7.3 Relations between Distributed Load, Shear and Moment Consider beam AD subjected to an arbitrary load w = w(x) and a series of concentrated forces and moments Distributed load assumed positive when loading acts downwards

7.3 Relations between Distributed Load, Shear and Moment A FBD diagram for a small segment of the beam having a length ∆x is chosen at point x along the beam which is not subjected to a concentrated force or couple moment Any results obtained will not apply at points of concentrated loadings

7.3 Relations between Distributed Load, Shear and Moment The internal shear force and bending moments shown on the FBD are assumed to act in the positive sense Both the shear force and moment acting on the right-hand face must be increased by a small, finite amount in order to keep the segment in equilibrium

7.3 Relations between Distributed Load, Shear and Moment The distributed loading has been replaced by a resultant force ∆F = w(x) ∆x that acts at a fractional distance k (∆x) from the right end, where 0 < k <1

7.3 Relations between Distributed Load, Shear and Moment Slope of the = Negative of shear diagram distributed load intensity Slope of = Shear moment diagram

7.3 Relations between Distributed Load, Shear and Moment At a specified point in a beam, the slope of the shear diagram is equal to the intensity of the distributed load Slope of the moment diagram = shear If the shear is equal to zero, dM/dx = 0, a point of zero shear corresponds to a point of maximum (or possibly minimum) moment w (x) dx and V dx represent differential area under the distributed loading and shear diagrams

7.3 Relations between Distributed Load, Shear and Moment Change in = Area under shear shear diagram moment shear diagram

7.3 Relations between Distributed Load, Shear and Moment Change in shear between points B and C is equal to the negative of the area under the distributed-loading curve between these points Change in moment between B and C is equal to the area under the shear diagram within region BC The equations so not apply at points where concentrated force or couple moment acts

7.3 Relations between Distributed Load, Shear and Moment Force FBD of a small segment of the beam Change in shear is negative thus the shear will jump downwards when F acts downwards on the beam

7.3 Relations between Distributed Load, Shear and Moment Force FBD of a small segment of the beam located at the couple moment Change in moment is positive or the moment diagram will jump upwards MO is clockwise

7.3 Relations between Distributed Load, Shear and Moment Example 7.9 Draw the shear and moment diagrams for the beam.

7.3 Relations between Distributed Load, Shear and Moment Solution Support Reactions FBD of the beam

7.3 Relations between Distributed Load, Shear and Moment Solution Shear Diagram V = +1000 at x = 0 V = 0 at x = 2 Since dV/dx = -w = -500, a straight negative sloping line connects the end points

7.3 Relations between Distributed Load, Shear and Moment Solution Moment Diagram M = -1000 at x = 0 M = 0 at x = 2 dM/dx = V, positive yet linearly decreasing from dM/dx = 1000 at x = 0 to dM/dx = 0 at x = 2

7.3 Relations between Distributed Load, Shear and Moment Example 7.10 Draw the shear and moment diagrams for the cantilevered beam.

7.3 Relations between Distributed Load, Shear and Moment Solution Support Reactions FBD of the beam

7.3 Relations between Distributed Load, Shear and Moment Solution At the ends of the beams, when x = 0, V = +1080 when x = 2, V = +600 Uniform load is downwards and slope of the shear diagram is constant dV/dx = -w = - 400 for 0 ≤ x ≤ 1.2 The above represents a change in shear

7.3 Relations between Distributed Load, Shear and Moment Solution Also, by Method of Sections, for equilibrium, Change in shear = area under the load diagram at x = 1.2, V = +600

7.3 Relations between Distributed Load, Shear and Moment Solution Since the load between 1.2 ≤ x ≤ 2, w = 0, slope dV/dx = 0, at x = 2, V = +600 Shear Diagram

7.3 Relations between Distributed Load, Shear and Moment Solution At the ends of the beams, when x = 0, M = -1588 when x = 2, M = -100 Each value of shear gives the slope of the moment diagram since dM/dx = V at x = 0, dM/dx = +1080 at x = 1.2, dM/dx = +600 For 0 ≤ x ≤ 1.2, values of the shear diagram are positive but linearly increasing

7.3 Relations between Distributed Load, Shear and Moment Solution Moment diagram is parabolic with a linearly decreasing positive slope Moment Diagram

7.3 Relations between Distributed Load, Shear and Moment Solution Magnitude of moment at x = 1.2 = -580 Trapezoidal area under the shear diagram = change in moment

7.3 Relations between Distributed Load, Shear and Moment Solution By Method of Sections, at x = 1.2, M = -580 Moment diagram has a constant slope for 1.2 ≤ x ≤ 2 since dM/dx = V = +600 Hence, at x = 2, M = -100

7.3 Relations between Distributed Load, Shear and Moment Example 7.11 Draw the shear and moment diagrams for the shaft. The support at A is a thrust bearing and the support at B is a journal bearing.

7.3 Relations between Distributed Load, Shear and Moment Solution Support Reactions FBD of the supports

7.3 Relations between Distributed Load, Shear and Moment Solution At the ends of the beams, when x = 0, V = +3.5 when x = 8, V = -3.5 Shear Diagram

7.3 Relations between Distributed Load, Shear and Moment Solution No distributed load on the shaft, slope dV/dx = -w = 0 Discontinuity or “jump” of the shear diagram at each concentrated force Change in shear negative when the force acts downwards and positive when the force acts upwards 2 kN force at x = 2m changes the shear from 3.5kN to 1.5kN 3 kN force at x = 4m changes the shear from 1.5kN to -1.5kN

7.3 Relations between Distributed Load, Shear and Moment Solution By Method of Sections, x = 2m and V = 1.5kN

7.3 Relations between Distributed Load, Shear and Moment Solution At the ends of the beams, when x = 0, M = 0 when x = 8, M = 0 Moment Diagram

7.3 Relations between Distributed Load, Shear and Moment Solution Area under the shear diagram = change in moment Also, by Method of Sections,

7.3 Relations between Distributed Load, Shear and Moment Example 7.12 Draw the shear and moment diagrams for the beam.

7.3 Relations between Distributed Load, Shear and Moment View Free Body Diagram Solution Support Reactions FBD of the beam

7.3 Relations between Distributed Load, Shear and Moment Solution At A, reaction is up, vA = +100kN No load acts between A and C so shear remains constant, dV/dx = -w(x) = 0 600kN force acts downwards, so the shear jumps down 600kN from 100kN to -500kN at point B No jump occur at point D where the 4000kN.m coupe moment is applied since ∆V = 0

7.3 Relations between Distributed Load, Shear and Moment Solution Shear Diagram Slope of moment from A to C is constant since dM/dx = V = +100

7.3 Relations between Distributed Load, Shear and Moment Solution Moment Diagram

7.3 Relations between Distributed Load, Shear and Moment Solution Determine moment at C by Method of Sections where MC = +1000kN or by computing area under the moment ∆MAC = (100kN)(10m) = 1000kN

7.3 Relations between Distributed Load, Shear and Moment Solution Since MA = 0, MC = 0 + 1000kN.m = 1000kN.m From C to D, slope, dM/dx = V = -500 For area under the shear diagram between C and D, ∆MCD = (-500kN)(5m) = -2500kN, so that MD = MC + ∆MCD = 1000 – 2500 = -1500kN.m Jump at point D caused by concentrated couple moment of 4000kN.m Positive jump for clockwise couple moment

7.3 Relations between Distributed Load, Shear and Moment Solution At x = 15m, MD = - 1500 + 4000 = 2500kN.m Also, by Method of Sections, from point D, slope dM/dx = -500 is maintained until the diagram closes to zero at B