Q. Cantilever beam of length 4.2 m is loaded. Draw the shear force and bending moment diagram. SOLUTION: Total load= 30 + 20 + 12 * 3 = 86kN S.F.D ( SHEAR FORCE DIAGRAM) FA= 20kN FB = 86 kN (30+20+12×3= 86 kN) FC = 86kN(30 + 20 + 12 × 3 = 86kN) Max … Read more
Q. A cantilever beam of span 8 m carries an udl of 6 kN/m over entire span. In addition, it also carries three point loads of 10 kN, 20 kN and 15 kN at 2 m, 5 m and 7 m respectively from its fixed end. Draw S.F and B.M diagrams. SOLUTION: Total load = … Read more
Q. cantilever beam of span 6 m. A udl of 4 kN/m is acting upto distance of 3 m from the fixed end. Two point loads of 6 kN 8 kN are acting at 1 m and 2 m respectively from the free end. draw SFD and BMD SOLUTION: Tota load = 4 × 3 … Read more
Q. cantilever beam of span 6 m carrying udl of 2 kN/m over a complete span. Two point loads 10 kN and 20 kN are acting at 0 m and 4 m from the free end. Draw S.FD and BMD. SOLUTION: Total load =20 + 10 + 2×6 = 42 kN S.F.D ( SHEAR FORCE … Read more
Q. A cantilever beam of span 4 m carries a udl of 10 kN/m upto a distance of 2 m from fixed end and two point loads of 20 kN and 30 kN placed at a distance of 0 m and 1m from the free end respectively. Draw SFD and BMD. SOLUTION: Total load (10×2) … Read more
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
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
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