44 Mechanics of Materials §3.2 5m 20 kN 30 kN UIO kN -2m。 m 4 m R 12m Fig.3.4. Fig.3.4 carrying concentrated loads only.(The term simply supported means that the beam can be assumed to rest on knife-edges or roller supports and is free to bend at the supports without any restraint.) The values of the reactions at the ends of the beam may be calculated by applying normal equilibrium conditions,ie.by taking moments about F. Thus R4×12=(10×10)+(20×6)+(30×2)-(20×8)=120 R=10kN For vertical equilibrium total force up total load down R4+RF=10+20+30-20=40 R=30kN At this stage it is advisable to check the value of Rr by taking moments about A. Summing up the forces on either side of X-X we have the result shown in Fig.3.5.Using the sign convention listed above,the shear force at X-X is therefore +20kN,i.e.the resultant force at X-X tending to shear the beam is 20kN. 20 kN kN 20kN 30kN 30 kN 20 kN Fig.3.5.Total S.F.at X-X. Similarly,Fig.3.6 shows the summation of the moments of the forces at X-X,the resultant B.M.being 40 kN m. In practice only one side of the section is normally considered and the summations involved can often be completed by mental arithmetic.The complete S.F.and B.M.diagrams for the beam are shown in Fig.3.7,and the B.M.values used to construct the diagram are derived on page 45.44 Mechanics of Materials $3.2 Fig. 3.4. Fig. 3.4 carrying concentrated loads only. (The term simply supported means that the beam can be assumed to rest on knife-edges or roller supports and is free to bend at the supports without any restraint.) The values of the reactions at the ends of the beam may be calculated by applying normal equilibrium conditions, i.e. by taking moments about F. Thus RA x 12 = (10 x 10) + (20 x 6) + (30 x 2) - (20 x 8) = 120 RA = 10 kN For vertical equilibrium total force up = total load down RA+RF = 10+20+30-20 = 40 RF= 3OkN At this stage it is advisable to check the value of RF by taking moments about A. Summing up the forces on either side of X-X we have the result shown in Fig. 3.5. Using the sign convention listed above, the shear force at X-X is therefore +20kN, Le. the resultant force at X-X tending to shear the beam is 20 kN. IO 1* kN X I N ImkN 30kN 1~111 x)kN/ X 30kN i I , 20kN 20kN I Fig. 3.5. Total S.F. at X-X. X Similarly, Fig. 3.6 shows the summation of the moments of the forces at X-X, the resultant B.M. being 40 kNm. In practice only one side of the section is normally considered and the summations involved can often be completed by mental arithmetic. The complete S.F. and B.M. diagrams for the beam are shown in Fig. 3.7, and the B.M. values used to construct the diagram are derived on page 45