Truss – AssumptionsSimple TrussMethod of Joints -TrussMethod of Joints -TrussTruss – Example ProblemTruss – Example ProblemTruss – Example ProblemTruss – Example ProblemTruss – Example ProblemExample ProblemTruss – Example ProblemTruss – Example ProblemTruss – Example ProblemTruss – Example ProblemTruss – Example ProblemTruss – Example ProblemTruss – Example ProblemTruss Truss ––Assumptions Assumptions There are four main assumptions made in the analysis of truss Truss members are connected together at their ends only.Truss are connected together by frictionless pins.The truss structure is loaded only at the joints.The weights of the members may be neglected.1234Simple TrussSimple TrussThe basic building block of a truss is a triangle. Large truss are constructed by attaching several triangles together A new triangle can be added truss by adding two members and a joint. A truss constructed in this fashion is known as a simple truss.Method of Joints Method of Joints --TrussTrussThe truss is made up of single bars, which are either in compression, tension or no-load. Themeans of solving force inside of the truss use equilibrium equations at a joint. This method is known as the method of joints.Method of Joints Method of Joints --TrussTrussThe method of joints uses the summation of forces at a joint to solve the force in themembers. It does not use the moment equilibrium equation to solve the problem. In a two dimensional set of equations,In three dimensions,xy0 0 FF==∑∑z0 F=∑Truss Truss ––Example ProblemExample ProblemDetermine the loads in each of the members by using the method of joints.Truss Truss ––Example ProblemExample ProblemDraw the free-body diagram. The summation of forces and moment about B result in() ( ) ( )xAxByAy AyABBAx00 10 kips 10 kips 20 kips0 5 ft 10 kips 10 ft 10 kips 20 ft60 kips60 kipsFRRFR RMRRR== +== − − ⇒ === − −⇒=⇒=−∑∑∑Truss Truss ––Example ProblemExample ProblemLook at Joint BxBCBBC BCyBABA0 60 kips 60 kips0 0 kipsFTRT TFTT== + = + ⇒ =−== ⇒ =∑∑Truss Truss ––Example ProblemExample ProblemLook at Joint D and find the angle 1oxDCDAyDA DADC5 ft.tan 14.0420 ft.0cos0 sin 10 kips 41.231 kips40 kipsFTTFT TTααα−====− −== − ⇒ ==−∑∑Truss Truss ––Example ProblemExample ProblemLook at Joint C and find the angle ()( )()()1oyCA CAxCDCA CBo5 ft.tan 26.56510 ft.0 sin 10 kips 22.361 kips0cos40 kips 22.361 kips cos 26.565 60 kips0FT TFTT Tβββ−==== − ⇒ === − −=− − −−=∑∑Example ProblemExample ProblemDetermine the forces in members FH, DH,EG and BE in the truss using the method of sections.Truss Truss ––Example ProblemExample ProblemDraw the free-body diagram. The summation of forces and moment about H result in() () () () ()xHxHxyHyIHIIHy0 3 kips 3 kips 3 kips 3 kips12 kips00 15 ft 3 kips 10 ft 3 kips 20 ft 3 kips 30 ft 3 kips 40 ft20 kips20 kipsFRRFRRMRRR==++++⇒=−== +== − − − −⇒=⇒=−∑∑∑Truss Truss ––Example ProblemExample ProblemDo a cut between BD and CETruss Truss ––Example ProblemExample ProblemTake moment about A ()() ()100ACECE10 fttan 53.137.5 ft0 cos 53.13 20 ft 3 kips 10 ft2.5 kipsMTTα−==== +⇒=−∑Truss Truss ––Example ProblemExample ProblemDo a cut between HD and GETruss Truss ––Example ProblemExample ProblemTake the moment about ITake the moment about D()()()IHDHD0 20 kips 15 ft 15 ft 3 kips 10 ft18 kipsMTT== − −⇒=∑()()()()D GEGE0 12 kips 20 ft 20 kips 15 ft 3 kips 10 ft 15 ft6 kipsMTT==− + + +⇒=−∑Truss Truss ––Example ProblemExample ProblemDo a cut between HD and HITruss Truss ––Example ProblemExample ProblemTake the sum of forces in y direction()()100yHF HDHF010 fttan 53.137.5 ft0 sin 53.13 20 kips20 kips 18 kips2.5 kipssin 53.13FT TTα−==== + −−⇒=
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