Shear and Bending moments

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Transcript of Shear and Bending moments

1. 8.2 Shear and Bending-Moment Diagrams: Equation Form 2. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 1 of 6 3 ft 5 ft 7 ft x A B 9 kip 6 kip 7 ft Fy = 0: RA 9 kip 6 kip + RB = 0 MA = 0: (9 kip)(3 ft) (6 kip)(3 ft + 5 ft) + RB(3 ft + 5 ft + 7 ft) = 0 Solving gives RA = 10 kip and RB = 5 kip 5 ft 9 kip Draw a free-body diagram and find the reactions. + 3 ftRA + 1 A 6 kip RB 1. Express the shear V and bending moment M as functions of x, the distance from the left end of the beam to an arbitrary point on the beam. Plot V and M vs. x. 3. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 2 of 6 x 9 kip RA = 10 kip A 6 kip RB = 5 kip B Pass a section through the beam at a point between the left end and the 9-kip force.2 7 ft3 ft 5 ft 0 < x < 3 ft Draw a free-body diagram of the portion of the beam to the left of the section and find V and M at the section. 3 RA = 10 kip A x M (sign convention: M positive counterclockwise, on right end of section) V (sign convention: V positive down, on right end of section) Fy = 0: 10 kip V = 0 Mx = 0: (10 kip)x + M = 0 Solving gives V = 10 kip and M = 10x kipft valid for 0 < x < 3 ft. ++ 4. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 3 of 6 x 9 kip RA = 10 kip A 6 kip RB = 5 kip B Pass a section through the beam at a point between the 9-kip force and the 6-kip force. 7 ft3 ft 5 ft 3 ft < x < 8 ft Draw a free-body diagram of the portion of the beam to the left of the section and find V and M at the section. RA = 10 kip A V x M Fy = 0: 10 kip 9 kip V = 0 Mx = 0: (10 kip)x + (9 kip)(x 3 ft) + M = 0 Solving gives V = 1 kip (3) M = (x + 27) kipft (4) valid for 3 ft < x < 8 ft. ++ 4 3 ft (x 3 ft) 5 9 kip 5. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 4 of 6 x 9 kip RA = 10 kip A 6 kip RB = 5 kip B Pass a section through the beam at a point between the 6-kip force and the right end of the beam. 7 ft3 ft 5 ft 8 ft < x < 15 ft Draw a free-body diagram of the portion of the beam to the left of the section and find V and M at the section. RA = 10 kip A V x M Fy = 0: 10 kip 9 kip 6 kip V = 0 Mx = 0: (10 kip)x + (9 kip)(x 3 ft) + (6 kip)(x 8 ft) + M = 0 Solving gives V = 5 kip (5) M = (5x + 75) kipft (6) valid for 8 ft < x < 15 ft. ++ 3 ft (x 8 ft) 6 7 5 ft (x 3 ft) 9 kip 6 kip 6. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 5 of 6 Collect the results from Eqs. 1-6: 0 < x < 3 ft V = 10 kip M = 10x kipft 3 ft < x < 8 ft V = 1 kip M = (x + 27) kipft 8 ft < x < 15 ft V = 5 kip M = (5x + 75) kipft 8 Ans. 7. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 1, page 6 of 6 6 kip9 kip B RB = 5 kipRA = 10 kip A 7 ft3 ft 5 ft Plot V and M versus x.9 V (kip) M (kipft) 10 1 x 30 35 x 5 8. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 2, page 1 of 3 14 m x A Fy = 0: RA 28 N = 0 MA = 0: MA (28 N)(7 m) + 20 Nm = 0 Solving gives RA = 28 N and MA = 176 Nm Draw a free-body diagram and find the reactions. ++ 1 2 N/m 20 Nm 14 m A 20 Nm 2 N/m RA MA = 7 m 14 m 2 Resultant = (2 N/m)(14 m) = 28 N A couple-moment reaction must always be included at a built-in end of a beam. 2 2. Express the shear V and bending moment M as functions of x, the distance from the left end of the beam to an arbitrary point on the beam. Plot V and M vs. x. 9. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 2, page 2 of 3 2 N/m Resultant = (2 N/m)(x) Pass a section through the beam at an arbitrary point (located by x) x 3 0 < x < 14 m MA = 176 Nm RA = 28 N Fy = 0: 28 N 2x V = 0 Mx = 0: 176 Nm 28x + ( )(2 N/m)(x) + M = 0 Solving gives V = (2x + 28) N Ans. M = (x2 + 28x 176) Nm Ans. valid for 0 < x < 14 m. Draw a free-body diagram of the portion of the beam to the left of the section and find V and M at the section. ++ x 4 MA = 176 Nm RA = 28 N V M 20 NmA A x 2 2 x 14 m 10. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 2, page 3 of 3 14 m 2 N/m Plot V and M versus x. MA = 176 Nm RA = 28 N 5 M (Nm) V (N) x x 28 20 176 20 NmA 11. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 1 of 6 Fy = 0: RA 100 lb + RB = 0 MA = 0: (100 lb)(7 ft + 5 ft) + RB(7 ft + 10 ft + 3 ft) = 0 Solving gives RA = 40 lb and RB = 60 lb Draw a free-body diagram and find the reactions. ++ 1 = 5 ft 10 ft 2 Resultant = (10 lb/ft)(10 ft) = 100 lb x A B 10 lb/ft A B RA RB 7 ft 3 ft 3 ft10 ft7 ft 10 ft 3. Express the shear V and bending moment M as functions of x, the distance from the left end of the beam to an arbitrary point on the beam. Plot V and M vs. x. 12. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 2 of 6 Draw a free-body diagram and find the reactions. 7 ft 3 ft A B 10 ft 10 lb/ft x 2 Pass a section through the beam at a point between the left end of the beam and the beginning of the distributed load. x RA = 40 lb Fy = 0: 40 lb V = 0 Mx = 0: (40 lb)x + M = 0 Solving gives V = 40 lb (1) M = (40x) lbft (2) valid for 0 < x < 7 ft. V + M + 3 RA = 40 lb RB = 60 lb 0 < x < 7 ft 13. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 3 of 6 Draw a free-body diagram of the portion of the beam to the left of the section and solve for V and M at the section. x RA = 40 lb Fy = 0: 40 lb (10 lb/ft)(x 7 ft) V = 0 Mx = 0: (40 lb)x + [(10 lb/ft)(x 7 ft)] ( ) + M = 0 Solving gives V = (10x + 110) lb (3) M= (5x2 + 110x 245) lbft (4) valid for 7 ft < x < 17 ft. V + M + Pass a section through the beam at a point between the beginning and end of the distributed load. 4 10 ft7 ft A RA = 40 lb x B 3 ft RB = 60 lb 10 lb/ft 5 2 x 7 ft Resultant = (10 lb/ft)(x 7 ft) 7 ft < x < 17 ft 7 ft x 7 ft 2 x 7 ft 14. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 4 of 6 Draw a free-body diagram of the portion of the beam to the left of the section and solve for V and M at the section. x RA = 40 lb V M Pass a section through the beam at a point between the right end of the distributed load and the right end of the beam. 10 ft7 ft A RA = 40 lb x B 3 ft RB = 60 lb 10 lb/ft 17 ft < x < 20 ft 7 ft x 17 ft 6 10 ft = 5 ft 2 10 ft Resultant = (10 lb/ft)(10 ft) = 100 lb 7 15. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 5 of 6 Fy = 0: 40 lb 100 lb V = 0 MA = 0: (40 lb)x + (100 lb)[(x 17 ft) + 5 ft] + M = 0 Solving gives V = 60 lb (5) M = (60x + 1200) lbft (6) valid for 17 ft < x < 20 ft. ++ 9 Collect the results from Eqs. 1-6: 0 < x < 7 ft V = 40 lb M = 40x lbft 7 ft < x < 17 ft V = (10x + 110) lb M = (5x2 + 110x 245) lbft 17 ft < x < 20 ft V = 60 lb M = (60x + 1200) lbft Ans. 8 16. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 3, page 6 of 6 7 ft 3 ft A B 10 ft 10 lb/ft Plot V and M versus x. RA = 40 lb RB = 60 lb 10 V (lb) M (lbft) 60 40 x x 280 360 180 40 60 17. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 1 of 6 4. Express the shear V and bending moment M as functions of x, the distance from the left end of the beam to an arbitrary point on the beam. Plot V and M vs. x. 3 ft 5 ft 7 ft x A B 7 ft Fy = 0: RA + RB = 0 MA = 0: RB(3 ft + 5 ft + 7 ft) 27 kipft 18 kipft = 0 Solving gives RA = 3 kip = 3 kip RB = 3 kip 5 ft Draw a free-body diagram and find the reactions. + 3 ftRA + 1 A RB 27 kipft 18 kipft 27 kipft 18 kipft 18. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 2 of 6 x RA = 3 kip A RB = 3 kip B Pass a section through the beam at a point between the left end and the 27 kipft moment couple. 2 0 < x < 3 ft Draw a free-body diagram of the portion of the beam to the left of the section and find V and M at the section. 3 A V x M Fy = 0: 3 kip V = 0 Mx = 0: (3 kip)x + M = 0 Solving gives V = 3 kip (1) M = 3x kipft (2) valid for 0 < x < 3 ft. ++ 27 kipft 18 kipft RA = 3 kip 7 ft5 ft3 ft 19. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 3 of 6 x RA = 3 kip A RB = 3 kip B Pass a section through the beam at a point between the 27 kipft and 18 kipft moment couples. 3 ft < x < 8 ft Draw a free-body diagram of the portion of the beam to the left of the section and find V and M at the section. A V x M Fy = 0: 3 kip V = 0 Mx = 0: (3 kip)x 27 kipft + M = 0 Solving gives V = 3 kip (3) M = (3x + 27) kipft (4) valid for 3 ft < x < 8 ft. ++ 27 kipft 18 kipft RA = 3 kip 27 kipft 4 5 3 ft x 3 ft 3 ft 7 ft5 ft 20. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 4 of 6 x RA = 3 kip A RB = 3 kip B Pass a section through the beam at a point between the 18 kipft moment couple and the right end of the beam. 7 ft3 ft 5 ft 3 ft < x < 8 ft Draw a free-body diagram of the portion of the beam to the left of the section and find V and M at the section. A V x M Fy = 0: 3 kip V = 0 Mx = 0: (3 kip)x 27 kipft 18 kipft + M = 0 Solving gives V = 3 kip (5) M = (3x + 45) kipft (6) valid for 8 ft < x < 15 ft.++ 27 kipft 18 kipft RA = 3 kip 27 kipft 3 ft 6 7 18 kipft 5 ft 21. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 5 of 6 8 Collect the results from Eqs. 1-6: 0 < x < 3 ft V = 3 kip M = 3x kipft 3 ft < x < 8 ft V = 3 kip M = (3x + 27) kipft 8 ft < x < 15 ft V = 3 kip M = (3x + 45) kipft Ans. 22. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 4, page 6 of 6 Plot V and M versus x.9 V (kip) M (kipft) 3 x x 18 21 3 18 kipft 7 ft3 ft A RA = 3 kip 5 ft 27 kipft RB = 3 kip B 9 3 23. 8.2 Shear and Bending-Moment Diagrams: Equation Form Example 5, page 1 of 6 5. Express the shear V and bending moment M in the horizontal portion ACDB of the beam as functions of x, the distance from the left end of the beam to an arbitrary point on the beam. Plot V and M versus x. x A B 4 kip Fy = 0: RA 4 kip 4 kip + RB = 0 MA = 0: (4 kip)(2 ft