BEAMS-SHEAR AND MOMENT.ppt

7/30/2019 BEAMS-SHEAR AND MOMENT.ppt

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1

BEAMS

SHEAR AND MOMENT


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2

Beam Shear

Shear and Moment

Diagrams

Vertical shear:

tendency for one partof a beam to move

vertically with respect

to an adjacent part


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3

Beam Shear

Magnitude (V) = sum of vertical forces oneither side of the section

can be determined at any section along the

length of the beamUpward forces (reactions) = positive

Downward forces (loads) = negative

Vertical Shear = reactionsloads

(to the left of the section)


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4

Beam Shear

Why?

necessary to know the maximum value ofthe shear

necessary to locate where the shear changesfrom positive to negative

where the shear passes through zero

Use of shear diagrams give a graphicalrepresentation of vertical shear throughoutthe length of a beam


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5

Beam Shear

Simple beam

Span = 20 feet

2 concentrated loads

Construct shear

diagram


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6

Beam ShearExample 1

1) Determine the reactions

Solving equation (3):


Figure6.7a =>

)'20()'161200()'68000(03

1200800002

01

2

..

1

2

..

1

RM

RRF

F

lblb

lblb

y

x

.

.

..

2

....

2

360,320

200,67

200,19000,48'20

lb

ft

ftlb

ftlbftlb

R

R

.

1

...

1

840,5

360,3200,1000,8

lb

lblblb

R

R


7/357

Beam ShearExample 1 (pg. 64)

Determine the shear at various

points along the beam

.

)18(

.

)8(

.

)1(

360,3200,1000,8480,5

160,2000,8480,5

480,50480,5

lb

x

lb

x

lb

x

V

V

V


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Beam ShearExample 1

Conclusions

max. vertical shear = 5,840 lb.

max. vertical shear occurs at greater

reaction and equals the greater reaction

(for simple spans)

shear changes sign under 8,000 lb. load

where max. bending occurs


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Beam ShearExample 2

Simple beam

Span = 20 feet

1 concentrated load

1 uniformly distr. load

Construct shear diagram,

designate maximum shear,

locate where shear passes

through zero


10/3510

Beam ShearExample 2

Determine the reactions



)'24()'166000()]'6)('12000,1[(03

000,6)'12000,1(02

01

2

..

1

2

..

1

RM

RRF

F

lbftlb

lbftlb

y

x

.

.

..

2

....

2

000,724

000,168

000,96000,72'24

lb

ft

ftlb

ftlbftlb

R

R

.

1

...

1

000,11

000,7000,6000,12

lb

lblblb

R

R


11/3511

Shear and Moment Diagrams


12/3512

Beam ShearExample 2



.)24(

.

)16(

.

)16(

.

)12(

.)2(

.

)1(

000,7000,6000,112000,11

000,7000,6000,112000,11

000,1)000,112(000,11

000,1)000,112(000,11

000,9)000,12(000,11

000,10)000,11(000,11

lbx

lb

x

lb

x

lb

x

lbx

lb

x

V

V

V

V

V

V


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Beam ShearExample 2

Conclusions

max. vertical shear = 11,000 lb.

at left reaction

shear passes through zero at some

point between the left end and theend of the distributed load

x = exact location from R1

at this location, V = 0

feetx

xV

11

)000,1(000,110


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Beam ShearExample 3

Simple beam withoverhanging ends Span = 32 feet


1 uniformly distr. load

acting over the entire beam

Construct shear diagram,designate maximum shear,locate where shear passes

through zero


15/3515

Beam ShearExample 3

)'20(4232)'32500()'28000,2()'6000,12()'4000,4(04

)'20(82

32)'32500()'24000,4()'14000,12()'8000,2(03

)'32500(000,4000,12000,202

01

1

.....

2

2

.....

1

2

...

1

.

RM

RM

RRF

F

ftlblblblb

ftlblblblb

ftlblblblb

y

x

.

.

..

2

........

2

800,18

20

000,376

000,128000,96000,168000,16'20

lb

ft

ftlb

ftlbftlbftlbftlb

R

R

...

1

........

1

200,15

'20

000,304

000,192000,56000,72000,16'20

lbftlb

ftlbftlbftlbftlb

R

R


16/3516

Determine the reactions




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Beam ShearExample 3



.......)32(

.......

)28(

......

)28(

......

)22(

.....

)22(

.....

)8(

....

)8(

000,4'32500000,12000,2800,18200,15

000,6'28500000,12000,2800,18200,15800,12'28500000,12000,2200,15

800,9'22500000,12000,2200,15

200,2'22500000,2200,15

200,9'8500000,2200,15

000,6'8500000,20

lbftlblblblblb

x

lbftlblblblblb

x

lbftlblblblb

x

lbftlblblblb

x

lbftlblblb

x

lbftlblblb

x

lbftlblb

x

V

VV

V

V

V

V


18/3518

Beam ShearExample 3

Conclusions

max. vertical shear = 12,800

lb.

disregard +/- notations

shear passes through zero at

three points

R1, R2, and under the 12,000lb.

load


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Bending Moment

Bending moment: tendency of a beam to bend dueto forces acting on it

Magnitude (M) = sum of moments of forces oneither side of the section can be determined at any section along the length of the

beam

Bending Moment = moments of reactionsmoments of loads

(to the left of the section)


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Bending Moment


21/3521

Bending MomentExample 1

Simple beam

span = 20 feet


shear diagram fromearlier

Construct momentdiagram


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22


1) Compute moments at

critical locations

under 8,000 lb. load &

1,200 lb. load

....)'16(

...

)'6(

440,13)'10000,8()'16840,5(

040,350)'6840,5(

ftlblblb

x

ftlblb

x

M

M


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Simple beam

Span = 20 feet

1 concentrated load

1 uniformly distr. Load

Shear diagram

Construct moment

diagram


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1) Compute moments at

critical locations

When x = 11 ft. and under

6,000 lb. load

...)'16(

...

)'11(

000,5642

1212000,1)'16000,11(

500,602

1111000,1)'11000,11(

ftlbftlblb

x

ftlbftlblb

x

M

M


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Negative Bending Moment

Previously, simple beams subjected to positive

bending moments only

moment diagrams on one side of the base line

concave upward (compression on top)

Overhanging ends create negative moments concave downward (compression on bottom)


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Negative Bending Moment

deflected shape has

inflection point

bending moment = 0

See example


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Negative Bending Moment -

Example Simple beam with

overhanging end on right

side

Span = 20

Overhang = 6

Uniformly distributed load

acting over entire span

Construct the shear and

moment diagram

Figure 6.12


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Example

)'20(

2

26)'26600(03

)'26600(02

01

2

.

1

2

.

1

RM

RRF

F

ftlb

ftlb

y

x

1) Determine the

reactions

- Solving equation (3):

- Solving equation (4):

.

.

..

2

..

2

140,1020

800,202

800,202'20

lb

ft

ftlb

ftlb

R

R

.

1

..

1

460,5

140,10600,15

lb

lblb

R

R


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Example

2) Determine the shear at

various points along

the beam and draw

the shear diagram

.

)20(

.

)20(

.

)10(

.

)1(

600,3)60020(140,10460,5

540,6)60020(460,5

540)60010(460,5

860,4)6001(460,5

lb

x

lb

x

lb

x

lb

x

V

V

V

V


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Example

3) Determine where the

shear is at a

maximum and where

it crosses zero max shear occurs at the right

reaction = 6,540 lb.

feetx

xV

1.9

)600(460,50


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Example

4) Determine the

moments that the

critical shear points

found in step 3) anddraw the moment

diagram

...)20(

...

)1.9(

800,10

2

20'20600)'20460,5(

843,242

1.9'1.9600)'1.9460,5(

ftlbftlblb

x

ftlbftlblb

x

M

M


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Example4) Find the location of the inflection

point (zero moment) and max.bending moment

since x cannot =0, then we use x=18.2

Max. bending moment =24,843 lb.-ft.

feetfeetx

x

xxxxM

xxM

xxxM ftlb

2.18;0600

460,55460

)300(2

)0)(300(4)460,5(460,5

460,5300300460,50

2600460,50

2600)460,5(0

2

22

2


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Rules of Thumb/Review

shear is dependent on the loads and reactions

when a reaction occurs; the shear jumps up by the

amount of the reaction

when a load occurs; the shear jumps down by theamount of the load

point loads create straight lines on shear diagrams

uniformly distributed loads create sloping lines of

shear diagrams


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Rules of Thumb/Review

moment is dependent upon the shear diagram

the area under the shear diagram = change in the

moment (i.e. Ashear diagram = M)

straight lines on shear diagrams create slopinglines on moment diagrams

sloping lines on shear diagrams create curves on

moment diagrams

positive shear = increasing slope

negative shear = decreasing slope


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Typical Loadings

In beam design, only

need to know:

reactions

max. shear max. bending moment

max. deflection

Documents

BEAMS-SHEAR AND MOMENT.ppt