Bending Shear and Moment Diagram, Graphical method to construct shear and moment diagram, Bending deformation of a straight member, The flexure formula

Shear and moment diagram Axial load diagram Torque diagram Both of these diagrams show the internal forces acting on the members. Similarly, the shear and moment diagrams show the internal shear and moment acting on the members

Type of Beams Simply Supported Beam Overhanging Beam Cantilever Beam Statically Determinate Simply Supported Beam Overhanging Beam Cantilever Beam

Propped Cantilever Beam Type of Beams Statically Indeterminate Continuous Beam Propped Cantilever Beam Fixed Beam

Example 1 A B M V F Equilibrium equation for 0  x  3m: * internal V and M should be assumed +ve

Shear Diagram M V F x Sign convention: V= -9kN Lecture 1

Shear Diagram F M=-9x Sign convention: M= -9x kNm X=0: M= 0 V = -9 kN F x M=-9x Sign convention: M= -9x kNm X=0: M= 0 X=3: M=-27kNm

At cross section A-A At section A-A V=9kN M=X X

Example 2 1) Find all the external forces

Moment equilibrium Force equilibrium Moment equilibrium Force equilibrium

Boundary cond for V and M M=5x M=10-5x

Solve it Draw the shear and moment diagrams for simply supported beam.

Distributed Load For calculation purposes, distributed load can be represented as a single load acting on the center point of the distributed area. Total force = area of distributed load (W : height and L: length) Point of action: center point of the area

Example

Example

Solve it Draw the shear and moment diagrams the beam:

Solving all the external loads Distributed load will be Solving the FBD

Boundary Condition Equilibrium eq

Boundary Condition Equilibrium eq

x=0 V= 12 kN x=4 V=-20 kN x=0 M= 0 kN x=4 V=-16 kN x=4 V= 16kN x=6 V= 0 kN x=4 V= -16kN x=6 V= 0 kN

Graph based on equations Straight horizontal line y = c y = mx + c y=3x + 3 y=-3x + 3 y = ax2 + bx +c y=3x2 + 3 y=-3x2 + 3

Graphical method Relationship between load and shear: Relationship between shear and bending moment: 26

Regions of distributed load: Dividing by Dx and taking the limit as Dxà0, the above two equations become: Regions of distributed load: Slope of moment diagram at each point = shear at each point Slope of shear diagram at each point =  distributed load intensity at each point 27

Example

= = The previous equations become: Area under shear diagram change in shear Area under distributed load change in moment = =

+ve area under shear diagram

Bending deformation of a straight member Observation: bottom line : longer top line: shorter Middle line: remain the same but rotate (neutral line)

Strain Before deformation After deformation, Dx has a radius of curvature r, with center of curvature at point O’ Similarly Therefore

Maximum strain will be -ve: compressive state +ve: tension

The Flexure Formula The location of neutral axis is when the resultant force of the tension and compression is equal to zero. Noting

Since , therefore Therefore, the neutral axis should be the centroidal axis

Maximum normal stress Normal stress at y distance

Line NA: neutral axis Red Line: max normal stress c = 60 mm Yellow Line: max compressive stress c = 60mm Line NA: neutral axis Red Line: Compressive stress y1 = 30 mm Yellow Line: Normal stress y2 = 50mm Refer to Example 6.11 pp 289

I: moment of inertial of the cross sectional area Find the stresses at A and B

I: moment of inertial of the cross sectional area Locate the centroid (coincide with neutral axis)

I: moment of inertial of the cross sectional area Profile I I about Centroidal axis I about Axis A-A using parallel axis theorem A A Profile II Total I * Example 6-12 to 6-14 (pp 290-292)

Solve it If the moment acting on the cross section of the beam is M = 6 kNm, determine the maximum bending stress on in the beam. Sketch a three dimensional of the stress distribution acting over the cross section If M = 6 kNm, determine the resultant force the bending stress produces on the top board A of the beam

Total Moment of Inertia Max Bending Stress at the top and bottom 1.45MPa Bottom of the flange 1.14MPa 6kNm

Resultant F = volume of the trapezoid 300 mm 40 mm 1.45MPa 1.14MPa

Solve it The shaft is supported by a smooth thrust load at A and smooth journal bearing at D. If the shaft has the cross section shown, determine the absolute maximum bending stress on the shaft

External Forces Draw the shear and moment diagram Absolute Bending Stress Mmax = 2.25kNm