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===Method of Joints===
===Method of Joints===
The principle of equilibrium may be used to solve the loads impinging upon a truss by analyzing the effect of each load at a single point on the truss.  Each load is broken down into its <math>\ x</math> and <math>\ y</math> component vectors and these are summed to equal zero at a particular point on the truss (e.g. '''A''').  Similarly, each load creates a moment rotating about that same point '''A''' and these may also be summed together to equal zero.  Therefore, given enough information about the loads applied, any of the other loads may be calculated.
The principle of equilibrium may be used to solve the loads impinging upon a ''massless'' truss by analyzing the effect of each load at a single point on the truss.  Each load is broken down into its <math>\ x</math> and <math>\ y</math> component vectors and these are summed to equal zero at a particular point on the truss (e.g. '''A''').  Similarly, each load creates a moment rotating about that same point '''A''' and these may also be summed together to equal zero.  Therefore, given enough information about the loads applied, any of the other loads may be calculated.


Therefore, the Method of Joints may also be applied to a truss joint to determine the force (compressive or tensile) in each member of the truss.  A place where the forces and members meet may become the joint to be examined ('''A'''), and the principle of equilibrium is applied, thus solving the question of where the forces are transmitted in that truss around that joint.
Therefore, the Method of Joints may also be applied to a truss joint to determine the force (compressive or tensile) in each member of the truss.  A place where the forces and members meet may become the joint to be examined ('''A'''), and the principle of equilibrium is applied, thus solving the question of where the forces are transmitted in that truss around that joint.
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'''Example:'''<BR>
'''Example:'''<BR>
Consider a truss bridge under loads <math>\vec F_1</math>, <math>\vec F_2</math> and <math>\vec F_L</math>.  What resultant forces <math>\vec R_1 </math> and <math>\vec R_2 </math> act upon the bottom the bridge if <math>\vec F_1 = 3N</math> and <math>\vec F_2 = 7N</math>?  Assume that <math>\vec R_1 </math> and <math>\vec R_2 </math> are <math>\ L</math> meters apart, and <math>\vec F_L</math> is <math>\frac{2}{3}L</math> from <math>\vec R_1 </math>.
Consider a truss bridge under loads <math>\vec F_1</math>, <math>\vec F_2</math> and <math>\vec F_L</math>.  What are the forces in each of the members of the loaded truss, under the loads applied?
 
Assume that <math>\vec F_1 = 3N</math>, <math>\vec F_2 = 7N</math> and <math>\vec F_L = 11N</math>.  Also assume that each member '''AB''', '''CE''' etc. is <math>\ 5</math> meters long and that <math>\vec F_L</math> is <math>\frac{2}{3}L</math> from <math>\vec R_1 </math>.
<BR>
<BR>
[[Image:LoadedTrussBridge.JPG|center|600px|thumb]]
[[Image:LoadedTrussBridge.JPG|center|600px|thumb]]
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<BR>
<BR>
'''Solution:'''<BR>
'''Solution:'''<BR>
Equilibrium dictates that  <math> \sum \vec F = 0 </math>  and  <math> \sum \vec M = 0 </math>.
First we must solve the external forces on the truss completely.  We do so by analyzing the free body diagram.  Equilibrium dictates that  <math> \sum \vec F = 0 </math>  and  <math> \sum \vec M = 0 </math>.


We may sum the forces around an arbitrary point '''A'''.  Lets select the location of <math>\vec F_1</math> for '''A'''...
''First'' we sum the ''moments'' about a particular point.  In this case, let's use '''A''':


There are no <math>\ x</math> components of the force vectors, so  <math> \sum \vec F_x \ = 0 </math>.
<math> \sum \vec M_A \ = \vec F_1*2L + \vec F_2*3L + \vec F_L*\frac{8L}{3} - \vec R_2*4L = 0</math>  
<BR>Or...<BR>
<math> \sum \vec M_A \ = 3N*10m + 7N*15m + 11N*\frac{8*5m}{3} - \vec R_2*20 = 0</math>  


However, <math> \sum \vec F_y \ = \vec R_1 + \vec R_2 - \vec F_1 - \vec F_2 - \vec F_L = 0 </math>.
<math> \vec R_2 = 14.1 N</math>.


Therefore, <math> \sum \vec F_y \ = \vec R_1 + \vec R_2 - 3N - 7N - \vec F_L = 0 </math>  and...


  <math> \vec R_1 + \vec R_2 = 10N + \vec F_L</math>   (1)
We may sum the forces around an arbitrary point '''A'''. There are no <math>\ x</math> components of the force vectors, so  <math> \sum \vec F_x \ = 0 </math>.


<math> \sum \vec F_{yA} \ = \vec R_1 + \vec R_2 - \vec F_1 - \vec F_2 - \vec F_L = 0 </math>.
<BR>Or...<BR>
<math> \sum \vec F_{yA} \ = \vec R_1 + 14.1N - 3N - 7N - 11N = 0 </math>  and...


We may then sum the moments around point '''A'''...
<math> \vec R_1 = 6.9N</math>


<math> \sum \vec M_A \ = \vec R_1*\frac{L}{2} - \vec R_2*\frac{L}{2} + \vec F_2*\frac{L}{4} + \vec F_L*\frac{L}{6} = 0</math>
[[Image:TrussJointA.JPG|left|200px|thumb]]
Secondly, we may begin to solve for the internal forces within each member by summing the forces around point '''A''' at the location of <math>\vec R_1</math>.


Therefore, <math> \sum \vec M_A \ = \vec R_1*\frac{L}{2} - \vec R_2*\frac{L}{2} + 7N*\frac{L}{4} + \vec F_L*\frac{L}{6} = 0</math>.  However, this leaves us with too many unknowns to solve independently.
<math> \sum \vec F_{yA} \ = \vec R_1 - \vec F_{AB} * cos{30} = 0 </math>
<BR>Or...<BR>
<math> \vec F_{AB} = \frac{\vec R_1}{cos{30}} = \frac{6.9N}{cos{30}} = 7.97N </math> ''(in Compression)''


 
<BR>
Alternately, we may sum the forces around an arbitrary point '''B''' at the location of <math>\vec F_2</math>. This has no effect on the force equilibrium, so next we sum the moments around point '''B'''...
<math> \sum \vec F_{xA} \ = \vec F_{AC} - \vec F_{AB} * sin{30} = 0 </math>
 
<BR>Or...<BR>
<math> \sum \vec M_B \ = \vec R_1*\frac{3L}{4} - \vec R_2*\frac{L}{4} + \vec F_1*\frac{L}{4} - \vec F_L*\frac{L}{12} = 0</math>
  <math> \vec F_{AC} = \vec F_{AB} * sin{30} = 7.97N * sin{30} = 3.99N </math> ''(in Tension)''
 
Substituting in the values we already have...
 
<math> \vec R_1*\frac{3L}{4} - \vec R_2*\frac{L}{4} + 3N*\frac{L}{4} - \vec F_L*\frac{L}{12} = 0</math>  or <math> \vec R_1*\frac{3L}{4} - \vec R_2*\frac{L}{4} + 3N*\frac{L}{4} = \vec F_L*\frac{L}{12}</math>
 
Re-arranging... 
 
<math> \vec F_L = 9\vec R_1 - 3\vec R_2 + 9N</math>   (2)
 
We can combine (2) with (1) from above and get
 
<math> \vec F_L = \vec R_1 + \vec R_2 - 10N = 9\vec R_1 - 3\vec R_2 + 9N</math>
 
or
 
<math>  \vec R_2 = 2\vec R_1 + \frac{19N}{4}</math>


===Method of Sections===
===Method of Sections===

Revision as of 23:19, 28 October 2007

Part of the Statics course offered by the Division of Applied Mechanics, School of Engineering and the Engineering and Technology Portal

Lecture

Structural engineering relies heavily on the strengths of materials and their ability to withstand forces of tension or compression. When used in conjunction with each other, as in the case of a truss, individual load bearing members both share and transmit loads, enabling the structure to accomplish much more than any individual member could alone.

Plane Trusses

One way of distributing a force across a large distance is by building a Plane Truss, which takes advantage of the principle of Equilibrium to translate forces along a system of interconnecting members.

A Warren Truss
A Pratt Truss


File:Simpletruss.PNG
A Simple Triangle Truss







The simplest truss is a triangle made of three points: A, B and C, and three members: AB, BC and CA. The method of distributing forces amongst many members relies on the engineer's ability to place a tensile or compressive force at any particular location. To properly sum forces at a particular point, one must be able to sum forces into that point and also away from it. Thus, AB and AC are in compression while BC is in tension.

Method of Joints

The principle of equilibrium may be used to solve the loads impinging upon a massless truss by analyzing the effect of each load at a single point on the truss. Each load is broken down into its <math>\ x</math> and <math>\ y</math> component vectors and these are summed to equal zero at a particular point on the truss (e.g. A). Similarly, each load creates a moment rotating about that same point A and these may also be summed together to equal zero. Therefore, given enough information about the loads applied, any of the other loads may be calculated.

Therefore, the Method of Joints may also be applied to a truss joint to determine the force (compressive or tensile) in each member of the truss. A place where the forces and members meet may become the joint to be examined (A), and the principle of equilibrium is applied, thus solving the question of where the forces are transmitted in that truss around that joint.


Example:
Consider a truss bridge under loads <math>\vec F_1</math>, <math>\vec F_2</math> and <math>\vec F_L</math>. What are the forces in each of the members of the loaded truss, under the loads applied?

Assume that <math>\vec F_1 = 3N</math>, <math>\vec F_2 = 7N</math> and <math>\vec F_L = 11N</math>. Also assume that each member AB, CE etc. is <math>\ 5</math> meters long and that <math>\vec F_L</math> is <math>\frac{2}{3}L</math> from <math>\vec R_1 </math>.

File:LoadedTrussBridge.JPG


Solution:
First we must solve the external forces on the truss completely. We do so by analyzing the free body diagram. Equilibrium dictates that <math> \sum \vec F = 0 </math> and <math> \sum \vec M = 0 </math>.

First we sum the moments about a particular point. In this case, let's use A:

<math> \sum \vec M_A \ = \vec F_1*2L + \vec F_2*3L + \vec F_L*\frac{8L}{3} - \vec R_2*4L = 0</math>
Or...
<math> \sum \vec M_A \ = 3N*10m + 7N*15m + 11N*\frac{8*5m}{3} - \vec R_2*20 = 0</math>

<math> \vec R_2 = 14.1 N</math>.  


We may sum the forces around an arbitrary point A. There are no <math>\ x</math> components of the force vectors, so <math> \sum \vec F_x \ = 0 </math>.

<math> \sum \vec F_{yA} \ = \vec R_1 + \vec R_2 - \vec F_1 - \vec F_2 - \vec F_L = 0 </math>.
Or...
<math> \sum \vec F_{yA} \ = \vec R_1 + 14.1N - 3N - 7N - 11N = 0 </math> and...

<math> \vec R_1 = 6.9N</math>
File:TrussJointA.JPG

Secondly, we may begin to solve for the internal forces within each member by summing the forces around point A at the location of <math>\vec R_1</math>.

<math> \sum \vec F_{yA} \ = \vec R_1 - \vec F_{AB} * cos{30} = 0 </math>
Or...

<math> \vec F_{AB} = \frac{\vec R_1}{cos{30}} = \frac{6.9N}{cos{30}} = 7.97N </math> (in Compression)


<math> \sum \vec F_{xA} \ = \vec F_{AC} - \vec F_{AB} * sin{30} = 0 </math>
Or...

<math> \vec F_{AC} = \vec F_{AB} * sin{30} = 7.97N * sin{30} = 3.99N </math> (in Tension)

Method of Sections

Space Trusses

Assignments

Activities:

Readings:

Study guide:

  1. Wikipedia article:Plane Truss
  2. Wikipedia article:Tension
  3. Wikipedia article:Compression
  4. Wikipedia article:Method of Joints
  5. Wikipedia article:Method of Sections