CHAPTER 8 BENDING MOMENT AND SHEAR FORCE DIAGRAMS EXERCISE 51, Page 121 1. 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: Resolving vertically gives: RR12+= 3 But as the beam is symmetrically loaded, RR12= Hence, R12= R = 1.5kN Bending moment expressions: Range AC At x, M = Rx1 = 1.5 x i.e. a straight line Range CB At x, M = R1 x−− 3(x 1) = 1.5 x – 3x + 3 i.e. M = - 1.5 x + 3 i.e. a straight line Shearing Force expressions: Range AC SF = + 1.5 kN Range CB SF = + 1.5 – 3 = - 1.5 kN The bending moment and shearing force diagrams are shown below. 146 © John Bird & Carl Ross Published by Taylor and Francis 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: R1 ×=× 3 41 4 Hence, R= = 1.333kN 1 3 Resolving vertically gives: RR12+= 4 Hence, R2 = 4−=− R1 4 1.333 = 2.667 kN Bending moment expressions: Range AC At x, M = Rx1 = 1.333 x i.e. a straight line Range CB At x, M = R1 x−− 4(x 2) = 1.333 x – 4x + 8 147 © John Bird & Carl Ross Published by Taylor and Francis i.e. M = - 2.667 x + 8 i.e. a straight line Shearing Force expressions: Range AC SF = R1 = + 1.333 kN Range CB SF = R1 - 4 = + 1.5 – 4 = - 2.667 kN The bending moment and shearing force diagrams are shown below. 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: R1 ×=×+× 3 12 41 6 Hence, R= = 2kN 1 3 Resolving vertically gives: R12+=+ R 14 Hence, R2 = 52− = 3 kN 148 © John Bird & Carl Ross Published by Taylor and Francis Bending moment expressions: Range AC At x, M = Rx1 = 2 x i.e. a straight line Range CB At x, M = R1 x−− 1(x 1) = 2 x – x + 1 i.e. M = x + 1 i.e. a straight line Range DB At x, M = R1 x− 1(x −− 1) 4(x − 2) = 2 x – x + 1 – 4x + 8 i.e. M = - 3x + 9 i.e. a straight line Shearing Force expressions: Range AC SF = R1 = + 2 kN Range CD SF = R1 - 1 = 2 – 1 = 1 kN Range DB SF = R1 - 1 - 4 = 2 – 1 – 4 = - 3 kN The bending moment and shearing force diagrams are shown below. 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. 149 © John Bird & Carl Ross Published by Taylor and Francis Taking moments about D gives: R1 ×+×=× 24014 4 Hence, R= = 2kN 1 2 Resolving vertically gives: R12+=+ R 14 Hence, R2 = 52− = 3 kN Bending moment expressions: Range CA At x, M = −1x = - x i.e. a straight line Range AB At x, M = −+1x R1 (x − 2) = -1 x + 2 x - 4 i.e. M = x - 4 i.e. a straight line Range DB At x, M = −+1x R12 (x −+ 2) R (x −− 4) 4(x − 4) = -1 x + 2x - 4 + 3x - 12 – 4x + 16 i.e. M = 0 Shearing Force expressions: Range CA SF = - 1 kN Range AB SF = - 1 + R1 = - 1 + 2 = + 1 kN Range DB SF = - 1 + 2 – 4 + 3 = 0 kN The bending moment and shearing force diagrams are shown below. 150 © John Bird & Carl Ross Published by Taylor and Francis 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: R1 ×+ 36 =× 42 86− Hence, R= = 0.667kN 1 3 Resolving vertically gives: RR12+= 4 Hence, R2 = 4− 0.667 = 3.333 kN Bending moment expressions: Range AC At x, M = Rx1 = 0.667 x i.e. a straight line Range CD At x, M = R1 x−− 4(x 1) = 0.667 x - 4 x + 4 i.e. M = - 3.333 x + 4 i.e. a straight line 151 © John Bird & Carl Ross Published by Taylor and Francis Range DB At x, M = R1 x− 4(x −+ 1) 6 = 0.667 x - 4x + 4 + 6 i.e. M = - 3.333 x + 10 i.e. a straight line Shearing Force expressions: Range AC SF = R1 = + 0.667 kN Range CD SF = R1 - 4 = 0.667 - 4 = - 3.333 kN Range DB SF = R1 – 4 = 0.667 – 4 = - 3.333 kN The bending moment and shearing force diagrams are shown below. 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: Range AC 152 © John Bird & Carl Ross Published by Taylor and Francis At x, M = - 4 x i.e. a straight line Range CB At x, M = - 4 x – 6(x – 2) = - 4 x - 6 x + 12 i.e. M = - 10 x + 12 i.e. a straight line Shearing Force expressions: Range AC SF = - 4 kN Range CB SF = - 4 - 6 = - 10 kN The bending moment and shearing force diagrams are shown below. 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: Range AC At x, M = - 2 x i.e. a straight line Range CB At x, M = - 2 x + 6 i.e. a straight line 153 © John Bird & Carl Ross Published by Taylor and Francis Shearing Force expressions: Range AC SF = - 2 kN Range CB SF = - 2 kN The bending moment and shearing force diagrams are shown below. 8. A 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 2 m from the left end, the second load is of magnitude 2 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 R A and R B : Taking moments about B in the diagram below gives: Clockwise moment = anticlockwise moment i.e. R A × 7 = 3 × 5 + 2 × 3 + 4 × 1 i.e. R A × 7 = 25 25 from which, R = = 3.57 kN A 7 Resolving vertically, R A + R B = 3 + 2 + 4 = 9 from which, RB = 9 – 3.57 = 5.43 kN 154 © John Bird & Carl Ross Published by Taylor and Francis Range AC At x, M = R A × x = 3.57 x i.e. a straight line (1) SF = + R A = 3.57 kN (2) Range CD At x, M = R A × x – 3(x – 2) = 3.57 x – 3 x + 6 i.e. M = 0.57 x + 6 i.e. a straight line (3) SF = R A - 3 = 3.57 – 3 = 0.57 kN (4) Range DE At x, M = R A × x – 3(x – 2) – 2(x – 4) = 3.57 x – 3 x + 6 – 2x + 8 i.e. M = - 1.43 x + 14 i.e. a straight line (5) SF = R A - 3 – 2 = 3.57 – 5 = - 1.43 kN (6) Range EB At x, M = R A × x – 3(x – 2) – 2(x – 4) - 4(x – 6) 155 © John Bird & Carl Ross Published by Taylor and Francis = 3.57 x – 3 x + 6 – 2x + 8 – 4 x + 24 i.e. M = - 5.43 x + 38 i.e. a straight line (7) SF = R A - 3 – 2 – 4 = 3.57 – 5 – 4 = - 5.43 kN (8) Plotting equations (1) to (8) results in the bending moment and shearing force diagrams shown above. Summarising, Bending moments (kN m): 0, 7.14, 8.28, 5.42, 0 Shearing forces (kN): 3.57, 3.57/ 0.57, 0.57/ - 1.43, - 1.43/ - 5.43, - 5.43 156 © John Bird & Carl Ross Published by Taylor and Francis EXERCISE 52, Page 124 1. 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. kN Total load = 6× 7m = 42 kN m By inspection, RR12= 42 Hence, R= R = = 21kN 122 Bending moment expression: x2 At x, M = Rx−× 6 1 2 i.e. M = 21 x - 3 x2 i.e. a parabola Shearing Force expression: At x, SF = R1 − 6x i.e. SF = 21 – 6x i.e. a straight line The bending moment and shearing force diagrams are shown below. 157 © John Bird & Carl Ross Published by Taylor and Francis 2. 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. kN Total load = 5× 12m = 60 kN m By inspection, RR12= 60 Hence, R= R = = 30kN 122 Bending moment expression: x2 At x, M = Rx−× 5 1 2 i.e. M = 30 x – 2.5 x2 i.e. a parabola Shearing Force expression: At x, SF = R1 − 5x i.e.