Structural Theory 1 (Part 2)

7/27/2019 Structural Theory 1 (Part 2)

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THEORY1 Structural Theory 1 Chapter 2

Reactions on a Three-Hinged Arch;

- a common problem in structural analysis is to find the hinge reactions on athree-hinged arch by algebraic or graphical methods. A three-hinged arch is a

structure composed of two curved segments, joined together by an internal hinge

and supported at two external hinges.

EX. Find the horiontal and vertical components of the three-hinge arch shown on

the figure.

!"#A$ % &'

()!*# " +!# " !(+# " +!) - y/ !)*# % &' y/ % ((.& 01

!"#.$ % &'

Ay/ !)*# 2 ()!) 2 +!(+# 2 !# 2 +!*# % &' Ay/ % (+.& 01

!"#3

$ % &'

()!# " +!*# 2 (+!()# " Ax/ !+# % &' Ax/ % ().& 01

!"# 3$ % &'

!*# " +!# 2 ((!()# " x/ !+# % &' x/ % ().& 01

!"# yF % &' 4 the left segment,

(+ 2 () 2 + " 3y/ % &' 3y/ % ).& 01

!"# xF % &' 4 the left segment,

() - 3x/ % &' 3x/ % ().& 01

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!"# yF % &' 4 the right segment,

-) 2 2 + " (( % &' o0

Engr. .R. !on"on (

+ 4 *.& m % )*.& m

() 01

01+ 01

+ 01

A

3

+.&m

3y/

Ay/

3x/

Ax/ y/

x/

3x/

3y/


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!"# xF % &' 4 the right segment,

() 2 () % &' o0

EX. Find the horiontal and vertical components of the three-hinge arch shown onthe figure.

!"# A$

% &' 4 the left segment,

&!*# - 3x/ !+# - 3y/ !# % &

3x/ " * 3y/ % (+&' e5n. (

!"#.

$ % &' 4 the right segment,

3x/ !*# - 3y/ !# 2 +&!*# % &

3x/ - ) 3y/ % +&' e5n. )

solving simultaneously' 3x/ % 6+.& 01

3y/ % -).& 01

!"# xF % &' 4 the left segment,

Ax/ - 6+ % &' Ax/ % 6+.& 01

!"# yF % &' 4 the left segment,

Ay/ - & - ) % &' Ay/ % ).& 01

!"# xF % &' 4 the right segment,

6+ - x/ % &' x/ % 6+.& 01

!"# yF % &' 4 the right segment,

y/ " ) 2 +& % &' y/ % 6.& 01

Engr. .R. !on"on )

A

3

* 4 *.& m % (+.& m

& 01 +& 01

*.&m

+.&m

Ay/

3y/

Ax/

3x/

x/

3x/

y/

3y/


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!"# xF % &'

6+ - 6+ % &' o0

!"# yF % &'

) " 6 2 & 2 +& % &' o0

#le$i%le Suspension Ca%les;

- in engineering structures, there are structures that are encountered which aremade of flexible cables or chains suspended between supports at the ends and

subjected to some 0ind of a distributed vertical load. 7he e5uilibrium shape

assumed by such a cable is, of course, a funicular curve for loading to which

the cable is subjected.

d %8,

a)

7 % &.6a)

)

d(+

a( +

9 % a

+

+ ...

a

d

:

)6+

a

d

6

-)

a

d

-

,(

+*)

where; d - sag

- uniform loada - span

8 - tension at midpoint

7 - tension at supports

9 - length of cable

EX. Each cable of a suspension bridge carries a horiontal load of (& 01hat is the length of the cable?

by using the formula,

7 % &.6a)

)

d(+

a(+

Engr. .R. !on"on

d

a

8

7


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% &.6!(!()

)

#()#!(+!

#(,&!( + % *@).@* 01

8 %d,

a)

%#()#!,!

#(,!(&! )

% :6.& 01

9 % a

+

+ ...

a

d

:

)6+

a

d

6

-)

a

d

-

,(

+*)

% !(

+

+

+*)

(,&

()

:

)6+

(,&

()

6

-)

(,&

()

-

,(

% ().(( m

by using the e5uilibrium e5uations,

!"# xF % &'

7 cos - 8 % & e5n. (

!"# yF % &'

7 sin - @&& % & e5n.)

dividing e5n. ) by e5n. (,

tan %8

@&&

but, tan %*6

(),

*6

()%

8

@&&' 8 % :6.& 01

7

6:.*+

*6- :6 % &' 7 % *@).@* 01

EX. A cable having a mass of (.6 01


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!"# xF % &'

7 cos - 8 % & e5n. (

!"# yF % &'

7 sin - (.6x % & e5n.)

dividing e5n. ) by e5n. (,

tan % 8x6.(

but, tan %x6.&

y,

x6.&

y%

8

x6.(' 8y % &.:6 )x

since, x % a, when y % )& m,

)&8 % &.:6 )a ,

similarly, on the right side,

(&8 % &.:6 )b ,

from the figure, a " b % &&

&& %:6.&

8)&"

:6.&

8(&' 8 % ((6.() 01

a %:6.&

#().((6,#!)&!% (:6.:* m

b %:6.&

#().((6,#!(&!% ()*.)+ m

s5uaring both e5n. ( and e5n. ),

)7

)cos % )8 e5n.

Engr. .R. !on"on 6

(& m

a b

& m

)&& m

3

A


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)7

)sin % ).)6 )x e5n. *

adding e5n. and e5n. *,

)7 !

)sin " )cos # % )8 " ).)6

)x

since the maximum tension occurs at the left support,

7 % )) #:*.(:6#!)6.)!#().((6,! + % ((:.:* 01

EX. A cable carrying )&& 1


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)a % !*&&&& 2 *&&a "

)a #

)a - +&&a " +&&&& % &' a % ()+.:@ m

b % :.)( m

38 % -&

(&.&

!()+.:@#)

% 6.6@ 01

!"# xF % &'

6.6@ -A8 % &' A8 % 6.6@ 01

!"# yF % &'

AB - &.)&!()+.:@# % &' AB % )6.+ 01

A7 % ))

#-+.)+!#[email protected]! + % 6@.:) 01

!"# xF % &'

.8 - 6.6@ % &' .8 % 6.6@ 01

!"# yF % &'

.B - &.)&!:.)(# % &'

.B % (*.+* 01

.7 % ))

#+*.(*!#[email protected]! + % 66.66 01

9 % a

+

+ ...a

d

:

)6+

a

d

6

-)

a

d

-

,(

+*)

for segment A3,

a

d%

#:@.()+#!)!

-&% &.((

a %)

#:@.()+#!)!% ()+.:@ m

(9 % !()+.:@# ( ) ( ) ( )

++ +*) ((,.&

:

)6+((,.&6

-)((,.&-

,(

% ((. m

for segment 3,

a

d%

#)(.:-#!)!

(&% &.&+

Engr. .R. !on"on :


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a %)

#)(.:-#!)!% :.)( m

)9 % !:.)(# ( ) ( ) ( )

++ +*) &+,.&

:

)6+&+,.&

6

-)&+,.&

-

,(

% :*.(( m

79 % (9 " )9

% ((. " :*.(( % )&6.*@ m

EX. For the cable shown, determine the reaction components, the tension at the

ends and the vertical displacement of .

ta0ing the whole structure,

!"#A

$ % &'

+&&!+# " *6&!((# - .B !(6# % &' .B % 6:&.& 01

!"#.$ % &'

AB !(6# 2 +&&!@# 2 *6&!*# % &' AB % *&.& 01

ta0ing the F of A3,

!"#3

$ % &'

*&!+# -A

8 !)# % &'A

8 % (**&.& 01

ta0ing the F of 3,

!"#3

$ % &'

.8 !)# " *6&!6# 2 6:&!@# % &' .8 % (**&.& 01

Engr. .R. !on"on

) m

*6& 01

+&& 01

+ m 6 m * m

3

A

yA8

) m

+&& 01

37

AB

F of A3

*6& 01

+&& 01

y) m

.8

A37

F of 3

*6& 01

y

(**& 01

6:& 01


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A7 % ))

#(**&!#*,&! + % (6(:.@ 01

.7 % ))

#(**&!#6:&! + % (6*.:( 01

!"# $

% &'

(**&!y# 2 6:&!*# % &' y % (.6 m

&o'ent (iagra' %y )arts;

- in order to apply the basic application of the theorems of structural analysisis to the necessity to compute easily and accurately the area under any part of

a moment diagram and also the moment of such an area about any axis. A method of

doing this from calculus is to integrate the two expressions'

dx$ and #dx$!x

between proper limits, noting that the bending moment must be expressed as afunction of x.

A method being used is by dividing moment diagrams into parts whose areas

and centroids are 0nown' this permits simple numerical calculations to replace

integrations. 7he first step is to learn how to draw moment effects of each

separate loading !hereafter-called moment diagram by parts# instead of a

conventional moment diagram.

7he construction of moment diagram by parts depends on two basic

principles'

- the resultant bending moment at any section caused by any load system is the

algebraic sum of the bending moments at that section caused by each load

acting separately. 7his statement is expressed algebraically by'

$ % 9$ % /$

where 9$ indicates the sum of the moments caused by all the forces to the

left of the section and/

$ is the sum of the moments caused by all the

forces to the right of the section.

- the moment effect of any single specified loading is always some variation of

the general e5uation'

y % n0x

EX. For the beam shown, draw the moment diagram by parts.

Engr. .R. !on"on @

*& 01

)*&)*&

& 01


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!"#A

$ % &'

&!*#!*# - ./ !+# % &' ./ % &.& 01

!"#.$ % &'

A/ !+# 2 &!*#!)# % &' A/ % *&.& 01

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!"# yF % &'

*& " & 2 &!*# % &' o0

EX. For the beam shown, draw the moment diagram by parts.

!"#A

$ % &'

)&!+#!6# - 3/ !+# % &' 3/ % (&&.& 01

!"#3

$ % &'

A/ !+# 2 )&!+#!(# % &' A/ % )&.& 01

3hec0'

!"# yF % &'

Engr. .R. !on"on (&

)& 01


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)& " (&& 2 )&!+# % &' o0

Engr. .R. !on"on ((

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Structural Theory 1 (Part 2)