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5.3 · Analyze a statically determinate frame

Learn to analyze a statically determinate frame through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Frames and Machines

Use free-body diagrams and equilibrium to find support reactions and member forces

A frame is an assembly of connected members that carries applied loads. To analyze one, choose a system, draw its free-body diagram, and apply planar equilibrium. The complete frame is often a useful first system: forces between its members are internal and do not appear on that diagram. To find a force at a connection, isolate the relevant member and show the connection force on its free-body diagram. A statically determinate frame has enough independent equilibrium equations to find its unknown reactions and member forces using equilibrium alone. This lesson uses ideal pins and rollers. A pin between members transmits force components, and the forces on the two connected members are equal and opposite.

What you will learn

  • Identify the members, supports, loads, and connections in a planar frame.
  • Choose a system and draw its free-body diagram.
  • Use force and moment equilibrium to determine unknown reactions in a statically determinate frame.
  • Isolate a member when needed and check its force and moment balance.

1. Choose the system and draw its free-body diagram

Before writing equations, identify the members, supports, applied loads, and internal pins. An ideal pin support prevents translation in two directions, so show horizontal and vertical reaction components. A roller on a horizontal surface supplies a vertical reaction. These supports exert forces in the directions they restrain; a roller resting on a horizontal surface pushes upward on the supported body.
For a free-body diagram of the complete frame, include applied loads and support reactions. Do not include forces between members, because they act within the chosen system. If you isolate one member, show all forces acting on it, including forces transmitted at its pin connections. On the other connected member's diagram, the force at that pin has the opposite direction.
A member is a two-force member only when forces act at two points and no other loads or couples act on it. If a load acts between its ends, include that load in the member's equilibrium equations; do not use the two-force-member model for it.
FC,1=−FC,2\mathbf{F}_{C,1}=-\mathbf{F}_{C,2}
  • Choose the complete frame or an individual member as the system before drawing its diagram.
  • Show external loads and support reactions on a complete-frame diagram.
  • Show pin forces on member diagrams; forces at a shared pin are equal and opposite.

2. Choose signs and apply equilibrium

Use one coordinate convention throughout a calculation. In the examples, right and up are positive, and counterclockwise moment is positive. A force creates a moment equal to its magnitude times the perpendicular distance from the moment centre to the force's line of action. Assign the moment sign from the direction the force tends to turn the body. For example, a leftward force above a point creates a counterclockwise moment.
A body in planar static equilibrium must have zero resultant horizontal force, zero resultant vertical force, and zero resultant moment about any point. These equations apply to the complete frame and to each isolated member. A convenient moment centre often removes unknown reactions from the moment equation: a force whose line of action passes through that point contributes no moment about it.
Assume directions for unknown reaction components when drawing the diagram. A negative result means the actual component points opposite to the assumed direction. For a roller on a horizontal surface, a calculated downward reaction means the stated support model cannot provide the required force. Recheck the loading or model rather than reporting that the roller pulls downward. Keep units consistent, such as kN for force and kN·m for moment.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • State the positive axes and moment direction before writing equations.
  • Use the perpendicular distance from the moment centre to the force's line of action.
  • Check whether the calculated reaction directions are possible for the stated supports.

3. Analyze the complete frame, then its members

Begin with the complete frame when its free-body diagram makes the external reactions easiest to find. Use moment equilibrium to remove reactions whose lines of action pass through the chosen moment centre. Then use force equilibrium for the remaining components. Solve symbolically before substituting values when practical, and retain units throughout.
After finding external reactions, isolate a member if you need a pin force or want to check member equilibrium. Include every applied load on that member, its support forces, and the forces at its pins. The member must satisfy force and moment equilibrium on its own. A force at a pin has zero moment about that same pin, but may have a moment about another point.
Finish by checking the complete frame's force and moment sums, and the isolated member's sums when applicable. If equilibrium does not determine all the unknowns, equilibrium alone is not sufficient for that problem; do not invent an extra relationship.
∑(r×F)z=0\sum (\mathbf{r}\times\mathbf{F})_z=0
  • Use the complete-frame diagram to find external reactions when possible.
  • A member free-body diagram must include every load and connection force acting on that member.
  • Verify both force balance and moment balance for each isolated system.

4. Read the result and verify it

Substitute the calculated reactions into the original equilibrium equations, not just the equation used to find the last unknown. Horizontal and vertical force sums have units of force; moment sums have units of force times distance. If the units do not match the equation, revisit the setup.
For members joined by a pin, the forces on the two member diagrams must be equal in magnitude and opposite in direction. This relation applies to the interaction at the connection. It does not mean pin forces belong on a free-body diagram of the complete frame, where they are internal.
FC, on member 1+FC, on member 2=0\mathbf{F}_{C,\,\mathrm{on\ member\ 1}}+\mathbf{F}_{C,\,\mathrm{on\ member\ 2}}=\mathbf{0}
  • Check force sums and moment sums independently.
  • Keep force and moment units distinct.
  • Use opposite pin-force directions on the two connected-member diagrams.

Worked example

1. Reactions on a loaded frame

A planar frame is supported by a pin at A and a roller at B, 4 m to the right of A. It carries a 2 kN downward load 1 m to the right of A and a 4 kN downward load 3 m to the right of A. Neglect member weight. Find the support reactions by treating the complete frame as one rigid body.
  1. Define the system and signs
    Take the complete frame as the system. Let the pin reactions be AxA_x and AyA_y, and the roller reaction be ByB_y. Choose right, up, and counterclockwise as positive. The roller reaction is assumed upward, as appropriate for a roller resting on a horizontal surface.
  2. Take moments about A
    The reactions at A pass through A, so they create no moment about A. The two downward loads create clockwise moments; the upward reaction at B creates a counterclockwise moment.
    4By−2(1)−4(3)=04B_y-2(1)-4(3)=0
  3. Find the vertical reaction at B
    Solve the moment equation for ByB_y. Its positive value confirms the assumed upward direction.
    By=3.5 kNB_y=3.5\text{ kN}
  4. Use force equilibrium
    There are no horizontal applied loads, so horizontal equilibrium gives zero horizontal reaction at A. The upward reactions must balance the total downward load.
    Ax=0,Ay+By−2−4=0,Ay=2.5 kNA_x=0,\quad A_y+B_y-2-4=0,\quad A_y=2.5\text{ kN}
  5. Verify the frame
    The horizontal forces sum to zero. The upward reactions total 6 kN, matching the downward loads. The moment sum about A is also zero.
    ∑Fx=0,∑Fy=2.5+3.5−2−4=0,∑MA=4(3.5)−2(1)−4(3)=0\sum F_x=0,\quad \sum F_y=2.5+3.5-2-4=0,\quad \sum M_A=4(3.5)-2(1)-4(3)=0
Answer: The reactions are Ax=0A_x=0, Ay=2.5 kNA_y=2.5\text{ kN} upward, and By=3.5 kNB_y=3.5\text{ kN} upward.
Check: The reactions total 6 kN upward, balancing the 6 kN downward load. About A, the roller contributes 14 kN·m counterclockwise, balancing the loads' combined 14 kN·m clockwise moment.

Worked example

2. A horizontal force above a support

A planar frame is pinned at A and supported by a roller at B, 3 m to the right of A. A 2 kN force acts leftward at a point 2 m above A. A 3 kN downward load acts 2 m to the right of A. Neglect member weight. Find the support reactions for the complete frame.
  1. Set up the complete-frame diagram
    Use the complete frame as the system. Let the pin reactions be AxA_x and AyA_y, and the roller reaction be ByB_y. Choose right, up, and counterclockwise as positive. The leftward force above A creates a counterclockwise moment about A.
  2. Take moments about A
    The roller reaction has a 3 m moment arm. The downward load creates a clockwise moment with a 2 m arm. The leftward force, 2 m above A, creates a counterclockwise moment.
    3By+2(2)−3(2)=03B_y+2(2)-3(2)=0
  3. Solve for the roller reaction
    The moment equation gives a positive reaction, consistent with the roller pushing upward.
    By=23 kNB_y=\frac{2}{3}\text{ kN}
  4. Find the pin reactions
    Horizontal equilibrium balances the 2 kN leftward load with a rightward reaction at A. Vertical equilibrium balances the 3 kN downward load with the two upward reactions.
    Ax=2 kN,Ay=3−23=73 kNA_x=2\text{ kN},\quad A_y=3-\frac{2}{3}=\frac{7}{3}\text{ kN}
  5. Verify forces and moments
    The horizontal reactions balance the horizontal load, and the vertical reactions balance the downward load. The moment contributions about A also cancel.
    ∑Fx=2−2=0,∑Fy=73+23−3=0,∑MA=3(23)+2(2)−3(2)=0\sum F_x=2-2=0,\quad \sum F_y=\frac{7}{3}+\frac{2}{3}-3=0,\quad \sum M_A=3\left(\frac{2}{3}\right)+2(2)-3(2)=0
Answer: The reactions are Ax=2 kNA_x=2\text{ kN} to the right, Ay=73 kNA_y=\frac{7}{3}\text{ kN} upward, and By=23 kNB_y=\frac{2}{3}\text{ kN} upward.
Check: The horizontal reaction balances the 2 kN leftward load, and the upward reactions total 3 kN. About A, the roller's 2 kN·m counterclockwise moment plus the horizontal force's 4 kN·m counterclockwise moment balances the load's 6 kN·m clockwise moment.

Worked example

3. Force transmitted at a pin

At pin C, member AC exerts a force on member CB of 4 kN left and 3 kN up. Determine the force exerted on AC by CB, including its magnitude.
  1. Reverse the pin-force components
    The forces exerted by the connected members on each other are equal and opposite. Thus the force on AC is 4 kN to the right and 3 kN down.
    FC, on AC=(4i−3j) kN\mathbf{F}_{C,\,\mathrm{on\ AC}}=(4\mathbf{i}-3\mathbf{j})\text{ kN}
  2. Find the force magnitude
    The horizontal and vertical components are perpendicular, so use the Pythagorean theorem. The magnitude has units of kN.
    FC=(4 kN)2+(3 kN)2=5 kNF_C=\sqrt{(4\text{ kN})^2+(3\text{ kN})^2}=5\text{ kN}
  3. Check the pin interaction
    Adding the forces on the two connected members gives zero in both components. This checks the pin-force pair; the internal pin forces are omitted from the complete-frame diagram.
    (4−4) kN=0,(3−3) kN=0(4-4)\text{ kN}=0,\quad (3-3)\text{ kN}=0
Answer: The force exerted on AC by CB is 4 kN to the right and 3 kN down, with magnitude 5 kN.
Check: The two member forces are opposite in both components, so their component sums are zero.

Common mistakes and how to avoid them

Including internal pin forces on the complete-frame free-body diagram.
Correction: Internal pin forces act between parts of the chosen system, so omit them when the complete frame is isolated.
Drawing a pin force in the same direction on both connected-member diagrams.
Correction: The two forces at a shared pin are equal in magnitude and opposite in direction.
Reporting a downward reaction from a roller on a horizontal surface without checking the model.
Correction: That roller can push upward, not pull downward. Recheck the loading and support model if equilibrium requires a downward force.
Assuming a force at a pin has zero moment about every point.
Correction: A force at a pin has zero moment about that pin, but it may have a nonzero moment about another point.
Treating a loaded member as a two-force member.
Correction: Use the two-force-member model only when forces act at two points and no other loads or couples act on the member.

Lesson summary

  • Define the system and draw its free-body diagram before writing equations.
  • Use whole-frame equilibrium for external reactions and member equilibrium when pin forces are needed.
  • Apply horizontal-force, vertical-force, and moment equilibrium with consistent signs and units.
  • Check support directions and verify force and moment balance for each isolated system.

Check your understanding

Question 1

A pin force on one member is 4 kN left and 2 kN up. What is the force on the connected member?
  1. 4 kN left and 2 kN up
  2. 4 kN right and 2 kN down
  3. 2 kN right and 4 kN down
  4. Zero
Show answer and explanation
4 kN right and 2 kN down
The force on the connected member is equal and opposite: 4 kN right and 2 kN down.

Question 2

A 7 kN force acts leftward at a point 2 m above a moment centre. With counterclockwise positive, what moment does it create?
  1. 14 kN·m counterclockwise
  2. 14 kN·m clockwise
  3. Zero
  4. 7 kN·m clockwise
Show answer and explanation
14 kN·m counterclockwise
A leftward force above the point tends to rotate counterclockwise. Its moment magnitude is force times perpendicular distance, giving 14 kN·m.

Question 3

Which forces belong on a free-body diagram of a complete frame?
  1. Only internal pin forces
  2. Applied external loads and support reactions
  3. Only forces on one selected member
  4. Support reactions and every internal pin-force pair
Show answer and explanation
Applied external loads and support reactions
The complete-frame diagram includes external loads and support reactions. Internal pin forces are omitted because they act between parts within the chosen system.

Key terms

Frame
A connected assembly of members that carries applied loads.
Support reaction
A force exerted by a support on the frame.
Internal pin force
A force transmitted between members connected by a pin.
Free-body diagram
A drawing of an isolated system showing the external forces and moments acting on it.
Statically determinate
A structure whose unknown reactions and member forces can be determined using equilibrium equations alone.

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