outline: sign convention again the relationship between...

Outline:

Sign Convention again

The relationship between Load, Shear and Bending

How to draw SFD and BMD without doing calculations

More examples

ENGR 1205 Chapter 5 - supplement 1

Internal forces at a specific point in a beam can be calculated by cutting the beam at that point

Cutting the beam at point A will expose the internal forces at A.

Generally there are 3 internal actions (2Forces + 1 Moment).

In beams where all forces are vertical we only have 2 internal actions.

◦ Vertical force Shear

◦ Moment Bending moment

8m

5 KN/m15 KN

422A

15 KN

18.7516.25

20 KN

YA

MA AM

AY

ENGR 1205

In North America, if the moment tends to cause the beam to curve upward it is positive; if the moment tends to cause it to curve downward it is negative.


The bending moment is positive if

◦ the upper side is in compression

◦ it holds water

◦ it’s a smiley face


A positive shear force at the cut causes the beam to rotate clockwise and matches the “smiley face” shape


Positive Shear and Moment on a whole Beam

Positive Shear and Moment on a cut-out Beam

Note: Coming from the left or the right makes a difference, so always start from left.

Diagrams illustrate the value of the internal forces (shear/ moment) that occur at each point along a structure (beam).

Additionally,


The shear diagram is the graphic representation of the shear force at successive points along the beam.

Upward acting forces are assumed positive and downward forces negative.

The shear force (V) at any point is equal to the algebraic sum of the external loads and reactions to the left of that point.

Since the entire beam must be in equilibrium (the sum of V=0), the shear diagram must close to zero at the right end.

Changes in loading cause changes in the slope of the shear curve.


The moment diagram is the graphical representation of the magnitude of the bending moment at successive points along the beam.

The bending moment for the moment diagram (M) at any point equals the sum of moments of the forces on the beam to the left about that point.

Since the entire beam is in equilibrium (Sum of M=0), the bending moment diagram must close to zero at the right side.


Cut the beam between each concentrated load.

For each section solve for the unknown shear force and bending moment

◦ Equation for each in terms of x (the distance along the beam

◦ Sub in endpoint values of x to get numerical values of V and M at each cut

Plot V (shear vs x) and M (moment vs x) along the beam


The “text” method works every time, but it is time consuming

You can draw the diagrams from the FBD if you know the relationships between w (load intensity), V (shear force) and M (bending moment).


dx

dMV

dx

dVw

the slope of the moment diagram at the given point is the shear at that point

the negative slope of the shear diagram at a given point equals the load at that point

◦ If there is no change in the load along the length under consideration, the shear curve is a straight horizontal line (or a curve of zero slope). The slope at any point is defined as the tangent to the curve at that point.

◦ If a concentrated load exists, then there is a vertical jump in the SFD

◦ If a load exists, and is uniformly distributed, the slope of the shear curve is constant and non-horizontal.

◦ If a load exists, and increases in magnitude over successive increments, the slope of the shear curve is positive (approaches the vertical); if the magnitude decreases, the slope of the shear curve is negative (approaches the horizontal).


If the slope of the SFD is zero, then the moment curve has a constant slope that is equal to the value of the shear for that increment.

If the slope of the SFD is positive, then the slope of the moment curve is getting steeper.

If the slope of the SFD is negative, then slope of the moment curve is getting flatter.

Changes in the shear diagram will produce changes in the shape of the moment curve.


The area of the shear diagram to the left or to the right of the section is equal to the moment at that section.

The slope of the moment diagram at a given point is the shear at that point.

The maximum moment occurs at the point of zero shears. When the shear (also the slope of the moment diagram) is zero, the tangent drawn to the moment diagram is horizontal.


Draw the shear and bending moment diagrams for the beam and loading shown and determine the location and magnitude of the maximum bending moment.


Draw the shear and bending moment diagrams for the beam and loading shown.


For each beam calculate the external reaction forces then draw the shear force and bending moment diagrams.


These slides were adapted from:

E. Tamer. Module 5: Area Moment of Inertia and Internal Forces. CIV100: Mechanics Lecture Notes. University of Toronto. http://www.civ.utoronto.ca/sect/coneng/i2c/civ100/old/fall08/Civ100-M5.pdf, Accessed November 25, 2011.


outline: sign convention again the relationship between...

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