A simply supported beam is subjected to a combination of loads as shown in figure. The point of contraflexure occurs at a point, whereĤ. The maximum bending moment occurs at a point where T aking section between B and A, at a distance x from C, the bending moment is: Taking section between C and B, bending moment at a distance x from end C, we haveĪt x = 1 m. To draw bending moment diagram we need bending moment at all salient points. (The sign is taken positive taken when the resultant force is in downward direction the RHS of the section). To draw the shear force diagram and bending moment diagram we need R A and R B.Īt x = 4 m F A = 4 – 8/3 = + 4/3 kN = + 1.33 kN Solution: Consider a section (X – X’) at a distance x from end C of the beam. Sketch the shear force and bending moment diagrams and find the position of point of contra-flexure. X = 2.5 m M A = –108 (2.5 – 0.9) = –172.8 kN mīMD is parabolic in nature from B to C and straight line from C to A.Ī simply supported beam overhanging on one side is subjected to a uniform distributed load of 1 kN/m. SFD will be triangular from B to C and a rectangle from C to A.īending moment between B and C M x = – (wx).x/2 = –wx 2/2 (The sign is taken to be positive because the resultant force is in downward direction on right hand side of the section).
Since, there is no load between points A and C for this region Fx remains constant.
Shear force = Total unbalanced vertical force on either side of the section.
Solution: Consider a section (X – X’) at a distance x from end B. 19.2 Simply supported beam carrying UDL upto some extent Draw the shear force and bending moment diagrams for the beam.įig. The shape of bending moment diagram is parabolic in shape from B to D, D to C, and, also C to A.Ī cantilever beam carries a uniform distributed load of 60 kN/m as shown in figure. (The sign of bending moment is taken to be negative because the load creates hogging). Solution: Consider a section (X – X’) at a distance x from section B.
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