Solve all 8 members in this truss using the Method of Joints (MoJ).

Truss Example: Method of Joints (MoJ)Thursday, 2 May 201311:30 AM

MoJ-Example Page 1

There are no obvious roller joints, but the force in member DF must be horizontal, so Reaction at F must also be horizontal.So we can treat as:Joint E acts like a Pin JointJoint F acts like a Roller Joint

Find Reactions1.Thursday, 2 May 201311:39 AM

MoJ-Example Page 2

Take moments for whole truss at joint E:

ME = 0 = - (20*2.4) + (RFx * 1.03923)RFx = 48/1.03923 RFx = 46.18804 kN

1(a). Take Moments at Joint E (pin joint)Thursday, 2 May 201311:40 AM

MoJ-Example Page 3

Sum of vertical forces:

Fy = 0 = -20 + ReyREy = 20 kN

Sum of horizontal forces:

Fx = 0 = - RFx - REx

= - - 46.18804REx = - RFx

REx = 46.188 kN (Positive - to the right)

1(b). Find Reactions at Joint EThursday, 2 May 201311:40 AM

MoJ-Example Page 4

Preparing for Method of Joints (MoJ):Now we need to find a joint that has no more than 2 unknowns.A = 2 unknown forcesB = 3 unknown forcesC = 4 unknown forcesD = 4 unknown forcesE = 2 unknown forcesF = 1 unknown forces

So we could start at Joints A, E or F. (Note: F is not suitable because it takes us to Joint D where there are still 3 unknowns). So we can start with A or E.Let's start with joint A

Method of Joints: Look for a Suitable Joint (to Start MoJ)2.Thursday, 2 May 201312:10 PM

MoJ-Example Page 5

Isolate Joint A and treat Joint A as a Concurrent Forcesproblem with 2 unknown forces. The BODY is the joint, so the FBD shows what the members are doing TO THE JOINT.

Have a guess at the force directions. If not sure, just make it tension (i.e. pulling the joint)

Now solve equlibrium by drawing a Force Polygon.Using CAD, start with known force 20kN @ 270o

Then something horizontal + something at 60o gets back to start point. FAC = 11.547 kN and FAB = 23.094 kNYou must keep track of directions here! We can see from this Force Polygon that we guessed FAB wrong.FAB is in compression (pushing on joint A)So final answer is;FAC = 11.547 kN tensionFAB = 23.094 kN compression

Remember: Solving by CAD is exactly the same as using the equilibrium conditions; (Solving mathematically)

2. Joint A (Equilibrium of Joint A)Thursday, 2 May 201312:10 PM

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FAC = 11.547 kN tensionFAB = 23.094 kN compression

The last step in completing Joint A is to consider the equilibrium of the 2 members: Member AC and member AB.Since all members in a pin-jointed truss have only 2 joints - which means 2 forces.To have equilibrium with 2 forces means they must be equal and opposite. Since member AC is in tension, it must pull on Joint C by exactly 11.547 kN.Since member AB is in compression, it must push on Joint B by an exactly 23.094 kN.

2. Joint A (Tension/compression of members)Thursday, 2 May 201312:10 PM

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FAC = 11.547 kN tension, so it must pull on Joint CFAB = 23.094 kN compression, it must push on Joint B

Now we have more information about two adjacent joints - B and C. Joint B was 3, now 2 unknown forces.So the next joint to solve is Joint B!

2. Joint A (Forces on adjacent joints)Thursday, 2 May 201312:10 PM

MoJ-Example Page 8

Isolate Joint B and solve Concurrent Forces. Have a guess at the force directions.

Now draw FP. Note that FAB which was @120o is now FBA @ 300o.All forces are 60o apart, so the FP is an equilateral triangle.

3. Joint B (Equilibrium of Joint B)Thursday, 2 May 201312:10 PM

MoJ-Example Page 9

Looks like we guessed the right arrow directions!

3. Joint B (Check directions from FP)Thursday, 2 May 201312:10 PM

MoJ-Example Page 10

Now do equilibrium of members.Member BC is in tension, so it pulls on Joint C.Member BD is in compression, so it pushes on Joint D.

Now we have more information about two adjacent joints - C and D. Joint C is now 2 unknown forces, so the next joint to solve is Joint C!

3. Joint B (Equilibrium of members)Thursday, 2 May 201312:10 PM

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Isolate Joint C and solve Concurrent Forces. Have a guess at the force directions. Draw FP. FCE = 34.641 kN FCD = 23.094 kN

4. Joint C (Equilibrium of joint)Thursday, 2 May 201312:10 PM

MoJ-Example Page 12

Transfer these forces back into trussFCE = 34.641 kN @ 0o (Tension)FCD = 23.094 kN @ 120o (Compression)

4. Joint C (Put joint forces into truss)Thursday, 2 May 201312:10 PM

MoJ-Example Page 13

Now do equilibrium of members.Member CE is in tension, so it pulls on Joint E.Member CD is in compression, so it pushes on Joint D.

Now we have more information about two adjacent joints - E and D. Joint D is now 2 unknown forces, so the next joint to solve is Joint D!

(Of course, we could also do Joint E too, but we will wait and see...)

4. Joint C (Equilibrium of members)Thursday, 2 May 201312:10 PM

MoJ-Example Page 14

Isolate Joint D and solve Concurrent Forces. Have a guess at the force directions. Draw FP.Start with known forces FDB and FDC.

FDE = 23.094 kN (pulling)FDF = 46.188 kN (pushing)

5. Joint D (Equilibrium of Joint D)Thursday, 2 May 201312:10 PM

MoJ-Example Page 15

Transfer these forces back into trussFDE = 23.094 kN (Tension)FDF = 46.188 kN (Compression)

5. Joint D (Transfer joint force into truss)Thursday, 2 May 201312:10 PM

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Now do equilibrium of members.Member DE is in tension, so it pulls on Joint E.Member DF is in compression, so it pushes on Joint F.

Finished!

5. Joint D (Equilibrium of members)Thursday, 2 May 201312:10 PM

MoJ-Example Page 17

Check equilibrium at Joint F:

46.188 - 46.188 = 0 Correct!

Check equilibrium at Joint E:(Force Polygon)Correct!

Final CheckThursday, 2 May 201312:10 PM

MoJ-Example Page 18