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beamanalysis.ppt
- 1. âą Structural memberthat carries a load that is applied transverse to its length âą Used in floors and roofs âą May be called floor joists, stringers, floor beams, or girders Beam Analysis
- 2. âą The loadsare initially applied to a building surface (floor or roof). âą Loads are transferred to beams which transfer the load to another building component. Chasing the Load
- 3. EXAMPLE Partial 2nd FloorFraming Plan Design Area
- 4. EXAMPLE Tributary or ContributingArea 6â8â Width Column B-3 Beam B.3 Girder 3BC
- 5. Static Equilibrium âą Thestate of an object in which the forces counteract each other so that the object remains stationary âą A beam must be in static equilibrium to successfully carry loads
- 6. Static Equilibrium âą Theloads applied to the beam (from the roof or floor) must be resisted by forces from the beam supports. âą The resisting forces are called reaction forces. Applied Load Reaction Force Reaction Force
- 7. Reaction Forces âą Reactionforces can be linear or rotational. â A linear reaction is often called a shear reaction (F or R). â A rotational reaction is often called a moment reaction (M). âą The reaction forces must balance the applied forces.
- 8. Beam Supports The methodof support dictates the types of reaction forces from the supporting members.
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- 10. Beam Types Fixed Moments ateach end Propped â Fixed at one end; supported at other Overhang
- 11. Simple Beams Applied Load BEAMDIAGRAM FREE BODY DIAGRAM Applied Load Note: When there is no applied horizontal load, you may ignore the horizontal reaction at the pinned connection.
- 12. Fundamental Principles of Equilibrium Thesum 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. ïœ ï„ y F 0 ïœ ï„ x F 0 ïœ ï„ p M 0
- 13. Moment âą A momentis created when a force tends to rotate an object. âą The magnitude of the moment is equal to the force times the perpendicular distance to the force (moment arm). F M M d F ïœ ï ï d moment arm ï ïœ
- 14. Calculating Reaction Forces Sketcha beam diagram.
- 15. Calculating Reaction Forces Sketcha free body diagram.
- 16. Calculating Reaction Forces Usethe equilibrium equations to find the magnitude of the reaction forces. â Horizontal Forces â Assume to the right is positive + ïœ ï„ x F 0
- 17. Calculating Reaction Forces âąVertical Forces âą Assume up is positive + Equivalent Concentrated Load Equivalent Concentrated Load ïœ ï„ y F 0
- 18. Calculating Reaction Forces âąMoments âą Assume counter clockwise rotation is positive + A B 0 = = 7700 lb 20 4000 6 13 000 10 0 0 ï ï ï ï ï ï« ï ïœ yB yA ( F ft ) ( lb ft ) ( , lb ft ) ( F ) 20 24 000 130 000 0 0 ï ï ï ï ï« ïœ yB ( ft )F , ft lb , ft lb 20 154 000 ïœ ï yB ( ft )F , ft lb 154 000 20 ï ïœ yB , ft lb F ft 7 700 ïœ yB F , lb
- 19. Calculating Reaction Forces âąNow that we know , we can use the previous equation to find . = 7700 lb 9300 lb = 0 =
- 20. Shear Diagram = 7700lb 9300 lb = 0 =
- 21. Moment Diagram 1400 215 650 ïœ ïœ lb ft lb x. ft Kink in moment curve
- 22. Moment Diagram 9300 lb= 0 = 4000 lb 2.15â M P ï« ï« ï ï ïœ . ft lb ft ( lb)( . ft ) ( )( M . ft ) ( ) ( lb)( . ft ) 815 2 4000 215 650 815 9300 815 0 45608 ïœ ïœ ï max M M ft lb
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- 25. Beam Analysis âą Example: simple beam with a uniform load, w1= 1090 lb/ft âą Span = 18 feet Test your understanding: Draw the shear and moment diagrams for this beam and loading condition.
- 26. Moment Shear Max. Moment =44,145l ft-lb Max. Shear = 9,810 lb Shear and Moment Diagrams