THE BASIC FUNDAMENTALS OF STATICSTHE BASIC FUNDAMENTALS OF STATICS

The physical laws used in this study that The physical laws used in this study that govern the action and reaction of forces on a govern the action and reaction of forces on a body include Sir Isaac Newton’s first and body include Sir Isaac Newton’s first and third Laws;third Laws;

First Law – First Law – A body at rest will remain at rest A body at rest will remain at rest and any body in motion will move uniformly and any body in motion will move uniformly in a straight line unless acted upon by a forcein a straight line unless acted upon by a force – which is a condition in structures called – which is a condition in structures called EQUILIBRIUM.EQUILIBRIUM.

Third Law – Third Law – For every force of action, there is For every force of action, there is a reaction that is equal in magnitude, a reaction that is equal in magnitude, opposite in direction, and has the same line opposite in direction, and has the same line of actionof action – which is the basic concept of – which is the basic concept of FORCE.FORCE.

Forces acting on a body generally Forces acting on a body generally cause two effects, and not necessarily cause two effects, and not necessarily simultaneously:simultaneously:

a) It will cause the body to move if it a) It will cause the body to move if it is at rest or change the motion of the is at rest or change the motion of the body if it is already in motion. body if it is already in motion.

b) It will cause a deformation of the b) It will cause a deformation of the body if the body is restrained.body if the body is restrained.

a horizontal beam will benda horizontal beam will benda member in compression will shortena member in compression will shortena member in tension will stretcha member in tension will stretcha member in shear will distorta member in shear will distort

A force is a A force is a QUANTITYQUANTITY, caused by an , caused by an object that has mass, or weight – or a force is object that has mass, or weight – or a force is caused due to a caused due to a RESISTANCERESISTANCE to a mass or to a mass or weight.weight.

Forces in mathematics are represented Forces in mathematics are represented by an illustration called a VECTOR, which is by an illustration called a VECTOR, which is simply a graphic representation, a line, that simply a graphic representation, a line, that has MAGNITUDE, or value, and a DIRECTION.has MAGNITUDE, or value, and a DIRECTION.

Use of vectors is a simple way to Use of vectors is a simple way to mathematically represent forces in order to mathematically represent forces in order to determine the solution to a structural determine the solution to a structural situation. situation.

Vectors can be used to represent Vectors can be used to represent distance, mass, or area, but their elementary distance, mass, or area, but their elementary use in structures is to represent forces. use in structures is to represent forces.

CHARACTERIZATION of a vector is evidenced by:CHARACTERIZATION of a vector is evidenced by:

1 MAGNITUDE – the quantity of a force, a 1 MAGNITUDE – the quantity of a force, a numerical measure of its intensity, usually in numerical measure of its intensity, usually in units of pounds or Newtons.units of pounds or Newtons.

2 DIRECTION – the line of action of a force, 2 DIRECTION – the line of action of a force, such as gravity pulling downward due to weight, such as gravity pulling downward due to weight, or resistance of a restraint pushing that weight or resistance of a restraint pushing that weight upward to negate actual movement and remain upward to negate actual movement and remain static.static.

3 POINT OF APPLICATION – a location that 3 POINT OF APPLICATION – a location that generally describes the origin of a force.generally describes the origin of a force.

A weight supported from a A weight supported from a ceiling by a string is ceiling by a string is represented by a vector, the represented by a vector, the line with the arrowhead that line with the arrowhead that defines directiondefines direction

A weight represented by a A weight represented by a vector, supported by a post, vector, supported by a post, using a vector to indicate using a vector to indicate resistance, the direction of resistance, the direction of each defined by the each defined by the arrowheads.arrowheads.

FORCE

FORCE

RESISTANCE

POST

The magnitude of the force The magnitude of the force shown as 100 pounds, and shown as 100 pounds, and the resistance shown as the resistance shown as 100 pounds, both directions 100 pounds, both directions defined by the arrowheads. defined by the arrowheads. Point of application is the Point of application is the small circle. When the small circle. When the forces are equal and forces are equal and opposite, the object is in opposite, the object is in EQUILIBRIUM.EQUILIBRIUM.FORCE

RESISTANCE

POST

100 pounds

RESISTANCE100 pounds

FORCE

100 pounds

100 pounds

In analysis of structural elements, it In analysis of structural elements, it is necessary to add the effects of two or is necessary to add the effects of two or more forces to determine the net effect as more forces to determine the net effect as though ONLY ONE FORCE produced the though ONLY ONE FORCE produced the same result. This one force is called the same result. This one force is called the RESULTANT of a series of forces. RESULTANT of a series of forces.

Parallelogram Law: Parallelogram Law: If two forces act If two forces act concurrently at a point, the resultant concurrently at a point, the resultant force can be represented by the diagonal force can be represented by the diagonal of the parallelogram formed by the sides, of the parallelogram formed by the sides, parallel and proportional to the two parallel and proportional to the two forces.forces.

Illustration: If an object Illustration: If an object is under the influence of is under the influence of two forces as shown two forces as shown that tend to drag it that tend to drag it along, and the along, and the magnitude of the two magnitude of the two forces are illustrated by forces are illustrated by the length of their lines, the length of their lines, then;then;

A Which direction will A Which direction will the object be moved?the object be moved?

B What is the B What is the magnitude of a single magnitude of a single force that will move the force that will move the object.object.

FOR

CE

2 FORCE 1

OBJECT

Consider two forces 1 & 2, Consider two forces 1 & 2, each represented by a each represented by a straight line vector that straight line vector that indicates their magnitudes indicates their magnitudes by the lengths, and by the lengths, and directions by the directions by the arrowheads.arrowheads.

The two forces, acting The two forces, acting together would have an together would have an influence on the object. influence on the object.

By the graphic By the graphic parallelogram method, the parallelogram method, the RESULTANT, of the two RESULTANT, of the two forces can be constructed as forces can be constructed as shown, its value indicated shown, its value indicated by the graphic length.by the graphic length.

The RESULTANT has the The RESULTANT has the same effect on the object as same effect on the object as do forces 1 & 2. do forces 1 & 2.

FORCE 1

FO

RC

E 2 R

ES

ULT

AN

T O

F F

OR

CE

S 1

& 2

OBJECT

Obviously, if the numerical value of the Obviously, if the numerical value of the vectors is known, and the directions defined by vectors is known, and the directions defined by angles with some reference, the magnitude of the angles with some reference, the magnitude of the Resultant can be found using trigonometry – by the Resultant can be found using trigonometry – by the solution of the sides of a triangle.solution of the sides of a triangle.

It can also be found graphically if the diagram It can also be found graphically if the diagram is drawn accurately to scale. For instance this is drawn accurately to scale. For instance this illustration was done by a cadd drawing, and the illustration was done by a cadd drawing, and the system is accurate enough that the lengths of the system is accurate enough that the lengths of the lines can simply be measured. Doing this on the lines can simply be measured. Doing this on the drawing, F-1 measures 13.31 units at an angle of drawing, F-1 measures 13.31 units at an angle of 48 degrees, and F-2 measures 8.57 units at an 48 degrees, and F-2 measures 8.57 units at an angle of 115 degrees. The Resultant measures angle of 115 degrees. The Resultant measures 18.45 units at an angle of 78 degrees. The units 18.45 units at an angle of 78 degrees. The units can be in inches, feet, pounds, Newtons, or any can be in inches, feet, pounds, Newtons, or any other system of units, and the relationship between other system of units, and the relationship between the forces and the resultant will remain the same.the forces and the resultant will remain the same.

If the units are in pounds, realize that the If the units are in pounds, realize that the single 18.45 pound force has the same influence on single 18.45 pound force has the same influence on the object as do the two forces of 13.31 and 8.57 the object as do the two forces of 13.31 and 8.57 pound forces.pound forces.

On a cadd system, On a cadd system, the angles are the angles are measured from the measured from the right side X axis of right side X axis of the Cartesian the Cartesian Coordinates, in a Coordinates, in a counter – clockwise counter – clockwise direction.direction.

FORCE 1

FO

RC

E 2 R

ES

ULT

AN

T O

F F

OR

CE

S 1

& 2

OBJECT

X AXIS

Y A

XIS

48 DE

G

78 DE

G

115 DEG

In the study of statics, the relationship of In the study of statics, the relationship of forces will be significant in determining forces will be significant in determining structural qualities required to resist forces structural qualities required to resist forces that occur in the built environment. that occur in the built environment.

The X and Y axes of the Cartesian The X and Y axes of the Cartesian Coordinates system will become the Coordinates system will become the reference for forces that have directions reference for forces that have directions upward, downward, or at any angle.upward, downward, or at any angle.

Forces that are not parallel to either the X Forces that are not parallel to either the X and Y axes, are the RESULTANT of and Y axes, are the RESULTANT of component forces that are parallel and component forces that are parallel and perpendicular to those axes. perpendicular to those axes.

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

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

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

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

So it is in structure; Roofing pushes So it is in structure; Roofing pushes down on roof deck, which pushes down on down on roof deck, which pushes down on joists, which push down on beams, which joists, which push down on beams, which push down on columns, which push down on push down on columns, which push down on foundations, which push down on earth - - - foundations, which push down on earth - - - each because of gravity, and each successive each because of gravity, and each successive element pushes upward the same amount of element pushes upward the same amount of the sum of the weights above.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 downward, while actually, those items of structure don’t DO anything. They simply structure don’t DO anything. They simply have weight that reacts to the force of have weight that reacts to the force of gravity . . .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 So man, places these items in opposition to gravity in order to create those things we to gravity in order to create those things we call enclosed space, which we as architects call enclosed space, which we as architects refer to asrefer 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 maintaining that space according to the integrity of their strength.integrity 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 structural members, it may be necessary to illustrate the subject as a the subject as a

FREE – BODY DIAGRAMFREE – BODY DIAGRAM

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

And further, any object can be divided into parts, And further, any object can be divided into parts, whether separated at connections, or whether whether separated at connections, or whether members are shown to be cut at some point or plane - members are shown to be cut at some point or plane - - - and each part can be illustrated as an object - - and each part can be illustrated as an object suspended in space if the forces that hold it in suspended in space if the forces that hold it in equilibrium are shown.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 calculated, a Free-Body DiagramFree-Body Diagram simplifies the simplifies the calculation of the resisting forces that hold it in calculation of the resisting forces that hold it in equilibrium.equilibrium.

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

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

B C

A

weight of objects above

A

beam

B

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

And it also may And it also may

be easy to realize, be easy to realize, that if the load on that if the load on the beam is at the the beam is at the center, that A and B center, that A and B are the same, 25 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 Consider that the action of tightening or loosening a bolt with a wrench, the or loosening a bolt with a wrench, the force applied at the handle of the wrench force applied at the handle of the wrench creates a MOMENT about the pivot point – creates a MOMENT about the pivot point – the head of the bolt – and it is this the head of the bolt – and it is this MOMENT that acts to turn the bolt to MOMENT that acts to turn the bolt to make it tight or loose.make 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 For any body to remain in equilibrium, two things must be realized in this demonstration of things must be realized in this demonstration of forces on a free body diagram; forces on a free body diagram; ONE, vertical ONE, vertical forces in the upward direction must equal vertical forces in the upward direction must equal vertical forces in the downward directionforces in the downward direction, or, or

The sum of all vertical forces must = The sum of all vertical forces must = zero; zero;

If vertical forces are represented by “y”, then If vertical forces are represented by “y”, then Summation of FSummation of 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 must be equal and oppositeand opposite to moment caused by Forces that to moment caused by Forces that tend to cause counter-clockwise rotation about tend to cause counter-clockwise rotation about that same pointthat 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 The same is true that for horizontal forces to the right must be equal to the forces to the right must be equal to the horizontal forces to the left, or if horizontal horizontal forces to the left, or if horizontal forces are represented by “x”, thenforces are represented by “x”, then

The sum of FThe sum of Fxx = 0 = 0

Mathematically, it is convenient to Mathematically, it is convenient to adopt a sign convention for forces and adopt a sign convention for forces and moment, and the standard is set thus:moment, and the 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 a point that would represent a pivot if all the forces had a tendency to rotate about that forces had a tendency to rotate about that point -point - say at A, and sum the moments created about say at A, and sum the moments created about that point by the force at B and the 50 lb force. that point by the force at B and the 50 lb force. Realize that the force of 25 lbs at A will not Realize that the force of 25 lbs at A will not create any moment about A, since the distance create any moment about A, since the distance equals zero.equals zero.

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

- 250 + 250 = 0, so the amount of moment - 250 + 250 = 0, so the amount of moment created by the two forces cancel to zero.created 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 Realize also that the force of 25 lbs at B will

not create any moment about B, since the not create any moment about B, since the distance equals zero.distance equals zero.

Use Use BB as a point of rotation: + (50 x 5) - (25 x as a point of rotation: + (50 x 5) - (25 x 10) = 010) = 0

+ 250 - 250 = 0, so the amount of moment + 250 - 250 = 0, so the amount of moment created by the two forces cancel to zero.created by the two forces cancel to zero.Note the results are identical, except the Note the results are identical, except the signs are opposite.signs are 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 When there are two unknowns, such as the reactions at A and B, pick a point of one of the reactions at A and B, pick a point of one of the unknowns and sum the moments about that unknowns and sum the moments about that point, so there will be only one unknown.point, so there will be only one 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 By = 20 lbslbs

And since the sum of the forces in the vertical And since the sum of the forces in the vertical direction must equal zero, thendirection must 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 There is no moment created at B by the load at B, since the distance equals zero.at B, since the distance equals zero.

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

then solving for Athen solving for Ay ; y ; 10 A10 Ayy = 300, and = 300, and AAyy = = 30 lbs30 lbs

Consequently, since the sum of the forces in Consequently, since the sum of the forces in the vertical direction must equal zero, thenthe vertical direction must equal zero, thenBByy = 50 – 30 = 20 lbs = 50 – 30 = 20 lbs

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

50 lb

beam

B

NOW CONSIDER THIS EXERCISE NOW CONSIDER THIS EXERCISE

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

beam

A 3'-0"

60 lb

12'-0"9'-0" B

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

Since there are two unknowns, sum the Since there are two unknowns, sum the moments created by the 3 loads shown moments created by the 3 loads shown about point A ; 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 = = 00

- 1000 – 3000 – 7200 + 24B- 1000 – 3000 – 7200 + 24By y = 0= 0 24By = 7200 + 3000 + 24By = 7200 + 3000 +

10001000

24By = 11,200; By = 24By = 11,200; By = 11,20011,200 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 CONSIDER THISNOW CONSIDER THIS EXERCISE TWO: EXERCISE TWO: For the beam loaded as shown, calculate For the beam loaded as shown, calculate the magnitude of the reactions at points A the magnitude of the reactions at points A and B.and B.

Take moments about point A: Take moments about point A: -200x7 – 500x14 + By x 20 = 0 -200x7 – 500x14 + By x 20 = 0 So By = 1400+7000/20 and By = 420; Ay So By = 1400+7000/20 and By = 420; Ay

= 280= 280

A 6'

beam

20'-0"7' 7'

200 lb 500 lb

B

Consider a beam loaded as shown, with one Consider a beam loaded as shown, with one end that projects over its support. end that projects over its support. Calculate the magnitude of the reactions. Calculate the magnitude of the reactions. The procedure is the same as previous, The procedure is the same as 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, First, consider that there are two unknowns, the reactions at A and B. Select one of the the reactions at A and B. Select one of the points and sum the moments about that points and sum the moments about that point that produce rotation. Select point that produce rotation. Select AA, and , and remember the reaction force at A will not remember the reaction force at A will not produce any moment at A, because the produce any moment at A, because the distance is zero. Begin at the left - - -distance is zero. Begin at the left - - -

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

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

25 B25 By y = 9600= 9600`̀

BBy y = 9600 = 384 = 9600 = 384 lblb

2525

5'

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 +(400 x 30’) + (600 x 12’) + (200 x 6’) – 25 AAy 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 Remember, the cantilever will not present a problem of confusion present a problem of confusion IFIF you you remember the proper sign convention as the remember the proper sign convention as the forces have a tendency to rotate about the forces have a tendency to rotate about the point you choose.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 convention in order if you place a fingertip at the point of rotation, and then visualize at the point of rotation, and then visualize how the forces would have a TENDENCY to how the forces would have a TENDENCY to rotate about your finger.rotate about your 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 use judgment and assume a direction - - -judgment and assume a direction - - - If If you assume the wrong direction, it will be you assume the wrong direction, it will be evident if your answer turns out to be a evident if your answer turns out to be a negative number. negative number.