3rd lecture shear and moment diagram for determinate beam

Theory of Structures3rd Lecture (9/10/2012): Analysis of Determinate

Beams;

Shear & Moment Diagrams

Dr Hussein M. Al.Khuzaie; [email protected] 1

Beams

• Members that are slender and support loads applied perpendicular to their longitudinal axis.


Span, L

Distributed Load, w(x) Concentrated Load, P

Longitudinal

Axis

Types of Beams

• Depends on the support configuration


M

Fv

FHFixed

FV FV

FH

Pin

Roller

PinRoller

FVFV

FH

Statically Indeterminate Beams

• Can you guess how we find the “extra” reactions?


Continuous Beam

Propped Cantilever

Beam

Internal Loadings developed in a structure member

• in order to properly design structural components, we must know the distribution of the internal forces within the component.

• In this lecture, we will determine the normal (axial) force, shear, and moment at a point in a structural component.

• Internal loads at a point within a structural component. For a coplanar structure, the internal load at a specified point will consist of a normal (Axial) force, N, a shear force, V, and a bending moment, M.

Dr Hussein M. Al.Khuzaie; [email protected]


Positive normal (Axial) force, N, tends to elongate the

components. Again, note that the normal (axial) forces

act in opposite directions on either side of the cut.

Positive shear, V, tends to rotate the component

clockwise. Note that the shear is in opposite directions

on either side of a cut through the component.

Positive moment, M, tends to deform the component into

a dish-shaped configuration such that it would hold

water. Again, note that the moment acts in opposite

directions on either side of the cut.

Internal Reactions in Beams

• At any cut in a beam, there are 3 possible internal reactions required for equilibrium: – Normal (Axial) force,

– Shear force,

– Bending moment.


L

P

a b


• At any cut in a beam, there are 3 possible internal reactions required for equilibrium: – normal force,

– shear force,

– bending moment.

Dr Hussein M. Al.Khuzaie;

[email protected]

Pb/L

x

Left Side of Cut

V

M

N

Positive Directions

Shown!!!


• At any cut in a beam, there are 3 possible internal reactions required for equilibrium: – normal force,

– shear force,

– bending moment.


Pa/L

L - x

Right Side of CutVM

N

Positive Directions

Shown!!!

Finding Internal Reactions

• Pick left side of the cut:– Find the sum of all the vertical forces to the left of

the cut, including V. Solve for shear, V.

– Find the sum of all the horizontal forces to the left of the cut, including N. Solve for axial force, N. It’s usually, but not always, 0.

– Sum the moments of all the forces to the left of the cut about the point of the cut. Include M. Solve for bending moment, M

• Pick the right side of the cut:– Same as above, except to the right of the cut.


Example: Find the internal reactions at points indicated. All axial force reactions are zero. Points are 2-ft apart.


20 ft

P = 20 kips

12 kips8 kips12 ft

1

7

10

6

2 3 94 5 8

Point 6 is just left of P and Point 7 is just right of P.

20 ft

P = 20 kips

12 kips8 kips12 ft

1

7

10

6

2 3 94 5 8

V(kips)

M(ft-kips)

8 kips

-12 kips

96

4864

4872

24

80

1632

x

xDr Hussein M. Al.Khuzaie; [email protected] 13

20 ft

P = 20 kips

12 kips8 kips12 ft

V(kips)

M(ft-kips)

8 kips

-12 kips

96 ft-kips

x

x

V & M Diagrams

What is the slope

of this line?

a

b

c

96 ft-kips/12’ = 8 kips

What is the slope

of this line?-12 kips


20 ft

P = 20 kips

12 kips8 kips12 ft

V(kips)

M(ft-kips)

8 kips

-12 kips

96 ft-kips

x

x

V & M Diagrams

a

b

c

What is the area of

the blue rectangle?

96 ft-kipsWhat is the area of

the green rectangle?

-96 ft-kips


Draw Some Conclusions

• The magnitude of the shear at a point equals the slope of the moment diagram at that point.

• The area under the shear diagram between two points equals the change in moments between those two points.

• At points where the shear is zero, the moment is a local maximum or minimum.


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dxxVxM

dxxwxV

functionloadthexw

)()(

)()(

)(


The Relationship Between Load, Shear and

Bending Moment

Shear Force & Bending Moment Diagram

Structure under loads

wdx

dV

Vdx

dM

Bending moment diagram

Shear force diagram

Slope of V = load

Slope of M = V


Load

0 Constant Linear

Shear

Constant Linear Parabolic

Moment

Linear Parabolic Cubic


Common Relationships

Load

0 0 Constant

Shear

Constant Constant Linear

Moment

Linear Linear Parabolic


Common Relationships

M

Example: Draw Shear & Moment diagrams for the following beam


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3 m 1 m1 m

12 kN 8 kNA C

B

D

RA = 7 kN RC = 13 kN

3 m 1 m1 m

12 kNA C

B

D

V(kN)

M(kN-m)

7

-5

8

8 kN

7

-15

8

7

-82.4 m



23

Because the form of the loading does not

change anywhere along the beam, single

equations will suffice for moment and shear:

∑Fy=0; 75-10*x-20/(2*9)*x^2

∑M=0; 75*x-5*x^2-(10/27)*x^3

The reactions are VL = 75kN, VR =

105kN, and HL = 0.

75 kN 105 kN


-105

-85

-65

-45

-25

-5

15

35

55

75

0 1 2 3 4 5 6 7 8 9

Shear Diagram

0

20

40

60

80

100

120

140

160

180

200

220

0 1 2 3 4 5 6 7 8 9

Shear and

moment are

plotted :

Dr Hussein M. Al.Khuzaie; husseinma@coe-

muth.org25

Examples

Dr Hussein M. Al.Khuzaie; husseinma@coe-

muth.org 26


4- Develop equations for shear and moment as a function

of position for the following structural components. Plot

the functions.

9 m


Any questions?

3rd lecture shear and moment diagram for determinate beam

Engineering