Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips

Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

Truss AnalysisEngineering MechanicsStructural MechanicsMechanics of Materials

The Method of Joints is a simple method used to determine the forces acting on individual members of a truss. a step-by-step guide on how to apply the method to solve for forces at joints A, B, C, H, F, and G.

What you will learn

- How does the Method of Joints work for determining forces in trusses?
- Can the Method of Joints be used to find the forces in all members of a truss?
- What are the steps to apply the Method of Joints to solve for forces at a joint?

- What is the difference between the Method of Joints and the Method of Sections in truss analysis?
- How do you determine the direction of unknown forces using the Method of Joints?
- What is the role of equilibrium equations in the Method of Joints?

Typology: Lecture notes

2021/2022

1 / 67

Download Method of Joints: Determining Forces in Trusses and more Lecture notes Acting in PDF only on Docsity! Method of Joints ◮ The method of joints is one of the simplest methods for determining the force acting on the individual members of a truss because it only involves two force equilibrium equations. Method of Joints ◮ The method of joints is one of the simplest methods for determining the force acting on the individual members of a truss because it only involves two force equilibrium equations. ◮ Since only two equations are involved, only two unknowns can be solved for at a time. Therefore, you need to solve the joints in a certain order. That is, you need to work from the sides towards the center of the truss. Method of Joints Let’s start with joint A. We begin by drawing all the forces that act on the bolt at joint A. b A 120N 15N FAB FAC Note that FAB points towards the joint. This is because FAB and the 15 N force are the only vertical forces. Therefore FAB must point downwards to balance the 15 N force pointing up. Method of Joints Let’s start with joint A. We begin by drawing all the forces that act on the bolt at joint A. b A 120N 15N FAB FAC Note that FAB points towards the joint. This is because FAB and the 15 N force are the only vertical forces. Therefore FAB must point downwards to balance the 15 N force pointing up. This helps us determine the direction of FAC . FAC must point away from the joint since that is the only way to balance the forces in the horizontal direction. Method of Joints - Direction of Forces ◮ Determining the direction of the unknown forces is a good idea. You can still solve the problem if you choose the directions incorrect, but you will get negative values for the force. Therefore, it is best to figure this out when possible. In a few cases, it is not possible so you have to guess. Method of Joints - Direction of Forces ◮ Determining the direction of the unknown forces is a good idea. You can still solve the problem if you choose the directions incorrect, but you will get negative values for the force. Therefore, it is best to figure this out when possible. In a few cases, it is not possible so you have to guess. ◮ To get the directions for the unknown forces correct, you need to proceed in the correct order. Look back at the unknown forces, FAB and FAC . They both have horizontal components, but only FAB has a vertical component. ◮ Since only one of the unknowns, FAB has a vertical component, figure out the direction of FAB first. To do this, balance out the forces in the vertical direction. Since the only other force in the vertical direction is the 15 N force pointing up, FAB must point down, or towards the joint. ◮ Once you know the direction of FAB , you can see that both the 120 N force and FAB point to the left, so FAC must point to the right, or away from the joint. Interpreting the directions of forces in truss members ◮ When a force points toward the joint, the member is said to be in compression. If the force points away from the joint, the member is said to be in tension. Interpreting the directions of forces in truss members ◮ When a force points toward the joint, the member is said to be in compression. If the force points away from the joint, the member is said to be in tension. ◮ It is often important to know whether a truss member is in tension or in compression because some building materials have different strengths in compression versus tension. Method of Joints - Solving for forces at joint A Start by considering the horizontal and vertical forces. In the horizontal direction, you have two unknowns, FAB and FAC . In the vertical direction, you only have one unknown, FAB . Therefore, start with balancing the forces in the vertical direction. To break FAB into components, consider the right triangle that is described by the dimensions. b A 120N 15N FAB FAC 4 3 √ 32 + 42 = 5 5 vert. comp. → 3 5 hor. comp. → 4 5 ↑ + ∑ Fy = +15N − 3 5 FAB = 0 Method of Joints - Solving for forces at joint A Start by considering the horizontal and vertical forces. In the horizontal direction, you have two unknowns, FAB and FAC . In the vertical direction, you only have one unknown, FAB . Therefore, start with balancing the forces in the vertical direction. To break FAB into components, consider the right triangle that is described by the dimensions. b A 120N 15N FAB FAC 4 3 √ 32 + 42 = 5 5 vert. comp. → 3 5 hor. comp. → 4 5 ↑ + ∑ Fy = +15N − 3 5 FAB = 0 FAB = 5 3 (15N) = 25N (compression) Method of Joints - Solving for forces at joint A Now we can solve for FAC by summing forces in the horizontal direction. b A 120N 15N FAB = 25N FAC 4 3 √ 32 + 42 = 5 5 vert. comp. → 3 5 hor. comp. → 4 5 Method of Joints - joint B Now we can proceed to joint B. Since FAB has been determined, joint B has only two unknowns, FBC and FBD . b B120 N FAB 25 N FBC FBD Method of Joints - joint B Now we can proceed to joint B. Since FAB has been determined, joint B has only two unknowns, FBC and FBD . b B120 N FAB 25 N FBC FBD Here it is straightforward to figure out the directions of the unknowns. Since FBC is the only unknown force with a vertical component, it must point down (away from the joint) to balance the vertical component of FAB . Method of Joints - joint B Now we can proceed to joint B. Since FAB has been determined, joint B has only two unknowns, FBC and FBD . b B120 N FAB 25 N FBC FBD Here it is straightforward to figure out the directions of the unknowns. Since FBC is the only unknown force with a vertical component, it must point down (away from the joint) to balance the vertical component of FAB . Similarly, FBD must point left, towards the joint, to balance the 120 N force and the horizontal component of FAB . Method of Joints - joint B b B120 N FAB 25 N FBC FBD 3 4 5 ↑ + ∑ Fy = + 3 5 FAB − FBC = 0 FBC = 3 5 (25N) = 15N (tension) Method of Joints - joint B b B120 N FAB 25 N FBC FBD 3 4 5 ↑ + ∑ Fy = + 3 5 FAB − FBC = 0 FBC = 3 5 (25N) = 15N (tension) → + ∑ Fx = +120N + 4 5 FAB − FBD = 0 Method of Joints - joint B b B120 N FAB 25 N FBC FBD 3 4 5 ↑ + ∑ Fy = + 3 5 FAB − FBC = 0 FBC = 3 5 (25N) = 15N (tension) → + ∑ Fx = +120N + 4 5 FAB − FBD = 0 FBD = +120N + 4 5 (25N) = 140N (compression) Method of Joints - joint C Now that we have solved joints A and B, we can proceed to joint C. Since FAC and FBC have been determined, joint C has only two unknowns, FCD and FCE . b C FAC 140 N FBC 15 N FCD FCE Since FCD is the only unknown force with a vertical component, balancing forces in the vertical direction shows that FCD must point down, and toward the joint. Once we know that FCD points to the left, FCE must point right, away from the joint, to balance the forces in the horizontal direction. Method of Joints - joint C b C FAC 140 N FBC 15 N FCD FCE 4 35 Balancing forces, we get: Method of Joints - joint C b C FAC 140 N FBC 15 N FCD FCE 4 35 Balancing forces, we get: ↑ + ∑ Fy = +FBC − 3 5 FCD = 0 Method of Joints - joint C b C FAC 140 N FBC 15 N FCD FCE 4 35 Balancing forces, we get: ↑ + ∑ Fy = +FBC − 3 5 FCD = 0 FCD = 5 3 (15N) = 25N (compression) → + ∑ Fx = −FAC − 4 5 FCD + FCE = 0 FCE = 140N + 4 5 (25N) = 160N (tension) Method of Joints - joint H Now that we have gotten close to the center, we should move to the opposite side of the truss, namely joint H. b H 135 N FFH FGH Method of Joints - joint H Now that we have gotten close to the center, we should move to the opposite side of the truss, namely joint H. b H 135 N FFH FGH Since FFH is the only unknown force that has a vertical component, balancing the vertical forces means that FFH must point down, towards the joint. Method of Joints - joint H b H 135 N FFH FGH 4 3 5 Balancing forces gives: ↑ + ∑ Fy = +135N − 3 5 FFH = 0 Method of Joints - joint H b H 135 N FFH FGH 4 3 5 Balancing forces gives: ↑ + ∑ Fy = +135N − 3 5 FFH = 0 FFH = 5 3 (135N) = 225N (compression) Method of Joints - joint H b H 135 N FFH FGH 4 3 5 Balancing forces gives: ↑ + ∑ Fy = +135N − 3 5 FFH = 0 FFH = 5 3 (135N) = 225N (compression) → + ∑ Fx = 4 5 FFH − FGH = 0 Method of Joints - joint F Now we can move in to joint F. b F 150 N FFH 225 N FFG FDF Since FDF is the only unknown force with a horizontal component, balancing forces in the horizontal direction means that FDF must point to the right, towards the joint. Method of Joints - joint F Now we can move in to joint F. b F 150 N FFH 225 N FFG FDF Since FDF is the only unknown force with a horizontal component, balancing forces in the horizontal direction means that FDF must point to the right, towards the joint. FFG is the only unknown force that has a vertical component. To balance the vertical forces means requires some calculation. The vertical component of FFH is 3 5 (225N) or 135N. Since 135N is smaller than the 150N external force, FFG must point up, towards the joint. Method of Joints - joint F Now we can move in to joint F. b F 150 N FFH 225 N FFG FDF Since FDF is the only unknown force with a horizontal component, balancing forces in the horizontal direction means that FDF must point to the right, towards the joint. FFG is the only unknown force that has a vertical component. To balance the vertical forces means requires some calculation. The vertical component of FFH is 3 5 (225N) or 135N. Since 135N is smaller than the 150N external force, FFG must point up, towards the joint. Method of Joints - joint F b F 150 N FFH 225 N FFG FDF 4 3 5 Balancing forces gives: ↑ + ∑ Fy = −150N + 3 5 (225N) + FFG = 0 FFG = 150N − 3 5 (225N) = 15N (compression) Method of Joints - joint F b F 150 N FFH 225 N FFG FDF 4 3 5 Balancing forces gives: ↑ + ∑ Fy = −150N + 3 5 (225N) + FFG = 0 FFG = 150N − 3 5 (225N) = 15N (compression) → + ∑ Fx = +FDF − 4 5 FFH = 0 Method of Joints - joint F b F 150 N FFH 225 N FFG FDF 4 3 5 Balancing forces gives: ↑ + ∑ Fy = −150N + 3 5 (225N) + FFG = 0 FFG = 150N − 3 5 (225N) = 15N (compression) → + ∑ Fx = +FDF − 4 5 FFH = 0 FDF = 4 5 (225N) = 180N (compression) Method of Joints - joint G Now we can move in to joint G. b G FGH 180 N FFG 15 N FDG FEG Since FDG is the only unknown force with a vertical component, balancing forces in the vertical direction means that FDG must point up, away from the joint. Solving for the direction of FEG is a little tricky. Balancing forces in the horizontal direction requires balancing FGH , the horizontal component of FDG and FEG . Although FDG is unknown at this point, FDG only has to balance out the 15 N force in the vertical direction. Therefore the horizontal component of FDG is unlikely to be larger than FGH . So, FEG should point to the left, away from the joint. Method of Joints - joint G b G FGH 180 N FFG 15 N FDG FEG 4 3 5 Balancing forces gives: Method of Joints - joint G b G FGH 180 N FFG 15 N FDG FEG 4 3 5 Balancing forces gives: ↑ + ∑ Fy = + 3 5 FDG − FFG = 0 Method of Joints - joint G b G FGH 180 N FFG 15 N FDG FEG 4 3 5 Balancing forces gives: ↑ + ∑ Fy = + 3 5 FDG − FFG = 0 FDG = 5 3 (15N) = 25N (tension) → + ∑ Fx = −FEG − 4 5 FDG + FGH = 0 FEG = 180N − 4 5 (25N) = 160N (tension) Method of Joints - joint E Now that we have done the joints from the sides, let’s move to the center with joint E. bFCE 160 N FEG 160 N FDE Method of Joints - joint E Now that we have done the joints from the sides, let’s move to the center with joint E. bFCE 160 N FEG 160 N FDE Although FDE is shown pointing up, there really is no vertical force to balance FDE . Method of Joints - joint D ◮ We have actually solved for the forces in all the members at this point, so there is no need to solve for anything at joint D. Method of Joints - joint D ◮ We have actually solved for the forces in all the members at this point, so there is no need to solve for anything at joint D. ◮ Member FDE is a zero-force member. In reality, member DE actually would have some small force acting on it to maintain the shape of the truss. But, the magnitude of this force would be small compared to truss members that are not zero-force members. Method of Joints - joint D ◮ We have actually solved for the forces in all the members at this point, so there is no need to solve for anything at joint D. ◮ Member FDE is a zero-force member. In reality, member DE actually would have some small force acting on it to maintain the shape of the truss. But, the magnitude of this force would be small compared to truss members that are not zero-force members. ◮ If you wanted to check your work, you could verify that the forces acting at D do balance.