1 BEAMS SHEAR AND MOMENT. 2 Beam Shear Shear and Moment Diagrams Vertical shear: tendency for one...

1

BEAMS

SHEAR AND MOMENT

2

Beam Shear

Shear and Moment Diagrams

Vertical shear: tendency for one part of a beam to move vertically with respect to an adjacent part

3

Beam Shear

Magnitude (V) = sum of vertical forces on either side of the section– can be determined at any section along the

length of the beam Upward forces (reactions) = positive Downward forces (loads) = negative

Vertical Shear = reactions – loads(to the left of the section)

4

Beam Shear

Why? necessary to know the maximum value of

the shear– necessary to locate where the shear changes

from positive to negative• where the shear passes through zero

Use of shear diagrams give a graphical representation of vertical shear throughout the length of a beam

5

Beam Shear –

Simple beam – Span = 20 feet

– 2 concentrated loads

Construct shear diagram

6

Beam Shear – Example 1

1) Determine the reactions

Solving equation (3):


Figure 6.7a =>

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

1200800002

01

2..

1

2..

1

RM

RRF

F

lblb

lblby

x

..

..

2

....2

360,320

200,67

200,19000,48'20

lbft

ftlb

ftlbftlb

R

R

.

1

...1

840,5

360,3200,1000,8lb

lblblb

R

R

7

Beam Shear – Example 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

lbx

lbx

lbx

V

V

V

8


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

9


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


Determine the reactions



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

000,6)'12000,1(02

01

2..

1

2..

1

RM

RRF

F

lbftlb

lbftlby

x

..

..

2

....2

000,724

000,168

000,96000,72'24

lbft

ftlb

ftlbftlb

R

R

.

1

...1

000,11

000,7000,6000,12lb

lblblb

R

R

11

Shear and Moment Diagrams

12



.

)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

lbx

lbx

lbx

lbx

lbx

V

V

V

V

V

V

13


Conclusions– max. vertical shear = 11,000 lb.

• at left reaction

– shear passes through zero at some point between the left end and the end of the distributed load• x = exact location from R1

– at this location, V = 0

feetx

xV

11

)000,1(000,110

14


Simple beam with overhanging ends– Span = 32 feet– 3 concentrated loads– 1 uniformly distr. load

acting over the entire beam

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

15


)'20(42

32)'32500()'28000,2()'6000,12()'4000,4(04

)'20(8232)'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

ftlblblblby

x

..

..

2

........2

800,1820

000,376

000,128000,96000,168000,16'20

lbft

ftlb

ftlbftlbftlbftlb

R

R

...

1

........1

200,15'20

000,304

000,192000,56000,72000,16'20

lbftlb

ftlbftlbftlbftlb

R

R

16

Determine the reactions



17



.......

)32(

.......)28(

......)28(

......)22(

.....)22(

.....)8(

....)8(

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

000,6'28500000,12000,2800,18200,15

800,12'28500000,12000,2200,15

800,9'22500000,12000,2200,15

200,2'22500000,2200,15

200,9'8500000,2200,15

000,6'8500000,20

lbftlblblblblbx

lbftlblblblblbx

lbftlblblblbx

lbftlblblblbx

lbftlblblbx

lbftlblblbx

lbftlblbx

V

V

V

V

V

V

V

18


Conclusions

– max. vertical shear = 12,800 lb.• disregard +/- notations

– shear passes through zero at three points • R1, R2, and under the 12,000lb.

load

19

Bending Moment Bending moment: tendency of a beam to bend due

to forces acting on it

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

beam

Bending Moment = moments of reactions – moments of loads

• (to the left of the section)

20

Bending Moment

21

Bending Moment – Example 1 Simple beam

– span = 20 feet– 2 concentrated loads– shear diagram from

earlier

Construct moment diagram

22

Bending Moment – Example 1

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

ftlblbx

M

M

23

Bending Moment – Example 2 Simple beam

– Span = 20 feet– 1 concentrated load– 1 uniformly distr. Load– Shear diagram

Construct moment diagram

24

Bending Moment – Example 2

1) Compute moments at critical locations

– When x = 11 ft. and under 6,000 lb. load

...

)'16(

...)'11(

000,56421212000,1)'16000,11(

500,6021111000,1)'11000,11(

ftlbftlblbx

ftlbftlblbx

M

M

25

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)

26

Negative Bending Moment

deflected shape has inflection point– bending moment = 0

See example

27

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

28


)'20(226)'26600(03

)'26600(02

01

2.

1

2.

1

RM

RRF

F

ftlb

ftlby

x

1) Determine the reactions

- Solving equation (3):

- Solving equation (4):

..

..

2

..2

140,1020

800,202

800,202'20

lbft

ftlb

ftlb

R

R

.

1

..1

460,5

140,10600,15lb

lblb

R

R

29


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

lbx

lbx

lbx

lbx

V

V

V

V

30


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

31


4) Determine the moments that the critical shear points found in step 3) and draw the moment diagram

...

)20(

...)1.9(

800,10220'20600)'20460,5(

843,2421.9'1.9600)'1.9460,5(

ftlbftlblbx

ftlbftlblbx

M

M

32


4) 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

33

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 the amount of the load

point loads create straight lines on shear diagrams uniformly distributed loads create sloping lines of

shear diagrams

34

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 = ΔM)

straight lines on shear diagrams create sloping lines on moment diagrams

sloping lines on shear diagrams create curves on moment diagrams

positive shear = increasing slope negative shear = decreasing slope

35

Typical Loadings

In beam design, only need to know:– reactions

– max. shear

– max. bending moment

– max. deflection

1 BEAMS SHEAR AND MOMENT. 2 Beam Shear Shear and Moment Diagrams Vertical shear: tendency for one...

Documents

Transcript of 1 BEAMS SHEAR AND MOMENT. 2 Beam Shear Shear and Moment Diagrams Vertical shear: tendency for one...