Draft

13.2CurvedSpaceStructures 221

withrespectto thexandyaxisareBPandABrespectively. Applyingthreeequations

of equilibriumwe obtain

F

A

z 

Z

=

=0

wRd= 0 ) F

A

z =wR (13.23-a)

M

A

x

Z

=

=0

(wRd)(Rsin) = 0 ) M

A

x = 2wR

2

(13.23-b)

M

A

y 

Z

=

=0

(wRd)R(1cos) = 0 ) M

A

y =wR

2

 (13.23-c)

II InternalForcesaredeterminednext

1. ShearForce:

(+ 6 ) Fz= 0)V

Z



0

wRd= 0) V=wr (13.24)

2. BendingMoment:

MR= 0)M

Z



0

(wRd)(Rsin ) = 0) M=wR

2

(1cos) (13.25)

3. Torsion:

MT= 0)+

Z



0

(wRd)R(1cos ) = 0) T=wR

2

(sin) (13.26)

III De
ection aredeterminedlastwe assumea rectangularcross-sectionof widthbandheight

d= 2banda Poisson'sratio= 0:3.

1. Notingthatthemember willbe subjectedto both
   exuralandtorsionaldeforma-

tions,we seekto determinethetwo stinesses.

2. The
   exuralstinessEIis givenbyEI=E

bd

3

12

=E

b(2b)

3

12

=

2 Eb

4

3

=: 667 Eb

4

.

3. Thetorsionalstinessof solidrectangularsectionsJ=kb

3

dwherebis theshorter

sideof thesection,dthelonger,andka factorequalto .229for

d

b

= 2. Hence

G=

E

2(1+)

=

E

2(1+:3)

=: 385 E, andGJ= (: 385 E)(: 229 b

4

) =: 176 Eb

4

.

4. Consideringboth
   exuralandtorsionaldeformations,andreplacingdxbyrd:

P


|{z}

W



=

Z



0

M

M

EIz

Rd

| {z }

Flexure

• 
  

Z



0

T

T

GJ

Rd

| {z }

Torsion

| {z }

U



(13.27)

wheretherealmoments weregivenabove.

5. Assuminga unitvirtualdownwardforceP= 1, we have

M = Rsin (13.28-a)

T = R(1cos) (13.28-b)