applied forces and stress

7/29/2019 Applied Forces and Stress

1/23

Systems of Forces and Moments

Equivalent Force Systems

Shear and Bending Moment Diagrams

Dynamics and You


2/23

2


Two force and moment systems (1 and 2) are

equivalent if they make equivalent contributions to

the equilibrium equations. In other words,

F F

M M

1 2

1 2P P

P P P

Fr

MrP

M F3

F2F1

EquivalentQ

Fr

dEquivalent


3/23

3

Equivalent Force Systems:

Example

Find the point force that is equivalent to the

distributed load shown

in.67.6

lbs.750

lbs.-in.5000

Therefore,

isoriginthefromlocatedbemustforceresultantheat which tdistancethe

Handbook,EngineerABDRin the3.6.2.2sectiontoAccording

lbs.-in.500015)(

isoriginabout themomentresultantThe

lbs.75015)(

islineaalongondistributiforcethisofresultantforceThe

10

0

2

10

0

10

0

10

0

d

F

Md

dxxdxxwxM

dxxdxxwF

r

r

r

r

wx x( ) = 15

10 in.

Fr

6.7 in.


4/23

4

Typical Equivalent Force Systems

xc

L

w x( )

wo

xc

L

F

wl

F

F

w x( )

w x( )

wl

xc

L

w x w=

F wL=

xcL

2---=

w x w1 w0

L-------------------

x w0+=

Fw1 w0+

2--------------------

L=

xc

2w 1 w0+ L

3 w1 w 0+ --------------------------------=

w x w 1x

L---=

F2

3--- w1L=

xc3

5

---L=


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5


Example

Example: Represent the loaded structure with one force and

one moment vector at point A.

10 in

5 in

500 lbs

300 lbs

200 lbs

1800 in lbs

750 lbs

10 in

5 in

500 lbs

300 lbs

200 lbs

150 lbs/in

1800 in lbs

A


6/23

6


Example

Example: Represent the loaded structure with one force and

one moment vector at point A.

10 in

5 in

500 lbs

300 lbs

200 lbs

1800 in lbs

750 lbs

10 in

5 in

500 lbs

300 lbs 200 lbs

1800 in lbs

A

750 lbs

5000 in lbs

1000 in lbs


7/23

7


Example

10 in

5 in

500 lbs

300 lbs 200 lbs

1800 in lbs

A

750 lbs

5000 in lbs

1000 in lbsA

500 lbs

250 lbs

5800 in lbs

A

559 lbs

5800 in lbs

27 degrees

OR

We are able to do this under the assumption the

truss structure is infinitely stiff


8/23

8


Graphical representation of the shear and bending

loads along the length of a beam.

By applying equilibrium equations to the sectionedstructure, one obtains the axial, shear, and bending

moment loads as functions of the X-coordinate

Discontinuous at points where loading abruptly

changes

(therefore section between points where loading is

discontinuous)


9/23

9


Oursign convention will be asfollows:

Positive shear on the positive

face

Positive moment on the

positive face

Distributed loads are positive

in the positive y-direction

M

V

V+dV

M+dM

dx

O

w(x)

y

x

+ V + M


10/23

10


This beam is loaded with a

uniform distributed load.

Cut out a section of this beam and replace the

removed portions with shear forces and bending

moments.

Now, sum the forces and moments:

F V V dV w dx

dV w dx wdV

dx

M M V dx w dxdx

M dM

dM V dx V dM

dx

y

O

0

20

Differential relation-

ships between M, V,

and w.

w(x)

Ry

x

y

(a)

M

V

V+dV

M+dM

dx

O

(b)

w(x)

y

x

Rx

MA

dxxx+dx


11/23

11


Integral Relationships

GivendV

dx

w and dM

dx

V

dV x

dxdx w x dx C V x w x dx C

dM x

dxdx V x dx C M x V x dx C

:

( )

The integral relationships are then found by integrating both sides of the equations.

1 1

2 2


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12


Solution Process

Solve for reaction loads at the supports

Divide beam into sections based on changes in applied load

Make cuts inside each section and replace discarded piece

of beam with a shear force and a bending moment. Obeysign conventions

Solve for shear force and bending moment within each

section

Summation of forces and moments (preferred)

Using integral / differential relationships (more difficult)

Take into account first order discontinuities caused by

concentrated loads and moments

Graph the results


13/23

13

Shear / Bending Moment: Example

10 5 5 5 15

Graph the shear and bending moment diagram for this

beam using the summation of forces method. Go one

section at a time.


14/23

14


The first step is to solve for the reaction loads. To do this,

we need a free body diagram.

lbs3.3453lbs3.2203

lbsin3.86333in25

0in40lbs3000in25 in3

123lbs1250lbsin2000in5lbs500

lbs1250

0

0lbs3000lbs1250lbs500

yy

y

yA

yy

xx

yyy

BA

B

BM

BA

AF

BAF

Notice how Ivealready calculated

the equivalent

concentrated loads

for the distributed

loads.


15/23

15


Section 1.

F lbslbs

inin

lbs

lbs inlbs

inin in

in lbs

y

2203 3 50 0

50 2203 3

2203 3 502

0

25 2203 32

.

( ) .

.

( ) .

x V

V x x

M M x xx

M x x x

O

For 0 < x < 10 in


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16


Section 2:

F V

V x

M x x M

M x x

y

O

22033 50 10 0

17033

22033 50 10 5 0

17033 2500

.

( ) .

.

( ) .

lbslbs

inin

lbs

lbs inlbs

inin in in

in lbs

For 10 in < x < 15 in


17/23

17


Section 3:

F V

V x

M x x M

M x x

y

O

2203 3 50 10 0

17033

2203 3 50 10 5 2000 0

1703 3 4500

.

( ) .

.

( ) .

lbslbs

inin

lbs

lbs inlbs

inin in in in lbs

in lbs

For 15 in < x < 20 in


18/23

18


Section 4:

F x x V

V x x x

M x x

x x x M

M x x x x

y

O

2203 3 50 101

220 100 20 0

50 2000 18296 7

2203 3 50 10 5 2000

1

320

1

220 100 20 0

50

31000 18296 7

413500

3

2

3 2

.

( ) .

.

( ) .

lbslbs

inin in in

lbs

in

lbs

lbs inlbs

inin in in in lbs

in in in inlbs

in

in lbs

For 20 in < x < 25 in


19/23

19


Section 5:

F V

V x lbs

M x x

x x M

M x x

y

O

22033 50 101

2500 5 34533 0

3000

22033 50 10 5 2000

1

2500 5 23

1

334533 25 0

3000 120000

. .

( )

.

.

( )

lbslbs

inin

lbs

inin lbs

lbs inlbs

inin in in in lbs

lbs

inin in in lbs in in

in lbs

For 25 in < x < 40 in


20/23

20


50 lbs/in 3000 lbs500 lbs/in

10" 5" 5" 5" 15"

3000

453

1703

2203

+

-

19533

28050

30050

3856644992

V(x)lbs

M(x)in lbs

-

+

1 2 3 5

2000 in-lbs

BA


21/23

21

Dynamic Forces

ABDR avoids dynamic analysis because of the

complexities involved

Assume all dynamic forces act about the center of

gravity of the part Form:

n is referred to as the load factor. For steady state

flight, the load factor is 1.0

F Wa

gW n where n

a

ggrav

1 1


22/23

22

Dynamic Forces

Calculate the load, P, at the two attach

points for steady level flight and a 6.0

g pull-up.

ote: JP8 fuel density is 6.7 lbs/gal

Weight at each attach point:

lbsWAP 12952

1107.6370

lbsP

lbsP

nWP

g

g

AP

77670.61295

12950.11295

0.6

0.1


23/23

23

Summary



Dynamics and You

applied forces and stress

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