Cantilever beam subjected to udl

Q. A cantilever beam 7 m long carriers a udl of 2 kN/m is acting at a length 3 m from free end; another udl of 4 kN/m is acting at a length of 3 m from Fixed end. Draw SFD and BMD.

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SOLUTION: Total load = 4×3 + 2×3 = 18 kN S.F.D ( SHEAR FORCE DIAGRAM) FA=F{max}=18kN[4×3+2×3] FB=0 Fc = 6 kN FD = 6 kN B.M.D ( BENDING MOMENT DIAGRAM) MB=0 MC= -(2×3)3/2 = -9 kNm MD= -(2×3)(3/2+1) = -15kNm Mmax = MA = -(4×3)3/2-(2×3)(3/2+4) = – 18 – 33 = – 51 KNm Max … Read more

Q. A cantilever beam of 3 m span carries a udl of 10 kN/m over a span of 2 m from fixed end. Draw S.F.D and B.M.D

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SOLUTION: Total load = 10×2 = 20 kN S.F.D (SHEAR FORCE DIAGRAM) FB = Fc = 0 FA = 10×2 = 20 kN max S.F occurs at support A = 20 kN B.M.D ( BENDING MOMENT DIAGRAM) MB = MC = 0 MA = -10 x 2 x 2/2 = -20 kNm (or) MA = … Read more

Q. A cantilever beam of span 3 m. A udl of 2 kN/m is acting between 1 m to 3 m from fixed end. Draw SFD & BMD.

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SOLUTION: Total load = 2 * 2 = 4kN S.F.D (SHEAR FORCE DIAGRAM) F{A} = 4kN (same between A and C) F{C} = 4kN F{B} = 0kN Max S.F occurs at A = 4kN BMD( BENDING MOMENT DIAGRAM) M{B} = 0 M{C} = – (2 * 2) * 2/2 = – 4kNm M max =M … Read more