Draft

13.1Arches 217

Z



=0

wRd
R(cos cos) +M= 0 (13.12)

) M=wR

2





2

(1sin) + (



2

) cos



(13.13)

III De
ection aredeterminedlast

1. Therealcurvatureis obtainedby dividingthemoment byEI

=

M

EI

=

wR

2

EI





2

(1sin) + (



2

) cos



(13.14)

1. ThevirtualforceP willbe a unitverticalpoint in thedirectionof thedesired

de
ection,causinga virtualinternalmoment

M=

R

2

[1cossin] 0 



2

(13.15)

p

2. Hence,applicationof thevirtualworkequationyields:

1

|{z}

P

= 2

Z

2

=0

wR

2

EI





2

(1sin) + (



2

) cos



| {z }

M

EI

=

R

2

[1cossin]

| {z }

M

Rd

|{z}

dx

=

wR

4

16 EI

h

7 

2

 18  12

i

= : 0337

wR

4

EI

(13.16-a)

13.1.2 StaticallyIndeterminate

Example13-3: StaticallyIndeterminateArch, (Kinney1957)

Determinethevalueof thehorizontalreactioncomponent of theindicatedtwo-hingedsolid

ribarch, Fig. 13.8as causedby a concentratedverticalloadof 10k at thecenterlineof the

span.Considershearing,axial,and
exuralstrains.Assumethattheribis a W24x130witha

totalareaof 38.21in

2

, thatit hasa webareaof 13.70in

2

, a moment of inertiaequalto 4,000

in

4

,Eof 30,000k/in

2

, anda shearingmodulusGof 13,000k/in

2

.

Solution:

1. ConsiderthatendCis placedonrollers,as shownin Fig.??A unitctitioushorizontal

forceis appliedatC. Theaxialandshearingcomponents of thisctitiousforceandof

theverticalreactionatC, actingonany sectionin theright halfof therib,areshown

at theright endof theribin Fig. 13-7.