CHAPTER 8 BENDING MOMENT AND SHEAR FORCE DIAGRAMS

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1 CHPTE 8 BENDING MOMENT ND SHE FOCE DIGMS EXECISE 5, Page. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. To calculate the reactions: esolving vertically gives: + = 3 But as the beam is symmetrically loaded, = Hence, = =.5kN Bending moment expressions: ange C t x, M = x =.5 x i.e. a straight line ange CB t x, M = x 3(x ) =.5 x 3x + 3 i.e. M = -.5 x + 3 i.e. a straight line Shearing Force expressions: ange C ange CB SF = +.5 kn SF = = -.5 kn 46

2 . Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. Taking moments about B gives: 3= 4 4 Hence, = =.333kN 3 esolving vertically gives: + = 4 Hence, = 4 = =.667 kn Bending moment expressions: ange C t x, M = x =.333 x i.e. a straight line ange CB t x, M = x 4(x ) =.333 x 4x

3 i.e. M = x + 8 i.e. a straight line Shearing Force expressions: ange C SF = = kn ange CB SF = - 4 = = kn 3. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. Taking moments about B gives: 3= Hence, = = kn 3 esolving vertically gives: + = + 4 Hence, = 5 = 3 kn 48

4 Bending moment expressions: ange C t x, M = x = x i.e. a straight line ange CB t x, M = x (x ) = x x + i.e. M = x + i.e. a straight line ange DB t x, M = x (x ) 4(x ) = x x + 4x + 8 i.e. M = - 3x + 9 i.e. a straight line Shearing Force expressions: ange C SF = = + kn ange CD SF = - = = kn ange DB SF = = 4 = - 3 kn 4. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. 49

5 Taking moments about D gives: + 4 0= 4 4 Hence, = = kn esolving vertically gives: + = + 4 Hence, = 5 = 3 kn Bending moment expressions: ange C t x, M = x = - x i.e. a straight line ange B t x, M = x + (x ) = - x + x - 4 i.e. M = x - 4 i.e. a straight line ange DB t x, M = x + (x ) + (x 4) 4(x 4) i.e. M = 0 Shearing Force expressions: = - x + x x - 4x + 6 ange C SF = - kn ange B SF = - + = - + = + kn ange DB SF = = 0 kn 50

5. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams.

6 5. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. Taking moments about B gives: 3+ 6= 4 Hence, 8 6 = = kn 3 esolving vertically gives: + = 4 Hence, = = kn Bending moment expressions: ange C t x, M = x = x i.e. a straight line ange CD t x, M = x 4(x ) = x - 4 x + 4 i.e. M = x + 4 i.e. a straight line 5

7 ange DB t x, M = x 4(x ) + 6 = x - 4x i.e. M = x + 0 i.e. a straight line Shearing Force expressions: ange C SF = = kn ange CD SF = - 4 = = kn ange DB SF = 4 = = kn 6. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. Bending moment expressions: ange C 5

8 t x, M = - 4 x i.e. a straight line ange CB t x, M = - 4 x 6(x ) = - 4 x - 6 x + i.e. M = - 0 x + i.e. a straight line Shearing Force expressions: ange C ange CB SF = - 4 kn SF = = - 0 kn 7. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. Bending moment expressions: ange C t x, M = - x i.e. a straight line ange CB t x, M = - x + 6 i.e. a straight line 53

9 Shearing Force expressions: ange C ange CB SF = - kn SF = - kn 8. horizontal beam of negligible mass is of length 7 m. The beam is simply-supported at its ends and carries three vertical loads, pointing in a downward direction. The first load is of magnitude 3 kn and acts m from the left end, the second load is of magnitude kn and acts 4 m from the left end, and the third load is of magnitude 4 kn and acts 6 m from the left end. Calculate the bending moment and shearing force at the points of discontinuity, working from the left support to the right support. To determine the reactions and B : Taking moments about B in the diagram below gives: Clockwise moment = anticlockwise moment i.e. i.e. from which, esolving vertically, from which, 7 = = 5 + = 5 7 = 3.57 kn B = = 9 B = = 5.43 kn 54

10 ange C t x, M = x = 3.57 x i.e. a straight line () ange CD SF = + = 3.57 kn () t x, M = x 3(x ) = 3.57 x 3 x + 6 i.e. M = 0.57 x + 6 i.e. a straight line (3) ange DE SF = - 3 = = 0.57 kn (4) t x, M = x 3(x ) (x 4) = 3.57 x 3 x + 6 x + 8 i.e. M = -.43 x + 4 i.e. a straight line (5) ange EB SF = - 3 = = -.43 kn (6) t x, M = x 3(x ) (x 4) - 4(x 6) 55

11 = 3.57 x 3 x + 6 x x + 4 i.e. M = x + 38 i.e. a straight line (7) SF = = = kn (8) Plotting equations () to (8) results in the bending moment and shearing force diagrams shown above. Summarising, Bending moments (kn m): 0, 7.4, 8.8, 5.4, 0 Shearing forces (kn): 3.57, 3.57/ 0.57, 0.57/ -.43, -.43/ ,

12 EXECISE 5, Page 4. Determine expressions for the bending moment and shearing force distributions for the following simply supported beams; hence, plot the bending moment and shearing force diagrams. Total load = kn 6 7m = 4 kn m By inspection, = 4 Hence, = = = kn Bending moment expression: t x, M = x x 6 i.e. M = x - 3 x i.e. a parabola Shearing Force expression: t x, SF = 6x i.e. SF = 6x i.e. a straight line 57

13 . Determine expressions for the bending moment and shearing force distributions for the following simply supported beams; hence, plot the bending moment and shearing force diagrams. Total load = kn 5 m = 60 kn m By inspection, = 60 Hence, = = = 30kN Bending moment expression: t x, M = x x 5 i.e. M = 30 x.5 x i.e. a parabola Shearing Force expression: t x, SF = 5x i.e. SF = 30 5x i.e. a straight line 58

14 3. Determine expressions for the bending moment and shearing force distributions for the following cantilevers; hence, or otherwise, plot the bending moment and shearing force diagrams. Bending moment expression: t x, M = x 6 i.e. M = 3 x i.e. a parabola Shearing Force expression: t x, SF = - 6 x i.e. a straight line 4. Determine expressions for the bending moment and shearing force distributions for the following cantilevers; hence, or otherwise, plot the bending moment and shearing force diagrams. 59

15 Bending moment expression: t x, M = x 5 i.e. M =.5 x i.e. a parabola Shearing Force expression: t x, SF = - 5 x i.e. a straight line EXECISE 53, Page 4 nswers found from within the text of the chapter, pages to 3. EXECISE 54, Page 4. (b). (c) 3. (c) 4. (a) 5. (c) 6. (b) 60

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