Powerpoint Templates

Page 2
BENDING MOMENT
DIAGRAM

BMD show how the applied loads to a beam create
a moment variation along the length of the beam.
Mt

Mt

Bending Moment

Bending Moment = moments of reactions – moments of
loads
Distributed load acts downward on beam.
Internal shear force causes a clockwise rotation of the beam
segment; and the internal moment causes compression in the to...
SIGN CONVENTIONS

• A force that tends to bend the beam
downward is said to produce a
positive bending moment. A force
tha...
PROCEDURE

1.

Draw the free-body-diagram of the beam with sufficient room
under it for the shear and moment diagrams
(if ...
PROCEDURE
3. Draw the moment diagram below the shear diagram.
• The shear load is the slope of the moment and point moment...
Relations between Distributed
Load, Shear and Moment
Distributed Load
Slope of the shear
diagram
Slope of the
shear diagra...
Procedure for Analysis
Support Reactions
•
•

Find all reactive forces and couple moments acting on the beam
Resolve them ...
Shear & Moment Diagram
Common Relationships
0

Constant

Linear

Constant

Linear

Parabolic

Linear

Parabolic

Cubic

Load

Shear

Moment
Common Relationships
0

Constant

Constant

Constant

Linear

Linear

Load

0

Linear

Parabolic

M

Shear

Moment
Rules of Thumb/Review
• Moment is dependent upon the shear diagram
 the area under the shear diagram = change in the mome...
BEAMS IN BENDING
P1

w

P2

Ra

P3

Rb
Point of zero

SFD

Point of maximum

BMD
Point of contra-flexure

Point of maximum
Concentrated Load
•

Find reactions

•

Cut through beam to the left of the
load P (a distance x from the left
end), FBD
–...
Concentrated Load Moment
Diagram
Bending Moment Diagram
•

The bending moment in the left
side increases linearly from zer...
a

b

A

R Ay =

y

P

Example 1 :

C

P ⋅b
( a + b)

R By =

x
P

a

∑M

z

Mxz

A

P ⋅b
( a + b)

Qxy

=0 ;

+ Mxz + P (...
y

P

a

b

A

Pb
( a + b)

x

B

Pa
( a + b)

x

(i) When x ≤ a :
(ii) When x > a :
BMD:

C

Mxz =
Mxz

Mxz

Pab
( a + b)...
Example 2:

P

y

Mxz=P.L

A

x

B
L

RAy=P

P.L

Mxz
P

Mxz

Qxy

Mxz

Qxy

Q & M are POSITIVE

Qxy

x

P

Mxz
Qxy

∑F = ...
Q xy = P; Mxz = −P( L − x )

P

L

y

P.L
Mxz

A

B

x

Qxy

x
P

P
Qxy

Mxz

+ve

Shear Force
Diagram (SFD)

0

0
-P.L

-...
V & M Diagrams

8 kips

12 ft

P = 20 kips

12 kips

20 ft

8 kips

V

x

(kips)
What is the area of the
blue rectangle?
9...
BMD for simply supported beam
with UDL:
Parabolic, max moment at mid span of value WL2/8 ,
where w is the distributed load...
UNIFORM LOAD
•

The beam and its loading is
symmetric, the reactions are equal to
wL/2

•

The slope of the shear diagram ...
UNIFORM LOAD
•

Therefore, the shear force and bending
moment at a distance x from the left end
are:

dV
dM
=− ( x ),
w
=V...
Distributed Load w
per unit length

Example : Distributed Load
y
x

Mxz=

wL2

A

2

B
L
x

RAy=wL
wL2

Mxz

wx

2

Qxy

w...
⇒ Q xy = w ( L − x )

⇒ Mxz

wx 2 wL2
= wLx −
−
2
2

Mxz
BMD:

L

0
-ve
x
-wL2
2

wL2
=−
2

@ x = 0;

Mxz

@ x = L;

Mxz =...
Draw Some Conclusions

• The magnitude of the shear at a point equals the slope of the
moment diagram at that point.
• The...
10.01.03.029
10.01.03.029
10.01.03.029
10.01.03.029
10.01.03.029
10.01.03.029
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10.01.03.029

  1. 1. Powerpoint Templates Page 2
  2. 2. BENDING MOMENT DIAGRAM BMD show how the applied loads to a beam create a moment variation along the length of the beam.
  3. 3. Mt Mt Bending Moment Bending Moment = moments of reactions – moments of loads
  4. 4. Distributed load acts downward on beam.
  5. 5. Internal shear force causes a clockwise rotation of the beam segment; and the internal moment causes compression in the top fibers of the segment.
  6. 6. SIGN CONVENTIONS • A force that tends to bend the beam downward is said to produce a positive bending moment. A force that tends to shear the left portion of Positive Bending Negative Bending the beam upward with respect to the right portion is said to produce a positive shearing force. Positive Shear Negative Shear
  7. 7. PROCEDURE 1. Draw the free-body-diagram of the beam with sufficient room under it for the shear and moment diagrams (if needed, solve for support reactions first). 1. Draw the shear diagram under the free-body-diagram. • The change in shear ∆V equals the negative area under the distributed loading. • Label all the loads on the shear diagram ∆V = − ∫ w( x )dx
  8. 8. PROCEDURE 3. Draw the moment diagram below the shear diagram. • The shear load is the slope of the moment and point moments result in jumps in the moment diagram. • The area under the shear diagram equals the change in moment over the segment considered (up to any jumps due to point moments). • Label the value of the moment at all important points on the moment diagram. ∆M = ∫ V ( x )dx
  9. 9. Relations between Distributed Load, Shear and Moment Distributed Load Slope of the shear diagram Slope of the shear diagram dV = − w(x) dx dM =V dx Negative of distributed load intensity Shear moment diagram Change in shear ∆M BC = ∫ Vdx Area under shear diagram Change in moment ∆VBC = − ∫ w( x)dx Area under shear diagram
  10. 10. Procedure for Analysis Support Reactions • • Find all reactive forces and couple moments acting on the beam Resolve them into components Shear and Moment Reactions • • • • Specify separate coordinates x Section the beam perpendicular to its axis V obtained by summing the forces perpendicular to the beam M obtained by summing moments about the sectioned end Shear and Moment Reactions • • Plot (V versus x) and (M versus x) Convenient to plot the shear and the bending moment diagrams below the FBD of the beam
  11. 11. Shear & Moment Diagram
  12. 12. Common Relationships 0 Constant Linear Constant Linear Parabolic Linear Parabolic Cubic Load Shear Moment
  13. 13. Common Relationships 0 Constant Constant Constant Linear Linear Load 0 Linear Parabolic M Shear Moment
  14. 14. Rules of Thumb/Review • Moment is dependent upon the shear diagram  the area under the shear diagram = change in the moment (i.e. A shear diagram = ΔM) • Straight lines on shear diagrams create sloping lines on moment diagrams • Sloping lines on shear diagrams create curves on moment diagrams • Positive shear = increasing slope • Negative shear = decreasing slope
  15. 15. BEAMS IN BENDING P1 w P2 Ra P3 Rb Point of zero SFD Point of maximum BMD Point of contra-flexure Point of maximum
  16. 16. Concentrated Load • Find reactions • Cut through beam to the left of the load P (a distance x from the left end), FBD – Equilibrium yields V and M for the left side of the beam • Cut through the beam to the right of P, FBD – Equilibrium yields V and M for the right side of the beam.
  17. 17. Concentrated Load Moment Diagram Bending Moment Diagram • The bending moment in the left side increases linearly from zero at the support to P(ab/L) at the concentrated load x=a • In the right side, the bending moment is again a linear function of x, varying from P(ab/L) at x=a to zero at the support x=L. • The maximum bending moment is therefore P(ab/L), which occurs at the concentrated load. In this example a=b=L/2
  18. 18. a b A R Ay = y P Example 1 : C P ⋅b ( a + b) R By = x P a ∑M z Mxz A P ⋅b ( a + b) Qxy =0 ; + Mxz + P ( x − a ) ∴ Mxz Where Pb ( x) = 0 − ( a + b) Pb ( x ) − P ( x − a) = ( a + b) ( x − a) x B can only be +VE or ZERO. P⋅a ( a + b)
  19. 19. y P a b A Pb ( a + b) x B Pa ( a + b) x (i) When x ≤ a : (ii) When x > a : BMD: C Mxz = Mxz Mxz Pab ( a + b) Pb ( x ) − P ( x − a) ( a + b) 0 Pb ( x ) − P ( x − a) = ( a + b) Eq. 1 2 Eq. 2 +ve 0 A C 1 B
  20. 20. Example 2: P y Mxz=P.L A x B L RAy=P P.L Mxz P Mxz Qxy Mxz Qxy Q & M are POSITIVE Qxy x P Mxz Qxy ∑F = 0 ; ∑M = 0 ; y z ∴ Q xy = P ∴ Mxz = −P( L − x )
  21. 21. Q xy = P; Mxz = −P( L − x ) P L y P.L Mxz A B x Qxy x P P Qxy Mxz +ve Shear Force Diagram (SFD) 0 0 -P.L -ve Bending Moment Diagram (BMD)
  22. 22. V & M Diagrams 8 kips 12 ft P = 20 kips 12 kips 20 ft 8 kips V x (kips) What is the area of the blue rectangle? 96 ft-kips M (ft-kips) -12 kips 96 ft-kips b What is the area of the green rectangle? -96 ft-kips a c x
  23. 23. BMD for simply supported beam with UDL: Parabolic, max moment at mid span of value WL2/8 , where w is the distributed load and L the length of the beam.
  24. 24. UNIFORM LOAD • The beam and its loading is symmetric, the reactions are equal to wL/2 • The slope of the shear diagram at each point equals the negative distributed load intensity at each point dV = −w( x ) dx
  25. 25. UNIFORM LOAD • Therefore, the shear force and bending moment at a distance x from the left end are: dV dM =− ( x ), w =V dx dx • These equations are valid through the length of the beam and can be plotted as shear and bending moment diagrams.  The maximum value at the midpoint where dM = V = 0. dx  Mmax= wL2/8
  26. 26. Distributed Load w per unit length Example : Distributed Load y x Mxz= wL2 A 2 B L x RAy=wL wL2 Mxz wx 2 Qxy wL Mxz Qxy ∑F = 0 ; ∑M = 0 ; y z wL − wx − Q xy = 0 ⇒ Q xy = w ( L − x ) wL2 x − wL ( x ) + wx   = 0 Mxz + 2 2
  27. 27. ⇒ Q xy = w ( L − x ) ⇒ Mxz wx 2 wL2 = wLx − − 2 2 Mxz BMD: L 0 -ve x -wL2 2 wL2 =− 2 @ x = 0; Mxz @ x = L; Mxz = 0 L @x = ; 2 Mxz wL2 =− 8
  28. 28. 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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