• Applied loads result in internal forces consisting of a shear force (from the shear stress distribution) and a bending couple (from the normal stress

• Beams - structural members supporting loads at various points along the member.

• Transverse loadings of beams are classified as concentrated loads or distributed loads

• Applied loads result in internal forces consisting of a shear force (from the shear stress distribution) and a bending couple (from the normal stress

distribution)

Classification of Beam Supports

ANALYSIS OF BEAM AND FRAME STRUCTURES

In the case of trusses all bars were subjected to only axial loads. In beams and frames, this is not the case. Loads act everywhere therefore besides axial force, there is shear and bending moment.

They are not constant. Their value change along the axis of the member.

Internal force components

Flexural structures

V(x)=Shear

M(x)=Moment

N(x)=Axial

a

a Section at a

F

F F

For Axial Force (N) Tension is positive , Compression is negative

For Shear Force (V)

V

Clockwise +ve

For Moment (M) concave up (sagging) +ve , hogging -ve

M

Positive bending moment compresses the upper part of the beam and a negative bending moment compresses the lower part of the beam.

Sign conventions for stress resultants are called deformation sign conventions

because they are based upon how the material is deformed.

Wide-flange beam supported on a concrete wall and held down by anchor bolts that pass through slotted holes in the lower flange of the beam.

Beam to column connection in which the beam is attached to the column flange by bolted angles.

Metal pole welded to a base plate that is anchored to a concrete pier embedded deep in the ground.

Relationship Between , Load, Shear and Bending Moment

x x

M M+M

x

V+V V

q

C

Equilibrium of forces: The equilibrium of forces in the vertical direction in the segment shown of the member results in

x q V

V V

x q V



 

 











 ( ) 0

dx q

dV  

Equilibrium of moments: The equilibrium of moments around the centroid C for the section shown yields

) 2

2 ( 1

2 0 ) 2 (

V x V

M

V x x V

V M

M M





 

 

 





 









Taking the limit as gives

0

 x

dx V

dM 

Slope of shear diagram at a point =

intensity of distributed load at that point

Slope of bending moment diagram at a point = shear at that point

dx q

dV  

dx V

dM 

Load intensity

0 Constant Linear

Shear force

Constant Linear Parabolic

B.Moment

Linear Parabolic Cubic

Problem : Construct shear force and bending moment diagrams for the given system.

Show the critical values on the diagrams.

Problem : Plot shear force and bending moment diagrams for the given system. Show the critical values on the diagrams. Span length is 9 m.

Suggested Problem : Plot V and M diagrams.

Suggested Problem : a-) Compute the shear force (V) value and bending

moment (M) value at point C. b-) Plot V and M diagrams.

Determine the shear and moment in the beam shown in figure as a function of x.

Suggested Problem :