# Shear force and bending moment diagram for simply supported beam _1P

sushma chintaasst.prof at chaitanya bharathi institute of technology um chaitanya bharathi institute of technology
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### Shear force and bending moment diagram for simply supported beam _1P

• 2. Shear force and Bending moment diagram When designing a beam, we usually need to know how the shear forces and bending moments vary throughout the length of the beam. Of special importance are the maximum and minimum values of these quantities. Information of this kind is usually provided by graphs in which the shear force and bending moment are plotted as ordinates and the distance x along the axis of the beam is plotted as the abscissa. Such graphs are called shear-force and bending-moment diagrams.
• 3. Q1.Draw the shear force and bending moment diagram for a simple beam AB supporting a concentrated load P. The load P acts at distance โaโ from the left-hand support and distance โbโ from the right-hand support.
• 4. Free body diagram of given simple beam: C
• 5. Free body diagram of section1:
• 6. Free body diagram of section1: when x=0 section coincides with A, when x=a section coincides with C 0< ๐ฅ<a Section1-1 Shear force V= P๐ ๐ฟ Bending Moment M= P๐ ๐ฟ . ๐ฅ ๐ฅ=0 (at A) VA= P๐ ๐ฟ MA= P๐ ๐ฟ . 0= 0 ๐ฅ= a (at C) VC= P๐ ๐ฟ MC= P๐ ๐ฟ . a = Pa๐ ๐ฟ ๐บ๐๐ฒ = ๐ RA-V =0 ๐บ๐ ๐โ๐ = ๐ - RA . ๐ฅ + M =0 M= RA . ๐ฅ = P๐ ๐ฟ . ๐ฅ
• 7. Free body diagram of section2: when x=a section coincides with C, when x=L section coincides with B ๐บ๐๐ฒ = ๐ ๐บ๐ ๐โ๐ = ๐
• 8. Free body diagram of section2: when x=a section coincides with C, when x=L section coincides with B a< ๐ฅ<L Section2-2 Shear force V= - Pa ๐ฟ Bending Moment M= Pa ๐ฟ (๐ฟ โ ๐ฅ) ๐ฅ=a (at C) VC= - ๐๐ ๐ฟ MC= Pa ๐ฟ (๐ฟ โ a) = ๐a๐ ๐ฟ ๐ฅ= L (at B) VB= - ๐๐ ๐ฟ MB= Pa ๐ฟ (๐ฟ โ L) = 0 ๐บ๐๐ฒ = ๐ RA- P-V =0 ๐บ๐ ๐โ๐ = ๐ -RA . ๐ฅ +P (๐ฅ -a) + M =0 M= RA . ๐ฅ - P (๐ฅ -a) M= Pa ๐ฟ . ๐ฅ - P (๐ฅ -a) M= Pa ๐ฟ (๐ฟ โ ๐ฅ)
• 9. 0< ๐ฅ<a Section1-1 Shear force Bending Moment ๐ฅ=0 (at A) VA= P๐ ๐ฟ MA= 0 ๐ฅ= a (at C) VC= P๐ ๐ฟ MC= Pa๐ ๐ฟ a< ๐ฅ<L Section2-2 Shear force Bending Moment ๐ฅ=a (at C) VC= - ๐๐ ๐ฟ MC = ๐a๐ ๐ฟ ๐ฅ= L (at B) VB= - ๐๐ ๐ฟ MB= 0
• 10. 0< ๐ฅ<a Section1-1 Shear force Bending Moment ๐ฅ=0 (at A) VA= P๐ ๐ฟ MA= 0 ๐ฅ= a (at C) VC= P๐ ๐ฟ MC= Pa๐ ๐ฟ a< ๐ฅ<L Section2-2 Shear force Bending Moment ๐ฅ=a (at C) VC= - ๐๐ ๐ฟ MC = ๐a๐ ๐ฟ ๐ฅ= L (at B) VB= - ๐๐ ๐ฟ MB= 0 Shear-force and bending moment diagrams for a simple beam with a concentrated load
• 11. Q2. Draw the shear force and bending moment diagram for a simple beam AB supporting a uniformly distributed load of intensity โqโ through out the length of the beam.
• 12. Free body diagram of given simple beam: Because the beam and its loading are symmetric, we see immediately that each of the reactions (RA and RB) is equal to qL/2.
• 13. Free body diagram of given simple beam:
• 14. Free body diagram of section1: 0< ๐ฅ < L when x=0 section coincides with A, when x=L section coincides with B 0< ๐ฅ< L Section1- 1 Shear force V = ๐.๐ณ ๐ - q. ๐ Bending Moment M= ๐.๐ณ.๐ฅ ๐ - ๐.๐ฅ2 2 ๐ฅ=0 (at A) VA= ๐.๐ณ ๐ MA= ๐.๐ณ.0 ๐ - ๐.02 2 = 0 ๐ฅ= L (at B) VB= ๐.๐ณ ๐ โ q๐ฟ = โ ๐.๐ณ ๐ MB= ๐.๐ณ.๐ฟ ๐ - ๐.๐ฟ2 2 = 0 ๐ฅ= L/2 (at C) VC= ๐.๐ณ ๐ โ ๐.๐ณ ๐ =0 MC= ๐.๐ณ. ๐ฟ 2 ๐ - ๐.( ๐ณ ๐ ) ๐ ๐ = ๐๐ฟ2 8 ๐บ๐๐ฒ = ๐ RA-q . ๐ฅ - V =0 V = RA- q . ๐ฅ V = ๐.๐ณ ๐ - q. ๐ ๐บ๐ ๐โ๐ = ๐ - RA . ๐ฅ + q . ๐ฅ. ๐ฅ 2 + M =0 M= RA . ๐ฅ - q . ๐ฅ. ๐ฅ 2 = ๐.๐ณ ๐ . ๐ฅ - q . ๐ฅ. ๐ฅ 2 = ๐.๐ณ.๐ฅ ๐ - ๐.๐ฅ2 2
• 15. 0< ๐ฅ< L Section1- 1 Shear force V = ๐.๐ณ ๐ - q. ๐ Bending Moment M= ๐.๐ณ.๐ฅ ๐ - ๐.๐ ๐ ๐ ๐ฅ=0 (at A) VA= ๐.๐ณ ๐ MA= 0 ๐ฅ= L (at B) VB=โ ๐.๐ณ ๐ MB= 0 ๐ฅ= L/2 (at C) VC=0 MC= ๐๐ฟ2 8
• 16. Q3.A simply supported beam of length 6 m, carries point load of 3 kN and 6 kN at distances of 2 m and 4 m from the left end. Draw the shear force and bending moment diagrams for the beam.
• 17. Estimate the reactions at supports: ๐ด๐ญ๐ = 0 ๏RA -3 - 6 + RB =0 ๏RA + RB = 9 โ (1) ๐ด๐ด๐จ = 0 ๏-3 * 2 โ 6 * 4 + RB * 6 =0 ๏ RB * 6 = 3 * 2 + 6 * 4 ๏RB = 5 kN โ (2 ) Substitute 2 in (1) ๏RA = 9 - RB ๏RA = 9 โ 5 ๏RA = 4 kN
• 19. 4 kN
• 20. Free body diagram of section1: 0< ๐ฅ< 2 m when x=0 section coincides with A, when x=a section coincides with C 0< ๐ฅ<2 Section1-1 Shear force (๐๐) V= 4 Bending Moment (kN-m) M= 4 * ๐ฅ ๐ฅ=0 (at A) VA= 4 MA= 4 โ 0= 0 ๐ฅ= 2 m (at C) VC=4 MC=4 * 2= 8 ๐บ๐๐ฒ = ๐ RA-V =0 V =RA = 4 kN ๐บ๐ ๐โ๐ = ๐ - RA * ๐ฅ + M =0 M= RA * ๐ฅ = 4. ๐ฅ
• 21. 4 kN
• 22. Free body diagram of section2: when x=2 m section coincides with C, when x=4 m section coincides with D 2< ๐ฅ< 4 Section2-2 Shear force V= 1 kN Bending Moment(kN-m) M= ๐ฅ + 6 ๐ฅ=2 (at C) VC= 1 MC= 2+ 6= 8 ๐ฅ= 4 (at D) VD= 1 MD=4 + 6 = 10 ๐บ๐๐ฒ = ๐ RA- 3 -V =0 ๏V = RA- 3 ๏V = 4- 3 = 1kN ๐บ๐ ๐โ๐ = ๐ -RA . ๐ฅ +3 * (๐ฅ -2) + M =0 M= 4 . ๐ฅ โ 3 * (๐ฅ -2) M= 4 . ๐ฅ โ 3 * ๐ฅ + 6 M= ๐ฅ + 6 2 m < ๐ฅ< 4 m
• 24. ๐บ๐๐ฒ = ๐ Free body diagram of section 3: 4 m < ๐ฅ< 6 m when x=4 m section coincides with D, when x=6 m section coincides with B RA โ 3 โ 6 โ V =0 ๏V = RA- 9 ๏V = 4- 9 = - 5 kN ๐บ๐ ๐โ๐ = ๐ -RA . ๐ฅ +3 * (๐ฅ -2) + 6 * (๐ฅ -4) + M =0 M= 4 . ๐ฅ โ 3 * (๐ฅ -2) โ 6 * (๐ฅ -4) M= - 5 * ๐ฅ + 30 4 < ๐ฅ< 6 Section 3-3 Shear force V= -5 kN Bending Moment(kN-m) M= - 5 * ๐ฅ + 30 ๐ฅ=4 (at D) VD= -5 MD= - 5 * 4 + 30= 10 ๐ฅ=6 (at B) VB= -5 MB=- 5 * 6 + 30 = 0
• 25. 0< ๐ฅ<2 Section1-1 Shear force (๐๐) V= 4 Bending Moment M= 4 * ๐ฅ ๐ฅ=0 (at A) VA= 4 MA= 0 ๐ฅ= 2 m (at C) VC=4 MC= 8 2< ๐ฅ< 4 Section2-2 Shear force V= 1 kN Bending Moment M= ๐ฅ + 6 ๐ฅ=2 (at C) VC= 1 MC= 8 ๐ฅ= 4 (at D) VD= 1 MD= 10 4 < ๐ฅ< 6 Section 3-3 Shear force V= -5 kN Bending Moment M= - 5 * ๐ฅ + 30 ๐ฅ=4 (at D) VD= -5 MD=10 ๐ฅ=6 (at B) VB= -5 MB= 0 4 kN 5 kN
• 26. Q5. Draw the shear force and bending moment diagram for a simply supported beam of length 9 m and carrying a uniformly distributed load of 10 kN/m for a distance of 6 m from the left end. Also calculate the maximum B.M. on the section.
• 28. Estimate the reactions at supports: ๐ด๐ญ๐ = 0 ๏RA - 10 * 6 + RB =0 ๏RA + RB = 60 โ (1) ๐ด๐ด๐จ = 0 ๏- 10 *6 * 3 + RB * 9 =0 ๏ RB * 9 = 180 ๏RB = 20 kN โ (2 ) Substitute 2 in (1) ๏RA = 60 - RB ๏RA = 60 โ 20 ๏RA = 40 kN
• 30. Free body diagram of section1: 0< ๐ฅ< 6 m when x=0 section coincides with A, when x=6 section coincides with C 0< ๐ฅ<2 Section1-1 Shear force (๐๐) V=40 -10 *๐ฅ Bending Moment (kN-m) M= 40*๐ฅ - 10 * ๐ฅ2 2 ๐ฅ=0 (at A) VA=40 -10 *0 = 40 MA= 40*0 - 10 * 02 2 = 0 ๐ฅ= 6 m (at C) VC=40 โ10 โ6 = โ20 MC=40*6 - 10 * 62 2 = 60 ๐ฅ= 4 m (at D) V=0 B.M is maximum MD=40*4 - 10 * 42 2 = 80 ๐บ๐๐ฒ = ๐ RA-10 *๐ฅ - V =0 V = RA-10 *๐ฅ V = 40 -10 *๐ฅ ๐บ๐ ๐โ๐ = ๐ - RA * ๐ฅ + 10. * ๐ฅ. ๐ฅ 2 + M =0 M= 40 * ๐ฅ - 10* ๐ฅ โ ๐ฅ 2 = 40*๐ฅ - 10 * ๐ฅ2 2 40 kN V = 0 40 -10 *๐ฅ = 0 10 *๐ฅ = 40 ๐ฅ = 4 m
• 31. 20 kN
• 32. Free body diagram of section2: 0< ๐ฅ< 3 m when x=0 section coincides with B, when x=3 section coincides with C 0< ๐ฅ< 3 Section2-2 Shear force (๐๐) V=-20 Bending Moment (kN-m) M= 20 * ๐ฅ ๐ฅ=0 (at B) VB=-20 MB= 20*0 = 0 ๐ฅ= 3 m (at C) VC=โ20 MC=20*3 = 60 ๐บ๐๐ฒ = ๐ RB + V =0 V = - RB V = -20 kN ๐บ๐ ๐โ๐ = ๐ โ M + RB * ๐ฅ =0 M= 20 * ๐ฅ
• 33. 0< ๐ฅ<2 Section1-1 Shear force (๐๐) V=40 -10 *๐ฅ Bending Moment (kN-m) M= 40*๐ฅ - 10 * ๐ฅ2 2 ๐ฅ=0 (at A) VA=40 MA= 0 ๐ฅ= 6 m (at C) VC=โ20 MC= 60 ๐ฅ= 4 m (at D) V=0 B.M is maximum MD= 80 0< ๐ฅ< 3 Section2-2 Shear force (๐๐) V=-20 Bending Moment (kN-m) M= 20 * ๐ฅ ๐ฅ=0 (at B) VB= -20 MB= 0 ๐ฅ= 3 m (at C) VC=โ20 MC= 60
• 34. Q6. Draw the shear force and B.M. diagrams for a simply supported beam of length 8 m and carrying a uniformly distributed load of10 kN/m for a distance of 4 m as shown in Fig.
• 35. Q7. A beam AB of length L simply supported at the ends A and B and carrying a uniformly varying load from zero at end A to โqโ per unit length at B. Draw shear force and bending moment diagram.
• 36. Estimate the reactions at supports: ๐ด๐ญ๐ = 0 ๏RA - ๐ ๐ ๐๐ณ+ RB =0 ๏RA + RB = ๐ช๐ณ ๐ โ (1) ๐ด๐ด๐จ = 0 ๏- 1 2 ๐๐ฟ ( 2๐ฟ 3 ) + RB * L =0 ๏ RB * L = ( ๐ช๐ณ ๐ ๐ ) ๏RB = ๐ช๐ณ ๐ โ (2 ) Substitute 2 in (1) ๏RA = ๐ช๐ณ ๐ - RB ๏RA = ๐ช๐ณ ๐ - ๐ช๐ณ ๐ ๏RA = ๐ช๐ณ ๐
• 37. ๐ ๐ณ = ๐ ๐ ๐ => ๐ ๐= ๐.๐ ๐ณ
• 39. Free body diagram of section1: 0< ๐ฅ < L when x=0 section coincides with A, when x=a section coincides with C 0< ๐ฅ< L Section1-1 Shear force V= ๐ช๐ณ ๐ โ ๐.๐ ๐ ๐๐ณ Bending Moment ๐ด = ๐ช๐ณ ๐ . ๐ฅ - ๐.๐ ๐ ๐๐ณ ๐ฅ=0 (at A) VA= ๐ช๐ณ ๐ MA= ๐ช๐ณ ๐ . 0 - ๐.๐ ๐ ๐๐ณ = 0 ๐ฅ= L (at B) VB= ๐ช๐ณ ๐ โ ๐.๐ณ ๐ ๐๐ณ = โ ๐.๐ณ ๐ MB= ๐ช๐ณ ๐ . ๐ฟ - ๐.๐ณ ๐ ๐๐ณ = 0 ๐บ๐๐ฒ = ๐ RA- 1 2 . ๐.๐ ๐ณ ๐ฅ - V =0 V = RA- 1 2 . ๐.๐ ๐ณ ๐ฅ V = ๐ช๐ณ ๐ โ ๐.๐ ๐ ๐๐ณ ๐บ๐ ๐โ๐ = ๐ - ๐ช๐ณ ๐ . ๐ฅ + 1 2 . ๐.๐ ๐ณ ๐ฅ. ๐ฅ 3 + M =0 M= ๐ช๐ณ ๐ . ๐ฅ - ๐.๐ ๐ ๐๐ณ
• 40. Free body diagram of section1: 0< ๐ฅ < L when x=0 section coincides with A, when x=a section coincides with C 0< ๐ฅ< L Section1-1 Shear force V= ๐ช๐ณ ๐ โ ๐.๐ ๐ ๐๐ณ Bending Moment ๐ด = ๐ช๐ณ ๐ . ๐ฅ - ๐.๐ ๐ ๐๐ณ ๐ฅ=0 (at A) VA= ๐ช๐ณ ๐ MA= ๐ช๐ณ ๐ . 0 - ๐.๐ ๐ ๐๐ณ = 0 ๐ฅ= L (at B) VB= ๐ช๐ณ ๐ โ ๐.๐ณ ๐ ๐๐ณ = โ ๐.๐ณ ๐ MB= ๐ช๐ณ ๐ . ๐ฟ - ๐.๐ณ ๐ ๐๐ณ = 0 ๐ฅ= L/ 3 (at C) VC=0 MC= ๐ช๐ณ ๐ . ๐ณ ๐ - ๐.( ๐ณ ๐ ) ๐ ๐๐ณ = ๐๐ณ ๐ ๐ ๐ ๐บ๐๐ฒ = ๐ RA- 1 2 . ๐.๐ ๐ณ ๐ฅ - V =0 V = RA- 1 2 . ๐.๐ ๐ณ ๐ฅ V = ๐ช๐ณ ๐ โ ๐.๐ ๐ ๐๐ณ ๐บ๐ ๐โ๐ = ๐ - ๐ช๐ณ ๐ . ๐ฅ + 1 2 . ๐.๐ ๐ณ ๐ฅ. ๐ฅ 3 + M =0 M= ๐ช๐ณ ๐ . ๐ฅ - ๐.๐ ๐ ๐๐ณ
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