A Simplified Approach to Calculating Truss ForcesJames Nadir
A simplified approach to calculating truss forces. This presentation is built upon a wonderful presentation which comes from Fazirah Abdul Ghafar. This presentation uses Power Point Animations and is best viewed using Power Point Slide Show to present.
A Simplified Approach to Calculating Truss ForcesJames Nadir
A simplified approach to calculating truss forces. This presentation is built upon a wonderful presentation which comes from Fazirah Abdul Ghafar. This presentation uses Power Point Animations and is best viewed using Power Point Slide Show to present.
Concept of Particles and Free Body Diagram
Why FBD diagrams are used during the analysis?
It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.
Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.
Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.
A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.
Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.
Solutions manual for hydrologic analysis and design 4th edition by mc cuen ib...frazob
Solutions manual for hydrologic analysis and design 4th edition by mc cuen ibsn 9780134313122
download at: https://goo.gl/hA8T7p
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Concept of Particles and Free Body Diagram
Why FBD diagrams are used during the analysis?
It enables us to check the body for equilibrium.
By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body.
It helps to identify the forces and ensures the correct use of equation of equilibrium.
Note:
Reactions on two contacting bodies are equal and opposite on account of Newton's III Law.
The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces.
Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable.
Physical Meaning of Equilibrium and its essence in Structural Application
The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium.
A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium.
The rigid body in equilibrium means the body is stable.
Equilibrium means net force and net moment acting on the body is zero.
Essence in Structural Engineering
To find the unknown parameters such as reaction forces and moments induced by the body.
In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters.
For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.
Solutions manual for hydrologic analysis and design 4th edition by mc cuen ib...frazob
Solutions manual for hydrologic analysis and design 4th edition by mc cuen ibsn 9780134313122
download at: https://goo.gl/hA8T7p
people also search:
hydrologic analysis and design 4th edition pdf
hydrologic analysis and design 3rd edition pdf download
hydrologic analysis and design 4th edition pdf download
hydrologic analysis and design 4th edition solutions
hydrologic analysis and design pdf
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11 - 3
Experiment 11
Simple Harmonic Motion
Questions
How are swinging pendulums and masses on springs related? Why are these types of
problems so important in Physics? What is a spring’s force constant and how can you measure
it? What is linear regression? How do you use graphs to ascertain physical meaning from
equations? Again, how do you compare two numbers, which have errors?
Note: This week all students must write a very brief lab report during the lab period. It is
due at the end of the period. The explanation of the equations used, the introduction and the
conclusion are not necessary this week. The discussion section can be as little as three sentences
commenting on whether the two measurements of the spring constant are equivalent given the
propagated errors. This mini-lab report will be graded out of 50 points
Concept
When an object (of mass m) is suspended from the end of a spring, the spring will stretch
a distance x and the mass will come to equilibrium when the tension F in the spring balances the
weight of the body, when F = - kx = mg. This is known as Hooke's Law. k is the force constant
of the spring, and its units are Newtons / meter. This is the basis for Part 1.
In Part 2 the object hanging from the spring is allowed to oscillate after being displaced
down from its equilibrium position a distance -x. In this situation, Newton's Second Law gives
for the acceleration of the mass:
Fnet = m a or
The force of gravity can be omitted from this analysis because it only serves to move the
equilibrium position and doesn’t affect the oscillations. Acceleration is the second time-
derivative of x, so this last equation is a differential equation.
To solve: we make an educated guess:
Here A and w are constants yet to be determined. At t = 0 this solution gives x(t=0) = A,
which indicates that A is the initial distance the spring stretches before it oscillates. If friction is
negligible, the mass will continue to oscillate with amplitude A. Now, does this guess actually
solve the (differential) equation? A second time-derivative gives:
Comparing this equation to the original differential equation, the correct solution was
chosen if w2 = k / m. To understand w, consider the first derivative of the solution:
−kx = ma
a = −
k
m
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
x
d 2x
dt 2
= −
k
m
x x(t) = A cos(ωt)
d 2x(t)
dt 2
= −Aω2 cos(ωt) = −ω2x(t)
James Gering
Florida Institute of Technology
11 - 4
Integrating gives
We assume the object completes one oscillation in a certain period of time, T. This helps
set the limits of integration. Initially, we pull the object a distance A from equilibrium and
release it. So at t = 0 and x = A. (one.
2. READING QUIZ 1. In the method of sections, generally a “cut” passes through no more than _____ members in which the forces are unknown. A) 1 B) 2 C) 3 D) 4 2. If a simple truss member carries a tensile force of T along its length, then the internal force in the member is ______ . A) tensile with magnitude of T/2 B) compressive with magnitude of T/2 C) compressive with magnitude of T D) tensile with magnitude of T
3. APPLICATIONS Long trusses are often used to construct bridges. The method of joints requires that many joints be analyzed before we can determine the forces in the middle part of the truss. Is there another method to determine these forces directly?
4. THE METHOD OF SECTIONS In the method of sections, a truss is divided into two parts by taking an imaginary “cut” (shown here as a-a) through the truss. Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut member will also be either tensile or compressive with the same magnitude. This result is based on the equilibrium principle and Newton’s third law .
5. STEPS FOR ANALYSIS 1. Decide how you need to “cut” the truss. This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general). 2. Decide which side of the cut truss will be easier to work with (minimize the number of reactions you have to find). 3. If required , determine the necessary support reactions by drawing the FBD of the entire truss and applying the EofE.
6. PROCEDURE (continued) 4. Draw the FBD of the selected part of the cut truss. We need to indicate the unknown forces at the cut members. Initially we assume all the members are in tension, as we did when using the method of joints. Upon solving, if the answer is positive, the member is in tension as per our assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either tension or compression by inspection as was done in the figures above.)
7. PROCEDURE (continued) 5. Apply the equations of equilibrium (EofE) to the selected cut section of the truss to solve for the unknown member forces. Please note that in most cases it is possible to write one equation to solve for one unknown directly.
8. EXAMPLE Given : Loads as shown on the roof truss. Find : The force in members DE, DL, and ML. Plan: a) Take a cut through the members DE, DL, and ML. b) Work with the left part of the cut section. Why? c) Determine the support reaction at A. What are they? d) Apply the EofE to find the forces in DE, DL, and ML.
9. EXAMPLE (continued) Analyzing the entire truss, we get F X = A X = 0. By symmetry, the vertical support reactions are A Y = I Y = 36 kN + M D = – 36 (8) + 6 (8) + 12 (4) + F ML (5) = 0 F ML = 38 . 4 kN ( T )
10.
11. CONCEPT QUIZ 1. Can you determine the force in member ED by making the cut at section a-a? Explain your answer. A) No, there are 4 unknowns. B) Yes, using M D = 0 . C) Yes, using M E = 0 . D) Yes, using M B = 0 .
12. CONCEPT QUIZ 2. If you know F ED , how will you determine F EB ? A) By taking section b-b and using M E = 0 B) By taking section b-b, and using F X = 0 and F Y = 0 C) By taking section a-a and using M B = 0 D) By taking section a-a and using M D = 0
13. GROUP PROBLEM SOLVING Given : Loading on the truss as shown. Find : The force in members BC, BE, and EF. Plan: a) Take a cut through the members BC, BE, and EF. b) Analyze the top section (no support reactions!). c) Draw the FBD of the top section. d) Apply the equations of equilibrium such that every equation yields answer to one unknown.
14. SOLUTION + F X = 5 + 10 – F BE cos 45 º = 0 F BE = 21 . 2 kN (T) + M E = – 5(4) + F CB (4) = 0 F CB = 5 kN (T) + M B = – 5 (8) – 10 (4) – 5 (4) – F EF (4) = 0 F EF = – 25 kN or 25 kN (C)
15. ATTENTION QUIZ 1. As shown, a cut is made through members GH, BG and BC to determine the forces in them. Which section will you choose for analysis and why? A) Right, fewer calculations. B) Left, fewer calculations. C) Either right or left, same amount of work. D) None of the above, too many unknowns.
16. ATTENTION QUIZ 2. When determining the force in member HG in the previous question, which one equation of equilibrium is best to use? A) M H = 0 B) M G = 0 C) M B = 0 D) M C = 0