16.01 0 I LIE1 .1 99% Selesai + DF = 0.60IL DE = 1.020L OG = CG = R Joint D 182 Chapter 4 Structures PROBLEMS 4/4 Determine the force in each member of the truss. Note the presence of any zero-force members. Introductory Problems 5 kN 4/1 Determine the force in each member of the loaded truss. 3 m 12 m l m 2 m Problem 4/4 100 kg Problem 4/1 4/5 Calculate the forces in members AC, AD, and DE for the loaded truss. Restraining link BC is horizontal. 4/2 Determine the force in each member of the loaded truss. 3' 30 3' 60 E 600 1b Problem 4/5 100 kg 4/6 Calculate the force in each member of the loaded Problem 4/2 truss. 4/3 Determine the force in each member of the loaded D 2kN truss. 13 m 450 3 m 3 m C Problem 4/6 150 Problem 4/3 188 Chapter 4 Structures 4/4 Method of Sections When analyzing plane trusses by the method of joints, we need only two of the three equilibrium equations because the procedures involve concurrent forces at each joint. We can take advantage of the third or moment equation of equilibrium by selecting an entire section of the truss for the free body in equilibrium under the action of a nonconcur- rent system of forces. This method of sections has the basic advantage that the force in almost any desired member may be found directly from an analysis of a cartion which has out that member Thus it is not, nec- nem- TT V X 0 - 3, we s are pen- dent equilibrium relations. Editor teks Edit gambar Isi Formulir Anotasi O16.01 2 0 LTE1 .1 99% Selesai Section 2 From the free-body diagram of section 2, which now includes the known VDJ 3 value of CJ, a balance of moments about G is seen to eliminate DE and JK. 14.14 kN Thus, J [EMG = 0] 12DJ + 10(16) + 10(20) - 18.33(24) - 14.14(0.707)(12) = 0 18.33 kN DJ = 16.67 kN T Ans. 3 Observe that a section through mem- Again the moment of CJ is determined from its components considered to be act- bers CD, DJ, and DE could be taken ing at J. The answer for DJ is positive, so that the assumed tensile direction is which would cut only three unknown correct. members. However, since the forces An alternative approach to the entire problem is to utilize section 1 to deter- in these three members are all con- mine CD and then use the method of joints applied at D to determine DJ. current at D, a moment equation about D would yield no information about them. The remaining two force equations would not be sufficient to solve for the three unknowns. 192 Chapter 4 Structures PROBLEMS 4/34 Determine the force in member EF of the loaded symmetrical truss. Assume no horizontal reactions Introductory Problems at the supports A and C. 4/31 Determine the force in member CG. AWW 15 10' G 10' F 10' E 2 m 15 250 10 E B 5 kips 5 kips 5 kips Problem 4/34 Problem 4/31 4/35 Calculate the forces in members AB, BG, and GF. 4/32 Determine the forces in members CG and GH of the Solve for each force from an equilibrium equation symmetrically loaded truss. which contains that force as the only unknown. -3 m- 4 m 3 m- 6 kN D A 2.4m B 2.4 m C 2.4 m .D B 3 m 2 m 3 n 15 m 30) Problem 4/35 Problem 4/32 4/36 Determine the forces in members BC, CF, and EF of 4/33 Determine the force in member BE of the loaded truss. the loaded truss. D 6 kN - 30 2 m 5 kN 2 n 10 B 4 KN 2 m E T L 3 m Problem 4/36 Problem 4/33 TT V X 0 - Editor teks Edit gambar Isi Formulir Anotasi O