1. CTC / MTC 222 Strength of Materials
Chapter 6
Shear Forces and Bending Moments in Beams

2. Chapter Objectives
Draw the Free Body Diagram of a beam or portion of a beam
Use the Equations of Equilibrium to solve for the reactions of a stable, statically determinate beam
Calculate the shear and moment at any location in a beam
Use the relationships between load, shear and moment to construct the shear and moment diagrams

3. Review of Statics Types of supports Roller support Pinned support
Provides 1 reaction component - a force perpendicular to the surface the roller sits on
Pinned support
Provides 2 reaction components - forces perpendicular to each other (usually assumed to be horizontal and vertical)
Fixed support
Provides 3 reaction components – 2 perpendicular forces and a moment

4. Review of Statics Equilibrium – A balance of force and of moment
Statically Determinate – All forces can be determined using only the equations of equilibrium
Statically Indeterminate – There are more unknown forces than there are equations of equilibrium
Stable Structure
Equations of equilibrium are satisfied, and
Structure is restrained by its supports
In a statically determinate, stable structure the equations of equilibrium alone are sufficient to find all forces and moments in the structure

5. Review of Statics
If overall structure is in equilibrium, any portion of it is in equilibrium
General 3-Dimensional structure
6 degrees of freedom
3 translations and 3 rotations
6 equations of equilibrium
∑ FX = 0, ∑ FY = 0, ∑ FZ = 0
∑ MX = 0, ∑ MY = 0, ∑ MZ = 0
2-Dimensional structure with coplanar loads
3 degrees of freedom
2 translations and 1 rotation
3 equations of equilibrium
∑ FX = 0, ∑ FY = 0, ∑ MZ = 0

6. Internal Forces and Moments at a Point
Cut a section through member at the point
Internal forces and moments must be present to maintain equilibrium of the beam segment
For a 2-D structure with coplanar loads, there are two forces and a moment
Normal (or axial) force – N
Shear force – V
Bending Moment – M
Forces and moments are equal but in opposite direction on opposite sides of section cut

7. Internal Forces and Moments - Sign Convention
Positive values on left hand face at section cut
N - To the right
V - Downward
M – Counter-clockwise
Positive values are in opposite direction on right hand face

8. Internal Forces and Moments - Sign Convention
Same sign convention in other terms
Looking at a small segment of the member
Positive normal force N
Elongates segment  tension
Positive Shear force V
Rotates segment clockwise
Positive Moment M
Bends segment concave upward  compression on top

9. Internal Forces and Moments at a Point, O
Solve for reactions
Cut a section through member at point O
Draw FBD of either segment either side of section cut
Show N, V and M in their positive sense
Apply 3 equations of equilibrium
∑ FX = 0, ∑ FY = 0, ∑ MO = 0
Solve for N, V and M
If solution of EOE yields a positive solution, the quantity is positive; if it yields a negative solution, the quantity is negative

10. Shear and Moment Functions
Structural design requires knowledge of the variations in V and M
Axial force N must also be known, but usually doesn’t vary along the member
Cut a section through member at a distance x from some point
Draw FBD of either segment either side of section cut
Show V and M in their positive sense
Apply 2 equations of equilibrium
∑ FY = 0, ∑ MO = 0
Solve for V and M in terms of x, the distance along the member
The result is an expression for V and for M as a function of x

11. Shear and Moment Functions
This approach can be tedious, particularly for complicated loading
Shear and moment functions (or their slopes) will be discontinuous at points where:
The magnitude or type of distributed load changes
Concentrated loads are applied
Concentrated moments are applied
Each time one of these things occur, there is a new equation for V or M
Construction of V and M diagrams can be simplified by using the relationships between load, shear and moment

12. Relationships Between Load, Shear and Moment
Shear Diagram
Application of a downward concentrated load causes a downward jump in the shear diagram. An upward load causes an upward jump.
The slope of the shear diagram at a point (dV/dx) is equal to the (negative) intensity of the distributed load w(x) at the point.
The change in shear between any two points on a beam equals the (negative) area under the distributed loading diagram between the points.

13. Relationships Between Load, Shear and Moment
Moment Diagram
Application of a clockwise concentrated moment causes an upward jump in the moment diagram. A counter-clockwise moment causes a downward jump.
The slope of the moment diagram at a point (dM/dx) is equal to the intensity of the shear at the point.
The change in moment between any two points on a beam equals the area under the shear diagram between the points.

14. Relationships Between Load, Shear and Moment
A point of local maximum moment will occur at a point of zero shear.
The point of absolute maximum moment may occur at either end of the beam, at a point of zero shear or at a point of applied concentrated moment (or couple).
The moment and shear which occur at any point on the beam can be calculated from the free-body diagram of the beam either side of that point.

15. Constructing Shear and Moment Diagrams
Draw the FBD of the structure
Determine the reactions of the structure
Resolve all forces acting on the member into components parallel and perpendicular to the member’s axis
Construct the shear diagram
Establish the V and x axes
Determine and plot the shear at the start of the member
Use the relations between the load and the shear diagrams to determine the shear in the rest of the member.
Construct the moment diagram
Establish the M and x axes
Determine and plot the moment at the start of the member
Use the relations between the load, shear and moment diagrams to determine the moment in the rest of the member.