There are many structures whose individual members are made up of ?beams. Some examples include bridges, skyscrapers, artwork, signs, ?parking covers, and even furniture. These beams
There are many structures whose individual members are made up of beams. Some examples include bridges, skyscrapers, artwork, signs, parking covers, and even furniture. These beams are intentionally designed with different cross-sectional shapes to optimize both strength, and weight all depending on the type of loading condition they will be subject to. As an engineer, it is our responsibility to consider the loading conditions of the structures we design then to select from the available beam geometries a shape and size that will be more than strong enough to last.
Common Beam Cross-Sectional Shapes
Statics allows us to quantify the loading conditions and use the support types to understand the forces and moments a beam must be designed to hold. Each symbol of a support type correlates to real-life connections that are commonly used in structural design. In some cases, support types that become too constraining create a statically indeterminant situation where statics is not enough to solve for the forces and moments on a structure.
Structures and Members
In your original post, include the following:
- Find a real-world example where beams are used in a structure and identify/discuss the following as they apply to your example
- Beam type
- Support types
- Loading type
MET211 Statics
Structures & Members
Introduction
A two-force member has forces acting on each end. These forces line up with the member and cause either tension or compression.
A truss is formed if several two-force members are joined in one or more connected triangles. Each of the members is pinned at each end and, if carrying a load, is in either tension or compression.
The direction of the member indicates the direction of the tensile or compressive force in the member acting on the joint or pin. The ends of the members are pinned together to form a joint. A member in tension and pinned at a certain joint exerts a pull on the pin.
Each truss member is a two-force member if we neglect the weight of the member. This is a relatively safe assumption since the member weight is often small compared to the loads carried by the truss.
Method of Joints
The method of joints consists of a number of free-body diagrams of adjacent joints.
The first joint selected must have only two unknown forces and one or more known forces.
The unknown forces are determined by using ΣFx = 0 and ΣFy = 0. These newly found forces are used in the free-body diagram of an adjacent joint.
The load in each truss member is found by taking consecutive free-body diagrams of joints throughout the complete truss.
Method of Joints
Example:
Determine the load in each member of the truss shown
Solution:
Step 1. Solve for the external reactions RA and RH.
Step 2. Choose a pin or joint for the first free-body diagram.
Step 3. Draw a free-body diagram of A.
Step 4. Solve for AB
Step 5. Solve for AC
Step 6. Follow the value of a new known load, AB = 6.25, from pin A to an adjacent pin, B.
Step 7. Solve for BD
Step 8. Solve for BC
And so on…
Method of Joints
Step 1.
Solve for the external reactions RA and RH. The first step of the solution is one with which you are already familiar: Solve for the external reactions at points A and H. Taking moments about point H and solving for RA, we obtain:
Method of Joints
Step 1.
Solve for the external reactions RA and RH. The first step of the solution is one with which you are already familiar: Solve for the external reactions at points A and H. Taking moments about point H and solving for RA, we obtain:
Method of Joints
Step 2. Choose a pin or joint for the first free-body diagram.
Note that only four joints—A, D, G, and H—have known forces.
Joint D has four unknowns,
G has three unknowns,
Joints A and H each have two unknowns.
Thus, the first free-body diagram could be of joint A or H.
Let us arbitrarily choose joint A.
Assume member AB to be in compression and member AC to be in tension
Step 3. Draw a free-body diagram of A
Method of Joints
Step 4. Solve for AB
Step 5. Solve for AC
The compression or tension of a member should be indicated following the value with C or T, respectively
Method of Joints
Step 6. Follow the value of a new known load, AB = 6.25, from pin A to an adjacent pin, B.
Step 7. Solve for BD
Step 8. Solve for BC
Method of Joints
Step 9. Knowing the value of BC = 5, move from pin B to pin C and draw a free-body diagram
Step 10. Solve for CD
Step 11. Solve for CE
Method of Joints
Step 12. Reduce your possibility of error.
Go to joint H and work back toward the center of the truss
Step 13. Show RH = 7 kips on a free-body diagram of H
Step 14. Solve for GH, using ΣFy = 0
Step 15. Solve for EH, using ΣFx =0
Method of Joints
Step 16. Knowing GH = 8.75 kips C, move from pin H to a free-body diagram of G
Step 17. Solve for EG, using ΣFy = 0
Step 18. Solve for DG, using ΣFx = 0
Method of Joints
Step 19. Draw a free-body diagram of E showing CE = 7.5 kips, EG = 3 kips, and EH = 5.25 kips .
Step 20. Solve for DE, using ΣFy = 0
Step 21. Check accuracy of calculations, using ΣFx = 0
Method of Joints
Step 22. Label forces for all truss members
Method of Sections
The method of sections is used to solve for the force in a member near the middle of a truss.
The time-consuming method of joints is avoided.
The method of sections consists of cutting a truss into two sections by cutting through the truss where a member force is required; one section is discarded.
A free-body diagram of the remaining section is drawn. On this free-body diagram, we show a tensile or compressive force where each member was cut.
These are equivalent forces that have the same effect as the discarded section had. Suppose that a truss member, cut in two by the method of sections, had been in compression. The free-body diagram would show a force pushing on the remaining half of the member.
Only three members are usually cut at one time, although a partial solution is possible when four or more members are cut.
Method of Sections
Solve for the load in members CB, AB, and JK of the truss shown
Method of Sections
Solve for the load in members CB, AB, and JK of the truss shown
ΣFy = 0
ΣMC = 0
Method of Sections
Solve for the load in members CB, AB, and JK of the truss shown
ΣMD = 0
Method of Members
In the previous truss problems, the truss members were two-force members; there was a force acting at each end of the member; each member was in either tension or compression. The force of a member on a joint had the same direction as the slope of the member. For this reason, we could draw free-body diagrams of individual joints.
Free-body diagrams of joints cannot be used when the members have three or more forces acting on them. A three-force member may be subject to bending, and the force that it exerts on a joint no longer has the same slope or direction as the member.
Therefore, a free-body diagram is drawn of the member, not of the joint.
A frame consists of a number of members fastened together so that each member has two or more forces acting on it. Determining these forces consists of drawing free-body diagrams of individual members or of the complete frame.
Method of Members
Example
Note that the internal forces at B are in opposite directions in the two diagrams. If AB pushes down on CD, then CD pushes up on A
Method of Members
ΣMA = 0
ΣMD = 0
N
References
All figures and examples are taken from
APPLIED MECHANICS FOR ENGINEERING TECHNOLOGY, EIGHTH EDITION, Keith M. Walker, ISBN 978-0-13-172151-7
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