{"id":186,"date":"2021-03-01T09:00:50","date_gmt":"2021-03-01T14:00:50","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/statics\/?post_type=chapter&p=186"},"modified":"2025-08-01T16:42:38","modified_gmt":"2025-08-01T20:42:38","slug":"method-of-joints","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/statics\/chapter\/method-of-joints\/","title":{"raw":"5.2 Method of Joints","rendered":"5.2 Method of Joints"},"content":{"raw":"The method of joints<\/em><\/strong><\/span> is a form of particle analysis. After solving for the reaction forces, you solve for the unknown forces at each joint until you have found the value of each member. You start with your model:\r\n\r\n\"Truss\r\n\r\nConvert the constraints into reaction forces with the appropriate labels:\r\n\r\n\"FBD\r\n\r\n \r\n\r\nNow solve for the reaction forces (Rax\u00a0<\/sub> Ray<\/sub>\u00a0 Re<\/sub>) looking only at the external forces using the equilibrium equations for a rigid body:\r\n

[latex]\\sum F_x=0\\\\\\sum F_y=0\\\\\\sum M=0[\/latex]<\/p>\r\nAssuming the length of each member is L:\r\n

[latex]\\sum F_x=R_{ax} = 0, \\\\\\underline{R_{ax} = 0}[\/latex]<\/p>\r\n \r\n

[latex]\\sum F_y=R_{ay}+R_e \u2013 F_g - F_f= 0, \\\\R_{ay} +R_e = 150 lb[\/latex]<\/p>\r\n \r\n

[latex]\\sum M_a= -L*F_g \u2013 2L * F_f+3L*R_e = 0\\\\R_e = \\frac{100L + 100L}{3L}\\\\\\underline{R_e =66.7 lb}[\/latex]<\/p>\r\n \r\n

[latex]R_{ay} = 150 lb - 66.7 lb\\\\ \\underline{ R_{ay}= 83.3 lb }[\/latex]<\/p>\r\n \r\n\r\nNext, pick a joint where there are 2 or fewer unknown values, such as a or e. This is because you only have 2 equations available to find the unknowns: $latex \\sum F_x=0\u00a0 \\text{,\u00a0\u00a0 } \\sum F_y=0$. The following table shows the number of known and unknown forces at each joint.\r\n\r\n\r\n\r\n\r\n\r\n
Joint:<\/em><\/td>\r\na<\/em><\/td>\r\nb<\/em><\/td>\r\nc<\/em><\/td>\r\nd<\/em><\/td>\r\ne<\/em><\/td>\r\nf<\/em><\/td>\r\ng<\/em><\/td>\r\n<\/tr>\r\n
Known forces:<\/em><\/td>\r\n2<\/td>\r\n0<\/td>\r\n0<\/td>\r\n0<\/td>\r\n1<\/td>\r\n1<\/td>\r\n1<\/td>\r\n<\/tr>\r\n
Unknown forces:<\/em><\/td>\r\n2<\/td>\r\n3<\/td>\r\n4<\/td>\r\n3<\/td>\r\n2<\/td>\r\n4<\/td>\r\n4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nChoosing joint a (or e), do a particle analysis, assuming all of the members are in tension. That way, if the force is negative, that means it is in compression. Notice Rax<\/sub> has been excluded because it is equal to zero.\r\n\r\n\"Free-body\r\n\r\n \r\n

[latex]\\sum F_y=0\\\\R_{ay}+F_{ab}sin(60^\\circ) = 0\\\\F_{ab}=-\\frac{R_{ay}}{sin(60^\\circ)}=-\\frac{83.3 \\text{ lb}}{ 0.866} \\\\\\underline{F_{ab} = - 96.2 \\text{ lb}} \\text{(compression)}[\/latex]<\/p>\r\n \r\n

[latex]\\sum F_x=0\\\\F_{ag} + F_{ab}cos(60^\\circ) = 0 \\\\F_{ag} =- F_{ab}cos(60^\\circ) = - (-96.2 \\text{ lb}) * (0.5) \\\\ \\underline{F_{ag} = + 48.1 \\text{ lb}} \\text{(tension)}[\/latex]<\/p>\r\nNext, move to joint b because you now only have 2 unknowns at joint b (Fbc<\/sub> and Fbg<\/sub>).\r\n\r\nKeep analyzing joints until you've calculated the load in all members:\r\n\r\n\"Truss\r\n\r\n\r\n\r\n\r\n\r\n
Member<\/em><\/td>\r\nab<\/em><\/td>\r\nbc<\/em><\/td>\r\ncd<\/em><\/td>\r\nde<\/em><\/td>\r\nef<\/em><\/td>\r\nfg<\/em><\/td>\r\nag<\/em><\/td>\r\nbg\r\n<\/em><\/td>\r\ncg\r\n<\/em><\/td>\r\ncf\r\n<\/em><\/td>\r\ndf\r\n<\/em><\/td>\r\n<\/tr>\r\n
Force (lb)\r\n<\/em><\/td>\r\n96.2<\/td>\r\n96.2<\/td>\r\n77.0<\/td>\r\n77.0<\/td>\r\n38.5<\/td>\r\n86.6<\/td>\r\n48.1<\/td>\r\n96.2<\/td>\r\n19.3<\/td>\r\n19.3<\/td>\r\n77.0<\/td>\r\n<\/tr>\r\n
Tension or <\/em>Compression\r\n<\/em><\/td>\r\nC<\/td>\r\nC<\/td>\r\nC<\/td>\r\nC<\/td>\r\nT<\/td>\r\nT<\/td>\r\nT<\/td>\r\nT<\/td>\r\nT<\/td>\r\nC<\/td>\r\nT<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAnd that's it! If you don't specify compression or tension, you should use positive and negative to denote tension and compression, respectively.\r\n\r\n \r\n\r\nHere is a second explanation on how to solve using the method of joints:\r\n
\r\n\r\nThe method of joints<\/strong> is a process used to solve for the unknown forces acting on members of a truss<\/strong>. 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.<\/span>\r\n

Using This Method:<\/h2>\r\nThe process used in the method of joints is outlined below:\r\n
    \r\n \t
  1. 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 labelled with letters, and the joints will be labelled with numbers.\r\n\r\n[caption id=\"attachment_2752\" align=\"aligncenter\" width=\"457\"]\"Labeled The first step in the method of joints is to label each joint and each member.[\/caption]<\/li>\r\n \t
  2. 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.\r\n\r\n[caption id=\"attachment_2753\" align=\"aligncenter\" width=\"458\"]\"Truss Treat the entire truss as a rigid body and solve for the reaction forces supporting the truss structure.[\/caption]<\/li>\r\n \t
  3. 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:<\/span>\r\n