sfd presentation by 10.01.03.133

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Page 1: SFD PRESENTATION by 10.01.03.133

WELCOMETO

MY

Page 2: SFD PRESENTATION by 10.01.03.133

By-

Enamul Nasir

10.01.03.133

4th Year 2nd Semester

Dept. Of Civil Engineering

Ahsanullah University Of Science and Technology

Page 3: SFD PRESENTATION by 10.01.03.133

•What is “SHEAR FORCE”?

Shear force:

is the algebraic sum of the vertical

forces acting to the left or right of a cut section

along the span of the beam.

Page 4: SFD PRESENTATION by 10.01.03.133

•What is „SHEAR FORCE DIAGRAM‟?

Shear force diagram(SFD):

The graphical representation of shear force in which ordinates

represents shear force and the abscissa represents the position of the section is called

“shear force diagram” or it can be defined as “ A shear force diagram is the one which

shows the variation shear force along the length of a beam section.”

Page 5: SFD PRESENTATION by 10.01.03.133

What is the importance of SFD in case of designing structures?

Shear force diagrams are analytical tools used in

conjunction with structural analysis to help perform

structural design by determining the value of shear force at

a given point of a structural element such as a beam. These

diagrams can be used to easily determine the

type, size, and material of a member in a structure so that a

given set of loads can be supported without structural

failure.

Page 6: SFD PRESENTATION by 10.01.03.133

Sign convention for SFD drawing?

The normal convention used in most engineering applications

is to label a positive shear force one that spins an element

clockwise (up on the left, and down on the right).

Page 7: SFD PRESENTATION by 10.01.03.133

Also for simplification in solving many structuralproblems the shear force at a section will beconsidered positive when the resultant of the forces tothe left section is upwards, or to the right of thesection is downward.

Page 8: SFD PRESENTATION by 10.01.03.133

IMPORTANT POINTS FOR DRAWING SHEAR FORCE DIAGRAM

1. Consider the left or right portion of the section,

2. Add the forces ( including reactions ) normal to the beam on one

of the portion. [ If right portion of the section is chosen, a force on

the right portion acting downwards is positive while a force acting

upward is negative, if left section is chosen then sign convention

will be vice versa. ]

3. The positive values of shear force are plotted above base line, and

negative values below base line.

Page 9: SFD PRESENTATION by 10.01.03.133

4. The shear force diagram will increase or decrease suddenly by a vertical

straight line at a section where there is a vertical point load.

5. The shear force between any two vertical loads will be constant and

hence the shear force diagram between any two vertical loads will be

horizontal.

WHAT IS THE USE OF SFD?

To determine the internal shear forces at any given point on a rigid body.

Page 10: SFD PRESENTATION by 10.01.03.133

PRACTICAL APPLICATION OF SHEAR FORCE DIAGRAM

The figure below is referring the practical application ofshear force diagram in case of designing shearreinforcement of a beam.

Page 11: SFD PRESENTATION by 10.01.03.133

Some Examples of SFD

Simply Supported beam with UDL

Page 12: SFD PRESENTATION by 10.01.03.133

Simply supported beams with triangular load

Page 13: SFD PRESENTATION by 10.01.03.133

Cantilever Beam

Page 14: SFD PRESENTATION by 10.01.03.133

Cantilever beam carrying Three Loads

Page 15: SFD PRESENTATION by 10.01.03.133

Simply Supported beam carrying double point load and UDL

Page 16: SFD PRESENTATION by 10.01.03.133

A detail example of drawing SHEAR FORCE DIAGRAM

A simply supported beam overhanging on one side is subjected to a U.D.L. of 1 kN/m. Sketch

the shear force diagram.

Solution: Consider a section (X – X’) at a distance x from end C of the beam.

Page 17: SFD PRESENTATION by 10.01.03.133

To draw the Shear Force Diagram, RA and RB has to be defined,

By taking moment of all the forces about point A

∑MA =0

RB × 3 – w/2 × (4)2 = 0

RB = 1 × (4)2 / 2 × 3 = 8/3 kN

For static equilibrium;

∑Fy = 0

RA + RB – 1 × 4 = 0

RA = 4 – 8/3 = 4/3 kN

Page 18: SFD PRESENTATION by 10.01.03.133

Now, To draw SFD we need S.F. at all salient points:Taking a section between C and B, SF at a distance x from end C. we have,

Page 19: SFD PRESENTATION by 10.01.03.133

Fx = + w.x kN

At x = 0; FC = 0

x = 1 m; FB just right = 1 × 1 = + 1 kN

Now, again taking section between B and A, at a distance x from end C, the SF is:

Fx = x – 8/3

When, x = 1 m; FB = 1 – 8/3 = –5/3 kN = –1.67 kN

At x = 4 m; FA = 4 – 8/3 = + 4/3 kN = + 1.33 kN

Page 20: SFD PRESENTATION by 10.01.03.133

When The S.F. becomes zero;

Fx = x – 8/3 = 0

⇒ x = 2.67 m

(The sign is taken positive taken when the resultant

force is in downward direction on the RHS of the

section).

Page 21: SFD PRESENTATION by 10.01.03.133

Now the Final Shear Force Diagram

The SFD can be drawn by joining the ordinates at salient point by straight line, as

defined by governing equations.

Page 22: SFD PRESENTATION by 10.01.03.133

NECESSITY OF SFD IN CASE OF PRESTRESSED CONCRETE

Page 23: SFD PRESENTATION by 10.01.03.133

THANK YOU

ALL