction normal to its supporting surface.

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