ARCHITECTURE 2351ARCHITECTURE 2351

INTRODUCTION TOINTRODUCTION TOSTRUCTURE CONCEPTSSTRUCTURE CONCEPTS

Structural elements, whether rigid or flexible, Structural elements, whether rigid or flexible, have volume and mass, and are subject to the have volume and mass, and are subject to the forces of gravity.forces of gravity.

Most all structural elements are necessary, Most all structural elements are necessary, primarily because they are affected by the forces primarily because they are affected by the forces of gravity, and must exist in order to restrain of gravity, and must exist in order to restrain themselves, and possibly some other structural themselves, and possibly some other structural element against such affects of gravity in order to element against such affects of gravity in order to remain at rest.remain at rest.

A body at rest is said to be in equilibrium.A body at rest is said to be in equilibrium.

A body in equilibrium acted upon by another A body in equilibrium acted upon by another body reacts with an equal magnitude and opposite body reacts with an equal magnitude and opposite action, called a reaction. action, called a reaction.

During equilibrium, a person sitting at rest During equilibrium, a person sitting at rest has weight due to gravity, which pushes down on has weight due to gravity, which pushes down on the chair in which that person is at rest. The chair the chair in which that person is at rest. The chair pushes upward an equal and opposite amount. pushes upward an equal and opposite amount. The person and the chair have weight, which The person and the chair have weight, which pushes down on the floor, which in turn pushes pushes down on the floor, which in turn pushes upward the same amount of weight of the chair upward the same amount of weight of the chair and the person.and the person.

So it is in structure; Roofing pushes down So it is in structure; Roofing pushes down on roof deck, which pushes down on joists, which on roof deck, which pushes down on joists, which push down on beams, which push down on push down on beams, which push down on columns, which push down on foundations, which columns, which push down on foundations, which push down on earth - - - each because of gravity, push down on earth - - - each because of gravity, and each successive element pushes upward the and each successive element pushes upward the same amount of the sum of the weights above.same amount of the sum of the weights above.

We think in terms of these items pushing We think in terms of these items pushing downward, while actually, those items of structure downward, while actually, those items of structure don’t DO anything. They simply have weight that don’t DO anything. They simply have weight that reacts to the force of gravity . . .reacts to the force of gravity . . .

and gravity has a natural tendency to and gravity has a natural tendency to

pullpull them downward toward the earth. them downward toward the earth.

So man, places these items in opposition to So man, places these items in opposition to gravity in order to create those things we call gravity in order to create those things we call enclosed space, which we as architects refer to asenclosed space, which we as architects refer to as

The Built EnvironmentThe Built Environment

And those elements are successful in And those elements are successful in maintaining that space according to the integrity maintaining that space according to the integrity of their strength.of their strength.

beam

earth

foundation

column3

earth, the earth pushes upward - the foundtions push downward on the

foundations, the foundations pushthe columns push downward on the

on the columns, the columns pushthe ends of the beam push downward

the earth supports itself.

upward.

upward

upward an equalthe string pulls

downwardweight pulls

weight of objects above

a block is supportedby a string and doesnot move, yet gravityacts to pull the weight

downward

weight

1string

amount

chair push downwarddue to weight andgravity

a person in a

weight on beam pushes downward, the beam pushes upward

B C

2A

the floor pushesupward the sameamount

if equilibrium occursthe forces at B & Cmust equal the forceat A

OBJECTS IN EQUILIBRIUM

In order to simplify the graphical analysis of In order to simplify the graphical analysis of structural members, it may be necessary to illustrate the structural members, it may be necessary to illustrate the subject as a subject as a

FREE – BODY DIAGRAMFREE – BODY DIAGRAM

which means that any body can be illustrated as an object which means that any body can be illustrated as an object suspended in equilibrium as long as the forces that act to suspended in equilibrium as long as the forces that act to hold it in equilibrium are shown. hold it in equilibrium are shown.

And further, any object can be divided into parts, whether And further, any object can be divided into parts, whether separated at connections, or whether members are shown separated at connections, or whether members are shown to be cut at some point or plane - - - and each part can be to be cut at some point or plane - - - and each part can be illustrated as an object suspended in equilibrium if the illustrated as an object suspended in equilibrium if the forces that hold it in equilibrium are shown.forces that hold it in equilibrium are shown.

In the analysis of structure, after all acting forces are In the analysis of structure, after all acting forces are calculated, a Free-Body Diagram simplifies the calculation calculated, a Free-Body Diagram simplifies the calculation of the resisting forces that hold it in equilibrium.of the resisting forces that hold it in equilibrium.

A A free body diagramfree body diagram is a is a graphic of an object under the graphic of an object under the influence of forces with all influence of forces with all restraints removed and replaced by restraints removed and replaced by forces that hold the body in forces that hold the body in equilibrium.equilibrium.

Free body diagrams are Free body diagrams are useful in isolating a structural useful in isolating a structural member for analysis. member for analysis. B C

A

weight of objects above

A

beam

B

Consider this Consider this free body diagram of a free body diagram of a beam with a load beam with a load located directly at its located directly at its center. It is easy to center. It is easy to realize that the sum of realize that the sum of the upward forces at A the upward forces at A and B must equal the and B must equal the 50 lbs. 50 lbs.

And it also may And it also may

be easy to realize, that be easy to realize, that if the load on the beam if the load on the beam is at the center, that A is at the center, that A and B are the same, 25 and B are the same, 25 lbs. each.lbs. each.

beam

A B

50 lb

If the diagram is inverted, If the diagram is inverted, nothing changes in the nothing changes in the amount of the loads, and it amount of the loads, and it is easy to see that A and B is easy to see that A and B must be equal if the beam is must be equal if the beam is to remain balanced at its to remain balanced at its center. center.

Imagine the beam is 10’ long so the Imagine the beam is 10’ long so the distance from A to the 50 lb load is 5’, and is distance from A to the 50 lb load is 5’, and is the same for B. The point at the 50 lb load the same for B. The point at the 50 lb load is a fulcrum, or a point where the beam is a fulcrum, or a point where the beam could rotate. The force at A tends to rotate could rotate. The force at A tends to rotate downward, or counterclockwise about the downward, or counterclockwise about the 50 lb load. Force B tends to rotate 50 lb load. Force B tends to rotate downward, or clockwise about the 50 lb downward, or clockwise about the 50 lb load. load.

In other words A and B oppose each In other words A and B oppose each other in opposite directions of rotation, and other in opposite directions of rotation, and must be equal if the distances are the same.must be equal if the distances are the same.

A

beam

50 lb

B

Forces that tend to Forces that tend to rotate about a certain rotate about a certain point create a point create a MOMENTMOMENT with respect to that center with respect to that center point, equal in magnitude point, equal in magnitude to the force times the to the force times the perpendicular distance perpendicular distance from the center point to from the center point to the line of action of the the line of action of the force. force. MOMENT equals MOMENT equals Force times perpendicular Force times perpendicular distance.distance.

A

beam

50 lb

B

Consider that the action of tightening or Consider that the action of tightening or loosening a bolt with a wrench, the force applied loosening a bolt with a wrench, the force applied at the handle of the wrench creates a MOMENT at the handle of the wrench creates a MOMENT about the pivot point – the head of the bolt – and it about the pivot point – the head of the bolt – and it is this MOMENT that acts to turn the bolt to make is this MOMENT that acts to turn the bolt to make it tight or loose.it tight or loose.

FORCE

DISTANCE PERPENDICULAR TO LINE OF ACTION OF FORCE

pounds

feetIn this case, the amount of moment created by the FORCE is with respect to the PIVOT CENTER of the bolt. The FORCE has a TENDENCY to rotate about the PIVOT CENTER.

Moment equals FORCE times the perpendicular DISTANCE of the line of action of the force to the pivot center

Pivot center

Forces that tend to rotate Forces that tend to rotate about a certain point create a about a certain point create a MOMENTMOMENT with respect to that with respect to that center point, equal in center point, equal in magnitude to the force times magnitude to the force times the perpendicular distance the perpendicular distance from the center point to the from the center point to the line of action of the force. line of action of the force. MOMENT equals Force x MOMENT equals Force x perpendicular distance.perpendicular distance.

A

beam

50 lb

B

So, for the beam as loaded, at its center, the So, for the beam as loaded, at its center, the force at A and at B must be 25 pounds each. force at A and at B must be 25 pounds each. The force at A creates 25lbs x 5ft = 1255 ft-lbs of The force at A creates 25lbs x 5ft = 1255 ft-lbs of moment moment with respect to the point of the 50 lb with respect to the point of the 50 lb loadload, which is equal and opposite the moment , which is equal and opposite the moment created by the load at B - created by the load at B - and since the forces and since the forces create equal and opposite moments, they have a create equal and opposite moments, they have a tendency to BEND the beam.tendency to BEND the beam.

For any body to remain in equilibrium, two things For any body to remain in equilibrium, two things must be realized in this demonstration of forces on a free must be realized in this demonstration of forces on a free body diagram; body diagram; ONE, vertical forces in the upward ONE, vertical forces in the upward direction must equal vertical forces in the downward direction must equal vertical forces in the downward directiondirection, or, or

The sum of all vertical forces must = zero; The sum of all vertical forces must = zero; If vertical forces are represented by “y”, then Summation If vertical forces are represented by “y”, then Summation

of Fof Fyy = 0 = 0

Moment caused by Forces that tend to cause Moment caused by Forces that tend to cause clockwise rotation about any point clockwise rotation about any point must be equal and must be equal and oppositeopposite to moment caused by Forces that tend to cause to moment caused by Forces that tend to cause counter-clockwise rotation about that same pointcounter-clockwise rotation about that same point, or , or

The sum of ALL moments with respect to the The sum of ALL moments with respect to the same point must = zerosame point must = zero

Summation of MSummation of M0 0 must equal 0 must equal 0

The same is true that for horizontal forces to The same is true that for horizontal forces to the right must be equal to the horizontal forces to the right must be equal to the horizontal forces to the left, or if horizontal forces are represented by the left, or if horizontal forces are represented by “x”, then“x”, then

The sum of FThe sum of Fxx = 0 = 0

Mathematically, it is convenient to adopt a Mathematically, it is convenient to adopt a sign convention for forces and moment, and the sign convention for forces and moment, and the standard is set thus:standard is set thus:

Vertical forces upward are + Vertical forces upward are + (positive)(positive)Vertical forces downward are - Vertical forces downward are - (negative)(negative)Moments counter-clockwise are + Moments counter-clockwise are + (positive)(positive)Moments clockwise are - Moments clockwise are - (negative)(negative)Horizontal forces to the right are + Horizontal forces to the right are + (positive)(positive)Horizontal forces to the left are - Horizontal forces to the left are - (negative)(negative)

To demonstrate the To demonstrate the sum sum of momentsof moments, , assume thatassume thatBoth A and B equal 25 lbs.Both A and B equal 25 lbs.

Then pick a point of rotation,Then pick a point of rotation,a point that would represent a pivot if all the forces a point that would represent a pivot if all the forces had a tendency to rotate about that point -had a tendency to rotate about that point - say at A, and sum the moments created about that say at A, and sum the moments created about that point by the force at B and the 50 lb force. Realize point by the force at B and the 50 lb force. Realize that the force of 25 lbs at A will not create any that the force of 25 lbs at A will not create any moment about A, since the distance equals zero.moment about A, since the distance equals zero.

Use Use AA as a point of rotation: - (50 x 5) + (25 x 10) = 0 as a point of rotation: - (50 x 5) + (25 x 10) = 0

- 250 + 250 = 0, so the amount of moment created - 250 + 250 = 0, so the amount of moment created by the two forces cancel to zero.by the two forces cancel to zero.

A

10'-0"

50 lb

5'-0"

beam

5'-0" B

Next, pick B as the pointNext, pick B as the pointrotation, and sum therotation, and sum themoments created by themoments created by the50 lb force and the force50 lb force and the forceof 25 lb at A:of 25 lb at A:

Realize also that the force of 25 lbs at B will not Realize also that the force of 25 lbs at B will not

create any moment about B, since the distance create any moment about B, since the distance equals zero.equals zero.

Use Use BB as a point of rotation: + (50 x 5) - (25 x 10) = 0 as a point of rotation: + (50 x 5) - (25 x 10) = 0

+ 250 - 250 = 0, so the amount of moment created + 250 - 250 = 0, so the amount of moment created by the two forces cancel to zero.by the two forces cancel to zero.Note the results are identical, except the signs are Note the results are identical, except the signs are opposite.opposite.

A

10'-0"

50 lb

5'-0"

beam

5'-0" B

Next, arbitrarily pick a pointNext, arbitrarily pick a point that is 3’ to the left of B, that is 3’ to the left of B, and call it point “C”and call it point “C”

Sum the moments of all Sum the moments of all the forces about point C . . .the forces about point C . . .

*

5'

A

5'

*

2' 3'

50 lb

B

25 lb25 lb

(A) (B)

- 25 x 7 + 50 x 2 + 25 x 3 = 0

- 175 + 100 + 75 = 0 so, - 175 + 175 = 0

and no matter where rotational points are selected on the beam, the MOMENTS will sum to 0. And that is because the beam, loaded, with

reactions as shown remains in equilibrium.

C

But what if the load on a But what if the load on a beam is not at its center?beam is not at its center?How can the reactions atHow can the reactions at

A and B be found?A and B be found?

When there are two unknowns, such as the reactions at A When there are two unknowns, such as the reactions at A and B, pick a point of one of the unknowns and sum the and B, pick a point of one of the unknowns and sum the moments about that point, so there will be only one moments about that point, so there will be only one unknown.unknown.

Choose the point at Choose the point at AA: write the equation,: write the equation,- (50 x 4) + (By x 10) = 0- (50 x 4) + (By x 10) = 0, and , and – 200 + 10By = 0 ,– 200 + 10By = 0 ,

then solving for By ; 10 By = 200, and then solving for By ; 10 By = 200, and By = 20 lbsBy = 20 lbs

And since the sum of the forces in the vertical direction And since the sum of the forces in the vertical direction must equal zero, thenmust equal zero, then

Ay = 50 – 20 = 30 lbs Ay = 50 – 20 = 30 lbs

10'-0"6'-0"A 4'-0"

50 lb

beam

B

And to demonstrate that theAnd to demonstrate that the30 lb load at A is accurate,30 lb load at A is accurate,Sum the moments about Sum the moments about Point B:Point B:

There is no moment created at B by the load at B, There is no moment created at B by the load at B, since the distance equals zero.since the distance equals zero.

Choose the point at B, and solve for the unknown at Choose the point at B, and solve for the unknown at A: write the equation,A: write the equation,+ (50 x 6) - (A+ (50 x 6) - (Ayy x 10) = 0 x 10) = 0, and , and + 300 - 10B+ 300 - 10Byy = 0 , = 0 ,

then solving for Athen solving for Ay ; y ; 10 A10 Ayy = 300, and = 300, and AAyy = 30 lbs = 30 lbs

Consequently, since the sum of the forces in the Consequently, since the sum of the forces in the vertical direction must equal zero, thenvertical direction must equal zero, then

BByy = 50 – 30 = 20 lbs = 50 – 30 = 20 lbs

10'-0"6'-0"A 4'-0"

50 lb

beam

B

NOW TAKE A SHEET OF PAPER AND DO THISNOW TAKE A SHEET OF PAPER AND DO THISEXERCISE ONEEXERCISE ONE

For the beam loaded as shown, find the value of the For the beam loaded as shown, find the value of the reactions at A and Breactions at A and B

beam

A 3'-0"

60 lb

12'-0"9'-0" B

The process of finding the reactions to a The process of finding the reactions to a beam is the same for one with multiple loads as in beam is the same for one with multiple loads as in the previous examples. Consider a beam loaded the previous examples. Consider a beam loaded as shown and calculate the reactions. as shown and calculate the reactions.

Since there are two unknowns, sum the moments Since there are two unknowns, sum the moments created by the 3 loads shown about point A ; created by the 3 loads shown about point A ;

200 lb

A 5' 6'

beam

24'-0"5' 8'

300 lb 400 lb

B

The force at A will not create a moment at A The force at A will not create a moment at A because the dimension equals zero . . .because the dimension equals zero . . .

- (200 x 5) – (300 x 10) – (400 x 18) + 24B- (200 x 5) – (300 x 10) – (400 x 18) + 24By y = 0= 0- 1000 – 3000 – 7200 + 24B- 1000 – 3000 – 7200 + 24By y = 0= 0

24B24By y = 7200 + 3000 + 1000= 7200 + 3000 + 1000

24B24By y = 11,200; B= 11,200; By y = 11,200= 11,200 2424

BBy y = 466.67 lbs= 466.67 lbs Then sum the vertical loads to find AThen sum the vertical loads to find Ayy ; ;

AAy y = - 200 – 300 – 400 + 466.67 = 433.33 lbs= - 200 – 300 – 400 + 466.67 = 433.33 lbs

200 lb

A 5' 6'

beam

24'-0"5' 8'

300 lb 400 lb

B

NOW ON THE SAME SHEET OF PAPER DO THISNOW ON THE SAME SHEET OF PAPER DO THISEXERCISE TWO: EXERCISE TWO:

For the beam loaded as shown, calculate the For the beam loaded as shown, calculate the magnitude of the reactions at points A and B.magnitude of the reactions at points A and B.

A 6'

beam

20'-0"7' 7'

200 lb 500 lb

B

Consider a beam loaded as shown, with one end that Consider a beam loaded as shown, with one end that projects over its support. Calculate the magnitude projects over its support. Calculate the magnitude of the reactions. The procedure is the same as of the reactions. The procedure is the same as previous, previous, paying careful attention to sign paying careful attention to sign convention.convention.

5'

A

400 lbs 200 lbs

25'

beam

13' 6'

600 lbs

6' B

First, consider that there are two unknowns, the First, consider that there are two unknowns, the reactions at A and B. Select one of the points and reactions at A and B. Select one of the points and sum the moments about that point that produce sum the moments about that point that produce rotation. Select rotation. Select AA, and remember the reaction , and remember the reaction force at A will not produce any moment at A, force at A will not produce any moment at A, because the distance is zero. Begin at the left - - -because the distance is zero. Begin at the left - - -

+(+(400400 x 5’) –( x 5’) –(600 600 x 13’) –(x 13’) –(200200 x 19’) + 25 x B x 19’) + 25 x Byy = 0 = 0

+ + 20002000 – – 78007800 – – 38003800 + 25 B + 25 By y = 0 ;= 0 ;

25 B25 By y = 9600= 9600`̀

BBy y = 9600 = 384 lb= 9600 = 384 lb 25255'

A

400 lbs 200 lbs

25'

beam

13' 6'

600 lbs

6' B

To find the reaction at A, To find the reaction at A, find the algebraic sumfind the algebraic sumof all known vertical of all known vertical loads:loads:

+ 384 – 400 – 600 – 200 + A+ 384 – 400 – 600 – 200 + Ayy = 0, and = 0, and AAyy = = 816 lb816 lb

Or, take moments about B to find AOr, take moments about B to find Ay y ::

+(400 x 30’) + (600 x 12’) + (200 x 6’) – 25 A+(400 x 30’) + (600 x 12’) + (200 x 6’) – 25 Ay y = 0 = 0+12,000 + 7,200 + 1,200 - 25 A+12,000 + 7,200 + 1,200 - 25 Ay y = 0 = 0

25 A25 Ay y = 20,400 ; A = 20,400 ; Ay y = 20,400 = 816 lb = 20,400 = 816 lb 25 25

5'

A

400 lbs 200 lbs

25'

beam

13' 6'

600 lbs

6' B

384 lb

Remember, the cantilever will not present a Remember, the cantilever will not present a problem of confusion problem of confusion IFIF you remember the proper you remember the proper sign convention as the forces have a tendency to sign convention as the forces have a tendency to rotate about the point you choose.rotate about the point you choose.

Sometimes it helps to keep the sign Sometimes it helps to keep the sign convention in order if you place a fingertip at the convention in order if you place a fingertip at the point of rotation, and then visualize how the forces point of rotation, and then visualize how the forces would have a TENDENCY to rotate about your would have a TENDENCY to rotate about your finger.finger.

Consider finally, if you don’t know the Consider finally, if you don’t know the direction of an unknown force, direction of an unknown force, use judgment and use judgment and assume a direction - - -assume a direction - - - If you assume the wrong If you assume the wrong direction, it will be evident if your answer turns out direction, it will be evident if your answer turns out to be a negative number. to be a negative number.