Problems on simply supported beams

Problems on simply supported beams, concentrated load, uniformly distributed load

1. Problems on Simply supported Beam Mrs. Venkata Sushma Chinta

2. Q1. 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.

3. Estimate the reactions at supports: = 0 R A -3 - 6 + R B =0 R A + R B = 9     = 0 -3 * 2 – 6 * 4 + R B * 6 =0 R B * 6 = 3 * 2 + 6 * 4 R B = 5 kN       R A = 9 - R B R A = 9 – 5 R A = 4 kN

4. Free body diagram of section1: 0< < 2 m when x=0 section coincides with A, when x=a section coincides with C   0< <2 Section1-1 Shear force ( ) V= Bending Moment (kN-m) M= 4 * =0 (at A) V A = M A = 0= 0 = 2 m (at C) V C = M C =4 * 2 =   R A -V =0 V =R A = 4 kN   R A * + M =0 M= R A * = 4.  

5. Free body diagram of section2: when x=2 m section coincides with C, when x=4 m section coincides with D 2< < 4 Section2-2 Shear force V= 1 kN Bending Moment(kN-m) M= + 6 =2 (at C) V C = 1 M C = 2+ 6= 8 = 4 (at D) V D = 1 M D =4 + 6 = 10 Shear force V= 1 kN V C = 1 M C = 2+ 6= 8 V D = 1 M D =4 + 6 = 10   R A - 3 -V =0 V = R A - 3 V = 4- 3 = 1kN   -R A . +3 * ( -2) + M =0 M= 4 . – 3 * ( -2) M= 4 . – 3 * + 6 M= + 6   2 m < < 4 m  

6.   Free body diagram of section 3: 4 m < < 6 m when x=4 m section coincides with D, when x=6 m section coincides with B   R A – 3 – 6 – V =0 V = R A - 9 V = 4- 9 = - 5 kN   -R A . +3 * ( -2) + 6 * ( -4) + M =0 M= 4 . – 3 * ( -2) – 6 * ( -4) M= - 5 * + 30   4 < < 6 Section 3-3 Shear force V= -5 kN Bending Moment(kN-m) M= - 5 * + 30 =4 (at D) V D = -5 M D = - 5 * + 30= 10 =6 (at B) V B = -5 M B = - 5 * 6 + 30 = 0 Shear force V= -5 kN V D = -5 V B = -5 M B = - 5 * 6 + 30 = 0

7.   Free body diagram of section 3: 0 m < < 2m when x=4 m section coincides with D, when x=6 m section coincides with B   R B +V =0 V = - R B V =- 5 kN   R B . - M =0 M= R B . M= 5 *   4 < < 6 Section 3-3 Shear force V= -5 kN Bending Moment(kN-m) M= 5 * =2 (at D) V D = -5 M D = 5 * = 10 =0 (at B) V B = -5 M B = 5 * 0= 0 Shear force V= -5 kN V D = -5 V B = -5 M B = 5 * 0= 0 =5k N

8. 0< <2 Section1-1 Shear force ( ) V= Bending Moment M= 4 * =0 (at A) V A = M A = 0 = 2 m (at C) V C = M C = M A = 0 2< < 4 Section2-2 Shear force V= 1 kN Bending Moment M= + 6 =2 (at C) V C = 1 M C = 8 = 4 (at D) V D = 1 M D = 10 Shear force V= 1 kN V C = 1 M C = 8 V D = 1 M D = 10 4 < < 6 Section 3-3 Shear force V= -5 kN Bending Moment M= - 5 * + 30 =4 (at D) V D = -5 M D = 10 =6 (at B) V B = -5 M B = 0 Shear force V= -5 kN V D = -5 M D = 10 V B = -5 M B = 0 4 kN 5 kN

9. Q2. 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 .

10. Estimate the reactions at supports: = 0 R A - 10 * 6 + R B =0 R A + R B = 60     = 0 - 10 *6 * 3 + R B * 9 =0 R B * 9 = 180 R B = 20 kN       R A = 60 - R B R A = 60 – 20 R A = 40 kN

11. Free body diagram of section1: 0< < 6 m when x=0 section coincides with A, when x=6 section coincides with C   0< <2 Section1-1 Shear force ( ) V= 40 -10 * Bending Moment (kN-m) M= 40 * - 10 * =0 (at A) V A =40 -10 *0 = 40 M A = 40 * - 10 * = 0 = 6 m (at C) V C = M C = 40 * - 10 * = 60 = 4 m (at D) V=0 B.M is maximum M D = 40 * - 10 * = 80 V A =40 -10 *0 = 40 V=0   R A -10 * - V =0 V = R A -10 * V = 40 -10 *     R A * + 10. * . + M =0 M= 40 * - 10* = 40 * - 10 *   40 kN V = 0 40 -10 * 10 * = 40 = 4 m  

12. Free body diagram of section2: 0< < 3 m when x=0 section coincides with B, when x=3 section coincides with C   0< < 3 Section2-2 Shear force ( ) V= -20 Bending Moment (kN-m) M= 20 * =0 (at B) V B =-20 M B = 20 * = 0 = 3 m (at C) V C = M C = 20 *3 = 60 V B =-20 M C = 20 *3 = 60   R B + V =0 V = - R B V = -20 kN   M + R B * =0 M= 20 *  

13. 0< <2 Section1-1 Shear force ( ) V= 40 -10 * Bending Moment (kN-m) M= 40 * - 10 * =0 (at A) V A = 40 M A = 0 = 6 m (at C) V C = M C = 60 = 4 m (at D) V=0 B.M is maximum M D = 80 V A = 40 M A = 0 M C = 60 V=0 B.M is maximum M D = 80 0< < 3 Section2-2 Shear force ( ) V= -20 Bending Moment (kN-m) M= 20 * =0 (at B) V B = -20 M B = = 3 m (at C) V C = M C = 60 V B = -20 M B = M C = 60

14. Q3. 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.