CLASS DAY 24CLASS DAY 24

STRUCTURES 2STRUCTURES 2

THE CONCEPT OF COMPONENTSTHE CONCEPT OF COMPONENTSComponents are the parts of a Whole Components are the parts of a Whole DistanceDistance, or a Whole , or a Whole ForceForce, that help to , that help to

define the define the magnitude and directionmagnitude and direction of that of that Distance or ForceDistance or Force

Distances and Forces can be reduced to their Distances and Forces can be reduced to their components to aid in the analysis of components to aid in the analysis of

equilibrium, and in like manner, equilibrium, and in like manner, components can be used to identify components can be used to identify

magnitude and direction of distance and magnitude and direction of distance and force. force.

A component will be either VERTICAL or A component will be either VERTICAL or HORIZONTALHORIZONTAL

Cartesian Coordinates consist of a pair of Cartesian Coordinates consist of a pair of axes 90 degrees apart, used as reference for axes 90 degrees apart, used as reference for identifying identifying components components of forces that of forces that are not are not parallel or parallel or perpendicular to either perpendicular to either the vertical or horizontalthe vertical or horizontal

A vector whose directionA vector whose directionis at an angle with theis at an angle with thereference axes will have reference axes will have COMPONENTS – one for COMPONENTS – one for each of the reference each of the reference

axes, which will define axes, which will define their relationship with their relationship with the vector the vector

X X

Y

Y

0

Imagine the shape as a box, and the green Imagine the shape as a box, and the green line is a diagonal of the box, and represents line is a diagonal of the box, and represents a vector, so that two sides of the box are a vector, so that two sides of the box are represented by the represented by the yellow and cyan lines. yellow and cyan lines. The The two lines are two lines are COMPONENTS of the COMPONENTS of the

green line, just as two green line, just as two sides of a box are sides of a box are

COMPONENTS of the COMPONENTS of the diagonal of a box.diagonal of a box.

Just as two sides of a Just as two sides of a box box have a mathematical have a mathematical

relationship with relationship with the the diagonal of the diagonal of the box, so box, so do the do the COMPONENTS of a COMPONENTS of a vector have a vector have a

mathematical relationship mathematical relationship to the value of the to the value of the

vector.vector.

X X

Y

Y

0

Now consider a box whose Now consider a box whose side dimensions are 4’ and side dimensions are 4’ and 3’. This automatically 3’. This automatically makes the direction of the makes the direction of the diagonal at a slope of 4 to 3diagonal at a slope of 4 to 3

In other words the diagonal In other words the diagonal is at an angle defined by 3 is at an angle defined by 3 units vertical for each 4 units vertical for each 4 units horizontal. These are units horizontal. These are COMPONENTS of the COMPONENTS of the diagonal.diagonal.

The length of the diagonal The length of the diagonal can be found by the can be found by the Pythagorean solution of a Pythagorean solution of a right triangle, which right triangle, which defines the diagonal as the defines the diagonal as the square root of the sums of square root of the sums of the squares of the two the squares of the two sides. In this case, the sides. In this case, the diagonal is 5’.diagonal is 5’.

3

4

This says that for any vector whose slope is known (the relationship of the components) the magnitude of the vector can be found if the value of the components are known – the proportional relationships are the same.

F = force

in m

ember

X component of force = F

x = Ff

Ff =y

=F fx

=F fy

Y c

ompo

nent

of f

orce

= F

x

(

(

(

( )

y

3

4

553

54

35

54

LIKEWISE, FORCES CAN HAVE COMPONENTS

X component of force = F

F = force

in m

ember

f F=x

=f Fy

=F xf

=F yf

Y c

ompo

nent

of f

orce

= F

outer lines =

force

triangle

H

H

X

inner lines = slope triangle

x

Y

H

y(

x(

H

( x

H )(y

y

FOR ANY FORCE WHOSE SLOPE ANGLE IS KNOWN

COMPONENTS OF COMPONENTS OF DISTANCEDISTANCE

If one were to fly directly If one were to fly directly from Lubbock to Oklahoma from Lubbock to Oklahoma City, a distance of 300 City, a distance of 300 miles, the path would be at miles, the path would be at a direction of 30 degrees above an East/West a direction of 30 degrees above an East/West

linelinethrough Lubbock. In other words, through Lubbock. In other words, Oklahoma Oklahoma

City isCity isNortheast of Lubbock.Northeast of Lubbock.

To traverse the 300 miles, part of the path To traverse the 300 miles, part of the path would be north, at the same time part of it is would be north, at the same time part of it is east. Because of the 30 degree angle of the east. Because of the 30 degree angle of the direct route, Oklahoma City is 150 miles direct route, Oklahoma City is 150 miles NORTH of Lubbock, and it is also 269.8 miles NORTH of Lubbock, and it is also 269.8 miles EAST of Lubbock. EAST of Lubbock. The NORTH and EAST The NORTH and EAST distances are COMPONENTS of the 300 mile distances are COMPONENTS of the 300 mile direct route.direct route.

30

300 miles

Lubbock

150

mile

s

NORTH

0

EAST259.8 miles

CityOklahoma

In this example, the component distances In this example, the component distances are calculated by trigonometry, since the are calculated by trigonometry, since the north and east distances form a 30 degree north and east distances form a 30 degree right triangle, with a hypotenuse of 300 right triangle, with a hypotenuse of 300 miles.miles.

The short leg (north) is simply 300 x sin 30 The short leg (north) is simply 300 x sin 30 = 300 x .5 = 150 miles.= 300 x .5 = 150 miles.

The long leg (east) is 300 x cos 30 = 300 The long leg (east) is 300 x cos 30 = 300 x .866 = 259.8 miles.x .866 = 259.8 miles.

In a similar manner, COMPONENTS of In a similar manner, COMPONENTS of forces can be calculated. But it is necessary forces can be calculated. But it is necessary to adopt a standard for sign convention and to adopt a standard for sign convention and direction. The simple Cartesian Coordinates direction. The simple Cartesian Coordinates system is valuable in determining system is valuable in determining components in terms of X and Y values.components in terms of X and Y values.

Cartesian Coordinates consist of a pair of Cartesian Coordinates consist of a pair of axes 90 degrees apart, used as reference axes 90 degrees apart, used as reference for identifying for identifying components of forces that components of forces that

are not parallel or are not parallel or perpendicular to either perpendicular to either

the vertical or horizontal the vertical or horizontal

Values on the X Values on the X plane to plane to the right of 0 the right of 0 are are positive, and positive, and values values on the X on the X plane to plane to the left of 0 the left of 0 are are negative.negative.

Values on the Y plane Values on the Y plane above 0 are above 0 are

positive, and positive, and values on values on the Y plane the Y plane below 0 are below 0 are negative.negative.

X X

Y

Y

0

FINDING COMPONENTS OF A FORCE USING FINDING COMPONENTS OF A FORCE USING A SLOPE TRIANGLEA SLOPE TRIANGLE - - - - - -

If the If the directiondirection of a Force has a of a Force has a known relationship with both the X and Y known relationship with both the X and Y axes, such that the force acts in a axes, such that the force acts in a direction of direction of Y distanceY distance for for a given X a given X amount of distanceamount of distance, the X and Y , the X and Y components of the force become the value components of the force become the value that defines the SLOPE of the Force - - - that defines the SLOPE of the Force - - - the measure of the direction the FORCE the measure of the direction the FORCE varies from the X and Y axes. varies from the X and Y axes.

With the Y distance and X distances With the Y distance and X distances as legs of a right triangle, the hypotenuse as legs of a right triangle, the hypotenuse can be found by the Pythagorean can be found by the Pythagorean Theorem, and the resulting triangle Theorem, and the resulting triangle becomes the becomes the Slope TriangleSlope Triangle that defines that defines the direction of the Force. the direction of the Force.

A force with a magnitude of A force with a magnitude of 50 pounds lies in the XY 50 pounds lies in the XY plane, and the slope of plane, and the slope of the line that defines the the line that defines the force is Y = 3 units for force is Y = 3 units for each X = 4 units. The 3, each X = 4 units. The 3, 4 slope will form a right 4 slope will form a right triangle whose triangle whose hypotenuse = 5 units. hypotenuse = 5 units.

Imagine the line of force of Imagine the line of force of 50 is the hypotenuse of a 50 is the hypotenuse of a right triangle whose X right triangle whose X and Y components form a and Y components form a similar right triangle.similar right triangle.

The COMPONENT values The COMPONENT values are directly proportional are directly proportional to the legs of the to the legs of the Slope Slope TriangleTriangle, and may be , and may be calculated by simple calculated by simple proportion:proportion:

X

35

4

Force =50 pounds

Say the Say the Y componentY component of Force of Force is proportional to the is proportional to the 33 of of the slope triangle; and the the slope triangle; and the X componentX component of the Force is of the Force is proportional to the proportional to the 44 of the of the slope triangle; so it follows slope triangle; so it follows that the that the FORCEFORCE must be must be proportional to the proportional to the 55 of the of the slope triangle. slope triangle.

So, Y = Force ; cross So, Y = Force ; cross multiply, multiply,

3 53 5

And 5Y = 50 x 3, and Y = 3 x And 5Y = 50 x 3, and Y = 3 x 5050

55And And Y = 30 lbY = 30 lb,,

So the Y COMPONENT of the So the Y COMPONENT of the 50 pounds is 30 pounds.50 pounds is 30 pounds.

X

35

4

Force =50 pounds

Likewise, the Likewise, the X componentX component of the force of of the force of 5050 lb is lb is proportional to the proportional to the 44 of of the slope triangle;the slope triangle;

So, X = Force ; cross So, X = Force ; cross multiply,multiply,

4 54 5

And 5X = 50 x 4 ; and X = And 5X = 50 x 4 ; and X = 4 x 504 x 50

5 5

And And X = 40 lbX = 40 lb , ,

So, the X COMPONENT of So, the X COMPONENT of the 50 pounds is 40 the 50 pounds is 40 poundspounds

X

35

4

Force =50 pounds

If the components of a force can be If the components of a force can be determined from the magnitude and determined from the magnitude and direction of the force, then it follows that direction of the force, then it follows that if the magnitude of a force is not knownif the magnitude of a force is not known, , but but either the X or Y component can be either the X or Y component can be foundfound, then by the known direction of the , then by the known direction of the force, force, its magnitude can be found.its magnitude can be found.

Example: From the previous exercise, Example: From the previous exercise, since the Y component of the 50 lbs since the Y component of the 50 lbs equals 3 / 5 of the Force, then the force equals 3 / 5 of the Force, then the force must equal 5 / 3 times the component.must equal 5 / 3 times the component.

5 / 3 x 30 = 150 / 3 = 50 lbs.5 / 3 x 30 = 150 / 3 = 50 lbs.Since the X component of the 50 lbs Since the X component of the 50 lbs equals 4 / 5 of the Force, then the force equals 4 / 5 of the Force, then the force must equal 5 / 4 times the component.must equal 5 / 4 times the component.

5 / 4 x 40 = 200 / 4 = 50 lbs.5 / 4 x 40 = 200 / 4 = 50 lbs.

CONSIDER THIS EXERCISE CONSIDER THIS EXERCISE

11 A force of 60 lbs is at a direction such that A force of 60 lbs is at a direction such that the y direction of the slope triangle is 5, and the the y direction of the slope triangle is 5, and the x direction of the slope triangle is 12, then the x direction of the slope triangle is 12, then the hypotenuse is 13.hypotenuse is 13.

Find the X and Y Components of the force.Find the X and Y Components of the force.

22 If the Y component of a force of the same If the Y component of a force of the same slope as above equals 50 pounds, what is the slope as above equals 50 pounds, what is the amount of the Force?amount of the Force?

Y =

X =

5

12

13

Remember the Remember the Y componentY component of force is of force is proportional to the proportional to the 55 in the force triangle, the in the force triangle, the same way the same way the X componentX component of the force is of the force is proportional to the proportional to the 1212 in the force triangle – and in the force triangle – and the same way the the same way the 60 lb force60 lb force is proportional to is proportional to the hypotenuse, or the hypotenuse, or 13 13 in the force triangle.in the force triangle.

A really handy realization . . . So it follows A really handy realization . . . So it follows thatthat

6060//13 = 13 = Y componentY component // 5 5 , and, and Y component = Y component = 55//1313 x x 6060

Y component = 23.07 lbY component = 23.07 lb

and and 6060//1313 = = X componentX component / / 1212 , and , and

X componentX component = = 1212//1313 x x 6060 = = 55.38 lb55.38 lb

Consider the assembly Consider the assembly shown:shown:

A cable, AC supports a rigid A cable, AC supports a rigid bar, AB, which holds a 1200 bar, AB, which holds a 1200 lb load at end A, attached lb load at end A, attached to a wall at B and C.to a wall at B and C.

Find the tensile FORCE in the Find the tensile FORCE in the cable, and the horizontal cable, and the horizontal FORCE exerted at B.FORCE exerted at B.

Consider that the force in AC Consider that the force in AC must be pure tension, as must be pure tension, as the cable is flexible. Also the cable is flexible. Also consider that the force in consider that the force in the bottom member, AB the bottom member, AB must be in pure must be in pure compression, since the compression, since the member is horizontal, and member is horizontal, and no other forces occur on no other forces occur on the member.the member.

First, Show the assembly as a First, Show the assembly as a free body diagram - - -free body diagram - - -

1200 lbs

A

8'-0"

6'-0

"

B

C

3

4

5

Realize that since the cable is slanted with respect to vertical and horizontal – the force in the cable must have vertical and horizontal components.

Also realize the rigid member AB is horizontal and does not have components – only an axial force.

A free body diagram shows A free body diagram shows the assembly free of its the assembly free of its support, replaced by the support, replaced by the forces that hold it in forces that hold it in equilibrium.equilibrium.

The two forces that support The two forces that support the assembly are the the assembly are the Resisting Force at BResisting Force at B, and , and the the Resisting Force at CResisting Force at C, , which is equal and which is equal and opposite to the tension in opposite to the tension in the cable AC. the cable AC.

The force at B has no The force at B has no components in the XY components in the XY plane because it is plane because it is parallel with the X axis.parallel with the X axis.

The force at C has a The force at C has a direction as shown by the direction as shown by the slope triangle, and the slope triangle, and the force has an force has an X X componentcomponent and a and a Y Y component.component.

Realize that all forces in the X direction must = 0, as must all forces in the Y direction also = 0 – in order for the assembly to remain in equilibrium.

53

1200 lbs

A

8'-0"

4

B FORCEResisting

FORCEResisting

C

AC y

AC x

6'-0

"

Since the X and Y forces Since the X and Y forces at C are components at C are components of the Resisting force of the Resisting force at C, the resisting at C, the resisting force can be replaced force can be replaced by the components.by the components.

Note that there are only Note that there are only two forces in the Y two forces in the Y direction, one of which direction, one of which is 1200 lb. downward, is 1200 lb. downward, So the Component ACSo the Component ACy y must be 1200 lb. must be 1200 lb. upward.upward.

Note also there are only Note also there are only two forces in the X two forces in the X direction, neither of direction, neither of which are known. which are known.

Realize that all forces in the X direction must = 0, as must all forces in the Y direction also = 0 – in order for the assembly to remain in equilibrium.

1200 lbs

8'-0"

A

35

4

AC

C

y

ResistingFORCEB

xAC

6'-0

"

ResistingFORCE

Since ACSince ACyy = 1200 lb, then = 1200 lb, then the resisting force at C the resisting force at C can be found by the can be found by the slope triangle, since it is slope triangle, since it is the same configuration the same configuration as the assembly. By as the assembly. By proportion:proportion:

1200 = Resisting force C 1200 = Resisting force C ;;

3 53 5

Cross multiply: Cross multiply: 3 x Force C = 1200 x 5 ;3 x Force C = 1200 x 5 ;

Force C = 1200 x 5 ; andForce C = 1200 x 5 ; and 33

Force C = 2000 lb, which Force C = 2000 lb, which is the amount of tension is the amount of tension in the cable AC.in the cable AC.

Realize that all forces in the X direction must = 0, as must all forces in the Y direction also = 0 – in order for the assembly to remain in equilibrium.

8'-0"

5

1200 lbs

A

3

4

6'-0

"

FORCEResisting

B

FORCEResisting

C

AC y

AC x

Since the force at C = Since the force at C = 2000 lb, then the 2000 lb, then the horizontal component horizontal component can be found by the can be found by the slope triangle.slope triangle.

ACx = 4 x 2000 = ACx = 4 x 2000 = 8,000 ;8,000 ; 5 55 5

And ACx = 1,600 lb to And ACx = 1,600 lb to the rightthe right

Since ACx = 1,600 lb, Since ACx = 1,600 lb, then the resisting then the resisting force at B must be force at B must be 1,600 lb, to the left in 1,600 lb, to the left in order to maintain order to maintain equilibrium.equilibrium.

Realize that all forces in the X direction must = 0, as must all forces in the Y direction also = 0 – in order for the assembly to remain in equilibrium.

8'-0"

A

3

4

5

FORCEB

AC

Resisting

6'-0

"

FORCEResisting

C

yAC

x

Another way to calculate Another way to calculate the tensile force in AC is the tensile force in AC is by summation of by summation of moments created by the moments created by the external loads.external loads.

Since there are two Since there are two unknowns, the force at C unknowns, the force at C and the force at B, use and the force at B, use point B as the moment point B as the moment center. The resisting center. The resisting force at B will not cause force at B will not cause moment about B, since moment about B, since the distance equals 0.the distance equals 0.

But, perpendicular But, perpendicular distance, BD, from B to distance, BD, from B to the line of action of the the line of action of the tensile force in AC (shown tensile force in AC (shown by the dashed line) will by the dashed line) will have to be found. have to be found.

1200 lbs

A

8'-0"

3

4

5

ResistingFORCEB

6'-0

"

C

ResistingFORCE

D

Triangle BCD is the same proportion as the slope triangle. So BD / 4 = 6 / 5 ;

BD = 4/5 x 6 = 4.8 ft.

With BD = 4.8 ft. ;With BD = 4.8 ft. ;

Take moments about point Take moments about point B :B :

The line of action of the The line of action of the resisting force AC rotates resisting force AC rotates clockwise about B.clockwise about B.

The line of action of the The line of action of the 1200 lb load rotates 1200 lb load rotates counter clockwise about B, counter clockwise about B, soso

- (AC x 4.8) + (1200 x 8) = - (AC x 4.8) + (1200 x 8) = 00

4.8 AC = 9600, and4.8 AC = 9600, and

AC = 2000 lbAC = 2000 lb

1200 lbs

A

8'-0"

3

4

5

ResistingFORCEB

6'-0

"

C

ResistingFORCE

D

Triangle BCD is the same proportion as the slope triangle. So BD / 4 = 6 / 5 ;

BD = 4/5 x 6 = 4.8 ft.

EXAMPLE PROBLEM EXAMPLE PROBLEM TWOTWO

Two cords are Two cords are fastened to a ceiling fastened to a ceiling that support a 500 that support a 500 lb weight.lb weight.

Find the amount of Find the amount of tension in the two tension in the two cords, AC, and BC.cords, AC, and BC.

First show the First show the assembly as a free assembly as a free body diagram, body diagram, replacing the replacing the restraints at A and B restraints at A and B with forces that will with forces that will hold it in hold it in equilibrium.equilibrium.

5'-0

"

A B

12'-0"C

6.66'

500 lb

ceiling

Realize that the two vertical Y components at A and B must be upward and total 500 lbs.

Also realize that the horizontal components at A and B must be the same magnitude, but opposite in direction.

By observation one can By observation one can say that Asay that Ayy + B + Byy = 500, = 500,

and that Aand that Axx = B = Bxx..

If point A or point B is If point A or point B is used to sum moments, used to sum moments, the horizontal the horizontal components Acomponents Axx and B and Bxx will not cause rotation will not cause rotation about either point, about either point, since the distance = 0. since the distance = 0.

So if moments are So if moments are summed about point summed about point A, then only one A, then only one unknown remains, unknown remains, which is Bwhich is Byy , since A , since Ayy will not cause rotation will not cause rotation about A since the about A since the distance = zero.distance = zero.

500 lb

ceiling

12'-0"

A

6.66'C

5'-0

"

B

Observation reveals that the shortest solution is to find at least one of the components, then solve for the tension in both cords by what is known above.

As a starting place, select As a starting place, select point A, and sum the point A, and sum the moments of the forces moments of the forces that can cause rotation, that can cause rotation, REPLACING the forces REPLACING the forces with their components with their components

Write the equation:Write the equation:

-(500 x 12) + (B-(500 x 12) + (Byy x18.66) x18.66) = 0= 0

AndAnd- 6,000 + 18.66 B- 6,000 + 18.66 Byy = 0 = 0

BBy y = 6,000 = 321.54 lb.= 6,000 = 321.54 lb. 18.6618.66

Since Since BBy y + A+ Ayy = 500, then= 500, then

AAy y = 500 – 321.54 = = 500 – 321.54 = 178.46178.46

500 lb

ceiling

12'-0"

A

6.66'C

5'-0

"

B

With both component forces known, the tension in both cords can be calculated, BUT, slope of the forces must be known.

The slopes are given by the dimensions, but the length of the cords, (the hypotenuse of the slope triangles) must be calculated - - -

By the Theorem of Pythagoras,By the Theorem of Pythagoras,The length of the cords are thus:The length of the cords are thus:

AC = (5x5) + (12x12) = 13’AC = (5x5) + (12x12) = 13’

BC = (5x5) + (6.66x6.66) = 8.33’BC = (5x5) + (6.66x6.66) = 8.33’

And, since the forces and the components are directly And, since the forces and the components are directly proportional to the components of the force triangle:proportional to the components of the force triangle:

AC is proportional to 13’ as Ay is proportional to 5’, AC is proportional to 13’ as Ay is proportional to 5’, soso

AC = 13 x 178.46 = 464 lb , and sinceAC = 13 x 178.46 = 464 lb , and since 55

BC is proportional to 8.33’ as By is proportional to 5’, BC is proportional to 8.33’ as By is proportional to 5’, thenthen

BC = 8.33 x 321.54 = 535.69 lbBC = 8.33 x 321.54 = 535.69 lb 55

500 lb

ceiling

12'-0"

A

6.66'C

5'-0

"

B

To check the calculations, the statement was made To check the calculations, the statement was made at the beginning that Ax must equal Bx.at the beginning that Ax must equal Bx.

So Ax is proportional to 12’ as AC is proportional to So Ax is proportional to 12’ as AC is proportional to 13’, so13’, so

Ax = 12 x 464 = Ax = 12 x 464 = 428.30 lb428.30 lb , and , and 1313

Bx is proportional to 6.66’ as BC is Bx is proportional to 6.66’ as BC is proportional to 8.33’proportional to 8.33’so,so,

Bx = 6.66 x 535.69 = Bx = 6.66 x 535.69 = 428.30 lb428.30 lb 8.338.33

Some minute discrepancy in calculations will Some minute discrepancy in calculations will occur because of rounding off numbers to only two occur because of rounding off numbers to only two place accuracyplace accuracy