beam analysis

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Beam Analysis We require from buildings two kinds of goodness: first, the doing their practical duty well: then that they be graceful and pleasing in doing it. -John Ruskin

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Beam Analysis. We require from buildings two kinds of goodness: first, the doing their practical duty well: then that they be graceful and pleasing in doing it. -John Ruskin. Beam. Structural member that carries a load that is applied transverse to its length Used in floors and roofs - PowerPoint PPT Presentation

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Page 1: Beam Analysis

Beam Analysis

We require from buildings two kinds of goodness: first, the doing their practical duty well: then that they be graceful and pleasing

in doing it. -John Ruskin

Page 2: Beam Analysis

Beam

• Structural member that carries a load that is applied transverse to its length

• Used in floors and roofs

• May be called floor joists, stringers, floor beams, or girders

Page 3: Beam Analysis

• The loads are initially applied to a building surface (floor or roof).

• Loads are transferred to beams which transfer the load to another building component.

Gird

er

Beam Gird

er

Beam

Chasing the Load

Page 4: Beam Analysis

Static Equilibrium

• The state of an object in which the forces counteract each other so that the object remains stationary

• A beam must be in static equilibrium to successfully carry loads

Page 5: Beam Analysis

Static Equilibrium

• The loads applied to the beam (from the roof or floor) must be resisted by forces from the beam supports.

• The resisting forces are called reaction forces.

Applied Load

Reaction Force

Reaction Force

Page 6: Beam Analysis

Reaction Forces

• Reaction forces can be linear or rotational.– A linear reaction is often called a shear reaction (F or R).– A rotational reaction is often called a moment reaction

(M).

• The reaction forces must balance the applied forces.

Page 7: Beam Analysis

Beam Supports

The method of support dictates the types of reaction forces from the supporting members.

Page 8: Beam Analysis

Beam Types

Simple

Continuous

Cantilever Moment

(fixed at one end)

Page 9: Beam Analysis

Beam Types

Fixed

Moments at each end

Propped – Fixed at one end; supported at other

Overhang

Page 10: Beam Analysis

Simple Beams

Applied Load

BEAM DIAGRAM

FREE BODY DIAGRAM

Applied Load

Note: When there is no applied horizontal load, you may ignore the horizontal reaction at the pinned connection.

Page 11: Beam Analysis

Fundamental Principles of Equilibrium

The sum of all vertical forces acting on a body must equal zero.

The sum of all horizontal forces acting on a body must equal zero.

The sum of all moments (about any point) acting on a body must equal zero.

yF 0

xF 0

pM 0

Page 12: Beam Analysis

Moment

• A moment is created when a force tends to rotate an object.

• The magnitude of the moment is equal to the force times the perpendicular distance to the force (moment arm).

F

M

M dF d moment arm

Page 13: Beam Analysis

Calculating Reaction Forces

Sketch a beam diagram.

Page 14: Beam Analysis

Calculating Reaction Forces

Sketch a free body diagram.

Page 15: Beam Analysis

Calculating Reaction Forces

Use the equilibrium equations to find the magnitude of the reaction forces.– Horizontal Forces– Assume to the right is positive

+

xF 0

Page 16: Beam Analysis

Calculating Reaction Forces

• Vertical Forces

• Assume up is positive +

Equivalent Concentrated Load

Equivalent Concentrated Load

yF 0

Page 17: Beam Analysis

Calculating Reaction Forces

• Moments• Assume counter clockwise rotation is positive

+

A B

0 == 7700 lb

20 4000 6 13 000 10 0 0 yB yA( F ft ) ( lb ft ) ( , lb ft ) ( F )

20 24 000 130 000 0 0 yB( ft )F , ft lb , ft lb

20 154 000 yB( ft )F , ft lb

154 000

20

yB

, ft lbF

ft

7 700yBF , lb

Page 18: Beam Analysis

Calculating Reaction Forces

• Now that we know , we can use the previous equation to find .

= 7700 lb9300 lb =

0 =

Page 19: Beam Analysis

Shear Diagram

= 7700 lb9300 lb =

0 =

Shear at a point along the beam is equal to the reactions (upward) minus the applied loads (downward) to the left of that point.

Page 20: Beam Analysis

Moment Diagram

1400215

650

lbft

lbx . ft

Kink in moment curve

Page 21: Beam Analysis

Moment Diagram

9300 lb =

0 =

4000 lb

2.15’

MP

. ftlbft( lb )( . ft ) ( )(M . ft ) ( ) ( lb )( . ft )815

24000 215 650 815 9300 815 0

45608 maxMM ft lb

Page 22: Beam Analysis

Moment Diagram

Page 23: Beam Analysis

Moment Diagram

A

B

C

= 2.15 ft

Page 24: Beam Analysis

Beam Analysis

• Example : simple beam with a uniform load, w1= 1090 lb/ft

• Span = 18 feet

Test your understanding: Draw the shear and moment diagrams for this beam and loading condition.

Page 25: Beam Analysis

Moment

Shear

Max. Moment = 44,145l ft-lb Max. Shear = 9,810 lb

Shear and Moment Diagrams