EGR 236Lecture07StaticallyIndeterminate

8/12/2019 EGR 236Lecture07StaticallyIndeterminate

1/11


2/11

)ndeter$inate Loads

A staticall deter$inatestructure is one where the equations of statics are

sufficient to calculate all the support reactions.

Known: P

n!nowns: "Ax# "Ay# "$%quations:

0x

F =0

yF =

0M=

& un!nowns ' & equations: Statically determinate.

A staticall indeter$inatestructure has extra constraints. (he equations of

static equilibrium are not enou)h to find all the support reactions.

Known: P

n!nowns: "Ax# "Ay# "$x# "$y%quations:

0xF = 0yF = 0M=

* un!nowns + & equations

(his is an example of a statically indeterminate structure.

P

P

P

5

5

P

5

5


3/11

Identify each of the followin) structures as statically determinate ,S.-.#

statically indeterminate ,S.I.# or insufficiently constrained ,I./..

Process!

P

P

P

P

P

P

S. -.# S. I.# or I. /.

S. -.# S. I.# or I. /.

S. -.# S. I.# or I. /.

S. -.# S. I.# or I. /.

S. -.# S. I.# or I. /.S. -.# S. I.# or I. /.


4/11

In order to sol0e statically indeterminate problems# in addition to the static

equations# load deformation equations need to be found. 1ou will need one

additional load2deformation equation for each extra support reaction.

(he process for sol0in) a staticall indeter$inate pro/le$can be identified

by the followin) steps:

3 -raw the "$-

4 5ote the number of un!nowns

& 6rite out the static equilibrium equations

* Identify how many load2deformation equations are required.

7 6rite load2deformation equations for each extra un!nown.

8 Sol0e the set of equations.

Ea$ple 1!A structure made of a solid steel rod and a hollow brass cylinder is fixed

between two immo0able supports. (he plate between the materials is loaded

with 799 !5 as shown. -etermine the stress in each material and the

displacement of the plate at /.

%steel ' 499 Pa

%brass' 399 Pa

Solution:

3 -raw the "$-

4 5ote number

of un!nowns:

& 6rite outStatic %quilibrium

%quation,s

4m

3.7m

799 !5

Steel

$rass

-Steel

'79 mm

-$rass ;

' 399 mm

-$rass I

' 79 mm


5/11

Ea$ple 1 continued!


6/11

Ea$ple 2!

A &9 !ip load is applied

to the end of the ri)id

beam A-.

a determine the axial

stress set up in the

aluminum and stainless

steel wires.

b determine the

displacement of end -.

Solution:

Ass' 3.


7/11

Ea$ple 3!

(he column is constructed from hi)h stren)th concrete and six A2&8 steeel

reinforcin) rods. If it is sub=ected to an axial force of &9 !ip# determine the

required diameter of each rod so that 3>* of the load is carried by the concrete

and &>* of the load by the steel.

%st' 4?999!si %con'*499!si

Solution:

@ in dia

& ft

&9 !ip


8/11

Ea$ple 1 Solution!

A structure made of a solid steel rod and a hollow brass cylinder is fixed

between two immo0able supports. (he plate between the materials is loaded

with 799 !5 as shown. -etermine the stress in each material and the

displacement of the plate at /.

%steel ' 499 Pa

%brass' 399 Pa

Solution:

3 -raw the "$-

4 5ote number

of un!nowns: 4 un!nowns:

& 6rite out

Static %quilibrium

%quation,s

0F=500 0steel brassF F kN =

* Identify number of load2deformation equations needed.

number of un!nowns 2 number of equation ' additional equations needed

4 2 3 ' 3

7 6rite out load2deformation equation,s

%lon)ation of the steel ' compression of the brass

steel brass =

or

%lon)ation of steel %lon)ation of brass ' 5o o0erall chan)e of len)th

0steel brass + =

where: and

steel steelsteel

steel steel

F L

E A = brass brassbrass

brass brass

F L

E A =

4m

3.7m

799 !5

Steel

$rass

-Steel

'79 mm

-$rass ;

' 399 mm

-$rass I

' 79 mm

799 !5

"steel

"$rass


9/11

8 Sol0e:

500 0steel brassF F kN = ,%q. 3

0steel steel brass brass

steel steel brass brass

F L F L

E A E A

+ = ,%q. 4

where Lst' 4 m Lbr' 3.7 m

%st' 499 Pa %br' 399 Pa

2 22 3(50 ) 1963 0.001963

4 4steel

D mmA mm m

= = = =

2 2 2 22 3( ) ((100 ) (50 ) ) 5890 0.005890

4 4O I

brass

D D mm mmA mm m

= = = =

therefore:

2 2

(2 ) (1.5 )

(200 )(1963 ) (100 )(5890 )steel brass

F m F m

GPa mm GPa mm

=

2

2

1.5 200 19630.5

2.0 100 5890steel brass brass

m GPa mmF F F

m GPa mm= =

puttin) into %q 3

( 0.5 ) 500 0brass brassF F kN = 1.5 500brassF kN=

500

3331.5

brass

kNF kN= =

0.5 0.5(333 ) 167steel brassF F kN kN = = =

so2

2 9 2

(167 )(2 ) 1000 1000

1(200 )(1963 ) 10 /steel steel

steel

steel steel

F L kN m N GPa mm Pa

E A kN mGPa mm Pa N m

= =

0.000849 0.849steel m mm = = 0.849brass steel mm = =


10/11

Ea$ple 2 Solution!

A &9 !ip load is applied

to the end of the ri)idbeam A-.

a determine the axial

stress set up in thealuminum and stainless

steel wires.

b determine thedisplacement of end -.

Solution: from table: %al' 39999 !si

%ss' 4?999 !si

"rom static equilibrium:

0A

M =0 (3 ) (8 ) (30 )(10 )B CF ft F ft kip ft = + 3 8 3000B CF F kip+ =(wo un!nowns but only 3 equationB..3 deformation equation will be needed.

Assumin) a ri)id beam:"rom similar trian)les:

CB D

AB AC ADL L L

= =

3 8 10CB D

= =

from axial deformation:

B ss

B ss ss

F L

E A =

and

C al

C al al

F L

E A =

therefore:

1 1

3 8B ss C al

ss ss al al

F L F L

E A E A=

3

8al ss ss

B C

ss al al

L E AF F

L E A=

3 4 29000 2.25

1.868 3 10000 1.75

B C cF F F

= =

Sol0in):

1.86B cF F= and 3 8 3000B CF F kip+ =

3(1.86 ) 8 3000c CF F kip+ =

13.6 3000cF kip= 221cF kip= 1.86(221 ) 411B cF kip kip= =

2 2(411 )(3 )

0.0243(29000 )(1.75 ) /

B

kip ft ksi ft

ksi in kip in = =

so10 10

(0.0243 ) 0.0813 3

D B ft ft = = =

Ass' 3.


11/11

Ea$ple 3 Solution!

(he column is constructed from hi)h stren)th concrete and six a2&8 steeel

reinforcin) rods. If it is sub=ected to an axial force of &9 !ip# determine the

required diameter of each rod so that 3>* of the load is carried by the concrete

and &>* of the load by the steel.

%st' 4?999!si %con'*499!si

Solution:

"$-: ,4 un!nowns 0F=

30 0st conP P + + =

;nly 3 equation and 4 un!nowns:

Load2deformation relationship:

3

4stP P=

1

4conP P= con st =

where

con concon

c on c on

P L

E A = st st st

st st

P L

E A =

Sol0e:

con con st st

con con st st

P L P L

E A E A

= con st st st const con con

E P LA A

E P L

=

4200 0.75

0.43429000 0.25

st con con

ksi P LA A A

ksi P L= =

since

2

64

st

dA

=

2 2

64 4

con

D dA

=

then

2 2 2

6 0.434 64 4 4

d D d =

2 2 213.82 6d D d=

2 219.82d D=

1

0.22419.82

d D D= =

0.224(8 ) 1.797d in in= =

@ in dia

& ft

&9 !ip

Pcon

Pst

&9 !ip

Documents

EGR 236Lecture07StaticallyIndeterminate