Structures: Difference between revisions

Latest revision as of 03:58, 8 June 2009

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, compression & shear. 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 the 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

A 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> \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)

Each successive member may be analyzed for its internal forces, be they compressive or tensile in a similar manner. For a successful analysis of every member of the truss, it is imperative that every external force be included. Therefore it is advised that the engineer begin analysis by the method of joints at each joint where external forces have been resolved and work his or her way inward towards the center.

Method of Sections

If every joint in a structure does not need to be analyzed, yet the internal forces of some members need to be resolved, a potential shortcut method is the method of sections. The method of sections involves cutting a truss into a smaller part, thus changing the internal forces in the members in question to external forces, which may easily be solved by the method of Equilibrium.

File:LoadedTrussSection.JPG

Example:
If the truss shown here were to need analysis for the internal forces on one particular member, EF, the method of sections could be used. Here we use the previous values for <math>\vec F_1 = 3N</math> and <math>\vec R_2 = 14N</math>. Also assume that each member is <math>\ 5</math> meters long.

Solution:
We solve the external forces on the truss completely. Equilibrium dictates that <math> \sum \vec F = 0 </math> and <math> \sum \vec M = 0 </math>.

Once again, we first sum the moments about a particular point. In this case, let's use I:

<math> \vec F_{EH} = 17.3 N</math> (in tension).

Space Trusses

File:SpaceTruss.png

A Triangle Space Truss

A Space Truss is a 3D plane truss. Made up of similar structural forms, the space truss is a much more realistic form of truss thatn the Plane Truss. Calculation and resolution of loads, external and internal forces is performed the same way, however, care must be made to ensure that all three dimensions are correctly examined and properly resolved. Once again <math> \sum \vec F = 0 </math> and <math> \sum \vec M = 0 </math>.

Assignments

Activities:

Create an activity

Readings:

Peruse the appropriate sections of Wikibooks:Statics
Introduction to Structural Design, Virginia Tech.

Study guide:

Wikipedia article:Plane Truss
Wikipedia article:Tension
Wikipedia article:Compression
Wikipedia article:Method of Joints
Wikipedia article:Method of Sections
Wikipedia article:Space Truss

@@ Line 1: / Line 1: @@
-'''Part of the [[Topic:Statics|Statics]] course offered by the ''[[Topic:Applied Mechanics|Division of Applied Mechanics]]'', ''[[School:Engineering|School of Engineering]]'' and the ''[[Portal:Engineering and Technology|Engineering and Technology Portal]]'''''
+'''Part of the [[Statics|Statics]] course offered by the ''[[Applied Mechanics|Division of Applied Mechanics]]'', ''[[Engineering|School of Engineering]]'' and the ''[[Portal:Engineering and Technology|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.
+Structural engineering relies heavily on the strengths of materials and their ability to withstand forces of tension, compression & shear.  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.
+{{ntnes}}
 ===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.
+One way of distributing the 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.
 [[Image:Warren truss.PNG|thumb|400px|left|A Warren Truss]][[Image:Pratt truss.PNG|thumb|400px|right|A Pratt Truss]]
 <BR>
+[[Image:Simpletruss.PNG|thumb|200px|left|A Simple Triangle Truss]]
 <BR>
 <BR>
 <BR>
 <BR>
-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.
+<BR>
-[[Image:Simpletruss.PNG|thumb|400px|left|A Simple Triangle Truss]]
+<BR>
+A 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.
+<BR>
+<BR>
 ===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.
-===Method of Sections===
+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.
-===Space Trusses===
-Newton's Second Law for a static system (Equilibrium):
+'''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 are the forces in each of the members of the loaded truss, under the loads applied?
- <math> \sum \vec F \ = M * \vec a = 0 </math>
+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>.
-The sum of the forces is equal to the mass times the acceleration. The 2nd Law tells us that if the object or system is motionless, the acceleration is equal to zero. Therefore the sum of the vector forces must be equal to zero.
-'''Example:'''
-[[Image:Equilibrium.JPG|left|250px|thumb]]
-Consider a table with four legs and a 200 kg object resting motionless in the center of the table.  What force acts upon the bottom of each table leg <math> \vec F_L </math>?
-The force upward on the legs is found by balancing these forces in the equation given by Newton's Law.  Gravity is acting downward on the 200 kg object and the 100 kg table.  Therefore, we may substitute the acceleration due to gravity on Earth <math> \vec g </math> which is <math> 9.81 \frac{m}{s^2} </math>, for <math> \vec a </math>.
-<BR>
-<BR>
-Substitute in known values...
-<math> \sum \vec F_y = M * \vec a = 0 = - (200 kg * 9.81 \frac{m}{s^2}) - (100 kg * 9.81 \frac{m}{s^2}) + (F_L * 4) </math>
-<BR>
-<BR>
-And simplifying...
-<BR>
 <BR>
-<math> \vec F_L = \frac{200 kg * 9.81 \frac{m}{s^2} + 100 kg * 9.81 \frac{m}{s^2}}{4} </math>
+[[Image:LoadedTrussBridge.JPG|center|600px|thumb]]
-The unit <math> \frac{kg.m}{s^2} </math> is equivalent to the unit of force called a Newton <i>'''N'''</i>.  Thus when multiplying through, the kilograms cancel out, and we may solve for <math> \vec F_L </math>...
-<BR>
 <BR>
-<math> \vec F_L = \frac{1962  N + 981  N}{4} = 735.75  N </math>
+'''Solution:'''<BR>
+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>.
-This means that each leg exerts a downward force of 735.75 Newtons on the floor, and the floor simultaneously exerts the same force upward on each table leg.
+''First'' we sum the ''moments'' about a particular point.  In this case, let's use '''A''':
-Notice that it is critical that a consistent sign convention be followed throughout the entire analysis effort.  The sign convention is typically chosen in more complex problems to ease the total amount of algebra necessary to analyze the equations in the <math> x </math>, <math> y </math> and <math> z </math> axis.  We chose positive values to mean upward forces and negative values to mean downward forces, but we could have also used the opposite, provided we were consistent.
+<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>
-===Rotational Equilibrium===
+ <math> \vec R_2 = 14.1 N</math>.
-Lesson: In Statics the Sum of the Torques (Moments) is Equal to Zero.
+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>.
-The sum of all rotational forces, or torques, denoted by the capital Greek letter <i>tau ('''<math>\Tau </math>''')</i>, is also zero.  Commonly used units for torque are <i>Foot•pounds ('''ft•lb''')</i> and <i>Newton•meters ('''N•m''')</i>.
+<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...
-Newton's Second Law (applied to torques):
+ <math> \vec R_1 = 6.9N</math>
- <math>\sum \tau = \sum (\vec \omega * I) = (\vec \omega_1 * I_1) + (\vec \omega_2 * I_2) + (\vec \omega_3 * I_3) + ... + (\vec \omega_n * I_n)</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>.
-The sum of the torques is equal to the rotational mass or <i> moment of inertia ('''I''')</i> times the angular acceleration, denoted by the lower case Greek letter <i>omega ('''<math>\omega</math>''')</i>. The 2nd Law tells us that if the object or system is motionless, the angular acceleration is equal to zero. Therefore the sum of the vector torques must be equal to zero.
+<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)''
-Torques may also be calculated as forces times distances:
+<BR>
+<math> \sum \vec F_{xA} \ = \vec F_{AC} - \vec F_{AB} * sin{30} = 0  </math>
+<BR>Or...<BR>
+ <math> \vec F_{AC} = \vec F_{AB} * sin{30} = 7.97N * sin{30} = 3.99N </math> ''(in Tension)''
-  <math>\sum \vec \tau = \sum (\vec F * d) = \vec F_1 * d_1 + \vec F_2 * d_2 + \vec F_3 * d_3 + ... + \vec F_n * d_n </math>
+Each successive member may be analyzed for its internal forces, be they ''compressive'' or ''tensile'' in a similar manner.  For a successful analysis of every member of the truss, it is imperative that every external force be included.  Therefore it is advised that the engineer begin analysis by the method of joints at each joint where external forces have been resolved and work his or her way inward towards the center.
+===Method of Sections===
+If every joint in a structure does not need to be analyzed, yet the internal forces of some members need to be resolved, a potential shortcut method is the method of sections.  The method of sections involves cutting a truss into a smaller part, thus changing the internal forces in the members in question to external forces, which may easily be solved by the method of Equilibrium.
-'''Example:'''
+[[Image:LoadedTrussSection.JPG|200px|thumb|left]]
-[[image:LeverExample.png|thumb|200px|left]]
+'''Example:'''<BR>
-Consider a massless lever with two weights attached and a single massless support:
+If the truss shown here were to need analysis for the internal forces on one particular member, '''EF''', the method of sections could be used.  Here we use the previous values for <math>\vec F_1 = 3N</math>  and  <math>\vec R_2 = 14N</math>.  Also assume that each member is <math>\ 5</math> meters long.
-<BR>
-<BR>
-''Weight of Object 1, <math>\vec F_1</math> = 10 lb''<br>
-''Distance of Object 1 from fulcrum = 10 ft''<br>
-''Distance of support from fulcrum  = 7 ft''<br>
-''Weight of Object 2, <math>\vec F_2</math> = 80 lb''<br>
-''Distance of Object 2 from fulcrum = 4 ft''<br>
-<BR>
-What forces act upon the massless lever?''
-<BR>
-<BR>
-First let's write the static torque equation for the system.  Forces 1 and 2 are actually the weights of the two objects...
-<math>\sum \tau = \sum (\vec F * d) = \vec F_1 * d_1 + \vec F_2 * d_2 + \vec F_S * d_S = 0 </math>
+'''Solution:'''<BR>
-<BR>
+We solve the external forces on the truss completely.  Equilibrium dictates that <math> \sum \vec F = 0 </math>  and  <math> \sum \vec M = 0 </math>.
-<BR>
-Substitute in known values...
-<math>\sum \tau = \sum (\vec F * d) = (-10 lb)*(10 ft) + (-80 lb)*(4 ft) + \vec F_S * (7 ft) = 0 </math>
+Once again, we first sum the ''moments'' about a particular point.  In this case, let's use '''I''':
-<BR>
-<BR>
-Simplify...
-<math>\sum \tau = \sum (\vec F * d) = (-100 ft.lb.) + \vec F_S * (7 ft) + (-320 ft.lb.) = 0 </math>
+<math> \sum \vec M_I \ = -\vec F_1*L + \vec F_{EH}*cos{30} = 0</math>
-<BR>
+<BR>Or...<BR>
-<BR>
+<math> \sum \vec M_I \ = -3N*5m + \vec F_{EH}*cos{30} = 0</math>
-And solve for <math>\vec F_S</math> ...
-<math>\vec F_S * (7 ft) \ = (420 ft.lb.) </math> and therefore <math>\vec F_S = (420 ft.lb.) / (7 ft) \ = 60 lb.</math>
+ <math> \vec F_{EH} = 17.3 N</math> (in tension).
-<BR>
-<BR>
-Thus, a force of 60 pounds acts upward on the lever from the support in order to balance the torques on the lever.  Since the total weight on the lever is 90 pounds and 60 of those pounds are countered by the support, the remaining 30 pounds must act upward on the lever at the fulcrum in the lower, left corner.
-Note that we didn't need to use the fulcrum as the point from which all distances are measured, we could have chosen any point along the lever.  However, as with forces, a consistent sign convention must be used.  In this case, positive may be used for clockwise (CW) torques measured from one point of view, with negative torques used for counterclockwise (CCW) torques measured from the same POV.  The opposite system could also be used, so long as we were consistent.
-Also note that the table example used previously stated that the object was in the center of the table.  If not, then balancing the torques (in two directions) would result in more force supported by some table legs and less by others.  If the object was directly over one of the legs, for example, then it's entire weight would be supported by that leg, in addition to one-fourth of the table's weight.  The other legs would then only support one-fourth of the table's weight each.
+===Space Trusses===
+[[Image:SpaceTruss.png|thumb|300px|left|A Triangle Space Truss]]
+A Space Truss is a 3D plane truss.  Made up of similar structural forms, the space truss is a much more realistic form of truss thatn the Plane Truss.  Calculation and resolution of loads, external and internal forces is performed the same way, however, care must be made to ensure that all three dimensions are correctly examined and properly resolved.  Once again  <math> \sum \vec F = 0 </math>  and  <math> \sum \vec M = 0 </math>.
+<BR><BR><BR><BR><BR><BR><BR><BR>
 ==Assignments==
 '''Activities:'''
-* Create an [[activity]]
+* Create an [[Structures/activity|activity]]
 '''Readings:'''
 * Peruse the appropriate sections of [[Wikibooks:Statics]]
+* [http://urban.arch.virginia.edu/%7Ekm6e/arch324/highlights/home.html Introduction to Structural Design, Virginia Tech.]
 '''Study guide:'''
-# Wikipedia article:[[w:Plane Truss|Plane Truss]]
+# Wikipedia article:[[Wikipedia:Plane Truss|Plane Truss]]
-# Wikipedia article:[[w:Tension|Tension]]
+# Wikipedia article:[[Wikipedia:Tension|Tension]]
-# Wikipedia article:[[w:Compression|Compression]]
+# Wikipedia article:[[Wikipedia:Compression|Compression]]
-# Wikipedia article:[[w:Gravitational_acceleration|Gravitational Acceleration]]
+# Wikipedia article:[[Wikipedia:Method of Joints|Method of Joints]]
-# Wikipedia article:[[w:Newton|Newton]]
+# Wikipedia article:[[Wikipedia:Method of Sections|Method of Sections]]
-# Wikipedia article:[[w:Torque|Torque]]
+# Wikipedia article:[[Wikipedia:Space Truss|Space Truss]]
-# Wikipedia article:[[w:Moment of inertia|Moment of inertia]]
-# Wikipedia article:[[w:Angular acceleration|Angular acceleration]]
 [[Category:Statics]]
 [[Category:Advanced Classical Mechanics]]
-[[Category:Applied Mechanics]]
+[[Category:Applied mechanics]]
-[[Category:Mechanical Engineering]]
 [[Category:Engineering]]

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