Question

The beam is loaded as shown in the diagram below. The beam is uniformly loaded at 3 kN/m for the length of 4 m from B. The beam also has two point loads, 4 KN at 2 m from A and 3 KN at 3 m from B. 2 KN 3 KN 3KN/m A 2 m 2 m 11 m 3 m Fig. Q2 Draw a shear force and bending moment diagram. Also determine the location of maximum bending moment and calculate maximum bending moment. - b) Draw shear force and bending moment diagrams for a simply supported beam loaded as shown in figure Q3 below. Locate clearly the point of maximum bending moment. 2 KN 5 KN 2KN/m 2KN/m B. A 1.5m 1.5m 1.5m 1.5m 1 Fia.Q3

Answer 1

a 2 KN 3KN E 3KNIM B А C Эm .am 2m RB RA about A = Zero Sum of vertical forces - = Zero moment Sum of RA+Rg = 2 +3+3x4 RB X 8 2x2 + 3x5 + 3x4x (4+2) Ra + 1 + 315 = ? RB 11.375 KN RA = 5,625 EN Bending moment at a 11. 375 X2 = 22.75 KNM = RAX2 RAX4-2X2 = 14375x4-4 = 41-5 KNM Bending moment at D Bending moment at E RB X 3 3x3* RAX2 = 5-625 X2 11-25 KNM Bending moment at C = D = RAX4-2x2 = 5-625 X4-4 = 1875 kNm Bending moment at R$* 3–3*3*2*= 11-375*3 - 27 = 20.625 Bending moment at E KNM 3KN 2 KN | 8kim A 3m 1m RA = 5. 6s EN RO She 11.375 5.625 kn 18-625 KN 0-625 KN A B с shear force diagram 2.375 KN X = 3 where 11.375KN Maximum bending moment is at Shear force changes sign Bending moment = B17 mare = 20.62 S KNm 11.375x3-2 169SKAL 11.95 KNM 20.625 BMman A C Bending moment diagram Location of Bending moment 3m from right end B manimum = n av comme SKN 2 KN 2KNIM 2Knim AX B 1.5m-* 1.5m 1-5m -(.5m RA = 6:815 KN RB = 9.125 KN Sum of moments at A = 0 RB X 6 = 2 x to 5 + 5x 4.5 + 2x 1052 + 2x3*(1-5+3) RB = 9.125 KN Sum of Vertical forces = 0 RA+RB = 2 + 2x1.5 + 5 + 2x3 RA 6.875 KN RA + q: 195 = 16 Bending moment at c RAX1.5 - 2x los 8 96 km RBX1.5 Bending moment at E = 요. 2x 15 = 11.43 Kum 2 x1•5 – 2x(1-5+ +") x1.5 RAX3 moment at D Bending 10.875 KNM UJ 6.875 3-875 (-875 1•125 Shear force diagram 9.125 11.754 KNM 10.87 SKUM 11.43 KNM 8.06 2.0695 Bending moment diagrams Shearr force = 0 Manimum bending moment at SFX = 0 29.125 - 2XX - 5 X = 2.0625 Location Op marimum bending moment is at From night end B x = 2.0625 BMmar x=2062S 9.125 X 2.0625 - 2x 2.0625² -5(2.0625 -1-5) 2 11.754 KNM Scanned with CamScanner