Chapt 04b beams

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7.3 Relations between Distributed Load, Shear and Moment

Distributed Load• 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 upwards



Distributed Load• 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

• The internal shear force and bending moments assumed positive sense



Distributed Load• Distributed loading has been replaced by a resultant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

2)()(

0)()(;0)(

0)()(;0

xkxwxVMMMxkxxwMxVM

xxwVVVxxwVFy

(1)

(2)Divide both equations (1) and (2) by xLet x 0



Distributed Load

VdxdM

)(xwdxdV

Slope of the shear diagram

Negative of distributed load intensity

Slope of shear diagram

Shear moment diagram

VdxM BC

dxxwVBC )(

Change in shear Area under shear diagram

Change in moment Area under shear diagram



Force and Couple Moment• FBD of a small segment of the beam. Change in shear is

negative (or along the direction of positive sign convention. On the shear diagram, the shear value actually increases)

• FBD of a small segment of the beam located at the couple moment. Change in moment is positive


Example 7.9

Draw the shear and moment diagrams for the overhang beam.


Solution

The support reactions are shown.

Shear DiagramShear of –2 kN at end A of the beam is at x = 0.

Positive jump of 10 kN at x = 4 m due to the force.

Moment Diagram mkN 8420

04

MMM

xx

Engineering

Chapt 04b beams