STRENGTH OF MATERIAL

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.

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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.

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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

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

Q. A cantilever beam of span 3 m. carrier a udl of 400 N/m over its entire span. Sketch the B.M.D and state the position and magnitud, of maximum bending moment.

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SOLUTION: S.F.D: SHEAR FORCE DIAGRAM FB = 0 FA = 400×3 = 1200 N max S.F. occurs at A = 1200 N BMD ( BENDING MOMENT DIAGRAM) MB = 0 ( why because It’s free End ) MA = -(400×3) × 3/2 = -1800 Nm Max. B.M occurs at fixed end = 1800 Nm (or) … Read more

Q. A cantilever beam 6 m long carries four point loads of 5 kN, 10 kN, 15 kN and 20 kN are acting at 2 m, 3 m, 4 m and 6 m respectively from fixed end. Draw SFD and B.M.D.

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SOLUTION; Total load = 5 + 10 + 15 + 20 = 50 kN SHEAR FORCE DIAGRAM F A =F max = 50 kN (5 + 10 + 15 + 20) (same in between A &E) F{B} = (just left of B)= 20kN F{c} = (just left ofC )=35 kN FC = (just right of … Read more

Q.Q. A cantilever beam 3 m long carries THREE point loads of 2 kN, 4 kN, and 1kN are acting at 1 m, 2 m, and 3 m respectively from fixed end. Draw SFD and B.M.D.

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Q.Q. A cantilever beam 3 m long carries THREE point loads of 2 kN, 4 kN, and 1kN are acting at 1 m, 2 m, and 3 m respectively from fixed end. Draw SFD and B.M.D. SOLUTION TOTAL LOAD =2+4+1=7N SHEAR FORCE DIAGRAM FB = 1N FD = 5N (4+1=5N) FC = 7N (2+4+1=7N) MAX … Read more

Q. A cantilever 5m long carries three point loads of 20 kN, 30 kN and 15 kN at 1 m, 2.5 m and 4 m respectively from free end. Draw SF and BM diagrama. Calculate S.F. and B.M at 4.5 m from free end.

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SOLUTION: Total load = 15+ 30 + 20 = 65 kN SHEAR FORCE DIAGRAM FA = 65 kN (same between AC) Fc =(just left of B) = 65 kN Fc =(just right of B) 50 kN FD= (just left of D) = 50 kN FD =(just right of D) = 20 kN FE =(just left … Read more

Q.A cantilever AB is 6 m long and is fixed at A. It carries point loads of 10 kN, 8 kN and 12 kN at 2 m, 5 m and 6 m respectively from the fixed end. Draw SFD and state max SF

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SOLUTION: SHEAR FORCE DIAGRAM FA =30 kN (same in between AC) FC (just left of C) = 30 kN FC (just right of C) = 20 kN FD (just left of D) = 20 kN FD (just right of D) = 12 kN FB =(just left of B) = 12 kN FB (just right of … Read more