Problems on simply supported beams (udl , uvl and couple)

(udl , uvl and couple)

1. Problems on Simply supported beam subjected to UDL, UVL and couples By Mrs. Venkata Sushma Chinta

2. Q5. 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.

3. Q6. 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 ‘q’ per unit length at B. Draw shear force and bending moment diagram.

4. Estimate the reactions at supports: = 0 R A - + R B =0 R A + R B =     = 0 - ( ) + R B * L =0 R B * L = ( ) R B =       R A = - R B R A = - R A =  

5. => =  

6. Free body diagram of section1: 0< < L when x=0 section coincides with A, when x=L section coincides with B  

7. Free body diagram of section1: 0< < L when x=0 section coincides with A, when x=L section coincides with B   0< < L Section1-1 Shear force V= Bending Moment . - =0 (at A) V A = M A = . - = 0 = L (at B) V B = M B = . - = 0   R A - - V =0 V = R A - V =     . + . + M =0 M= . -  

8. Free body diagram of section1: 0< < L when x=0 section coincides with A, when x=L section coincides with B   0< < L Section1-1 Shear force V= Bending Moment . - =0 (at A) V A = M A = . - = 0 = L (at B) V B = M B = . - = 0 = L/ (at C) V C = M C = . - = V C = V =   M= . -  

9. 0< < L Section1-1 Shear force V= Bending Moment . - =0 (at A) V A = M A = = L (at B) V B M B = = L/ (at C) V C = M C = M A = M B = V C =

10. Q6. A simply supported beam of length 5 m carries a uniformly increasing load of 800 N/m run at one end to 1600 N/m run at the other end. Draw the S.F. and B.M. diagrams for the beam. Also calculate the position and magnitude of maximum bending moment

11. Estimate the reactions at supports: = 0 R A – 4000-2000+ R B =0 R A + R B =     = 0 - – 2000 ( )+ R B * 5 =0 R B * 5 =16666.7 R B =       R A = - R B R A = R A =   Then load on beam due to uniformly distributed load of 800 N/m = 800 x 5 = 4000 N Load on beam due to triangular loading = x 800x 5 =2000 N  

12. Consider any section 1-1at a-distance x from A. Rate of loading at the section 1-1 =Length CE =CD +DE = 800 + 160.   C E D D C E

13. 2666.7 N Then load on beam due to uniformly distributed load of 800 N/m = 800 . Load on beam due to triangular loading = 160 . . = 80. 2     R A -800 . - 80. 2 - V =0 V = 2666.7 -800 . - 80. 2     R A + 800 . . + 80. 2 . + M =0 M= R A - 800 . . - 80. 2 . M= - 400 . - 80.   Free body diagram of section1: 0< < 5m when x=0 section coincides with A, when x=5m section coincides with B  

14. 2666.7 N Free body diagram of section1: 0< < 5m when x=0 section coincides with A, when x=5m section coincides with B   0< < 5 Section1-1 Shear force V = 2666.7 -800 . - 80. 2 Bending Moment M= - 400 . - 80. =0 (at A) V A = = 2666.7 M A = *0 – 400* - 80* = 0 = 5 (at B) V B = 3333.3 M B = *5 – 400* - 80* = 0 = (at C) V C = M C = V C = M C =

15. 2666.7 N To find maximum bending moment make V=0 V = 2666.7 -800 . - 80. 2   M= - 400 . - 80.  

16. 2666.7 N To find maximum bending moment make V=0 0< <5 Section1-1 Shear force V = 2666.7 -800 . - 80. 2 Bending Moment M= - 400 . - 80. =0 (at A) V A = = 2667.7 M A = *0 – 400* - 80* = 0 = L (at B) V B = 3333.3 M B = *5 – 400* - 80* = 0 = 2.637 m (at C) V C = M C = *2.637 – 400* - 80* =3761.59 N-m V C = 2666.7 -800 . - 80. 2 =0 = 2.637m  

17. 0< < 5 Section 1-1 Shear force Bending Moment =0 (at A) V A = 2667.7 M A = = L (at B) V B = 3333.3 M B = = 2.637 m (at C) V C = M C = 3761.59 N-m Shear force Bending Moment V A = 2667.7 M A = M B = V C = M C = 3761.59 N-m

18. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR BEAMS SUBJECTED TO COUPLES Q7. A simply supported beam AB of length 6 m is hinged at A and B. It is subjected to a clockwise couple of24 kN-m at a distance of 2 m from the left end A. Draw the S.F. and B.M. diagrams.

19. Free body diagram of section1: 0< < 2 when x=0 section coincides with A, when x=2 section coincides with C   0< < 2 Section1-1 Shear force V = -4 Bending Moment M= -4 =0 (at A) V A = -4 M A =-4(0)=0 = 2 (at C) V C = -4 M C =-4(2)= -8 Shear force V = -4 V A = -4 M A =-4(0)=0 V C = -4 M C =-4(2)= -8

20. Free body diagram of section1: 2< < 6 when x=2 section coincides with C, when x=6 section coincides with B   2< < 6 Section2-2 Shear force V = -4 Bending Moment M= 24-4 =2 (at C) V C = -4 M C = 24-4 = 6 (at B) V B = -4 M B = 24-4 Shear force V = -4 V C = -4 V B = -4

21. 0< < 2 Section1-1 Shear force V = -4 Bending Moment M= -4 =0 (at A) V A = -4 M A =-4(0)=0 = 2 (at C) V C = -4 M C =-4(2)= -8 Shear force V = -4 V A = -4 M A =-4(0)=0 V C = -4 M C =-4(2)= -8 2< < 6 Section2-2 Shear force V = -4 Bending Moment M= 24-4 =2 (at C) V C = -4 M C = 24-4 = 6 (at B) V B = -4 M B =- 24-4 Shear force V = -4 V C = -4 V B = -4

22. Q8. A beam 10 m long and simply supported at each end, has a uniformly distributed load of 1000 N/m extending from the left end up to the center of the beam. There is also an anti-clockwise couple of 15000 Nm at a distance of 2.5 m from the right end. Draw the S.F. and B.M. diagrams

23. Q8. A beam 10 m long and simply supported at each end, has a uniformly distributed load of 1000 N/m extending from the left end up to the center of the beam. There is also an anti-clockwise couple of 15000 Nm at a distance of 2.5 m from the right end. Draw the S.F. and B.M. diagrams

24. 0< < 5 Section1-1 Shear force V =5250-1000 Bending Moment M= 5250 – 500 2 =0 (at A) V A =5250-1000 M A = 5250 – 500 2 =0 = 5 (at C) V C =5250-1000 M C = 5250 – 500 2 =13750

25. 2.5< < 5 Section2-2 Shear force V =250 Bending Moment M= 15000- 250 =2.5 (at D) V D =250 M D = 15000- 250 14735 = 5 (at C) V C =250 M C = 15000- 250 13750 Shear force V =250 V D =250 V C =250

26. 0< < 2.5 Section3-3 Shear force V =250 Bending Moment M= -250 =2.5 (at D) V D =250 M D =-250*2.5=-625 = 0 (at B) V B =250 M B =-250*0= 0 Shear force V =250 V D =250 M D =-250*2.5=-625 V B =250 M B =-250*0= 0

27. Section1-1 Shear force Bending Moment =0 (at A) V A =5250 M A = = 5 (at C) V C =250 M C =13750 Section1-1 Shear force Bending Moment V A =5250 M A = V C =250 M C =13750 Section2-2 Shear force Bending Moment =2.5 (at D) V D =250 M D 14735 = 5 (at C) V C =250 M C = 13750 Section2-2 Shear force Bending Moment V D =250 V C =250 M C = 13750 0< < 2.5 Section3-3 Shear force V =250 Bending Moment =2.5 (at D) V D =250 M D =-625 = 0 (at B) V B =250 M B = 0 Shear force V =250 Bending Moment V D =250 M D =-625 V B =250 M B = 0