5.4: Method of Joints

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The method of joints is a process used to solve for the unknown forces acting on members of a truss. The method centers on the joints or connection points between the members, and it is usually the fastest and easiest way to solve for all the unknown forces in a truss structure.

Using the Method of Joints:

The process used in the method of joints is outlined below.

In the beginning, it is usually useful to label the members and the joints in your truss. This will help you keep everything organized and consistent in later analysis. In this book, the members will be labeled with letters and the joints will be labeled with numbers.

Side view of a bridge with a 20-meter-long span, bisected into sections B (left) and E (right) at point 3 by the lower end of a 5-meter-long vertical beam C. The left end of B (point 1) is connected to the top end of C (point 2) by a diagonal beam A. The right end of E (point 4) is connected to point 2 by a diagonal beam D. Point 3 experiences a downwards force of 6 kN. Points 1 and 4 are held off the ground by a pin joint and a roller joint respectively. — Figure \(\PageIndex{1}\): The first step in the method of joints is to label each joint and each member.

Treating the entire truss structure as a rigid body, draw a free body diagram, write out the equilibrium equations, and solve for the external reacting forces acting on the truss structure. This analysis should not differ from the analysis of a single rigid body.

The section's Figure 1 above is shown with the reaction forces upon it: point 1 experiences a reaction force with both horizontal (x) and vertical (y) components; point 4 experiences a reaction force with only a y component. — Figure \(\PageIndex{2}\): Treat the entire truss as a rigid body and solve for the reaction forces supporting the truss structure.

Assume there is a pin or some other small amount of material at each of the connection points between the members. Next you will draw a free body diagram for each connection point. Remember to include:

Any external reaction or load forces that may be acting at that joint.
A normal force for each two force member connected to that joint. Remember that for a two force member, the force will be acting along the line between the two connection points on the member. We will also need to guess if it will be a tensile or a compressive force. An incorrect guess now though will simply lead to a negative solution later on. A common strategy then is to assume all forces are tensile, then later in the solution any positive forces will be tensile forces and any negative forces will be compressive forces.
Label each force in the diagram. Include any known magnitudes and directions and provide variable names for each unknown.

Free body diagrams of points 1, 2, 3 and 4 from Figures 1 and 2 above. Points 1 and 4 each experience an upward force of 3 kN; point 3 experiences a downwards force of 6 kN. In addition, each point experiences a tension force for each member that is attached to it. — Figure \(\PageIndex{3}\): Drawing a free body diagram of each joint, we draw in the known forces as well as tensile forces from each two-force member.

Write out the equilibrium equations for each of the joints. You should treat the joints as particles, so there will be force equations but no moment equations. With either two (for 2D problems) or three (for 3D problems) equations for each joint; this should give you a large number of equations.
- In planar trusses, the sum of the forces in the \(x\) direction will be zero and the sum of the forces in the \(y\) direction will be zero for each of the joints. \[ \sum \vec{F} = 0 \] \[ \sum F_x = 0 \, ; \,\,\, \sum F_y = 0 \]
- In space trusses, the sum of the forces in the \(x\) direction will be zero, the sum of the forces in the \(y\) direction will be zero, and the sum of forces in the \(z\) direction will be zero for each of the joints. \[ \sum \vec{F} = 0 \] \[ \sum F_x = 0 \, ; \,\,\, \sum F_y = 0 \, ; \,\,\, \sum F_z = 0 \]
Finally, solve the equilibrium equations for the unknowns. You can do this algebraically, solving for one variable at a time, or you can use matrix equations to solve for everything at once. If you assumed that all forces were tensile earlier, remember that negative answers indicate compressive forces in the members.

Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source: https://youtu.be/B8SEG7xPI-o.

Example \(\PageIndex{1}\)

Find the force acting in each of the members in the truss bridge shown below. Remember to specify if each member is in tension or compression.

A truss bridge spanning 30 meters; the leftmost end, point A, rests on the ground on a pin joint and the rightmost end, point F, rests on the ground on a roller joint. A and F are each endpoints of 10-meter-long members, joined by another 10-meter horizontal member with left endpoint B and right endpoint D. Member BD is the topmost edge of a rectangle whose other sides are formed by the members BC, CE, and ED. Points A and C are connected by a member that makes a 20° angle with the horizontal, and so are points E and F. A downwards force of 60 kN is applied at point B, and a downwards force of 80 kN is applied at point D. — Figure \(\PageIndex{4}\): problem diagram for Example \(\PageIndex{1}\). A truss bridge represented as a 2D plane truss, with a standard-orientation \(xy\)-coordinate system.

Solution: Video \(\PageIndex{2}\): Worked solution to example problem \(\PageIndex{1}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/vowewkEdTzw.

Example \(\PageIndex{2}\)

Find the force acting in each of the members of the truss shown below. Remember to specify if each member is in tension or compression.

A truss consists of a horizontal section 12 feet long, which is attached to a wall at its left end by a pin joint and consists of 2 6-foot members; a vertical member 6 feet long, attached at its bottom end to the midpoint of the horizontal members; two diagonal members attaching the top end of said vertical member to the left and right ends of the 12-foot span; a second 6-foot vertical member attached at the left end of the 12-foot span and extending downwards so its lower end is attached to the wall with a roller joint; and a diagonal member connecting the roller-joint end with the midpoint of the 12-foot span. A downwards force of 500 lbs is applied at the right end of the horizontal span. — Figure \(\PageIndex{5}\): problem diagram for Example \(\PageIndex{2}\). A plane truss mounted on a wall, with a standard-orientation \(xy\)-coordinate system.

Solution: Video \(\PageIndex{3}\): Worked solution to example problem \(\PageIndex{2}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/IxnClZ-ppjM.

Example \(\PageIndex{3}\)

Find the force acting in each of the members of the truss shown below. Remember to specify if each member is in tension or compression.

In the yz-plane (the plane of the screen), point C is at the origin, point A is 1 meter above C, and point D is 3 meters to the right of C, with all 3 points joined by members to form a right triangle. On the x axis, pointing out of the screen, point B is 2 meters forwards of C; B is joined by members to points A, C and D. At point D, a force of 600 N is applied down (in the negative z-direction) and out of the screen, making a 30° angle with the yz-plane. The whole truss is supported by a ball-and-socket joint, whose base points in the negative y-direction, at point A. — Figure \(\PageIndex{6}\): problem diagram for Example \(\PageIndex{3}\). A space truss supported by a single ball-and-socket joint, oriented on a 3D coordinate system with the \(yz\)-plane in the plane of the screen and the \(x\)-axis pointing out of the screen.

Solution: Video \(\PageIndex{4}\): Worked solution to example problem \(\PageIndex{3}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/sDKESSbufEk.