Lecture 1_SF and BM
-
Upload
mohammed-olaad -
Category
Documents
-
view
68 -
download
0
Transcript of Lecture 1_SF and BM
www.derby.ac.uk
5ME517 Structural Analysis
and Materials
Beam Bending – Shear Force and Bending Moment
Dr Amal Oraifige [email protected]
www.derby.ac.uk
Structural Analysis Structural analysis comprises a set of
physical laws and mathematics calculations that are required to study and predict the behaviour of different structures.
The structures can be described as engineering objects that will be analysed based upon their ability to withstand loads; they commonly include buildings, bridges, aircraft, ships and cars.
www.derby.ac.uk
Structural Analysis The primary goal of structural analysis is
the computation of deformations, internal forces, and stresses.
It is a method to drive the engineering design process or prove the soundness and safety of a design.
www.derby.ac.uk
Structural Beams A structural beam is a component used to
add strength to any structure or design.
Manufactured of steel, concrete or wood, structural beams are typically used to give support and strength through applications.
Steel “I” cross section beams are the most common type of beams used, while wooden beams are typically used in home/buildings construction.
www.derby.ac.uk
Structural Beams
www.derby.ac.uk
Beam Bending Beam bending theory is used to analyse
the soundness of structural beams on the way they are fixed and loaded.
The beams that will be analysed for this
module are either simply supported or cantilever.
www.derby.ac.uk
Point Loads • The perpendicular forces that are
acting on the beam are called Point Loads.
• A point load is shown as a single arrow and can act at different points.
www.derby.ac.uk
Uniformly Distributed Load (UDL) The beam can also carry a UDL, which is shown
as a series of arrows.
They can represent the weight of the beam per meter (N/m) or the weight of another object resting on the beam.
www.derby.ac.uk
If the UDL is distributed on the entire length, then for any given length x meters, the overall UDL (w) is (wx) in N and this is assumed to act at the centre of that length.
Uniformly Distributed Load (UDL)
www.derby.ac.uk
Examples of Loaded Beams
www.derby.ac.uk
Reaction Forces The loads that acting on the beam
forces the supporting points to produce reverse reaction forces
to make sure that the beam stays in a state of equilibrium.
www.derby.ac.uk
Reaction Forces So for the body to stay in a state of
equilibrium……..
Upwards Forces = Downwards Forces
For Simply Supported beam ONLY:
ΣMoments about any point on the beam must = 0
www.derby.ac.uk
Example 1 Calculate the Reaction Forces for the
loaded beam shown below.
www.derby.ac.uk
Shear Force (SF) The forces on a beam produce shearing at all sections
along the length. To the left, Up is positive and Down is negative. To the right, Up is negative and Down is positive. Or Clockwise is positive and Counter-Clock wise is
negative.
www.derby.ac.uk
In the case of uniformly distributed loads. The shear force increases proportional to the length. The shear force increases by the value of the UDL for every meter length.
Uniformly Distributed Load (UDL)
www.derby.ac.uk
Example 1 Calculate the shear forces and draw the
diagram.
www.derby.ac.uk
Example 1 – SF Diagram
www.derby.ac.uk
Moment of force about a point:
M = FD (Nm)
F
D
M = Moment
F = Force
D = Distance
Bending Moment (BM)
www.derby.ac.uk
The UDL is usually given as ‘w’ N/m. If the UDL is distributed along the entire
length, then for any length x:
M = wx*x/2 = wx2/2
Uniformly Distributed Load (UDL)
www.derby.ac.uk
Example 1 Calculate the bending moments and draw
the diagram.
www.derby.ac.uk
Calculate the BM at each point load (either left or right of point load, whatever's the easiest).
Plot the BM values and join with lines. Simply Supported: BM is zero at each end of beam.
Typical Cantilever: Max BM at the fixed end.
BM Diagram
www.derby.ac.uk
Note that BM at each end of the beam should equal to zero, the diagram looks like this:
Example 1 – BM Diagram
www.derby.ac.uk
Example 1 – SF and BM Diagram
www.derby.ac.uk
SF and BM Diagrams It can be shown wherever on the SF diagram
the value goes to zero, BM reaches a peak.
On more complex beams, the SF diagram might pass through the zero value several times, which results in several peak points for BM.
www.derby.ac.uk
Example 2 For the cantilever beam shown below: Calculate the reaction force. Calculate the shear forces and draw the
diagram. Calculate the bending moments and draw the
diagram.
www.derby.ac.uk
Example 2 – SF Diagram
www.derby.ac.uk
Example 2 – BM Diagram
www.derby.ac.uk
Example 3 For the simply supported beam shown below: Calculate the reaction forces. Calculate the shear forces and draw the diagram. Calculate the bending moments and draw the
diagram.
www.derby.ac.uk
Example 3 – SF Diagram
www.derby.ac.uk
Example 3 – BM diagram
www.derby.ac.uk
Example 3 – SF and BM diagram
www.derby.ac.uk
Example 3 – SF and BM diagram
Due to the complex loads acting on the beam, the SF diagram passes through the zero value three times, which results in three peak points for BM.