1. Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Beam Analysis
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
The students on the beam represent the load applied to the beam. The load is applied transversely, or perpendicularly, to the length of the beam.
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2. Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Beam Analysis
Beams are designed to carry the Shear and Bending Moment caused by the design loads
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3. Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Chasing the Load
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.
Girder
Beam
The load (dead, live, snow, wind, etc.) is initially applied to a building surface, such as a wall or a floor.
The load is then transferred through that component (the wall or floor) to the nearest beams.
The beams transfer the load to the structural components that supports its ends which may be a girder as indicated by the arrows. The beam can also transfer loads to a column or a load-bearing wall.
The load continues to be transferred through structural components until it reaches the supporting soil.
In this structural steel frame, a concentrated load is applied to an elevated floor.
The floor transfers the load to adjacent beams.
The beams carry the load to larger beams called girders.
These girders will then transfer the load to the supporting columns, etc.
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4. EXAMPLE Partial 2nd Floor Framing Plan
Design Area

5. EXAMPLE Tributary or Contributing Area
6’8” Width
Beam B.3
Column B-3
Girder 3BC

6. Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
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
Obviously we do not want the beams in a building to constantly move. Yes, the beam will deflect when a load is applied, but it will quickly return to a static condition – that is, it will stop moving.
In order to be static, a beam must be in equilibrium – all of the forces acting on the beam must counteract and balance each other.
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7. Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
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
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8. Reaction Forces Reaction forces can be linear or rotational.
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
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.
In order to design a beam, we must calculate the magnitude of the reaction forces.
The reaction forces will also indicate the types and magnitude of forces that will be transferred to the supporting members.
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9. Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Beam Supports
The method of support dictates the types of reaction forces from the supporting members.
At a roller support, the beam can rotate at the support and, in this case, move horizontally. A roller support can only react with a single linear reaction force in the direction of the support.
A pin connection will hold the beam stationary both vertically and horizontally but will allow the beam to rotate. This results in horizontal (Fx) and vertical (y) linear reaction forces from the support, but no rotational reaction.
A fixed support will not allow the beam to move in any way. The resulting reaction forces include a horizontal force (Fx), a vertical force (Fy), and a moment reaction, M.
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10. Beam Types Simple Continuous Cantilever Moment (fixed at one end)
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Beam Types
Simple Continuous Cantilever Moment (fixed at one end)
Beams come in many shapes and sizes and can be classified according to how they are supported.
A simple beam is supported by a pinned and a roller support.
A continuous beam is supported over several supports.
A cantilever beam is supported by one fixed connection.
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11. Beam Types Fixed Moments at each end
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Beam Types
Fixed
Moments at each end
Propped – Fixed at one end; supported at other
Overhang
A fixed beam is supported by a fixed connection at each end.
A propped beam is fixed at one end and supported by a pin or roller at the other end.
An overhang beam is similar to a simple beam except the beam extends over one of the supports.
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12. Simple Beams Applied Load Applied Load BEAM DIAGRAM FREE BODY DIAGRAM
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Simple Beams
Applied Load
BEAM DIAGRAM
Applied Load
In this class, we will primarily deal with simple beams which are beams that possess a single span, one pinned connection, and one roller connection.
The beam diagram and free body diagram for a simple beam are shown here.
The applied load can be concentrated, uniform, or a combination of both. A concentrated load is shown here.
Since the horizontal reaction at the pinned connection will always be zero when there is no applied horizontal load, you may draw the free body diagram without the horizontal reaction.
FREE BODY DIAGRAM
Note: When there is no applied horizontal load, you may ignore the horizontal reaction at the pinned connection.
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13. Fundamental Principles of Equilibrium
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
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.
In order for a beam to be in static equilibrium, the principles of equilibrium must hold true.
A moment is created when a force tends to rotate and object. It is equal to the force times the perpendicular distance to the force.
We will use these principles to calculate the beam reactions.
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14. Moment A moment is created when a force tends to rotate an object.
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
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
The perpendicular distance from the point to the force is often called the moment arm.
[click] to show formula
[click] to apply load and show moment arm on diagram M
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15. Calculating Reaction Forces
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Calculating Reaction Forces
Sketch a beam diagram.
A beam diagram shows
The beam and its span
The applied loads, including magnitude and location
The support conditions
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16. Calculating Reaction Forces
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Calculating Reaction Forces
Sketch a free body diagram.
A free body diagram shows
The beam
The applied load magnitudes and locations
The reaction forces based on the type of support – we will calculate the magnitude of the reaction forces to add to the free body diagram
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17. Calculating Reaction Forces
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Calculating Reaction Forces
Use the equilibrium equations to find the magnitude of the reaction forces.
Horizontal Forces
Assume to the right is positive +
Because there are no horizontal applied forces, there will be no horizontal reaction force.
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18. Calculating Reaction Forces
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Calculating Reaction Forces
Vertical Forces
Assume up is positive +
Equivalent Concentrated Load
Both the reaction forces are acting up and are therefore positive.
The applied forces are acting downward and are therefore negative.
[click] The uniform load must be converted to an equivalent concentrated load that takes into account the entire force.
[click]The equivalent concentrated load is assumed to act at the center of the uniform load application and represents the uniform load in the calculation.
This equation has two unknowns. We cannot solve a single equation with two unknowns. We need another equation.
Equivalent Concentrated Load
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19. Calculating Reaction Forces
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Calculating Reaction Forces
Moments
Assume counter clockwise rotation is positive +
The best point about which to sum moments is one of the support locations. You can choose any support location.
We will sum moments about point A in this example.
Since FyA acts through the point A, it has a zero moment arm.
The 4000 lb applied concentrated load and the equivalent concentrated load (from the uniform load) both cause clockwise rotation about point A; therefore, the moment caused by these forces is considered positive. Remember that the equivalent load (which takes the place of the uniform load) is assumed to act at the middle of the uniform load location.
The moment caused by FyB is negative, because FyB would cause a counterclockwise rotation about point A. A B
0 =
= 7700 lb
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20. Calculating Reaction Forces
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Calculating Reaction Forces
Now that we know , we can use the previous equation to find
= 7700 lb
9300 lb =
0 =
Substitute 7700 lbs for FyB.
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21. Shear Diagram 0 = = 7700 lb 9300 lb = Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Shear Diagram
= 7700 lb
9300 lb =
0 =
Because the applied loads do not occur at a support, shear forces are induced in the beam – that is, the particles in the beam try to slide against each other because the applied forces are pushing down and the reaction forces are pushing up.
The shear forces get larger as you get closer to the support.
A shear diagram is a way to represent the shear force at every point along the beam. It is really just a bookkeeping method to keep track of the vertical forces. [click]
(Walk through the creation of the shear diagram by pointing to the diagram as you discuss its creation.)
Start at the left end (point A) and plot the magnitude of the vertical reaction at that point.
The internal shear represented on the shear diagram at the left end of the beam is equal to the end reaction at the left end of the beam. In this diagram one square is approximately equal to 300 pounds vertically, although it is not critical that you sketch it perfectly to scale.
For uniform loads, the slope of the graph is equal to the magnitude of the uniform load. In this example, the slope decreases (goes down) at 650 pounds per foot of length of the beam. Therefore the shear diagram decreases at 650 lb/ft as you move to the right. For this diagram, the horizontal scale is 2 ft per square. Although, again, it is not critical that the diagram be sketched perfectly to scale.
If no uniform load is present, the shear diagram will look like a horizontal line except at concentrated loads.
For concentrated loads, the graph jumps an amount equal to the magnitude of the concentrated load. In this example, the graph jumps down 4000 pounds at the location of the concentrated load.
The shear diagram will continue to decrease at 650 lb/ft across the entire span until you reach the right support.
If completed correctly,
the magnitude of the internal shear in the diagram at the right end of the beam should be equal (and opposite to) the magnitude of the reaction force at the right end of the beam. In this case, pounds is equal and opposite to the 7700 pound FyB. The 7700 lbs up brings the shear back to zero.
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22. Moment Diagram Kink in moment curve Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Moment Diagram
Kink in moment
curve
The moment diagram represents the magnitude of the moment at every point along the beam.
[Point out the appropriate points on the moment diagram as you discuss it.]
For pinned and roller connections, we know there is no moment reaction. Therefore, the moment at a pinned or roller connection is always zero. Note that the moment at each end is zero here.
The moment diagram for a beam that has a uniform load is essentially parabolic. When a concentrated load is applied there is a slight kink in the diagram that is difficult to represent.
The moment diagram for a beam loaded only with concentrated loads is composed of straight lines. The slope of a line is equal to the magnitude of the shear at that point.
The maximum moment always occurs at a point of zero shear.
In this example, we need to know the distance x in order to determine where the maximum moment will occur. Because we know that the shear at 6 ft from the left end of the beam is 1400 lb (lower value), we can find x by determining how much length is required in order for the 650 lb/ft uniform load to counteract the 1400 pound force. [click] (See the calculation on the slide). Therefore, x = 2.15 feet.
Now we can determine the maximum moment by finding the moment induced by all of the forces to the left of the point of zero shear.
[click] Note that there will be a slight kink in the moment diagram where there is a jump in the shear diagram.
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23. Moment Diagram 4000 lb M 0 = 9300 lb = 2.15’ P Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Moment Diagram
9300 lb =
0 =
4000 lb M P
2.15’
The moment at this point, P (not necessarily the load location), in the beam is calculated by summing the moments induced by the forces acting to the left of the point.
First, sketch a free body diagram for the portion of the beam to the left of the point of zero shear. Then sum the moments about this point.
M represents the moment at this point required to resist the forces.
Remember counterclockwise about point P is positive.
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24. Moment Diagram Beam Analysis Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Moment Diagram
Therefore, the maximum moment will occur at 8.15 feet from the left end of the beam and will have a magnitude of 45,608 ft-lb.
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25. Moment Diagram = 2.15 ft A B C Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Moment Diagram A B C
= 2.15 ft
Another way to calculate maximum moment is to find the area under the shear diagram to the left of the point of zero shear.
The slight difference between this value and the previous maximum moment is due to rounding the value for x.
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26. Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Beam Analysis
Example : simple beam with a uniform load, w1= 1090 lb/ft
Span = 18 feet
[Give students a few minutes to sketch the diagrams, then go to the next slide.]
Test your understanding: Draw the shear and moment diagrams for this beam and loading condition.
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27. Shear and Moment Diagrams
Beam Analysis
Civil Engineering and Architecture
Unit 3 – Lesson 3.2 – Structures
Shear and Moment Diagrams
Shear
If you perform the calculations outlined in this presentation you will find that the
the maximum moment is 44,145 ft-lb. The max shear is 9,810 lb.
Note that because there is not a concentrated applied load, there is no jump in the shear diagram.
Also, because the loading is uniformly distributed, the moment diagram is parabolic.
Moment
Max. Moment = 44,145l ft-lb Max. Shear = 9,810 lb
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