SHEAR AND BENDING MOMENT DIAGRAMSSHEAR AND BENDING MOMENT DIAGRAMS

IN HORIZONTAL BEAMS WITHIN HORIZONTAL BEAMS WITH

DOWNWARD VERTICAL LOADINGDOWNWARD VERTICAL LOADING

When a horizontal beam, supported at each end, is When a horizontal beam, supported at each end, is subjected to vertical loads along its length, internal subjected to vertical loads along its length, internal stress develops within the beam, which is the stress develops within the beam, which is the measure of the beam’s resistance to loading in order measure of the beam’s resistance to loading in order that equilibrium may be maintained. that equilibrium may be maintained.

The loads may be intermittent point loads, or they The loads may be intermittent point loads, or they may be distributed over either a partial or full length may be distributed over either a partial or full length of the beam.of the beam.

A beam made of elastic material, with vertical A beam made of elastic material, with vertical downward loading has a downward loading has a TENDENCYTENDENCY to bend to bend downward. Such downward bend causes three types downward. Such downward bend causes three types of stress develop within the beam:of stress develop within the beam:

Shear stress, which is variable along its length, Shear stress, which is variable along its length, but is but is maximum at the neutral axis. maximum at the neutral axis.

Compressive stress, also variable along its Compressive stress, also variable along its length, but length, but maximum at the top surface of maximum at the top surface of the beam’s cross the beam’s cross section. section.

Tensile stress, variable along its length, and Tensile stress, variable along its length, and maximum maximum at the bottom surface of the beam’s at the bottom surface of the beam’s cross section.cross section.

Elevation of concrete block wall with opening

200 lb

A 5' 6'

beam

24'-0"5' 8'

300 lb 400 lb

B

A beam loaded as shown, might have simple support at the ends, but for analysis, the assembly would be shown as a FREE BODY DIAGRAM as shown, where the supports are

replaced by FORCES that will hold the beam in

EQUILIBRIUM.

THE STRESS THAT CAUSES THE BEAM TO BENDI.E. THE TENSION AND THE COMPRESSION IS CALLED

THE DASHED LINE AT 0 STRESS DOES NOT CHANGE LENGTH

BENDING STRESS

d-1

supportTENSILE STRESS

d-2

0 STRESS

COMPRESSIVESTRESS

support

Because of vertical, downward loading, the elastic beam bends in the direction of the load:

The shape of the beam then assumes a portion of the circumference of a circle:

stresscompressive

MOMENTis called BENDING

axis; the resulting quantitydistance to the neutral

multiplied by the the SINGLE FORCE

single

forcecompressive

dis

tan

ce

tensile stress

to the top half of the beam,

be represented as a SINGLE

single

forcetensile

the NEUTRAL AXIS

plane of zero stress is called

centroid of the stress.FORCE located at the

the compressive stress can

the cross sectional areastress is multiplied byif the summation of the

The same is true for the tensile force

RESISTANCE TO BENDING MOMENT, SHEAR AND BENDING STRESSA BEAM BY ITS OWN PHYSICAL CHARACTERISTICS HAS A DETERMINABLE

BENDING MOMENT DEVELOPS WITHIN A BEAM, AS COMPRESSIVEAND TENSILE STRESS OCCURS DUE TO LOADING ON THE BEAM

forcecompressive

single

is the plane where SHEAR stress is maximumstress is called the NEUTRAL AXIS, butthe plane of zero tensile and compressive

force couple times the perpendicular distancestress, and is equal to the value of thewith the amount of tensile and compressivepoint along the length of the beam variesThe maximum bending moment at any

tensile stress

dis

tan

ce

singletensileforce

between the two forces.

stresscompressive

compressiveforce

single

forcetensilesingle

Since the FORCES caused bytensile and compressive stresspush in opposite directions, SHEAR stress develops withinthe beam, and is maximum atthe neutral axis.

NEUTRAL AXIS

CROSS SECTION OF BEAMshowing SHEAR stress,maximum at the NEUTRALAXIS, and zero at the top andbottom of the beam.

NEUTRAL AXIS

Because of opposite forcescaused by tensile and compressivestress, the material tends toslide past itself at the neutralaxis.

Remember that BENDING MOMENT is equal to a Remember that BENDING MOMENT is equal to a force times a distance, and units are foot-pounds, force times a distance, and units are foot-pounds, foot-kips, inch-pounds, etc., whatever the units of foot-kips, inch-pounds, etc., whatever the units of loading and distance on the beam.loading and distance on the beam.

The value of SHEAR may also vary along the The value of SHEAR may also vary along the length of the beam, depending upon the way loads length of the beam, depending upon the way loads are placed on the beam.are placed on the beam.

The values of both SHEAR and BENDING The values of both SHEAR and BENDING MOMENT can be demonstrated graphically, by MOMENT can be demonstrated graphically, by determining the difference between the vertical determining the difference between the vertical upward loads, and the vertical downward loads at upward loads, and the vertical downward loads at any point along the beam.any point along the beam.

First, draw a diagram (in reasonable First, draw a diagram (in reasonable proportions) of the beam as a free body diagram, proportions) of the beam as a free body diagram, showing all the vertical forces on the beam, upward showing all the vertical forces on the beam, upward or downward as applicable, with the beam supports or downward as applicable, with the beam supports shown as reaction forces.shown as reaction forces. Second, find the values of Second, find the values of the reaction forces - - -the reaction forces - - -

Calculate the values of theCalculate the values of theend reactions at A and B:end reactions at A and B:

To find B, sum the moments To find B, sum the moments about point A, and write:about point A, and write:

-(200x5) – (300x10) –(400x18) + By x 24 = 0-(200x5) – (300x10) –(400x18) + By x 24 = 01000 – 3000 – 7200 + 24 x By = 0 24 By = 11,2001000 – 3000 – 7200 + 24 x By = 0 24 By = 11,200By = 11,200 ; By = 11,200 ; By = 466.67 lbBy = 466.67 lb 2424

Then, by summation of vertical forces, Then, by summation of vertical forces,

Ay = 900 – 466.67 = 433.33 lbAy = 900 – 466.67 = 433.33 lb

200 lb

A 5' 6'

beam

24'-0"5' 8'

300 lb 400 lb

B

Show the calculated Show the calculated forces at forces at A and B.A and B.

Below the free body Below the free body diagram, draw a line diagram, draw a line that will be a reference that will be a reference of zero, for values. All of zero, for values. All numbers below the line numbers below the line will be negative, will be negative, and all above the line and all above the line will will be positive. be positive.

Then extend a vertical Then extend a vertical line downward from line downward from each load each load on the free body on the free body diagram, including the diagram, including the reaction forces.reaction forces.

200 lb

A 5' 6'

beam

24'-0"5' 8'

300 lb 400 lb

B

433.33 466.67

Beginning at the left, from Beginning at the left, from the reference line, the reference line, directly under point A, directly under point A, draw a line upward that draw a line upward that representsrepresentsthe reaction at A – an the reaction at A – an upward force of upward force of 433.33 lb. Select a 433.33 lb. Select a suitable scale for the suitable scale for the length of the values, or length of the values, or use judgmentuse judgmentin comparative in comparative proportion.proportion.

Then move from that point Then move from that point above the reference, to above the reference, to the right, and realize the right, and realize nothing happens to nothing happens to change the upward change the upward 433.33 value - - - 433.33 value - - - until you until you get to the 200 lb loadget to the 200 lb load – – which is downward. Draw which is downward. Draw a line that represents the a line that represents the 200 lb force downward, 200 lb force downward, and subtract from the and subtract from the 433.33 lb load.433.33 lb load.

200 lb

A 5' 6'

beam

24'-0"5' 8'

300 lb 400 lb

B

433.33 466.67

433.33 233.33

Then from that point, draw Then from that point, draw a horizontal line to the a horizontal line to the right and realize that right and realize that nothing happens to nothing happens to change the 233.33 – change the 233.33 – until you get to the 300 until you get to the 300 lb force, downward.lb force, downward.

From that point, draw a From that point, draw a line downward that line downward that represents the 300 lb represents the 300 lb downward – then downward – then calculate the sum of the calculate the sum of the forces, forces, + 233.33 – 300 = - 66.67, + 233.33 – 300 = - 66.67, and if the lines are and if the lines are drawn to a suitable drawn to a suitable scale, see that your line scale, see that your line goes BELOW the goes BELOW the reference line, which reference line, which indicates the negative indicates the negative value of -66.67 lb.value of -66.67 lb.

200 lb

A 5' 6'

beam

24'-0"5' 8'

300 lb 400 lb

B

433.33 466.67

433.33 233.33

- 66.67

From that point, draw a From that point, draw a line horizontal to the line horizontal to the right and realize that right and realize that nothing happens to nothing happens to change the – 66.67, change the – 66.67, until you get to the until you get to the downward 400 lb force. downward 400 lb force.

From that point, draw a From that point, draw a line downward that line downward that represents the 400 lb represents the 400 lb downward – then downward – then calculate the sum of the calculate the sum of the forces forces -66.67 – 400 = - 466.67, -66.67 – 400 = - 466.67, which then indicates which then indicates the negative value the negative value below the line of -below the line of -466.67 lb.466.67 lb.

200 lb

A 5' 6'

beam

24'-0"5' 8'

300 lb 400 lb

B

433.33 466.67

433.33 233.33

- 66.67

- 466.67

From that point, draw a From that point, draw a line horizontal to the line horizontal to the right and realize that right and realize that nothing happens to nothing happens to change the – 466.67 change the – 466.67 until you get to the until you get to the UPWARD reaction at B, UPWARD reaction at B, which is a +466.67 which is a +466.67 force. force.

From that point, draw a From that point, draw a line vertically upward line vertically upward that that represents the represents the +466.67 lb +466.67 lb reaction - - - and see reaction - - - and see that the sum of the that the sum of the forces forces equals zero, at the right equals zero, at the right end of the reference end of the reference line. line.

200 lb

A 5' 6'

beam

24'-0"5' 8'

300 lb 400 lb

B

433.33 466.67

433.33 233.33

- 66.67

- 466.67

0 0

This demonstrates graphically that This demonstrates graphically that all the all the downward forces must equal all the upward downward forces must equal all the upward forcesforces, and satisfies the equilibrium formula that , and satisfies the equilibrium formula that the sum of vertical forces = 0.the sum of vertical forces = 0.

Also, important to the graphic values of bending Also, important to the graphic values of bending moments, is that the moments, is that the sum of the AREAS above the sum of the AREAS above the reference linereference line of the shear diagram must equal of the shear diagram must equal the the sum of the AREAS below the reference linesum of the AREAS below the reference line. .

The diagram drawn below The diagram drawn below the free body diagram the free body diagram is a graphic illustration is a graphic illustration of the values of vertical of the values of vertical SHEAR across the SHEAR across the length of the beam.length of the beam.

It is called a SHEAR It is called a SHEAR DIAGRAMDIAGRAMof the beam loading.of the beam loading.

200 lb

A 5' 6'

beam

24'-0"5' 8'

300 lb 400 lb

B

433.33 466.67

433.33 233.33

- 66.67

- 466.67

0 0

Now take a clean sheet of paper and do this Now take a clean sheet of paper and do this exercise.exercise.With the reactions calculated, SKETCH A SHEAR With the reactions calculated, SKETCH A SHEAR DIAGRAM.DIAGRAM.

Leave space below your diagram to draw a moment Leave space below your diagram to draw a moment diagram later.diagram later.

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

50 lb

beam

B By = 20 lb Ay = 30 lb

NOW ON THE NOW ON THE BACK SIDEBACK SIDE OF THAT SAME OF THAT SAME SHEET OF PAPER DO THISSHEET OF PAPER DO THISEXERCISE TWO: EXERCISE TWO: Sketch a shear diagram. Sketch a shear diagram.

Ay = 280 lb. By = Ay = 280 lb. By = 420 lb420 lb

A 6'

beam

20'-0"7' 7'

200 lb 500 lb

B

The shear diagram is a graphic illustration of the The shear diagram is a graphic illustration of the values of vertical SHEAR across the length of the values of vertical SHEAR across the length of the beam. The values vary because of the position of beam. The values vary because of the position of loads. It is easy to see that if one were to cut the loads. It is easy to see that if one were to cut the beam just to the right of the 200 lb load, then beam just to the right of the 200 lb load, then isolate the left side of the cut, that it would take a isolate the left side of the cut, that it would take a force of 233.33 lb to keep that part in equilibrium.force of 233.33 lb to keep that part in equilibrium.

Realize that any Realize that any portion of a body portion of a body may be isolated, may be isolated, provided the provided the supports for that supports for that portion of the portion of the body are replaced body are replaced with forces that with forces that will maintain will maintain equilibrium for equilibrium for the object.the object.

By = 466.67 lb5' 5' 8' 6'

233.33

referenceline

shear

433.33

-66.67

466.67

24'

Ay = 433.33 lb

below linenegative shear

positive shearabove line

Cut the beam just to the right of the 200 lb load Cut the beam just to the right of the 200 lb load and isolate the left side. Then sum the vertical and isolate the left side. Then sum the vertical forces and write:forces and write:

+ 433.33 lb - 200 lb - shear value = 0+ 433.33 lb - 200 lb - shear value = 0Shear value = 433.33 – 200 = 233.33 lb and Shear value = 433.33 – 200 = 233.33 lb and

must be must be downwarddownward

shear at the cut5'

a force of 233.33 lbis required here tohold the left part ofthe beam in equilibriumwhich is the value of

section

Ay = 433.33 lb

200 lb

cut

WITH THE SAME SECTION OF THE LEFT END OF WITH THE SAME SECTION OF THE LEFT END OF THE BEAM, CONSIDER THAT BENDING MOMENT IS THE BEAM, CONSIDER THAT BENDING MOMENT IS PRESENT IN THE BEAM AS WELL AS SHEAR. AT A PRESENT IN THE BEAM AS WELL AS SHEAR. AT A POINT JUST BELOW THE 200 LB LOAD, SUM THE POINT JUST BELOW THE 200 LB LOAD, SUM THE MOMENTS THERE AND WRITE:MOMENTS THERE AND WRITE:

433.33 x 5 = 2166.65 ft lbs of moment must 433.33 x 5 = 2166.65 ft lbs of moment must exist to hold the assembly in equilibrium. exist to hold the assembly in equilibrium. Realize that the 200 lb force will not cause Realize that the 200 lb force will not cause rotation about the moment point, because rotation about the moment point, because perpendicular distance is 0.perpendicular distance is 0.

200 lb

Ay = 433.33 lb5' point, counter-clockwise in

direction to hold the end ofthe beam in equilibrium

section

2166.65 must exist at thea moment of 433.33 x 5 =the 200 lb load, and realizethis point, just beneath sum the moments about

cut

NOW CUT A SECTION THROUGH THE BEAM JUST TO THE NOW CUT A SECTION THROUGH THE BEAM JUST TO THE RIGHT OF THE 300 LB LOAD - - -RIGHT OF THE 300 LB LOAD - - -

AND FIND THE FORCES REQURED FOR EQUILBRIUMTHE RIGHT END AWAY, ISOLATE THE LEFT ENDCUT A SECTION THROUGH THE BEAM, THROW

5'Ay = 433.33 lb

5'cutline

66.67

200 lb 300 lbHere, with the leftof the beam isolated,with 433.33 lb upand 500 lb down, ashear value of 66.67lb is required to holdthe assembly inequilibrium.

sdfgsdfg

Ay = 433.33 lb

200 lb

THEN WITH THE SAME SECTION, CONSIDER THEMOMENT VALUE. Select a point just beneath the 300lb load, sum the moments and write: -(433.33 x 10) + (200 x 5) = - 4333.3 + 1000 = 3333.3 ft lbs

5'

300 lb

point w

which means that amoment value of3333.3 ft lbs existsin the beam at this

5'

ONCE MORE, JUST TO THE RIGHT OF THE 400 LB ONCE MORE, JUST TO THE RIGHT OF THE 400 LB LOAD LOAD

300 lb

CUT A SECTION THROUGH THE BEAM, THROWTHE RIGHT END AWAY, ISOLATE THE LEFT ENDAND FIND THE FORCES REQURED FOR EQUILBRIUM

Here, with the left end of the beam isolated, with 433.33lb up and with 900 lb down, a shear value of 466.67 lbis required to hold the assembly in equilibrium.

Ay = 433.33 lb5' 5'

200 lb

8'

466.67 lb

400 lbcut line

- 7800 + 2600 + 2400 = - 2800 ft lb.

- (433.33 x 18) + (200 x 13) + (300 x 8) =

MOMENT VALUE. Select a point just beneath the 400

300 lb200 lb

5'Ay = 433.33 lb

5' 8'

lb. load, sum the moments, and write:

400 lb

of bending

exists atthis point

moment

2800 ft lb

THEN WITH THE SAME SECTION, CONSIDER THE

ONE CAN READILY SEE THAT A GREAT DEAL OF ONE CAN READILY SEE THAT A GREAT DEAL OF WORK IS REQUIRED TO FIND THE SHEAR AND WORK IS REQUIRED TO FIND THE SHEAR AND MOMENT VALUES AT EACH POINT, PARTICULARLY MOMENT VALUES AT EACH POINT, PARTICULARLY FOR A BEAM WITH A LARGE NUMBER OF LOADS. FOR A BEAM WITH A LARGE NUMBER OF LOADS. THE PROCESS IS EVEN MORE INVOLVED WHEN THE PROCESS IS EVEN MORE INVOLVED WHEN THE LOADS ARE DISTRIBUTED, RATHER THAN THE LOADS ARE DISTRIBUTED, RATHER THAN POINT LOADS.POINT LOADS.

WITH A SHEAR AND MOMENT DIAGRAM, ONE CAN WITH A SHEAR AND MOMENT DIAGRAM, ONE CAN SEE THE ENTIRE LENGTH OF THE BEAM, AND THE SEE THE ENTIRE LENGTH OF THE BEAM, AND THE VALUES OF SHEAR AND MOMENT AT EACH POINTVALUES OF SHEAR AND MOMENT AT EACH POINT

SINCE A RELATIONSHIP EXISTS BETWEEN SHEAR SINCE A RELATIONSHIP EXISTS BETWEEN SHEAR AND MOMENT, A GRAPHIC ILLUSTRATION OF THE AND MOMENT, A GRAPHIC ILLUSTRATION OF THE MOMENT VALUES CAN BE MADE FROM THE SHEAR MOMENT VALUES CAN BE MADE FROM THE SHEAR DIAGRAM.DIAGRAM.

Bending moment within the beam is created by the Bending moment within the beam is created by the loads, and the value also varies with the position loads, and the value also varies with the position and intensity of the loads.and intensity of the loads.

Since moment isSince moment isa force times aa force times adistance, thedistance, theunits of momentunits of momentare foot-poundsare foot-poundsin this example.in this example.

The area of theThe area of theshear diagram isshear diagram isalso in units ofalso in units offoot-pounds. foot-pounds.

Draw a line belowDraw a line belowThe shear diagram that represents a reference for The shear diagram that represents a reference for

moment.moment.

By = 466.67 lb5' 5' 8' 6'

233.33

referenceline

shear

433.33

-66.67

466.67

24'

Ay = 433.33 lb

below linenegative shear

positive shearabove line

Consider this statement and follow the Consider this statement and follow the procedure to construct and illustrate the moment procedure to construct and illustrate the moment diagram;diagram;

The The CHANGECHANGE in value of the in value of the bending moment bending moment between any two between any two pointspoints along the length of a beam is along the length of a beam is equal to the equal to the AREA OF THE SHEAR AREA OF THE SHEAR DIAGRAMDIAGRAM between the same two between the same two points.points.

To constructTo constructA MOMENTA MOMENTDIAGRAM,DIAGRAM,Begin at the leftBegin at the leftEnd of theEnd of thereference line,reference line,And realize . . .And realize . . .That the valueThat the valueOf BendingOf BendingMomentMomentMust begin at 0Must begin at 0andandEnd at 0End at 0

Then calculate the individual areas of the shear Then calculate the individual areas of the shear diagram, both above and below the reference diagram, both above and below the reference line.line.

By = 466.67 lb5' 5' 8' 6'

233.33

referenceline

shear

433.33

-66.67

466.67

24'

Ay = 433.33 lb

below linenegative shear

positive shearabove line

Moment reference line

By = 466.67 lb5' 5' 8' 6'

233.33

referenceline

shear

433.33

-66.67

466.67

24'

Ay = 433.33 lb

below linenegative shear

positive shearabove line2166.65 1166.65

533.36

2800

Beginning at the left end of the Moment reference Beginning at the left end of the Moment reference line, realize the area of the shear diagram for the line, realize the area of the shear diagram for the first 5’ is first 5’ is 2166.65 ft lb,2166.65 ft lb, which is the which is the CHANGECHANGE in the in the value of MOMENT. Draw a line from the beginning value of MOMENT. Draw a line from the beginning to a point above the reference that will represent to a point above the reference that will represent the value of the moment. the value of the moment. The line will be in an The line will be in an upward direction because the shear value is upward direction because the shear value is POSITIVEPOSITIVE

By = 466.67 lb5' 5' 8' 6'

233.33

referenceline

shear

433.33

-66.67

466.67

24'

Ay = 433.33 lb

below linenegative shear

positive shearabove line

2166.651166.65

533.36

2800

2166.65

Then, realize from the point of 2166.65, the Then, realize from the point of 2166.65, the CHANGECHANGE in inmoment for the next 5’ of beam is the area of the shearmoment for the next 5’ of beam is the area of the sheardiagram for that 5’ of beam,1166.65diagram for that 5’ of beam,1166.65. ADD this sum to the. ADD this sum to the2166.65 to get 3333.3. Continue the moment line 2166.65 to get 3333.3. Continue the moment line

UPWARD toUPWARD tothat point, because the shear is POSITIVE.that point, because the shear is POSITIVE.

By = 466.67 lb5' 5' 8' 6'

233.33

referenceline

shear

433.33

-66.67

466.67

24'

Ay = 433.33 lb

below linenegative shear

positive shearabove line

2166.651166.65

533.36

2800

2166.653333.3

Then, realize from that point of 3333.3, the CHANGE inThen, realize from that point of 3333.3, the CHANGE inmoment for the next 8’ of beam is the area of the shearmoment for the next 8’ of beam is the area of the sheardiagram. diagram. But note that the next segment of shear is But note that the next segment of shear is

below thebelow theline, line, NEGATIVENEGATIVE, so , so the moment line will continue DOWNthe moment line will continue DOWNWARDWARD ; ; SUBTRACT the SUBTRACT the 533.36533.36 from 3333.33 to get from 3333.33 to get 2800.2800.

By = 466.67 lb5' 5' 8' 6'

233.33

referenceline

shear

433.33

-66.67

466.67

24'

Ay = 433.33 lb

below linenegative shear

positive shearabove line

2166.651166.65

533.36

2800

2166.653333.3

2800

The last value of shear area is The last value of shear area is 28002800, which must be , which must be thatthat

amount since the last value of moment was also 2800, amount since the last value of moment was also 2800, in in

order that the last segment of line on the moment order that the last segment of line on the moment diagramdiagram

must return to zero on the reference line.must return to zero on the reference line. Moment Moment values must BEGIN with zero, and END with zero.values must BEGIN with zero, and END with zero.

By = 466.67 lb5' 5' 8' 6'

233.33

referenceline

shear

433.33

-66.67

466.67

24'

Ay = 433.33 lb

below linenegative shear

positive shearabove line

2166.651166.65

533.36

2800

2166.653333.3

2800

By = 466.67 lb5' 5' 8' 6'

233.33

referenceline

shear

433.33

-66.67

466.67

24'

Ay = 433.33 lb

below linenegative shear

positive shearabove line

2166.651166.65

533.36

2800

2166.653333.3

2800

The purpose of a shear and moment diagram is to determine the maximum values of shear and moment in a beam. These are two reactions to loading that must be known so that a beam of suitable material, strength and size can be chosen for the task.

FROM THE SHEAR AND MOMENT DIAGRAMS, ONE CAN SEE, the

Maximum Shear =

466.67 pounds

Maximum Bending Moment =

3333.3 foot pounds

NOW FOR THE TWO BEAMS FOR WHICH YOU DID NOW FOR THE TWO BEAMS FOR WHICH YOU DID SHEAR DIAGRAMS, SKETCH A MOMENT DIAGRAM SHEAR DIAGRAMS, SKETCH A MOMENT DIAGRAM FOR EACH ONE.FOR EACH ONE.

ON YOUR DRAWINGS, JUST BELOW THE SHEAR ON YOUR DRAWINGS, JUST BELOW THE SHEAR DIAGRAM, DRAW A LINE TO REPRESENT A DIAGRAM, DRAW A LINE TO REPRESENT A REFERENCE LINE FOR MOMENT.REFERENCE LINE FOR MOMENT.

THEN CALCULATE THE AREAS OF THE SHEAR THEN CALCULATE THE AREAS OF THE SHEAR DIAGRAM.DIAGRAM.

THEN, BEGINNING AT THE LEFT END OF THE THEN, BEGINNING AT THE LEFT END OF THE MOMENT REFERENCE LINE, DRAW A LINE FOR MOMENT REFERENCE LINE, DRAW A LINE FOR EACH SEGMENT THAT REPRESENTS SHEAR AREAEACH SEGMENT THAT REPRESENTS SHEAR AREA

REMEMBER, THE MOMENT LINES GO REMEMBER, THE MOMENT LINES GO UPWARD UPWARD FOR POSITIVEFOR POSITIVE SHEAR, AND SHEAR, AND DOWNWARD FOR DOWNWARD FOR NEGATIVENEGATIVE SHEAR SHEAR

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

50 lb

beam

B By = 20 lb Ay = 30 lb

0

30

- 20

AREA = 120

AREA = 120

SHEAR DIAGRAM

YOUR FIRST SHEAR DIAGRAM SHOULD LOOK LIKE THIS:

A 6'

beam

20'-0"7' 7'

200 lb 500 lb

B

By = 420Ay = 280

280

80

-420

AREA = 1960AREA = 560

AREA = 2520

SHEAR DIAGRAM

OBSERVE THAT A MAXIMUM MOMENT OCCURS WHERE THE SHEAR LINE CROSSES THE ZERO REFERENCE LINE

YOUR SECOND SHEAR DIAGRAM SHOULD LOOK LIKE THIS:

0