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5.5 · Determine internal pin forces between connected members

Learn to determine internal pin forces between connected members through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Frames and Machines

Isolate connected members, apply equilibrium, and interpret the force pair at a pin

A pin joining two members can transmit a force from one member to the other. In a planar model, that force can be represented by horizontal and vertical components. The force on one member and the force on the connected member are equal in magnitude and opposite in direction. The pin does not transfer a couple moment in this ideal model. To determine the force, isolate one member or the joint, draw every force acting on that isolated system, and apply equilibrium. The connection force is internal to the complete assembly, but it is external to either member once that member is isolated.

What you will learn

  • Explain how an ideal pin transfers force between connected members in planar statics.
  • Draw a free-body diagram for an isolated member or pin joint and choose consistent force directions.
  • Use equilibrium to determine internal pin-force components or forces in two-force members.
  • Interpret a negative assumed force and verify force and moment balance.

1. What the pin force represents

A pin joining two members allows them to exert forces on one another at the connection. In two dimensions, describe the force with components along chosen horizontal and vertical axes. If the force on member 1 is to the right and upward, the force on member 2 is to the left and downward. These forces act at the same pin and form an equal-and-opposite pair.
An ideal pin does not supply a couple moment to either member in this planar model. Do not draw a moment reaction at that pin. A different support elsewhere may have its own reaction components, depending on its model.
A member with only two pin connections and no other applied forces or couples is called a two-force member. Its end forces must be equal, opposite, and along the line between the pins. Otherwise, the forces would create a moment that could not be balanced. Use this rule only when the member meets those conditions.
F1→2=−F2→1\mathbf{F}_{1\to2}=-\mathbf{F}_{2\to1}
  • Isolate one member or one pin joint to expose the connection force.
  • Forces exerted on the two connected members at a pin are equal and opposite.
  • An ideal pin transmits force, not a couple moment.

2. Draw the free-body diagram and choose signs

Begin by defining the isolated system: one bar, bracket, or pin joint. Draw its free-body diagram and include only forces acting on that system. Replace a connection to another member by unknown force components unless the two-force-member condition lets you use a single force along the member.
Choose axes, commonly +x+x to the right and +y+y upward. Show an assumed direction for each unknown. That direction is a convenient starting choice; if the solution is negative, the actual direction is opposite. For a force of magnitude FF at angle θ\theta measured counterclockwise from +x+x, its components are Fx=Fcos⁡θF_x=F\cos\theta and Fy=Fsin⁡θF_y=F\sin\theta.
For a planar rigid member in equilibrium, apply two force balances and one moment balance. A force whose line of action passes through the moment centre contributes zero moment about that point. Taking moments about a pin often removes its unknown force components from the moment equation. Keep force units, such as newtons, separate from moment units, such as newton-metres.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • Draw forces acting on the isolated body, not forces acting on the complete assembly.
  • Use a consistent sign convention, such as counterclockwise moments positive.
  • A negative component means the actual component points opposite to its assumed direction.

3. Solve for forces between connected members

For the complete assembly, the equal-and-opposite forces at an internal pin cancel. That is why a whole-assembly free-body diagram usually does not reveal the internal pin force. Isolate one connected member instead: the force exerted by the other member then appears as an external force on the diagram.
In a multi-member problem, start with the member or joint that has the fewest unknowns. If one connected member is a two-force member, its force direction is known. The other member’s moment balance may then determine the force magnitude, followed by force balances for remaining components or support reactions.
If a member has an applied load between its pins, do not assume its pin force acts along its length. Keep unknown pin forces as separate components and use equilibrium to solve them. When switching to the other member, draw a new free-body diagram and reverse the connection force.
MO=xFy−yFxM_O=xF_y-yF_x
  • Internal pin forces cancel for the complete assembly but appear on an isolated member.
  • Use a two-force-member direction only when the member has no other applied force or couple.
  • Check force and moment equilibrium for each isolated rigid member.

4. Common mistakes to avoid

Do not draw both members’ forces at one pin on the same member’s free-body diagram. Each isolated member has only the force acting on it. Draw the opposite force when you isolate the other member.
Do not assign a moment reaction to an ideal pin. Also, do not treat every connected member as a two-force member: an applied load or couple on the member invalidates that condition.
A negative answer is not automatically an arithmetic error. It means the force acts opposite to the direction first assumed. Reverse its direction when describing the physical result, and retain the sign when checking the equations.
Finally, checking only vertical force balance is not enough. Independently check horizontal balance and moment balance as well.
  • Keep each free-body diagram tied to one clearly defined isolated system.
  • Do not assume a force direction along a member unless the two-force-member condition applies.
  • Verify all applicable force and moment equations.

Worked example

1. Symmetric two-force members meeting at a pin

Two straight members meet at pin OO. One extends up-left at 30∘30^\circ above the horizontal and the other extends up-right at 30∘30^\circ above the horizontal. A 600 N600\,\mathrm{N} load acts downward at OO. Find the force in each member and the force exerted on the left member by the pin.
  1. Model the connected members
    Each member has pin forces only at its ends and no other applied load, so each is a two-force member. Assume both members pull away from OO in tension along their axes.
  2. Balance horizontal forces
    For this symmetric geometry, the horizontal components cancel when the two member-force magnitudes are equal.
    −TLcos⁡30∘+TRcos⁡30∘=0  ⇒  TL=TR=T-T_L\cos 30^\circ+T_R\cos 30^\circ=0\;\Rightarrow\;T_L=T_R=T
  3. Balance vertical forces
    The upward components of the two member forces must balance the 600 N600\,\mathrm{N} downward load.
    2Tsin⁡30∘−600 N=0  ⇒  T=600 N2T\sin 30^\circ-600\,\mathrm{N}=0\;\Rightarrow\;T=600\,\mathrm{N}
  4. Interpret and check
    Both members are in tension with magnitude 600 N600\,\mathrm{N}. The force exerted by the pin on the left member is opposite to the force the left member exerts on the pin, so it points down-right along the member. All forces act at OO, so the pin’s moment balance about OO is also satisfied.
    ∑Fx=0,∑Fy=600sin⁡30∘+600sin⁡30∘−600=0\sum F_x=0,\quad \sum F_y=600\sin30^\circ+600\sin30^\circ-600=0
Answer: Each member carries 600 N600\,\mathrm{N} in tension. On the left member, the pin force is 600 N600\,\mathrm{N} directed down-right along the member.
Check: The horizontal components cancel because the geometry is symmetric and the magnitudes are equal. The total upward component is 600 N600\,\mathrm{N}, balancing the load.

Worked example

2. A loaded bar connected to a two-force link

A horizontal bar ABAB is pinned to a support at AA and connected at BB to a two-force link directed up-left in a 33-44-55 direction. The bar is 2.0 m2.0\,\mathrm{m} long and carries a 500 N500\,\mathrm{N} downward load at its midpoint. Determine the force at pin BB on the bar and the reaction at AA.
  1. Take moments about A
    The reaction components at AA have zero moment about AA. The link’s vertical component acts 2.0 m2.0\,\mathrm{m} from AA, and the load acts 1.0 m1.0\,\mathrm{m} from AA. Take counterclockwise moments as positive.
    2.0 m TBy−500 N(1.0 m)=0  ⇒  TBy=250 N2.0\,\mathrm{m}\,T_{By}-500\,\mathrm{N}(1.0\,\mathrm{m})=0\;\Rightarrow\;T_{By}=250\,\mathrm{N}
  2. Use the link direction
    The link direction has vertical-to-total ratio 3/53/5 and horizontal-to-total ratio 4/54/5. Its force on the bar points up-left, so its horizontal component is negative.
    TBy=35TB  ⇒  TB=416.7 N,TBx=−45TB=−333.3 NT_{By}=\frac{3}{5}T_B\;\Rightarrow\;T_B=416.7\,\mathrm{N},\quad T_{Bx}=-\frac{4}{5}T_B=-333.3\,\mathrm{N}
  3. Find the reaction at A
    Horizontal equilibrium makes the reaction at AA balance the link’s leftward component. Vertical equilibrium makes the reaction balance the remaining downward load after including the link’s upward component.
    Ax=333.3 N,Ay=250 NA_x=333.3\,\mathrm{N},\quad A_y=250\,\mathrm{N}
  4. Verify the bar
    The force sums vanish. About AA, the link creates a 500 N⋅m500\,\mathrm{N\cdot m} counterclockwise moment and the applied load creates a 500 N⋅m500\,\mathrm{N\cdot m} clockwise moment.
    ∑Fx=333.3−333.3=0,∑Fy=250+250−500=0,∑MA=250(2.0)−500(1.0)=0\sum F_x=333.3-333.3=0,\quad \sum F_y=250+250-500=0,\quad \sum M_A=250(2.0)-500(1.0)=0
Answer: The pin force on the bar at BB is 416.7 N416.7\,\mathrm{N} up-left, with components (−333.3,250) N(-333.3,250)\,\mathrm{N}. The reaction at AA is (333.3,250) N(333.3,250)\,\mathrm{N}. The force on the link at its pin is opposite to the force on the bar.
Check: Horizontal and vertical force sums and the moment sum about AA are zero using the displayed component precision.

Worked example

3. A pin joint with one member in compression

At pin OO, a 300 N300\,\mathrm{N} force acts right and a 200 N200\,\mathrm{N} force acts downward. Two two-force members connect to OO: member 1 runs up-left at 30∘30^\circ above the negative horizontal direction, and member 2 is vertical. Determine both member forces. Assume tension pulls away from OO along each member.
  1. Balance horizontal forces
    Only member 1 has a horizontal component. Its leftward component balances the 300 N300\,\mathrm{N} force to the right.
    −F1cos⁡30∘+300 N=0  ⇒  F1=346.410 N-F_1\cos30^\circ+300\,\mathrm{N}=0\;\Rightarrow\;F_1=346.410\,\mathrm{N}
  2. Balance vertical forces
    The assumed upward component of member 1 is insufficient to balance both downward loads. The resulting negative value means the assumed downward tension force in member 2 points the opposite way.
    F1sin⁡30∘−F2−200 N=0  ⇒  F2=−26.795 NF_1\sin30^\circ-F_2-200\,\mathrm{N}=0\;\Rightarrow\;F_2=-26.795\,\mathrm{N}
  3. Interpret and verify
    Member 1 is in tension at about 346.410 N346.410\,\mathrm{N}. Member 2 is in compression at about 26.795 N26.795\,\mathrm{N}; its actual force on OO is upward. Using the rounded values shown, the force residuals are approximately zero, within rounding. All forces act at OO, so their moment sum about OO is zero.
    ∑Fx≈−346.410cos⁡30∘+300=0,∑Fy≈346.410sin⁡30∘+26.795−200=0\sum F_x\approx-346.410\cos30^\circ+300=0,\quad \sum F_y\approx346.410\sin30^\circ+26.795-200=0
Answer: Member 1 is in tension at approximately 346.410 N346.410\,\mathrm{N}. Member 2 is in compression at approximately 26.795 N26.795\,\mathrm{N}. At each pin, the force on the connected member is opposite to the force exerted by the member on the pin.
Check: With unrounded values, both force balances are zero. With the displayed rounded values, both residuals are approximately zero. The joint moment balance is zero because the forces act at the joint.

Common mistakes and how to avoid them

Drawing a couple moment at an ideal pin.
Correction: Represent the pin interaction with force components only in this planar model.
Putting both equal-and-opposite connection forces on one member’s free-body diagram.
Correction: Show only the force acting on the isolated member; draw the opposite force when the other member is isolated.
Assuming every member’s force acts along its length.
Correction: That direction is justified only for a member with two pin forces and no other applied force or couple.
Treating a negative assumed force as an arithmetic error.
Correction: Reverse the assumed direction when reporting it; a negative tension result can indicate compression.

Lesson summary

  • An internal pin force becomes external when one connected member is isolated.
  • Forces at a pin on the two connected members are equal and opposite.
  • Use two force balances and, for a rigid member, a moment balance.
  • Use the two-force-member direction only when its conditions are met.
  • Verify the applicable force and moment equations and interpret negative results as reversed directions.

Check your understanding

Question 1

A pin exerts a force on member A directed right and upward. What direction does member A exert a force on the connected member?
  1. Right and upward
  2. Left and downward
  3. Right and downward
  4. A couple moment
Show answer and explanation
Left and downward
The forces the connected members exert on one another are equal and opposite.

Question 2

A member has pin connections at its ends and no other applied load or couple. In what direction must its end forces act?
  1. Perpendicular to the member
  2. Along the line between its pins
  3. At any angle if their magnitudes match
  4. Only vertically
Show answer and explanation
Along the line between its pins
For moment equilibrium, the equal-and-opposite end forces must act along the same line joining the pins.

Question 3

An assumed upward pin-force component solves to −40 N-40\,\mathrm{N}. What does this mean?
  1. The actual component is 40 N40\,\mathrm{N} downward
  2. The actual component is 40 N40\,\mathrm{N} upward
  3. The pin transfers a 40 N⋅m40\,\mathrm{N\cdot m} moment
  4. The member cannot be in equilibrium
Show answer and explanation
The actual component is 40 N40\,\mathrm{N} downward
The negative sign means the actual component points opposite the assumed upward direction.

Key terms

Internal pin force
The force one connected member exerts on another through their pin connection.
Free-body diagram
A drawing of an isolated body showing the external forces and moments acting on it.
Two-force member
A member acted on only by forces at two pin connections, with no other applied load or couple.
Action–reaction pair
Two equal and opposite forces exerted by connected bodies on one another.

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