Question

For the beams of problems 6.2-6.16, draw the shear force and bending moment
diagrams and find the maximum shear force, maximum bending moment and
point(s) of contraflexure (PCF).

7 kN 6 kN/m 4 kN/m B 2 m e C D E 24 kN 1,5m 7590.0.751 Figure 6.41

Answer 1

Drawing FBD of bean 4kn 7KN 4 kalm 6kolm THE DE >k lism - 0.75 0.7 Rema 24kN The reaction forces are as shown above first resolving the uniformly distrubuted loads to point loads. R = 4x(2+1.5) R = 4x3.5 R = 14 KN This will be acting at mid-point of AC e at 3.5= 1.75 (m) from A orc 2 Now, R. 6x2 R = 12 (kn) This will be acting at EF le at 2 - Im mid-point of from torf

Efxco Не со Etyce RE - 4-14-12-17 +24=0 Rf = 13 (KN) laking moment of faces at f. (AEME=0 (4 x 7) + (14x5.25) -(24 x 5) + (7x2.75) + (12x1) - MEO M = 12.75 (kN-m) Shear force diagian is. - 12+24 12 KN 12-14x1.5) = 6kN - 4kn 67 - kn -13 KN -4-(4x2) = -12 KN Maximum Taking Now for finding shea force is 13 kN at f only magnitude. drawing Bending moment diagram moments at the points. I

Moment at A -o (kN-m) Moment at B = (-4x2)-(4x2x2) 2 due to UDL - -8 - 8 = -16 (kN-m) c = -(4x3.5x3.5 + (24x1.5) Moment at - (4x3.8 E-2.5 (kN-m) Moment at D - (4x4.25) - [4 x 3.5*(3.5 +0.75 ) 7 + (24x 1.5 + 0.78)] 2 22 -17-(4x3.5x2.5) + (24x 2.25) 2 (knom) Moment at .E. -(4+5) - [4 x 3.5+ (3.5 +1.5) 7 +(24+(1.5 +1.5)] -(7x0.75) --20-(4x 3.5x3.25) + (24x3) -(7x0.75) 1.25 (kN-m) Moment at F. 1-(4x7) - (4x 3.5 × ( 3.5 + 1.5 +2) + (245) -(7x2.75)-(12x1) 1-28 -(4x3.5x5.25)+(24x5)-(7*2.75) -12 -12.75 (kN-m)

Decomes Bending moment diagian becomes. 2 1.25 A , a 16 -12-75 Point of contraffenuture are where Bending moment is zelo : Labelled points n4y are point of Contraflenue and also poist A. And maximan bending moment is 16 (kw.m) Considering only magnitude.
Using FBD and equation of equilibrium to find the shear force and bending moment diagram.