484
14-17.
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
Consider the arbitrary two-hinged arch shown. Discuss how you
Prob. 14-17
I
z
I
Al
would generate the influence line for the horizontal reaction.
Utilize the results
contained in Examples 14-10 and 14-11.
15
Engineering Theory of
an Arbitrary Member
15-1.
INTRODUCTION; GEOMETRICAL RELATIONS
In the first part of this chapter, we establish the governing
equations for a
member whose centroidal axis is an arbitrary space
curve. The formulation is
restricted to linear geometry and negligible warping
and is referred to as the
enUgineering theory. Examples illustrating the application
of the displacement
and force methods are presented. Next, we outline a
restrained warping for!nulation and apply it to a planar circular member.
Lastly, we cast the force
method for the engineering theory in matrix form and
develop the member
force-displacement relations which are required for
the analysis of a system
of member elements.
The geometrical relations for a member are derived
in Chapter 4. For
convenience, we summarize the differentiation formulas
here. Figure 15-1
shows the natural and local frames. They are related by
t, = t
t2 = os 07i + sin 4)/
/:3 = - sin ¢)ir + cos
(a)
){h
where
= q(s). Differentiating (a) and using the Frenet equations (4-20),
we obtain
dt
dA
[dS
dS- =
d 3
K cos i
(15-1)
dSd
a
K sin
0+ d
-(Tnd0* o+tt/'
Note that a is skew-symmetric.
485
2
486
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
SEC. 15-1.
INTRODUCTION; GEOMETRICAL RELATIONS
487
and the local vectors at Q are orthogonal when a2 = 0, which requires
3
\
a2 3 =
t
d95
dS
0
1
R,
(15-6)
It is reasonable to neglect y/R terms with respect to unity when the member
is thin, i.e., when the cross-sectional dimensions are small in comparison to
Centroidal axis
Y
Local reference directions
X3
as
Fig. 15-1. Natural and local reference frames for a member element.
The curvilinear coordinates' of a point, say Q, are taken as S and Y2, Y3.
Letting R be the position vector to Q (see Fig. 15-2),
(S) + YJ
Y22((S)
+
2 (S)
R
(S)
3t
3
Li
(15-2)
'1T
and differentiating, we find
X2
aR
-=
8y2
OR
t
(15-3))
X1
8y3
Fig. 15-2. Curvilinear directions.
The differential volume at Q is
d(vol.) = (1
=
1
2 a1 2 -
-
,n
R,
Rc and R t. We express d9/dS as
y 3 a13)dS dy 2 dy 3
(15-4)
dS dy 2 dy 3
do- = 0(
where y, is the coordinate of Q in the normal (i) direction and RC= 1/K is
the radius of curvature. Also,
O'
[2 = --Y 3 a2 3 = -Y3
3R ­
aS t, = y 2 a2 3
1-See Sec. 4-8.
+ dS
/1 + dk)
Y2 -R
(15-7)
where L is the total arc length and A is the total increment
in 9i. The non­
orthogonality due to
can be neglected when the member is only slightly
twisted, i.e., when
b
(15-5)
L
(15-8)
where b is a typical cross-sectional dimension. In what follows, we will assume
the member is thin, (15-8) is satisfied, and
defines the orientation of the
principal inertia directions.
7-1
II
ENGINEERING THEORY OF AN ARBITRARY MEMBER
488
CHAP. 15
r
SEC. 15-2.
FORCE-EQUILIBRIUM EQUATIONS
Example 15-1
489
tdS
The curvature and torsion for a circular helix are derived in Example 4-5:
I 1
1
_dlF dS
-F+
.K= - =
we
./=~
2iRII
-, -1~~2
tr
dS
R
where R is the radius of the base circle and H is the rise in one full revolution. The helix
is thin when b/R << 1,where b is a typical cross-sectional dimension.
F!~r·IlI!
· Zl-
2
ad'
Fig. 15-3. Differential element for equilibrium analysis.
~~~~~~~~~
'~~~~~~~~~~~~~~~~~~~~~~~~~
By definition, a member is planar if - = 0 and the normal direction (ii) is an axis of
symmetry for the cross section. We take the centroidal axis to be in the X-X plane
2
and define the sense of t2 according to t2 x 13 =
i3. The'angle q)is constant and equal to
either 0° (t2 = ii) or 180 ° (t2 = -fi). Only a1 2 is finite for a planar member:
a1
3
a2
=
where F = F1, F2, F3 } etc. The vector derivatives are
d F + dFT
dS
dS-t + FTat
dM+
dclMT
dS-: =-t + Mat
a1 2 -- = +K
R=-
Also,
rv~~~~~lml~~~~~la
(C·
'2~~~~~~~~~
Example
15-3
Consider the case
c' where the centroidal axis is straight and 4) varies linearly with S.
The member
Inember is said
sa- to be naturally twisted. Only a2 3 is finite for this case:
t x F+
0
_
_ 1 .
=
F 2[ 3
-
F3 2. =
a23
,
uosttutlng in (15--9), and noting that a" = -a,
librium equations:
IdF
~d/S-
= = const = k
If bk << 1,we can assume OR/OS
RIOS is orthogonal to
to t,2, tt3
3.
dF+
=
(15-9)
dM+
ii + ,+x F
6
dS
We express the force and moment vectors in terms of components referred
to the local frame,
M+
= MTt
b = bTt
m = ImTt
(b)
to theefollowing equi­
F3
O
+F
To establish the force-equilibrium
force-equilibrium equations, we consider the differential
differential
15-3. We use the same notation as for the planar case.
element shown in Fig. 15-3.
The vector
vector equil
equilibrium equations follow from the requirement that the resultant
The
equil
force and moment
morn( vectors must vanish:
mrn(
Flj
lead
0
n +
FORCE-EQUILIBRIUM
FORCE-EQUILIBRIUM EQUATINS
FORCE
EQUATIONS
F+ =
-,F 3, F,2}rt
aF + ) -0
dM
aM+
dS--
(a)
.
a12 = a1 3 = 0
15-2.
2
= FTt
(15-10)
dF
dS - a 12 F 2 - a1 3 F3 + i
= 0
dF2
dS + al 2 F1 - a2 3 F 3 + 2 = 0.
dF3
dS + al 3 F1 + a2 3 F2 + b3 = 0
dMdS
a2M2-a13 f
(15-11)
+
±l
= 0
dM2
aM
- a2 3 M3 + 2 - F3 = 0O
dS
dM 3
-dS- + a1 3 M1 + a2 3 M 2
m
n 3 + F2 = 0
When the member is planar, a3 = a
=
3
0 and the equations uncouple
23
naturally into two systems, one associated
with in-plane loading (b1 , b2, in 3,
It>rF
i
ENGINEERING THEORY OF AN ARBITRARY MEMBER
490
II
CHAP. 15
F1, F2, M 3) and the other with out-of-plane loading (b , n,, n , F , M,, M ).
2
3
2
The in-plane equations coincide with (14-14) when we3
set a 1 2 = 1/R and the
out-of-plane equations take the form
dF
-+
dS
dM 1
dS
dM 2
dS
+
I
R M
NEGLIGIBLE WARPING RESRAINT
z
=
= ut = equivalent rigid body translation
vector at the centroid
= j Tt= equivalent rigid body rotation vector
Juji
= O
w=
1R
(15-12)
+ m2 - F3 =
Cjt
0
FORCE-DISPLACEMENT RELATIONS-NEGLIGIBLE WARPING
RESTRAINT; PRINCIPLE OF VIRTUAL FORCES
f+(-
= AFd
t7i+ M
- au +
(15-13)
'F
i`
G,/k.~,\ -Mj
r/*
am
kT AM)dS =
(15-14)
di APi
-aco d
!·
81 -a2U
(0
cS-
au +
k
o
aS
k--d
dS
i
and write the one-dimensional principle of virtual forces as
dS = Js(eT AF
+ AMT(td
and substituting in (15-14) lead to the following force-displacement
relations:
k= {}=
" kjl
sdV *
-M1iJdS
(a)
-o,)dS
+ (02
du
1,2,3
+ )±
-+o2j
We consider the material to be elastic and define V*
as the complementary
energy per unit arc length. Since we are neglecting
warping restraint, V* is a
function only of F and M. We let
i=
(15-15)
The virtual system satisfies the equilibrium equations (15-5) identically
and
therefore is statically permissible. Evaluating C
dAP,
d1i Pi ~L
15-3.
//0
Now, we apply the principle of virtual forces to the element shown
in
Fig. 15-4. We define a and Was
b
SEC. 15-3.
l
a *
du,
aI1
dS
)3
+ 0
(15-16)
a1 2 u2
-
a13
a 3tU
3
duS
e
dS + -a23t13
a 2 tt1
@2
--&AM+ +
e3
dS
+2
dFV
2
;+ cS\/i
IdS 61).
k2 _
k i,*
do.,
3MI
dS
I
+ a1 3 1 1
- a12
*
OM2 =
V*
O=
M3
Fig. 15-4. Virtual force system.
du 3
-(8
c2F3
2
0)3
+ a2 3 U
2 + (0)2
-
13 : 3
33V
dS + a1 2 °)1 - a230)3
dco3
-lS
d
+
a30)1 + a23C02
Once P* is specified, the eft-hand terms can be expanded.
The form. of P*
depends on the material properties, the particular stress expansios
selected,
and the member geometry. In what follows, we consider the
material to be
linearly elastic and approximation P* with the complementary
energy function
CHAP. 15
ENGINEERING THEORY OF AN ARBITRARY MEMBER
Anto
°
* = F,e +
7+,
l.I1
l~ll
--
2G
2
F2
2GJ
3F
M M2 + kM
3
+ kM2 + 2EC 2
3
(15-17)
+ - -- I M
2E133
where
MT = M1 + F2 Y3 - F3
Y2,
73
2
- torsional moment with respect to
the shear center
k2
where MT = M
dA
15-4.
YA dA
1
Note that (15.-17) is based on taking a linear expansion for the normal stress, I
(a)
M3
M2
F
(a)
Y3- --- Y2
+ ---th
12i1
and
theory,
and using the shear stress distribution predicted by the engineering
"i.
where
t
a
=
tl
+
du +
MT
G.I2 = dS3
Mr
dS9,
-
2 F 3.
1
(15-19)
R
dS
M2
=k
Clj
is the unrestrained torsional distribution due t M anda
+
E/d =
dc
7
1
2 +
- W,
Note that flexure and twist are coupled, due to the
(b
DISPLACEMENT METHOD-CIRCULAR PLANAR MEMBER
Since the displacement method involves integrating the governing differential
equations, its application is restricted to simple geometries. In what follows,
to
we apply the displacement method to a circular planar member subjected
shear
the
and
constant
is
out-of-plane loading. We suppose the cross section
0
center coincides with the centroid. It is convenient to take the polar angle
below
summarized
are
equations
as the independent variable. The governing
and the notation is defined in Fig. 15-5.
Equilibrium Equations (see (15-12))
dF,
is the
- + Rh 3
member
are also neglecting the effct of curvature, i.e., we are considering the
to be thin. The approximate eordilacement relations for a linearly elastic
thin curve member are
du1 - 2 u - 013U3
0lF1
+-E- dS
MT -
F
e2
F3
dS
+ --- y 3
-GA2
MT
=
- -
e3 = -4
GJ
+ a1 2 11
dS- + a
dS
Y2
3
ul
-
+
a2 3
3 -
+
dM
d-
M2
+
Rm
l
= 0
M1 + Rn 2
-
RF
3
- 0
T
MTk,
d?i
dS
kM2
k
a1 2(02
d2
EI=2
dS
EI 3
dS
-
R dO
(15-18)
R I (d
a1 3 c03
+ a1 2 c01 - a 23 0)3
(a)
1 du3
F3
GA 3
MGJ
kl GJ
0
Force-DisplacementRelations (see (15-19))
03
a2 3 L2 + C02
=
Lie~~~~~~~~~~~a
we
distribution due o F2, F .3 In addition to these approximations
flexual
e, e= °
2
curvature, even when the shear center coincides with the centroid.
c0A
Y3F'
--
F
GA3 3
=- GJkl
= coordinates of the shear center with respect to
the centroid
-1y
2) dis­
C1),
placements. That is, an out-of-plane loading will produce only out-of-plane
displacements. The approximate force-displacement relations for out-of-plane
deformation for a thin planar member are
M
2GA1
2AE
When the member is planar, the shear center is on the Y2 axist and there is
no coupling between in-plane (u1, u 2, c%) and out-of-plane (u3,
case. which is developed in Sec. 12-3:
..
.
...- 1;omttor tne p
493
CIRCULAR PLANAR MEMBER
SEC. 15-4.
k2 = k
+
2
R -(
El, `( d
(b)
2
+
,)
t The shear center axis lies in the plane containing the centroidal axis, which, by definition, is
a plane of symmetry for the cross section.
--�---^-�l'��?-i�
494
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
SEC. 15-4.
Boundary Conditions
F3
or
U3
M1
or
or
oi, prescribed at each end (pts. A, B)
M2
CIRCULAR PLANAR MEMBER
(c)
d 2(01
-2
(02
+
(o1
0
Rk
°
R2
_;^
1ml
R
+ L
EI2
The solution of the equilibrium equations is quite straightforward.
We
integrate the first equation directly:
F3 = C 1 - R R0 b3 dO
0)2
(15-20)
M2 =-d--
F3 -
+
12
m,
(e)
d
sin O + C3 cos 0 + -
M
RF3
dO
GA3
RM
t
(f)
R,°2
El2
GJ
II
2
-
du3
(d)
)
We solve (d) for M, and determine M from (e). The
resulting expressions are
2
M = CcosO + C3 sinO + M,
M 2 = -C
do)l
(1 + c,)M 2
where c, is a dimensionless parameter,
The remaining two equations can be transformed to
LdO2 + M = R
495
The solution of the force-displacement
relations is also straightforward.
First, we transform (b) to
(15-21)
p+
where Ml, P is the particular solution of (d).
F3
in
(15-2
to
I
1
I
(15-22)
which is an indicator for torsional deformation. Solving the first equation
for
ol and then determining co2 and u3 from the second and third
equations lead
w)
=
C4 cos 0 + C sin0 + ol,p
)2
=
- C 4 sin 0 - C 5 cos
cos
++
co
RM
dU
d 3 R=
' P
LG4R,+
AJ
(15-23)
<11
where co, is the particular solution for o.
The complete solution involves six integration constants which
mined by enforcing the boundary conditions. The fbllowing examplesare deterillustrate
the application of the above equations.
Example 15-4
The member shown is fixed at A and subjected to a uniform distributed loading.
Taking
Fig. E15-4
b3 = const in (15-20), we obtain
Fig. 15-5. Notation for circular member.
F3 = C - Rb3 0
(a)
496
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
SEC. 15-4.
CIRCULAR PLANAR MEMBER
The equation for M1 reduces to
32 =
d2M ±M
+ 1 =R
c102
RC, - R2h
Sl0
S1
-l)
(b)
Then,
s
-
M1, = RF3 = RC 1 - R2 b30
(c)
=
U
3
and the solution for M1 and M2 follows from (15-21),
(d)
2
M2 - - C 2 Si l l 0 + C3 cos 0 - R b3
-
The boundary conditions at B require
at
0 rt
COS 0 +
{c[-1
0B
sinsin 0
I
cos(05
RC&A
2
)}
- ct sin
C
c + -E}1 + c, sin(0o
0)
(d)
sin 0
c
-
cos
sin(
+)
ii
cos OB- c sin
-
c, +
r
i
C4A2
+ RwA1(i - cos 0) -
A
3
R3E --
M, = C2 cos 0 + C3 sin 0 + RF 3
F3 - M1 = M2 =
+
497
-
merms intolving CO,, C)A2 and lA3 define the rigid body displacements due to support
movement. Als, terms involving c, are due to tist deformation. The rotations and
0 = 0(
C1 = R, 30
C2 = -R 2 b 3 sin n
C3 = R2 b3 cos 0
(e
Fig. E15-5A
F3
Replacing O0 - 0 by , the final solution is
F3 = R7b3t1
Ml = K2b3[
(f)I
-l- sin 1]
M2 = - R2 h3[1 - cos q
Example 15-5
F3
-
The force system due to the end action, IFB3, can be determiled by applying the equi­
librium conditions directly to the segment shown in Fig. EI5-5A. This leads to
F3 = F 3
M1 = F3R( - cos j)
FB3R[ - cos(0 - 0)]
(a)i
M2 = - FB3R sin j = - F,,R sin(0B -- 0)
We suppose there is no initial deformation. Using (a), the equation for 0o1 becomes
I
d2
-R
(1 + c,)sin(0 - 0)
(b)
translation at B are listed below:
o()
=
,At Cos 0B + wA2 sin 0B
- -
The particular solution of (b) is
Cl P =
Ei,
- --2E--(1 + C) [0 cos(0
2E1 2
(c)
---0)]
Using the above results and specializing (15-23) for this support condition lead to the
following expressions for the displacements:
Co1 = COA cos 0 +
A.42sin 0
+ E.- {[ 2
EI,
2
cos OB +
0oB2
-COA
83
1
sin 0 -
--
0 cos(0
- 0)}
2
cos O,
W42
{cfcCos
-1]
-
= TiA3 + RJ.4
1 (lI - cos 0) - RcI
+
C,
sin 0B +
+
R2
1 +
2
REn3
0
2I
- -
-- c,
sin2 0
42
sin B
cos Bsin 0B -
El,
GA 3R2 JJ1
2
c, sin 0B
(e)
SEC. 15-5.
ENGINEERING THEORY OF AN ARBITRARY MEMBER
498
CHAP. 15
To investigate the relative importance of the various deformation terms, we consider
the rectangular cross section shown in Fig. E15--5B. The cross-sectional properties aret
6 1
61
1
5A
k
A3
(f)
J =--d d3 (for d2 A d3)
3
2
Then,
EI2
£C-G J
El_
AIli""ll
E
IJ-U
Consider a closed circular ring (Fig. E15-6) subjected to a uniformly distributed twisting
moment. From symmetry, F 3 = 0 and M,, M2 are constant. Then, using (15-16), we find
d2 d}
12
Cl
M =O
M2 = RmI
E!l
= '6
dn~2-1
LP
G L4J LJ:2
d3)
z
(g)
(a)
The displacements follow from (15-18)
U3 =
'!)
R
GL--
-d-A
Since (d2 /R) 2 << 1,we see that it is reasonable to neglect transverse shear deformation. In
general, we cannot neglect twist deformation when the member is not shallow. For the
shallow case, we can neglect c, in the expressions for coB 2, UB3.
r
5 d2 d 3
499
FORCE METHOD-EXAMPLES
C1 =
)2 = 0
RM 2
El2
(b)
R2 171l
=
El2
The values of 4k and c, for d/d2 = 1, 2, 3 and v = 0.3 are tabulated below:
4k
1
2
3
1.54
3.8
1.69
2.75
3.16
Fig. E15-6
X2
c, = E1 2/GJ
(for = 0.3)
=
2
=0
7.4
Fig. E15-5B
Xi
113
15-5.
Y3
t The torsional constant for a rectangular cross section is developed in Sec. ll-3.
FORCE METHOD-EXAMPLES
In this section, we illustrate the application of the principle of virtual forces
to curved members. The steps involved are the same as for the prismatic or
planar case and therefore we will not reiterate them here. We restrict this
discussion to the case where the material is linearly elastic, the member is thin
and slightly twisted, and warping is neglected. The general form of the
expression for the displacement at an arbitrary point and the compatibility
equations corresponding to these restrictions (see (15-14), (15-17)) follow.
SEC. 15-5.
CHAP. 15
ENGINEERING THEORY OF AN ARBITRARY MEMBER
500
Displacement at Point Q
dQ
Fl
e +17
+ A
+
-R,Q3
A
F2, Q
Fig. E15-7A
(15-24)
Q + (k3±+f-M
'VI)M2
+
+(
3
rlirpcti-n
L.I~,$I
(a)
= force redundants
,
Princinl inprti
,jdS
Co, mpatibility Equations
-7. 1 -Z . -.Z,
Y2
MT, Q
F3, Q + (
+_
501
Now, we consider the problem of determining the translations of the centroid at B due
to the loading shown in Fig. E15-7A. It is convenient to work with translation components
shear
(vB2, vB3) referred to the basic frame, i.e., the X 2, X 3 directions. We suppose that the
\
+
FORCE METHOD-EXAMPLES
r
Fj = F. o +
I
X2 r
(b)
Mj = Mj,o +
Ri = Ri,o +
=-1
Ak
.fkjZj=
where
f,.: = fi. =
JKJ
1[- Fi,jF,,k
1
F ,
+
"
~ 3 3F3,
,'
F2'jF2'k
-t (',- 'CF,, A GA,
Lu/±
IJs
I,,
+1.
+ ZT-/MT, M,
uF
(15-25)
(k= 1,2,...,r)
k + -= M21J
I'
l
Id jM3
1 M
+ _/-,MS
M2,, El3
uentroilal axis
) F 3, k
=
1
Eik Rik ai
,ik dS
+
\ U.J /
M
M 2,
dQ
k
E'
-
(
M2 M2 , Q + 3 M3 M3 , Q) d
(b)
2F3
Ft =
The reduced form for out-of-plane deformation is obtained by setting
°
F2 = M3 = e +
center coincides with the centroid and transverse shcar deformation is negligible. Spe­
the centroid,
cializing (15-24), and noting that M = 0 for a transverse load applied at
to
reduces
the displacement expression
I
3 F2 -
MT = M 1 +
k (,
-I
= 0.
Example 15-7
the
Consider the nonprismatic member shown below. The centroidat axis is straight but
centroidal
the
with
coincide
to
XI
take
We
vary.
orientations of the principal inertia axes
left end (point A).
axis and X2, X3 to coincide with the principal inertia directions at the
t
.
,
t
vectors
unit
the
by
2 3
The principal inertia directions are defined
t2 = cos 012 + sin 13
2i,+ COs 4i1
t3 = -sin
3
=0
at
xl = 0
(a)
Force Systems
at B (Fig.
The moment vectors acting on a positive cross section due to P2, P3 applied
E15-7B) are
(MP = P2 (L - X1)13
(c)
(M)p, = -P 3(L - x 2
to the local frame.
To find M2, M3, we must determine the components of M with respect
E15-7C:
Fig.
from
follow
These
For P2,
M2 = P(L - x)sin
4)
M3 = P2(L - xl)cos (P
(d)
SEC. 15-5.
CHAP. 15
503
FORCE METHOD-EXAMPLES
ENGINEERING THEORY OF AN ARBITRARY MEMBER
502
Fig. E15-7B
c^"llule
:]--
0
We rework Example 15-6 with the force method. Using symmetry, we see that
2_
_.
2
M =0
(a)
M 2 = nt R
i2
r -x
in the direction of inm is desired. The virtual loading for this
Suppose the rotation
displacement is mn= + 1. Starting with
B
1
- -rAM, dS
dS
co, Amt
1
(b)
13
and substituting for M2 , we obtain
R2
p2 (L -
o
\
A)
i,
El 2
(c)
.,
_
Fig. E15-7C
Example 15-9
12
Consider the closed ring shown. Only Mt and M2 are finite for this loading. Also, the
behavior is symmetrical with respect to XY and we have to analyze only one half the ring.
M2 t­
Fig. E15-9
ISL
X2
2T
2
Z1 -
-P3(L - Xl )2
M3 t3
For P3,
4)
M2 = -P 3(L - x)cs
M3 =
The virtual-force system for
we obtain
i
2=
=EE
Determination of VB3
I 2,J are constant
+ 1. Introducing (d) in (b),
B2 corresponds to P2 =
/,1
r
,,r.2
r>:.
s
JoL'2
I I1,
x,)
I3
I |(L
{j
(f)
dNL
We take the torsional moment at 0 = 0 as the force redundant. The moment distributions
are
T
M = - sin0 - Z, cos 0 = M, + ZM,
Due to P2
The virtual-force system for VB3corresponds to P3 =
VB3 --
(e)
+ P (L - x)sin b
Determination of vB2 Due to P2
VJ32
---
+ 1.
Using (e) leads to
(g)
(---
+
(L
2
) sin 4)cos 4)clx
2
T
M2 = - cos 0 - Z1 sin 0 = M2, 0 + ZlM 2 .
2
----------------
^­
_ -2<0
2
/2
(a)
504
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
SEC. 15-5.
Specializing (15-25) for this problem,
fZ
1+
2
l
=
h 11_
A, = -2R
_~~+
GEF13
GJ
-,t2 lI
ft 1 =2R
fi2R
FORCE METHOD--EXAMPLES
Primay tructure505
1
=
El-
The primary structure is defined in Fig. El 5-0B:
]idO
(b)
Rl
RRI-
dO
/2L J +
1
R-A
R2 =
Z1
=
3FA 3
0
=
d, =
R3
Rl
3
,
=
(a)
and then substituting for MI, M2,
A1 ,
-RT
2
(CJ
¼)
= I2(I
-RT
sinocos
J, 1 GJ
E
sin 0 Cos 0 d
1
1'r. ,
: _ ,
V\GJ - _ --7 L2Ill
'IJ-r/2
2
k
GS L
Fig. E15-10B
(c)
....
Ei2J
......
and it follows that Z, = 0. We could have arrived at this result by noting that the behavior
is also symmetrical with respect to X2. This requires M2 to be an even function of 0.
The virtual-force system for o)At is T = + 1. Using (15-24) and (a) leads to
,/2[(T sin 0Osin
f2w4
-,t12
1 = 2R2GJ
2
RT
11
8 LGJ
~~~~~~~~~
X2
T cos OY\cosO
±
2fEI,)
f1,O
I
-.2 j
(d)
1
El2
I
Example 15-10
We analyze the planar circular member shown in Fig. EIS-10A. The loading is out-ofplane, and only F3, M1 , and M2 are finite. To simplify the algebra, we consider the shear
center to coincide with the centroid and neglect transverse shear deformation. It is con­
venient to take the reaction at B as the force redundant.
X2
Fig. E15-10A
A
A
V
iForce
nt
Analyses
The force solutions for the loadings shown in Fig. EiS-IOC are:
For P:
F3 ,0 = +P
'" ,o0= r[tL - cos(q - c)]
M2, 0 = - PR sin(qi - tc)
For Z = + 1:
.
CdFq
F3,
3,,
Y~~~~~~~
+s
=
+ I
)
Mi.· 1 = R(I - cos
21
(b)
-R-sinj
tt
-I
l
(c)
SEC. 15-6.
506
RESTRAINED WARPING FORMULATION
507
CHAP. 15
ENGINEERING THEORY OF AN ARBITRARY MEMBER
15-6.
Fig. E15-10C
RESTRAINED WARPING FORMULATION
In what follows, we consider the member to be thin and slightly twisted.
Referring to Fig. 15-2, these restrictions lead to
dR
dS
(15-26)
dS dy 2 dy 3
d(vol.)
Therefore, in analyzing the strain at Q (S, Y2, y3), we can treat the differential
line elements as if they were orthogonal. The approach followed for the pris­
matic case is also applicable here. One has only to work with stress and strain
t2,73) rather than the global frame.
measures referred to the local frame (,
Our formulation is based on Reissner's principle (13-33):
=+
prij d(surface area)] = 0
b[ft(Tr - bTfi - V*)d(vol.) a, ii = independent quantities
E =
(a)
(a)
p, b = prescribed forces
V* = V*(F) = complementary energy density
CoinpatibilitY Equation (15-25)
At
jTI
/,, R I i
A-£
G
(d)
+ Mug1 d1,
{:,
We start with the strain measures, :
Lo
­ -
Ri tai- R
We introduce expansions for ii, in terms of one-dimensional displacement
and force measures (functions of S) and integrate over the cross section. The
force-equilibrium equations follow from the stationary requirement with respect
to displacement measures.
+ (k°
d l
-±ia ) M, tl
s
Y2.
y; 3}. One, can show thatt
'
lt
we obtain the following expressions
Substituting for the internal force and reactions,
for ftt and A1:
(i
Lk-
Vi/
l
Al
= UB3 -
) ,0-
2c, sin
A3 + R()A2 sin
+ R
)dO
1 +
COS(O0 -
Ct
0 cos(OB
-
O
COs OB)
0o
where hiis the displacement vector for Q (S, y 1 , y2). We use the samse displace­
- sin 0 - sicn
(e)
=
0 sin 041
EI2
Note that we could have determined A and f,
FItl + 1n2 t2 + it31
u
+ co2Y 3 - (3Y2
+ JC
(15-28)
0 = 05(Y 2 , Y3)
Expanding
fJTF
using the results of Example 15-5. --
=
-(s2
(Y3 -3 3)
ui2 = t
tI3 = Us3 + (c1(y 2 - Y2)
sin 0c}
- 0) - cos
(15-27)
- + t3
t t
3'13
1
+ sin Bcos 0,C-- cos
+
1
Sill OCo
02St
_Vl
1
ment expansion as for the prismatic case:
k sin(0B -
PRII j
EI2
-
o0-
OB- RK(74(
0
2
I<-
-FSee Prob. 15-5.
dy 2 dy 3 = .f((-,
1
+ 12Y2 + O13y 1 3)dy 2 dy3
(a)
508
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
SEC. 15-6.
RESTRAINED WARPING FORNAI 11ATir-11
'JCv
-.
leads to
Jflor
and b2, b3 are due to self-equilibrating stress distributions
dy 2 dy 3 = Fle,1
ths case,
b t O
"''ess
adg disributions.t It is reasonable,
based on the primary flexural shear stress distributions.
Expanding the stationary requirement with respect to force measures yields
+ F2
3 e3 + MTkt + M 2k 2
F2 e
+ M3 k 3 + MRf + Mef,s
el, e2 ,,
M = ff 1 I0
MR =
ff[
k
(defined by (15-16))
3
(15-29)
the force-displacement relations,
dy2 dy3
1 2 (0,
+
2
a 12
) +
1 3 (0,
+
3
a1 3 )]dy2 dy
el
eI
3
The equilibrium equations consist of (15-11) and the equation due to
warping
*
a
= -I
0F
k2
eF
+
eb
JF=2 V
F2
+
k=
Now, we use the stress expansion developed for the prismatic case.
The
derivation is discussed in Sec. 13-5, so we only list the essential results here.
The normal stress is expressed as
+
M
-+
M2
y-
M3
-
13
12
V +
M,/
where ee
e
2
F2 +- I/ 3F3 + ,IM'-
+ //,MT
,
Example 15-11
kl
q
1
+-
GJ
(F
+
2
+
+
+
3
2E.
3
(15-35)
k 3 are defined by (15-16). The corresponding unrestrained
--
Fig. E15-11
I
B
i
=
elV*
OF*
('~*
M':..-
(1 + b) f
(V's are functions of Y2, Y3.) The corresponding complementary energy function
is
3
=
(15-32)
Mr = MUT + MT
V* = fV* dy 2 d
f
To investigate the influence of warping restraint, we consider a planar circular member
having a doubly synetrical cross section (Fig. E-1 1), clampedat one end and subjected
where 0 = - ", the St. Venant warping function referred to the shear center.
We write the transverse shear stress distribution as
a =
2
f,,~, amo
warping relations are (15-18).
(15-3l)
,,(/)
-OM~-
OMT
weighted with respect to ¢.
A
a V*
k3 = -
M,
(15-30)
which can be interpreted as the stress equilibrium equation for the t1 direction
Fl1
aV*
-­
restraint,
MR = M,,
509
.......
It is reasonable,
Al'
I
-j(Mr)+ Cr(M)2 (1533)
i
GJ1
(F~y~r + 23
(F 2Y3 r +- F3 Y2 )M
II
to a torsional moment at the other end. We neglect transverse shear deformation due
to restrained torsion. Theoverning euios
for this loading (see Sec. 15-4) follow.
Equilibrim Equations
Also, (15-32) satisfies (see(13-50))
JJ(U1 d,
2
2
dM =
+ U1330 3)dy2 dy3 - M'.
(b)
dOM
2
Finally, noting (b), we express MR as
MR = (1 + b2,)M
+ b2 F 2 + b3 F3
(15- 34)
where the b's involve the curvature (al 2,a 3). If the cross section is symmetrical,
A2 3 = Y3 r = Y2,- = b =
0
1 dMl
R dO
M = M + M~
,Al
i'i
isi
(c)
t See
I
Prob. 15-6.
(a)
510
ENGINEERING THEORY OF AN ARBITRARY MEMBER
SEC. 15-7.
CHAP. 15
COMPLETE END RESTRAINT
511
The rotation at B is
Force-DisplacemnentRelations
M2 = EI2k2
2 (d 2
=i
+ oi)
(0B
R--M
-
sin
cos O) + cK
E,.I, (if
(g)
M-- R dO
(1do
- R \-iJ
(R -(
f= -kl =
M
(b)
+
)
B Ii
K fj2_1 B+ sin O~
I=
B cos 0sco
t
-rI
2
j - o --+
2
cCs2
0, tanh 0
If we set
= GJk1
R=-
L
0 =0
= O
o
= o2 = f = O
M
= M
7
O
LU
(h)
.
and let 0,-0. (g) reduces to (13-57), the prismatic solution. The influence of warping
restraint depends on and 0,. Values of K vs. - for 0) = 7r/4, iT/2
are tabulated below:
On
Boundary Conditions
0
+
(c)
K
MOt = 0
I
=, + I
M2 = 0
for 0, =
1+7
2
One can write the equilibrium solution directly from the sketch:
M =
I cos(OB - 0)
(d)
M2 = M sin(0O - 0)
K
K°°,
We substitute for the moments in the force-displacement relations.
M1 = GJk1
Ej d2 k,
-
Idco~R
-R
k2-(r
2
2
-
M
E,.l
__
1
_ +_ 2
for 0 =
I + I lr
t+2+- 1+--
= M
nEcos(On - 0)
K,
for=
4
(i)
= -2
O, + sin 0, cos 0,)
2
d
(e)
1)+
sin(03 - 0)
and solve for k,, and then ol. The resulting expressions are
k, =
1
=
G.I
El 2
EI 4,
GJ
i.
for OB = /4
0B= r/2
1
5
10
0.179
0.786
0.907
0.500
0.96
0.99
We showed in Chapter 13 that
{cos(0B- 0) - cos 01 [cosh 70 - sinh 70 tanh R0,B]}
GJ + - 22
R
(f)
!
A= 0
(2)
(open section)
2= 0
()
(closed section)
(j)
where t is the wall thickness and h is a depth measure. Since A= R and R/h >> I for
a thin curved member, the influence of warping restraint is not as significant as for the
prismatic case.
1(EI2
W(ap
neglCe cr e)a
-
i s t2t n
J
0cos
+
a
n'
2 [sillh 7.0 - tanh 70B cosh 70 + cos 0 tanh MIJB]
Warping restraint is neglected by setting E, =
and
= c.
15-7.
MEMBER FORCE-DISPLACEMENT
END RESTRAINT
RELATIONS-COMPLETE
In the analysis of a member system, one needs the relations between the forces
and displacements at the ends of the member. For a truss, these equations
512
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
SEC. 15-7.
reduce to a single relation between the bar force and the elongation. Matrix
notation is particularly convenient for this derivation so we start by expressing
the principle of virtual forces and the complementary energy density in terms
of generalized force and deformation matrices.
Referring back to Sec. 15-3, we define
COMPLETE END RESTRAINT
F=>{F1 F2 M 3 }
Y3 = 0
1
-0
Ia F/* ,i
0
i
9=
-
Uj
F =
r
(15-36)
f T,~=
A
002____
0
and write the principle of virtual forces as
.s dV* dS
Note that we are working with M , not MT. We use the complementary energy
function for a thin slightly twisted member with negligile warping restraint
(i.e., (15-17)). With the above notation,
Eq. (15-17) => V*
(o)rT
g,,,r-g
Lgr
!
-I
(15-38)
ot_
0
I
l
-
T
AR = dAP
(15-41)
where d contains prescribed displacements and R are the corresponding re­
actions; d contains unknown displacements and A' are forces corresponding
to d. The virtual-force system (AP, AR, A.F) must satisfy the force-equilibrium
equations, (15-11). It is more convenient to generate , and R with the equili­
brium equations for a finite segment rather than attempt to solve (15-11).
Consider the arbitrary member shown in Fig. 15-6. Each end is completely
restrained against displacement. The positive sense of S is from A toward B.
where
Fgfm
A2 G
.fS(g + gjF-)TA57- S
7
A-.
g.7
+
(15-40)
I
for planar loading applied to a planarmember.
Finally, we substitute for a in (15-37)and distinguish between prescribed and
unknown displacements. The principle of virtual forces expands to
(15-37)
jgTsg Af dS = d AP
513
of g corresponding to the zero force measures. For example,
0
I
I
7i7 + y­w-;f-,d
gf =
j7S
Sym
0
gfm =
0
YJ
G Y2
Y2Y3
GJ .
+
A 3G
0
C?]
I
0
0
0
0
A
El2
Sym
GJ
B
C;.
1
0
1
,
i3
o
i
1
i2
El,
The force-deformation relation implied by (15-38) is
= g + gY-
n
member frame
(15-39)
We will use these general expressions for planar and out-of-plane deformation
as well as for the arbitrary case. One has only to delete the rows and columns
XIn
Fig. 15-6. Arbitrary curved member.
514
ENGINEERING THEORY OF AN ARBITRARY MEMBER
SEC. 15-7.
CHAP. 15
We suppose the geometry of the member is defined with respect to a basic
frame which we refer to as frame n, and take the end forces at B as the
force
redundants. Then, the primary structure consists of the member cantilevered
from A.
Throughout the remaining portion of the chapter, we will employ the
notation
for force and displacement transformations that is developed in Chapter
5. A
superscript n is used to denote a quantity referred to the basic frame.
When
no frame superscript is used, it is understood the quantity is referred
to the
local frame. For example, .rQ represents the internal force matrix at
point Q
referred to the local frame at Q. Note that ds'Q acts on the positive
face. The
force matrix for the negative face is - J9Q. The end forces at A, B are
denoted
by ,
f
and are related to the internal force matrices by
B
=
An =
+
=
-
'
o
=
B
-
COMPLETE END RESTRAINT
515
·~(2Q leads to
ryiB
*orn,
BAt,
, +
J-
T(s
+ gQ.-Q, o)dS
(15-45)
A
A
[ SL
'A
'
(
Q](QS) JB
BQ
The first term is due to rigid body motion of the member about A whereas
the second and third terms are due to deformation of the member. We define
'"as the member deformation matrix:
X
=
/duc
"B9l
,
;
"iil id bdy
albout.4
(15-46)
Ilotion
By definition,
We also define
nL"is
(15-42)
B
A
f11 =
)~~1i8fi
S(y
f
Also, the displacement matrix at point Q is written as 61Q.
equal to the sum of the second and third terms in (i 5-45).
(° 'Q0BQ+ QQ o )S = initial deformation matrix
TgQ_ 07-gQ
,
BgQe)dS = member flexibility matrix
(15-47)
and (15-45) reduces to
6
01Q
= {Ul, U2 ,
t3
(tl,(t)2,)3Q = {O}
(15-43)
Ad' = -
For this system, '270 and W'/[ are prescribed.
We determine til[/ for the primary structure, i.e. the member cantilevered
from A, due to displacement of A, temperature, loads applied along the
member,
and the end forces at B and then equate it to the actual @tX1].
The virtual-force
system is
AP = A¢"
AR = A7- = -XB AIB3
(a)
7-nq
7-n . = nq er,,
~'
Q BQ
q-XQ B 3,;
1 1B-;.i=
B
Also,
(b)
d-fall
Introducing (a), (b) in (15-41), we obtain
8SB
(Ae
=
T(,)A,
[,,n,
.SA
Next, we express
(c)
f
7-)''q,
+ T(
± g(Q?,i)dSj
,IRQ as
i
1,
I
Q =
'Q,
;
~
+ .O-n 6C
(15- 44)
where Q, o is the internal force matrix at Q due to the prescribed external
loading applied to the member cantilevered from A. Finally, substituting
for
J
n,
- o,,,
"' B
T...4
,tt
A
..
' 0
f
&YB
(15-48)
Equation (15.-48) is the force-displacement relation for an arbitrary member
with complete end restraint. It is analogous to the force-elongation relation
for the ideal truss element that we developed in Chapter 6.
The member flexibility matrix, f, is a dtiu'al property of the member
since
it depends only on the geometry and material properties. For simple members
such as a prismatic member or a planar circular member with constant cross
section, one can obtain the explicit form of f. When the geometry is complex,
one must generally resort to numerical integration such as described in Sec.
14-8 in order to determine f and '/'. This problem is discussed in the next
section. Finally, we point out that the general definitions off, /'0 are also valid
for in-plane or out-of-plane deformation of a planar member. One simply has
to use the appropriate forms for the various matrices.
Up to this point, we have considered only a simple member. Now suppose
the actual member consists of a set of members rigidly connected to each other
and the flexibility matrix for each member is known. We can obtain the total
flexibility matrix by compounding the flexibility matrices for the individual
elements. To illustrate the procedure, we consider two members, AA and
1
AB, shown in Fig. 15-7.
The matrix, f", contains the displacements at B due to the end forces at B
with A fixed:
q6/B
= f
77
(a)
Now, suppose point A is fixed. Then, the displacement at B due to the deforma­
tion of member A B is
Al
member
A,B
fBB-B
A=
(h)
kv/
516
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
SEC. 15-8.
where f',B is the flexibility matrix for member A 1B referred to frame n. The
additional displacement at B due to movement of A, is
(
/B)displacementat,
=
TBAi
I
(c)
41
GENERATION OF MEMBER MATRICES
It remains to determine qlA.
A
li
Ai
where
A
--- "Bn
X .k
A
' A,
i
A
-'A,
-
kBA
Fig. 15-7. Segmented member.
= f,
Ia
Note that only k" and
I
AAL"'s
+
BA,
AAi CtBA° )t'IB
== f
(15-53)
- , _1
AkBA
=
'BA
+ kBBLBn + kr
A.4
, ; + k"B'IB ±4 kA
4i
PA?
(15-54)
are required in order to evaluate kA and ka.
(e)
15-8.
(fA X
(15-52)
i
I
Finally, we have
0&B =
n. i
=
ii
I
and the resulting deformation of member Al is
1
BA -CA
With this notation, the force-displacement relations simplify to
(d)I
At, lr
(b)
B
A,k
-
TkAA
Bnk ZJ
The force system at A t due to the end forces at B is given by
A
'VBAX, k
BA
n~T
k"
-
T
fAA,
c+B
--
BA,97BtBA
B
O -
(k)A
nAB
)
AJIImemberAA,
-
where "A,
i represents the initial end forces. In order to express the equations
in a more compact form, we let
B
dAA=
'A.O
Substituting for O,"leads to
Al
2
517
When analyzing a system of members by the displacement method, expressions for the end forces in terms of the end displacements are required. In
addition to (15-50), we need an expression for ,.'FA' Now,
B
(15-- 49)
I
I
i
GENERATION OF MEMBER MATRICES
The member flexibility matrix is defined by
The end forces at B are found by inverting (15-48):
~7e= k"(i"
-
./'7)
n= k ka/li -kn / + k/ l - k
-
,riSA,.
f
O
Q, 0 + '}.
(b)
gl
(15-51) I
The second and third terms are the end forces at B due to end displacement at
B, A. Once ,'B is known, we can evaluate the interior force matrix at a point
using (15-44),
j'Q =
= snq 7Rnq"n.
and letting
(15-50)
The first term is due to-external load applied along the member and represents
the initial (or fixed-end) forces at B. For convenience, let
-B i = -k
(a)
Noting that
k" = (f")- = member stiffness matrix
(a)
Thus, the analysis of a completely restrained member reduces to a set of matrix
multiplications once the member stiffness and initial deformation matrices are
established.
I
=
f'
;
<rnq. Tg
nq
(15-55)
CAB (i
QTg2Q)dS
IBQ
(15--56)
we can write
If numerical integration is used, the values of the integral at intermediate points
along the centroidal axis as well as the total integral can be determined in the
same operation. This is desirable since, as we shall show later, the intermediate
values can be utilized to evaluate the initial deformation matrix.
We consider next the initial deformation matrix:
%n =
We transform ,g, and
I-q,,B<-
T(
+
g
o)S
(c
from the local frame to the basic frame, using (15-55)
518
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
SEC. 15-8.
GENERATION OF MEMBER MATRICES
519
and simplify the notation somewhat:
and
.~'Q = .. q Thy
Q
(d)
BQ
=
The contributions of temperature and external load are
(15-57)
()
lad =
gQ)dS
SB (XQn
SA
ypnq
_
n
_
(axe)
(15-59)
c ={ Tc
Normally, the external force quantities are referred to the basic frame for the
member, i.e., frame n. The initial force matrix at Q due to this loading is
given by
n
C
37"o,0 =0
:
S,
-< SO
SC
SC < S
rg )dC S
Qn,BQ
T
,,=
l
( CB'
C)
(ax)
F
-2/g2
,(fix/)
The translation and rotation transformation matrices are developed in Sees.
5-1, 5-2 and the form of g for a thin curved member is given by (15-38).
The local flexibility matrix g' is defined by (15-55). Using the above notation,
the expressions for the submatrices are
gl = R gl Ra
g2 = RITgl 2R(
g22 = R22R
Note that g 2 = 0 and g, g22 are diagonal matrices when the shear center
coincides with the centroid. If, in addition, axial and shear deformation are
We let
( x )
(f)
=
fan, Tgnn
=
fI
(
(n
To n
x'll )ets
(15-61)
.....
J12 = g 2 + X'
(15-62)
Also,
fn
(15-63)
JB
The determination of the member flexibility matrix reduces to evaluating
J defined by (15-61). One can work with unpartitioned matrices, i.e., X', g,
but it is more convenient to express the integrand in partitioned form. The
partitioning is consistent with the partitioning of :f into F, M. Since the
formulation is applicable for arbitrary deformation, it is desirable to maintain
this generality when expanding X, ,. in partitioned form. Therefore, we define
a as the row order of F and ji as the row order of M.
'122 =
(fix )
(g XBQ)
2
BQ g22
BQ
(15-68)
Next, we partition J consistent with
i:
(a x )
jp = SA
Sp i dS =
J,
1
(ax
12 J
IJP,
)
22
1
(15-69)
JSA
Finally, we partition f:
(axa)
(ax )
- 1 _kt
(15-64)
Continuing, we partition X, 9 and g symmetrically, consistent with (15-64),
+
g22
22
fn =
MJF
(15-67)
J22
(t x )
The submatrices follow from (15-65):
I11 - g 1 + g1 2 X'BQ -
With this notation, (f) simplifies to
(ax )
t_\
f12
The bracketed term is an intermediate value of the integral defining f". Finally,
we let
Jp
(15-66))
,
neglected, g,, = 0.
and introducing the above relations in (15-59) result in
J1/1
L912 (15-60)
(e)
CB
_ r[SC (XjB,
( 0z -BQ[is
B
[A
|Q iT
Sn
Writing
XCQ -B
x_ )]
(15-58)
Suppose there is an external force system applied at an intermediate point,
say C. Let Pc, Tc denote the force and moment matrices and >c the total
force matrix:
<Son , o = arn
CQ
C.
(15-65)
(e x c)
(f(x )
f = JB, ij = i
j dS
a
(15-70)
520
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
SEC. 15-9.
MEMBER MATRICES-PRISMATIC MEMBER
521
The initial deformation matrix due to an arbitrary loading at point C can be
determined with (15-62). Its partitioned form is
= (JCnCB)pn
(T)-n o
{vO
(ax a)
Jc 12XBC '
JcII
1
.C,
-
2JJC,
2XBC
(a xp)
JC. 12
I
(aX1)
22
. 0 -0
(/3
t MB2, WB2
(15-71)
t2
I)
12
B 1
where v', O0 denote the initial translation and rotation matrices.
The member stiffness matrix, k", is obtained by inverting fn. We write
(ax a)
xi
(ax )
kr
k" = (fn)--l
A
;-
H/
]=
3'
t3
(15-72)
l'B3 /
TB I , UB
1
3,3
(p x pi)
One can easily show that (we drop the frame superscript on fj' for convenience)
kl
LA
= (ft1 - f2f22f12)-
- 222tl
(15-73)
1f
2k
-I
Y3
= f21I
f
krl!/i
k22:
L - Fl.t2n12)
kn22
L
n
k[A = -.
A
n ,1
f
Lk 2
1-k2
k2A
Xn, {-k
-
k2
[
,
n
Xn.T
k"''"
=="--- A--
A
Xn
+]
_
r
j,XkA
]
-_-T-; =
++ "22212
nn
Ik
i
12
0
X=o
(15-74)
0
0
I
(L -° °0
! B
.. T.... t'…LA
C
-
(a)
Then, using g defined by (15-38), we obtain
L/AE
15-9.
dXiS.
Now,
=
Tl~n
-IIIUld
Fig. 15-8. Summary of notation for a prismatic member.
Once k" is known, the stiffness matrices k7,n, kA and k4 A can be generated.
Expanding (15-53) leads to the following partitioned forms:
nk[
b Lm
X2 , X3 are principal inertia directions.
0
0
MEMBER MATRICES-PRISMATIC MEMBER
In Chapter 12, we developed the governing equations for a prismatic member
and presented a number of examples which illustrate the displacement and
force methods of solution. Actually, we obtained the complete set of force­
displacement relations and also the initial end forces for concentrated and
uniform loading. Now, in this section, we generate the member flexibility
matrix using the matrix formulation. We also list for future reference the
various member stiffness matrices.
The notation is summarized in Fig. 15-8. For convenience, we drop the
frame reference superscript n, since the basic frame coincides with the local
frame, i.e., R" q = I. The positive sense of a displacement, external force, or
end forces coincides with the positive sense of the corresponding coordinate
axis.
Starting with (15-66), we have g7 = g, since Ra, R/ are identity matrices.
Once XBQ is assembled we can determine the submatrices of I from (15-68).
|A ---
1------1)
2
__7
­
(15-75)
IL/GJ
-
Sym
0
L/E1
2
0
0
L/E13
I
ENGINEERING THEORY OF AN ARBITRARY MEMBER
522
CHAP. 15
SEC. 15-9.
MEMBER MATRICES-PRISMATIC
MEMBER
523
The submatrices of k are generated with (1 5- 73), (15- 74) and are listed below
for reference. Transverse shear deformation is neglected by setting a2 = a3 = 0:
12EI2
a2
12EI3
-
GA 3 L2
GA2L2
12
I3-
+ a2
1
+
GJ
12E
b, = +
(3I
_ AE
O
I
2
+ x2)
(change sign of(1, 2), (1, 3)in k2 2 )
Finally, the fixed end forces due to a concentrated transverse force and a
uniform transverse loading are summarized below.
O
12EI3
kll
ConcentratedForcePc2
Sym
12EI*
Sy m
I
F
0
L
1
0
0
I
I
I
12EI*3X3
kt2 =
-t
L3
b
6EI
--
(.1.5-76)
6EI2X
2
L2
2
2
I
MA
= -MB
0
FA2 =
MA 3
613)
B2
I
-L |
2 + FB2 +
L
ConcentratedForcePc3
12EI2
GA 3 L2
6EIl
L2
L3
=
PC2-
(15-77)
- MB3
= -x 3 (XcPC 2 + FB 2 )
0
--
-
2
MB
(4 +
312EI
P
! 6EIX 3
L2
Sym
12I00-
"_c-1 a, 2
MB3 = Lc2.c(1
FB 2 = -()
E1
A
I
L2
0
(4 + a 2)
L
22 =
A2L3
GA 2 L 3
6EI:
,22I
12EI 2 2
L:3
12EI 3
a3
j
2
MB2
-LP
3
xc(l -
XC)
(
I---)
(change sign of(2, 3) and (3, 2) in k 12 )
bl
B=
6EI2X2
2
L2
6EI3.X3
L2
- 6E*- 2
L
-(2
-
2
- a2)
EI*
2
I
FB3 =
6EI'3 3
L2
MA
1
00
2
+ _ MB
2
MBI '= X2 (CPC3 + FB3 )
0
EI
I -(2 - a 3)--I
Li
-PC3(X)
I
= -
A3 =
-PC
111
3
JMA 2 = L (PC3
-FB3
+
3 -
1 M )
(15-78)
ENGINEERING THEORY OF AN ARBITRARY
524
MEMBER
CHAP. 15
SEC. 15-10.
0
0
Concentrated Torque Tc1
M1
= - Tc1 Xc
-Tc(lXc)
T
MAl
(15-79)
X
= XQ
0
R(1 - cos )
q
R, = R2
(15-80)
- MA2
=
M
-R(1 - cos
sin
0
R =
(a)
R sin i7]
Ra = R2q
(b)
-R(1 - cos al)
,Re = -Rsin l
desired,
for out-of-plane deformation. Since the complete flexibility matrix isconsider
3 as to
it is just as convenient to work with submatrices of order
separately the planar and out-of-plane cases.
b3 L
2
FBi3 = FA3 --
=
R = 1
.
Uniformly Distributed Load, b3
a,n
for planar deformation and
12
-
MAt = 0
M
j-R
I
= [R( - cos )
b2L
22
b2L2
FB 2 =FA2 -
M3 =
0
R sin
We use
Uniformly DistributedLoad, b2
MB3 = -MA3
525
THIN PLANAR CIRCULAR MEMBER
b3 L2
12
(15-81)
=M1-L
(15-82)
0
IM=
Uniformly Distributed Torque, nml
MA=
15-10.
I
M A
2
-
n1
MEMBER
MEMBER MATRlCES-THIN PLANAR CIRCULAR
matrices
In this section, we generate the flexibility and initial deformation
matrix
using
section,
cross
constant
of
member,
for a thin planar circular
for the
deformation
operations. We include extensional and transverse shear
illustraas
obtained
sake of generality. Some of the relations have already been
In particular, the reader
tive examples of the force and displacement methods.
and Examples
deformation,
planar
treats
should review Example 14-6, which
deformation.
15-4, 15-5, 15-10 for out-of-plane
Y and Y3 are
The notation is summarized in Fig. 15-9. By definition, 2
Y3 = O0i.e., the shear center lies in the plane con­
principal inertia axes and
the basic frame (frame n)
taining the centroidal axis. It is convenient to take
forms of R,, R,
to be parallel to the local frame at B. The three-dimensional
and XQBrare
cos
Rnq
bq
=
sin
-
R = R, = Rnq
1
-sin
cos -- == R7i
l
R2
12
- 3bt
\
F31IB1
fora member.
notation
of
circular
Summary
15-9.
Fig.
planar
member.
Fig. 15-9. Summary of notation for a planar circular
_0
(15-83)
We consider the member to be thin and use the local flexibility matrix
the member flexibility
defined by (15-38). Expanding (15-66), (15-68) leads to
matrix.
I
II
ENGINEERING THEORY OF AN ARBITRARY MEMBER
526
13
e
= AR
El 2
GA3 R 2
-
THIN PLANAR CIRCULAR MEMBER
527
We consider next the determination of the initial deformation
matrix due to
an arbitrary concentraited load at an interior
point, C. Now, the flexibility
matrix for the segment AC referred to the local friame
at C, which we denote
by fac, is known. We just have to change
OB to c and superscript b to c in
(15-84). When the external load is referred to the local
frame at C, the displace­
ment at C is given by
GA2R2
EI2
GJ
SEC. 15-10.
i
EI3
as
2
CHAP. 15
a1 = ae + as
a2 = ae ­ as
C
-= 2(1 + c,)
C2 =
(1 -
(a)
The displacement at B due to rigid body motion about
C is
(15-84)
)
li4 -
Bbc
(b=
hc, T)q c
(b)
Finally, we can write
C4 = C,
C I (
A)
(/)
t
;Ob
ii
Symmetrical
+ sin0,[-2
R
fb
=
+
B(1
+ a2)]
2
' '1fc.,l
+
fc)c
l X BC TRTfc.
TCfT, 1
/C.
T
L-C
(15-85)
The uncoupled expressions follow.
-
(I + a2 )sin 0 cos 0,}
2't
0o. =B " I
o
o
- 0B(
R2
0
0c
I
+ C4 )
- 2c, sin 0, - c sin
0
EI3z~~
l
'b
iI
R'
0Ocos 0,,}
+
in
+
"-
I
l
2
.(R - sin 0,)
-1
2
(1)3
(I
I +-
+
_2
a
0
IE13
R
- {-c 10B + C3 sin 0 B
+ c2 sin 0 cos 1B
J}
-
--C2 sin
- c3 ( -
2
On
Vlo 2 =
II,
B2
2
0
_
-
Symmetrical
i
-- (c,0B + c2 sin 03 cos 0n)
El2
0
0
B
1-
sin
ac
-sin Oc sin 0B} PC2
Ct
+
-E+, 2
la 1
ub
sin
c
(15-86)
%C
c 'S
OS 0)}
.
=
cos
)
cosq
+ sin t1: - sin
fr
O sin
_
1
El,insv,%
(1 - cos 0)
0
2
1+
+_2 Cos 0P)}C
1 - cos Oc
+
R
2
C
' . T
.,
PlanarLoading
_ I
=
(
B= (XcIrAT,
· cj
+ (R fc, 12 + X"BC R"TfAc. 22)Tf
=
(Rftc, T 2 )Pc + (R/TFC,
"
2)T
ob4
-{- --- (l + L,)
+
ff, = (f2
(');,.
I
1 - cos 01
sin 0
2
-
UBub
V0
3
,
0
IRO,
I
+ Eco
0
cos
s Oc
B
i Oc sin 0,
P
c
csin
P
R2
R"
'I 2
·-
E
00,3
··
~~··
- sin Oc)Pfcl
(cos)P +
+
ROc
- TC3
EI,
528
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
SEC. 15-10.
THIN PLANAR CIRCULAR MEMBER
Out-of-Plane Loading
VO, 3
c(c Ccos c 4- C4 ) - C2 sin Oc cos
EI2
UB3 =
529
0
+ c3 (- sin Oc - sin
+ER I-{-CO
+ El2
+ c2 sin
cj°c sin /c -
Oc cos OB + C3 sin
c2 sin Oc sin 0,
R2
ER
1=
,
{-BciC cos
i 2
1
+ c3 (sin
0
ii
i
0
i
i
I
i
TC
- C3(1- cos Oc)
B -
P
i
B
P
+ sin
2
0
Fn
TC2
I_
sin qc)
B
±
+ c2 sin 0 c cos OBIPC 3
+
R+
EI2
+
-
EI
'/c-
2
IC 10c sin qc +
C2
o0, = ob2
=
Cc
COS
+
-EI2
{-c 1
OB TI
sin Oc sin 0
lOc
C sin c + C3 (Cos 0O
- c sin Oc sin 0
+ EI
sin Oc COS
(15-87)
i
rTc2
-
Cos 'IC)
Pc3
Fig. 15-10. Notation for planar loading.
0c sin c + c2 sin Oc sin 0 TT
Oc cos q
+
2 sin
fc
cos O
sin
I'C2
When the loading is symmetrical, one can utilize symmetry to determine the
fixed end forces. The most convenient choice of unknowns is the internal forces
at the midpoint, i.e., 0 = 0 /2 ; F1 and M 3 are unknown for the planar case and
only M2 is unknown for the out-of-plane case. Explicit expressions for the
fixed end forces due to various loading conditions arc listed below.
(I - cos a) + -- 12
sa
V =
1-__.
2
.
..
+
I
Mc
--2
F[1 = -lA1 =
_ ___ +
F(
sin 2 a'
2-
sin
sin cc
1
)
(15-88)
V/
_ _.
C
cos a
---FZc
Planar Loading
Fig. 15-10 defines the notation for the planar case.
We consider two loadings: a concentrated radial force P applied at C, and
a uniform distributed radial load b2 applied per unit arc length over the entire
segment. The basic frame is chosen to utilize symmetry. We determine the
axial force and moment at C from the symmetry conditions u = o3 = 0.
PR fsM
f
i-MIJ=-MII_
==PR
_i­ n
2
in cc-
I - cosa
----cc
COS
\c-i -coScc
CASE 2 -UNIFORM DISTRIBUTED RADIAL. LOAD b2
FC2 = 0
CASE 1-CONCENTRATED RADIAL FORCE P
P
2
FC = 2 q1
a
a
sin 2 cc
-
a (+
--
a
2
sin a cos cc
CHAP. 15
PR
(15-89)
s
= --Rb 2(co c -aeq)
Fni = -F P
FB2 =
MB =
2
R­
= R22b
-M.A2 = MC2-
2
PR
sin a
2
P
- Rb 2 sin cc
AZ =
A
MB
(15-90),
= Mi* ='--- (1 - cos )
~
- 1 +--
MC = R2b 2ae
531
FLEXIBILITY MATRIX-CIRCULAR HELIX
SEC. 15-11.
ENGINEERING THEORY OF AN ARBITRARY MEMBER
530
()
-
cos
F, =
x
3
FA =
CASE 2-CONCENTRATED TORQUE T
Out-of-Plane Loading
Fc 3 = 0
T
Figure 15-11 defines the notation for the out-of-plane case.
We consider four loadings: a concentrated force P, and a couple T-both
applied at C; a uniform distributed force b3; and a uniform distributed couple
min. The bending moment at C is obtained using the symmetry condition
=
MC
0)2 = 0.
=
M
11,11
M
2
c2 sin 2cc
T
2 acc + c2 sin a cos
T
MBI 1 ­
2
(15-91)
MB 2 = -MA2 = MeC
-I---~f~
FB = F, = 0
CASE 3-UNIFORM DISTRIBUTED LOAD b3
Fc 3 = MCI = 0
M R2b ct(sin oa - a) + c2 sin c( - cos Ca)+ c3 (sin a =
ac, + c2 sin accos ac
M, = R2 b3 (sina
cos c)
(15-92)
- a cos oy)
2
MB2 =
MA 2 = MC2 - R b3 (c sin a - 1 + cos
B; = FA =- PRa
c)
CASE 4--UNIFORM DISTKRIBUTED COUPLE m,
FC3 = MCI = 0
2 1(c
MC2 = mr
M2
MB1
MB2
- sin a) + c2 sin c(cos a - 1)
c% + c2 sin accos cc
=
MA 2
(15.-93)
sin
-nf4
-mR
= M2
- nR(1 - cos x)
Fig. 15-11. Notation for out-of-plane loading.
CASE t-CONCENTRATED FORCE P
P
FC3-
0
MC = 0
PR C2 sin 2 c + c3(1 - cos c)
cc + c2 sin o cos
2
MC2-
15-11.
FLEXIBILITY MATRIX-CIRCULAR HELIX
In this section, we develop the flexibility matrix for a member whose centroidal
axis is a circular helix. The notation is shown in Fig. 15-12. The principal
inertia direction, Y2 , is considered to coincide with the normal direction, i.e.,
the inward radial direction, at each point. We also suppose the cross-sectional
properties are constant. For convenience, we summarize the geometrical
532
ENGINEERING THEORY OF AN ARBITRARY MEMBER
relations :t
X1
CHAP. 15
SEC. 15-11.
FLEXIBILITY MARIX-CIRCULAR
HELIX
533
n
-
K( cos Ut
R sin 0
CO
x3
dS = a dO
a=
[R2 + C2 )1/2 = constant
1
t = ft = - (- R sin Oi, + R cos 072 + Cl3)
X2
3/
X3"
4.
l
Zt
F'
B3
(a)
n = t2 = -cos 07, - sin 072
b = t3
-(Csin 01, - C cos 072 + R73)
R
--- sin 0
I
--cos 0
C
?t
RaR = R= q
0
0
- ill 0
I
C
R
---- cos 0 /
I
a
j c/_
-cosO
C
.
sin 0
Lc~
I -C(OB
i
)
XBQ = C(B -R(sin
=
- sin 0)
- 0)
R(sin 01, ­
sin 0)
- R(cos O, - cos
K(cos O, - cos 0)
R
X
2
0)
X , X2n, X3n-directions of basic frame
0
1
X' /
The steps involve only algebraic operations and integration. We first deter­
mine gi~ using (15-66), then Bqij from (15---68), and finally fi'j with (15-70). In
what follows, we assume the shear center coincides with the centroid and neglect
extensional and transverse shear deformation. With these restrictions,
gn
=
gl
g22
°
K
=
Tg2Rnq
(b)
12, r3-principal inertia directions
1
Fig. 15-12. Notation for circular helix.
Notation-DimesnsionlessParameters
1
R2 El2
GJ
a2 GJ
2
R
12
g2 =
O
=
a3
*22
=
12 =
\11 =-
C2
R
1 + a
22
XXBQ
t
I----a12 \\GJ
22
1X2BQ
(c)
a, a ­
a,
13
I2
[I3 - GJ j
Ra3
The flexibility matrix for a constant cross section is given below.
t See Examples 4-6 and 5-3.
±
(EI\,
RC I2
E13
and the expressions for *ij reduce to
a2 I3
2
C
a
tEI
E12
C2 12
2
+ a6
2
a6 + 3a 4
2
2
1 - a
2
a6
a
a8
12
a4 - a 6
2
a 6 - 3a 4
a
2
534
- ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
SEC. 15-12.
PARTIAL END RESTRAINT
Elements of f, 1
f26
11 1 Syrn
ffl =
f
fi
C2a
-
{a 7 0
+
f21 =
±
2
0 -
aEl
f31 = El
-
f22 = C2a
EI {a 7 0
10
O
sin OB
+
-
+ a
2 sin
-
sin O
a8
cos 0B(0B
2
0 r
sin
a5{03
03
+ e3
- a 0
sin 0
+ -sin
sin 0B
4(01
-
f 3 6 =- ,EI
cos 0B
+ cos
2al sin O3 -
[f 4
a, sin ORcos O,.
(15-94)
fl 5
161
4 = ECc -a 4 sin 01 + a7 01 + a8 sin n cos O]
f6
+
f2
as
sin On
_EI {1 -cos O
{sin
{a4X
0- 0
Ei
{0 sin 0 -
+ cos 01
MEMBER FORCE-DISPLACEMENT
RESTRAINT
f2 == f2 E
- 2
(0 a
±
+a
sin
sin 01 - a4 (1co s
cos)
RELATIONS-PARTIAL END
In Sec. 15-7, we considered an arbitrary member
which is completely re­
strained at both ends. This led to the definition of
the member flexibility matrix
and a set of equations relating the end forces and the
end displacements. Now,
when the member is only partially restrained, there
is
a
reduction in the number
of member force unknowns. For example, if there
is no restraint against rotation
at B, M = 0, and there are only a unknowns (where
a is the order of IFB), the
rotation oB at B has no effect on the end forces.
To handle the case of partial
restraint, we first determine the compatibility equations
corresponding to the
reduced set of force unknowns. Inverting these equations
and using the equi­
librium relations for the end forces results in force-displacement
relations which
are consistent with the displacement releases.
Let Z denote the force redundants. Normally,
one would work with the
primary structure corresponding to Z = 0. However,
the force at a point, say Q, in terms of the end forces suppose we first express
at B, using, as a primary
structure, the member cantileveredfiom A:
. IF
,3r
2= J .Ca
2
}
=
15-12.
sin 0,) + a,,O cos OB
f12 = f2 4 f2 s f2 6
Lf34 f35 f36
2
)
Elements of f22
OB)
cos 0,f) - a4 sin2 0R + a0
nsin0}
Elements of f~' 2
f5 = E
cos
-
- sin
cos 0B sin 0
+ a(OB ­
= El2
(a + a)0BEI23
sin0B
BCO
B
+ a8 sin O + a4 sin 0B cos 0O}
f33
2
{50
El
sin 01O
COS O - 2a 4 01 cos 013
+ CO2
s2
f3 = EIRCl
{ 2 0 sin 0
I
+
f35 =ia5(sin aB - HB cos 0)
cos0n
B COS
sin 0
Ras
1 + cos 0B) - -sin2 0nB
O COS 0 + (al - a4 )(l ­
+
f34 =
J331
f3 2
+ 2a4 sin
3
OB (+
0El
-
I
03 ±
-a7B +
I
f2 2
-
535
=
~iQ,
=
-
+~~Q­
~~T
Next, using the primary system corresponding to Z
= 0,
of the applied external load and the force redundants: we express
,
= EZ + G
(a)
B[
in terms
(15-95)
536
ENGINEERING THEORY OF AN ARBITRARY MEMBER
SEC. 15-12.
CHAP. 15
The elements of G are the end forces at B (for Z = 0) due to the applied external
loads. Note that G = 0 if Z contains only end forces at B.
Now, the principle of virtual forces requires
S,
A.Tr(&o + g)dS = AB[' TB + Ago'
a
PARTIAI mn arTDA^hXI'T
. ... I.... .
I, I.
" .
(b)
FA =
i
for any self-equilibrating virtual-force system. Taking the system due to AZ
results in the compatibility equations for Z. It is convenient to work first with
the virtual force system due to &ABE.Equation (b) reduces to
agn,""(¥,
,^ 0n)+Tmfn/)
~
Bu~~~
0
BB
=
~
At,, T'T(o
B
- A 'o, A Tn,
/w?
= AJ",'/
(c)
where F", fn are the initial deformation and flexibility matrices for thefitll end
restraintcase. Substituting for 3.P using (15-95), and requiring the resulting
expression to be satisfied for arbitrary AZ, we obtain
(ETrE)Z ± ET(i.'" + fnG) = ETak"
ET(q/7 -
oA4 rln) (15-96)
It should be noted that °gR, R'} are the displacements of the supports at B, A.
We suppose Z is of order q x 1, i.e., there are q force redundants. Also, we
let i be the row order of ,- (and O}).
i
iI
f
I
I
,
A,
(s15-97)
!BAB
ETfnE
-
(15-100)
Note that, for complete end restraint,
ZG=O
E=[i
O, 2
(15-101)
1
We will use (15-100) in Chapter 17 when we develop
the formulation for a
member system.
Continuing, we let
(15-102)
The force redundants are obtained by inverting (15-99):
Z
j- 4
- FT
f, reduced flexibility matrix (q x q)
z =- '~ + f"G
f z _ ET(11.
Eq'-)_I ~fi~ll
f.Z
)
T
= E (Ok "Ai
Ta
//f
_ %.lj
(axi1
~ =
537
At this point, we summarize the force-displacement relations for partial end
restraint:
Z = member force matrix
.' = EZ + G
k,Er(iql'
',
-t
(15-103)
Substituting for Z, the end forces at B are given by
(fix 1)
With this notation,
(e)
We defined k~ as the effective member stiffness matrix:
E isi x q
Gisi x 1
(d)
and (15-96) represents q equations. For convenience, we let
f- E Tf"E
(q x q)
n"' = t/' + fnG
(i x 1)
(15-98)
ke = Ek,.ET
(15-104)
= E(ETf",E)-T
In general, kj is singular when q < i, since E is only of rank q. Equation (e)
takes the form
]3
and the member force-deformation relations take the form
f Z =
ET(/r" - "/',
ET (in _
f,,
.4
BA
z)
o",
-/,),
'/ O, Z)A
+ k
SS B,
,i = -k
(15-99)
We refer to f, as the reduced flexibility matrix since, in general, q < i. Actually,
f, is the flexibility matrix for Z and it is positivedefinite since E must be of rank
q, i.e., the force systems corresponding to the redundants must be linearly
independent. Note that one can determine f,. directly by working with the
primary system corresponding to Z = 0. This is the normal approach. The
approach that we have followed is convenient when the member flexibility
matrix is known.
i -
"Y" TIWt,·
o, z +
(15-105)
= -k,-'" + (I - kf)G
The end forces at A are determined from (a):
jn
A
~"nn 4,1n
i.
-
n
A,0
BAO'
I,-nn,T)/n
sBA4e*BX
B
B i
Finally, we write the relations in the generalized form
=
YA
i
FA, i
±
Bc.B
+ kRBB
+
A
A nA
kAB 0'B + k AA,,(A
MZA
(15-106)
MEMBER
ENGINEERING THEORY OF AN ARBITRARY
538
539
rogam or cintiic ranla
sral laticBeas,
. Z: hinWaled
LASV,
7.PROBLEMS
ransla­
7. VLASOV, V. Z.: Thin alled Elastic Beams, Israel Program for Scientific
Washington, D.C., 1961.
tions, Office of Technical Services, U.S. Dept. ofCommerce,
Bars of Variable Section,"
8. BAZANT, Z. P.: "Nonuniform Torsion of Thin-walled
Publications, Zurich,
Engineering
Structural
and
Bridge
for
International Association
Vol. 25, 1965, pp. 17-39.
CHAP. 15
(15-107)
where
kM-== kn
kA
BA
- ke
== (kB,)
k
RBAke ',4BA
A
kA = - - XBAkA
expressions for the com­
Comparing (15-107) with (15-53), the corresponding
partitioned
PROBLEMS
k" by k in the
plete restraint case, we see that one has only to replace
i is different, however, due
B<,
forms for kBB, kBA, and kA. The equation for
to the presence of the G term.
Example 15-12
oI7=
0. We take Z = FR
at B. Then,
Suppose there is no restraint against rotation
(15-95).
with
G
E,
and generate
r
-I}
_
}
AG{
(a)
-
for a typical wide-flange section
15-1. Refer to Example 15-5. Determine c,t
relative importance of torsional
and a square single cell. Comment on the
involving c, in Equation (e)).
terms
deformation vs. bending deformation (i.e.,
members.
shallow
and
Distinguish between deep
rectangular cross section and kb
15-2. Refer to Example 15-7. Consider a
sketch. Evaluate VB2/! 3- ,;j and
varying linearly with x l, as shown in the
7;and a/b.
b~~~~rgProb
~~~,,j~~r~~)a~~~.~~~.
vB3S/\ 3 EI) for a range of
E
Prob. 15-2
stiffness matrices follow from (15-98),
For this case, G = 0. The reduced flexibility and
(b!
Fr- .
,v}
(15-102),
kr= f't'll
X2
(15-104):
and the effective stiffness matrix follows from
2= 12
(c)
0
0
'f't
k
15-
b
X3
13 - 12
Y3
(15-99)):
Finally, the force-displacement relations are (see
fitF,,
=
u-
-
X'
T
&i -
(d)
Y)
r
0", the relative rotation at B, There is
Note that premultiplication of I'" by ET eliminates
in this case; i.e., the support rotation
rotations
end
the
no compatibility requirement for
any member deformation.
at B, which we have defined as o), does not introduce
(in the direction of P)
15-3. Determine the reaction at B and translation
shear deformation.
at C for the member sketched. Neglect transverse
Prob. 15-3
PP
REFERENCES
for Elastic Beams and Shells." J. Eng.
REISSNEFR, E.: "Variational Considerations
February 1962.
Mech. Div. A.S.C.E., Vol. 88, No. EMI,
1967.
Frame Analysis, Wiley, New York,
WOODHEAD:
W.
R.
and
S.
A.
2. HALL,
Van Nostrand, New
Structures,
Fraamed
oJ
Analysis
3. GERE, J. M. and W. WEAVER:
1.
of Structwres, Prentice-Hall, 1966.
RUBINSTEIN, M. F.: Matrix Computer Analysis
Analysis, Pergamon Press, London,
LIVESLEY, R. K.: Matrix Methods of Structural
1964.
Triiger (Curved thin-walled beams),
6. DABROWSKI, R.: Gekriimmte diinnwandige
Springer-Verlag, Berlin, 1968.
4.
5.
Y3
I-
I
Vertical restraint at B
i
540
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
15-4. Repeat Prob. 15-3, considering complete fixity at B. Utilize symmetry with respect to point C.
15-5. Derive (15-27). Start with the definitions for the strain measures
(see Fig. 15-2),
PROBLEMS
iI
I
p=RRll
(
sin -12
as as
0
15-14. Starting with (15-87), develop expressions for the initial deforma­
tions due to an aribitrary distributed loading, b3 = b(O). Specialize for b
3
constant and verify (15-92).
15-15. Using the geometric relations and flexibility matrix for a circular
helix (constant cross section; Y, coincides with the normal direction) developed
in Sec. 15-11:
(a) Develop a matrix equation for the displacements at B due to a loading
referred to the global frame and applied at 0,. int: See (15-85).
(b) Evaluate u 3 for the loading and geometry shown.
OS Oy2
) 5I
11
I
neglect second-order terms, and note (15-26).
15-6. Summarize the governing equations for restrained torsion. Evaluate
b2 and b3 (see (15-34)) for a symmetrical wide-flange section and a symmetrical
rectangular closed cell. Comment on whether one can neglect these terms.
15-7. Refer to Example 15-11. Specialize the solution (Equations f) for
GB =
L >> 1. Verify that (g) reduces to the prismatic solution, (13-57),
when
541
Prob. 15-15
Y3,b
Y2 l
d
B -* 0.
15-8. Consider a member comprising of three segments. Assuming the
flexibility matrices for the segments are known, determine an expression for
the member flexibility matrix in terms of the segmental flexibility matrices.
Generalize for n segments.
15-9. Discuss how you would apply the numerical integration schemes
described in Sec. 14-8 to evaluate Jp, defined by (15-69).
15-10. Verify (15-73) and (15--74).
15-11. Determine the fixed end forces for the member shown, using (15-77)
and (15-79).
X2
X3
i
P
B
Prob. 15-11
X2
P
A
oD = r/2
c =R/2
G = E/2
'K
X3
f,
A1
15--16. Determine the reduced member flexibility matrix for no restraint
against rotation at an interior point P.
15--17. For the planar member shown, determine E and G corresponding to
15-12.
15-13.
Solve Prob. 15-3 using (15-84) and (15-87).
Verify (15-90) and (15-91). Apply them to Prob. 15-4.
Ze
reeases
forrotation
at
B and d
e
Then specialize for rotation releases at A, B and determine ke.
542
ENGINEERING THEORY OF AN ARBITRARY MEMBER
CHAP. 15
Prob. 15-17
X2
'B2
-FjT
(
rJ
Y.
-
MA\-
15-18. Determine E and G for(a) no restraint against translation in a particular direction at
B
(b) no restraint against rotation about a particular axis at B
Hint: Review Example 15-12.
Part IV
ANALYSIS OF A
MEMBER SYSTEM