424
RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER
CHAP. 13
Hint: One can write
=-if
fi3 =
2
I3
(X2 V / 2 r + X3 V2rldA
2 02 2/32
I3
14
Also show that
13 =
2/2
13-12. Specialize Equations (13-84) and (13-85) for the case where the
cross section is symmetrical with respect to the X 2 axis. Utilize
JjHe(X2, x3)HNo( 2, x 3)dA =
0
where He is an even function and Ho an odd function of x3. Evaluate the co­
efficients for the channel section of Example 13-5. Finally, specialize the
equations for a doubly symmetric section.
13-13. Specialize (13-88) for a doubly symmetrical cross section. Then
specialize further for negligible transverse shear deformation due to flexure and
warping. The symmetry reductions are
.X2 =
3
P2 =3
=
X-C2r=
0
=/3=t
14-1.
X3r == 0
1. The centroidal axis is a plane curve.
2. The plane containing the centroidal axis also contains one of the
principal inertia axes for the cross section.
3. The shear center axis coincides with or is parallel to the centroidal axis.
However, the present discussion will be limited to the case where the
shear center axis lies in the plane containing the centroidal axis.
'i = 17i = 0
13-14. Consider the two following problems involving doubly symmetric
cross section.
(a) Establish "linearized" incremental equations by operating on (13-88)
and retaining only linear ternms in the displacement increments.
Specialize for a doubly symmetric cross section (see Prob. 13-12).
(b) Consider the case where the cross section is doubly symmetric and the
initial state is pure compression (F t = - '). Determine the critical
load with respect to torsional buckling for the following boundary
conditions:
1.
o
2.
o=1
=f = 0
df
dX
--
0
=0
at
=
(unrestrained warping)
,L
Neutral equilibrium (buckling) is defined as the existence of a nontrivial
solution of the linearized incremental equations for the same external
load. One sets
U3 =
01 =
(02 =
X2
i
j
Yl
F1 = -P
t2 =
We consider the centroidal axis to be'defined with respect to a global reference
frame having directions X1 and X2. This is shown in Fig. 14-1. The orthogonal
unit vectors defining the orientation of the local fame (Y,, Y2 ) at a point are
denoted by Ft, 2, where t[ points in the positive tangent direction and t x t2 =
t3 . Item 2 requires Y2 to be a principal inertia axis for the cross section.
(restrained warping)
at x = 0, L
INTRODUCTION: GEOMETRICAL RELATIONS
A member is said to be planar if--
1/A 2 3 = 0
0
72 = 173 =
i
1
1I4
Planar Deformation of a
Planar Member
(03 =
.f
=
and determines the value of P for which a nontrivial solution which
satisfies the boundary conditions is possible. Employ the notation
introduced in Example 13-7.
13-15. Determine the form of V, the strain energy density function (strain
energy per unit length along the centroidal axis), expressed in terms of displace­
ments. Assume no initial strain but allow for geometric nonlinearity. Note
that V = V* when there is no initial strain.
A
I
12
I4
-l1
.I
Fig. 14-1. Geometrical notation for plane curve.
425
426
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
By definition, t
dr
t
=--
dS
dxl_
dS
=
dx 2
dx2
dS
dS
2
1
3,
t
dS
=1 dxi
I
(14-7)
R
o
- /Jo-v
1
a03
-- =
dt2
dS
=
R
--
t2
(14-3)
1
R
tl
where
1
R
dt l
dd2.
d-S-
-
=
t
x2
d2 x 2 dxl
dS + dS dS
According to this definition, R is negative when dil/dS points
in the negative
direction, e.g., for segment AB in Fig. 14-1. One could take
2 = ii, the unit
normal vector defined by
t2
I
dtl
dt
dS
(14-4)
x t2 =
3
but this choice is inconvenient
when
there is a reversal in curvature. Also, this definition degenerates at an inflection
point, i.e., when dF/dS = . If the sense of the curvature is
constant, one can
always orient the X 1 -X 2 frame so that 2 coincides with
hi,to avoid working
with a negative R.
To complete the geometrical treatment, we consider the general
parametric
representation for the curve defining the centroidal axis,
X
=
-
d2x
d2
d22
-d
d-
d-2 d;
dxl
A planar member subjected to in-plane forces (Xi-X plane
our notation)
will experience only in-plane deformation. In what 2follows, for
we develop the
governing equations for planar deformation of a arbitrary
member.
This formulation is restricted to the linear geometric case. planar
The two basic
solution procedures, namely, the displacement and force methods, are
described
and applied to a circular member.
We also present a simplified formulation (Marguerre's equations) which
is
valid for a shallow member. Finally, we include a discussion of numerical
integration techniques, since one must resort to numerical integration
when
the cross section is not constant.
14-2.
FORCE-EQUILIBRIUM EQUATIONS
The notation associated with a positive normal cross
dCSI
rather than according to t
V
d-x2-a
(14-2)
12
The differentiation formulas for the unit vectors are
dS
+ d-2 )
a-kTT
it follows that
dxl,
+
1
427
2, and 1/R in terms of v are
(14-1)
dSS 12
Since we are taking t2 according to t1 x t =
2
FORCE-EQUILIBRIUM EQUATInm_-n
increasing y. Using (14-6), the expressions for ti,
+ldS
dS
SEC. 14-2.
l(Y)
i.e., a cross
section whose outward normal points in the + S direction, issection,
shownin Fig. 14-2.
We use the same notation as for the prismatic case, except that now the vector
,,
4
Y3
-Y
t1-J1
x2 = x 2 (Y)
where y is a parameter. The differential arc length is related
to dy by
dS = +
dX )
+ (
2
12
dv =
dy
(14-6)
According to this definition, the +S sense coincides with
the direction of
t We summarize here for convenience the essential geometric relations
for a plane curve which
are developed in Chapter 4.
Fig. 14-2. Force and moment components
acting on a positive cross section.
PLANAR DEFORMATION OF A PLANAR MEMBER
428
CHAP. 14
SEC. 14-3.
components are with respect to the local frame (Y1, Y 2, Y 3) rathe r than the
basic frame (X1 , X 2, X 3). The cross-sectional properties are defineed by
A = ff dy
dY' dy =
=
13 = ff(y2)2 dA
dA
2
12 = |f'(y3)
b = bl/l + b 212
Introducing the component expansionsin (14-12), and using the differentiation
formulas for the unit vectors (.14-3), lead to the following scalar differential
equilibrium equations:
(14-10)
r+ = F,i F?+~72/'
F~
F)
-(14-11)
1
R
t- + b = 0
(14-14)
Note that the force-equilibrium equations are coupled due to the curvature.
The moment equilibrium equation has the saime form as for the prismatic case.
= M3
Note that 3 is constantfor a planar member.
I
t~
.
F2
dS
dF 2
F1
dS
R
dM
-- + F2 +m=O
dS
In this case, we work with reduced expressions for F'+and M, (see Fig. 14-3)
and drop the subscript on M 3:
X2
IF I
F_
1_
When the member is planar (Xl-X2 plane) and is -subjected to a planar
loading,
M+ = M3
(14-13)
tt- = mt 3
Since Y 2, Y3 pass through the centroid and are principal directions, it follows
that
fJy 2 cA =
dA
d'y3= J Y2Y3 dA = 0
(14-9)
F3 = M1 = M2 = 0
429
We expand b and i in terms of the unit vectors for the local frame:
(14-8)
1A
PRINCIPLE OF VIRTUAL FORCES
b S
.
r,+~= - `e+
YI
f
dSa
"2
+
S
+rlM+dS +...
dS
2
(L
-S+'"
rl 2 "
AL +di- (
t(S)
-,I -,I
vUU
-,j
aAI
'1
Fig. 14-4. Differential element for equilibrium analysis.
L'
The positive sense of the end forces is shown in Fig. 14-5. We work with
components referred to the local frame at each end. The end forces are related
to the stress resultants and stress couples by
AlV
Al
Fig. 14-3. Force and moment components in planar behavior.
F7Bj = FjISB
MB = MISB
To establish the force-equilibrium equations, we consider the differential
volume element shown in Fig. 14-4. We define b andin as the statically equiva­
lent external force and moment vectors per unit arc length acting at the centroid.
For equilibrium, the resultant force and moment vectors must vanish. These
conditions lead to the following vector differential equilibrium equations:
dF +
dS
dS
dS
+ i
± f x F+ = 0
1A = -MSA
14-3.
=
(14-12)
(14-15)
FAj = - FISA
j-
1, 2
FORCE-DISPLACEMENT RELATIONS; PRINCIPLE OF VIRTUAL
FORCES
We establish the force-displacement relations by applying the principal of
virtual forces to a differential element. The procedure is the same as for the
430
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
prismatic case described in Sec. 12-3, except that now we work with displacement components referred to the localframe at each point. We define 0 and Zo as
=
uji = equivalent1 rigid-body translation vector at the
centroid.
5 = Ycojtj = equivalent rigid-body rotation vector
For planar deformation, only ut, zl2 and
U3, wo1, and )2 can be deleted:
l
=
l)
= 03t
ltl
3 are
+
3
=
(14-16)
finite, and the terms involving
SEC. 14-3.
PRINCIPLE OF VIRTUAL FORCES
We define V* as the complementary energy per unit arc length. For planar
deformation, V* = 17* (F1, F2, M). One determines V* by taking expansions
for the stresses in terms of F t, F2 , M, substituting in the complementary energy
density, and integrating with respect to the cross-sectional coordinates Y2, Y3.
We will discuss the determination of V* later.
Specializing the three-dimensional principle of virtual forces for the onedimensional elastic case, and writing
el /*
2
t2
drV* = F- AF +
(14-17)
t3
The positive sense of the displacement components is shown in Fig. 14-6.
431
P*
8P*
o
AF 2 +
AM
(14-18)
= el AF1 + e2 AF 2 + k AM
lead to the one-dimensional form
fs(el AF 1 + e2 AF 2 + k AM)dS = Y di AP,
(14-19)
where di is a displacement measure and Pi is the force measure corresponding
to di. The virtual-force system (AFt, AF 2, AM, APi) must be statically permis­
sible, i.e., it must satisfy the one-dimensional equilibrium equations.
YB2
>>}MB
A+P,
-till.
± (Al)
dS(
-
)do
+
+d
-
-- F
FAt
dudS
t
it
du gdS
+
d dS 2
Y - .i dw dS
Fig. 14-5. Convention for end forces.
cS 2
Fig. 14-7. Virtual force system
x2
We apply (14-19) to the differential element shown in Fig. 14-7. The virtual
force system must satisfy the force-equilibrium equations (14-17),
tL2t2
d
­
dS
­
ta)
dS AM+ + t x A
dS
........
.
­
Evaluating
-+
E di APi,
X1
Fig. 14-6. Definition of displacement measures.
t "Equivalence" refers to work. See (12-8).
E
di
{d
+ dS
dS
dS
ki=s
(b)
= AF, [dul _ u)
+
F2 (12
+ Ul
o) + AM dol dS
dSf
432
PLANAR DEFORMATION OF A PLANAR MEMBER
and then substituting in (14-19) results in the following relations between the
force and displacement parameters:
OV*
duL
el = -- = ­
cF,
dS
aO* du2
e2
caP*
dU2
aF2
dS
d9* deo
k
do)
OM
dS
Q 1
=d-ydi
=
i
PRINCIPLE OF VIRTUAL FORCES
F,
rll
R
U
1
(14-20)
+ III
R
I
- a
dt2
+ Y3
d
v
dv
M
(14-22)
Y2
where I
I3. A logical choice for ulj (when the cross section is thin-walled)
is the distribution predicted by the engineering theory o flexural shear stress
distribution described in Sec. 11 -7:
ju=tq(F2)
+-±Y2
433
The most convenient choice for a1 is the linear expansion,t
112
We interpret e as an average extension, e2 as an average transverse shear
deformation, and k as a bending deformation. Actually, k is the relative rotation
of adjacent cross sections. In what follows, we discuss the determination of V*.
Consider the differential volume element shown in Fig. 14-8. The vector
defining the arc QQ is
UQQ = ddy
SEC. 14-3.
CHAP. 14
q = F2¢
(14-23)
where t denotes the local thickness, and q is the flexural shear flow due to F .
2
Both expansions satisfy (a).
X2
(a)
Noting that
dy
at
d{ 2
o -..
dy
Rt
(b)
dy
for a planar member, (a) can be written as
dS
2
=-Q( =
(1
dy
(_
dS .
(c)
By definition, V* is the complementary energy per unit length along the
centroidalaxis. Substituting for dS2 in the general definition, we obtain
ff
F* dS
11
V* dS2 y2, dy3
A I
Y2.Y3
U4
rV/*
=,
(14-21)
I-
dA
In what follows, we consider the material to be linearly elastic. The comple­
mentary energy density is given by
In general, V = V* (,
,
3). We select suitable expansions for the
1 22,
stress components in terms of F1, F2, M, expand V*, and integrate over the
cross section. The only restriction on the stress expansions is that they satisfy
the definition equations for the stress resultants and couples identically:
JfJ'11 dA = F
fSY :3
JSS,
2 dA
= F2
JfU 1 d
dA
M
-fY 2c, 1 1
11 dA = 0
ff(Y 2 71 3 - Y3 91 2 )dA = O
Fig. 14-8. Differential volume element.
=0
(a)
1
Y*I l+
11J
2E
-a,,2
1(2
­ 2G
2
+
+
1)
12
(a)
°
where 0 is the initial extensional strain. Substituting (a) in (14-21) and taking
the stresses according to (14-22), (14-23) results in the following expression
t This applies for a homogeneous beam. Composite beams are more conveniently treated with
the approach described in the next section.
434
e
V*eOF+
k° 0 M=
2AE
+
F-±F
1
AER
F,
M +
1 M
1
1
2
+ 2AE + 2GF
2GA 2
2E*
435
by neglecting y2/R with respect to unity in the expression for the differential
arc length, i.e., by taking
dS2 . dS
(14-27)
Va* ,ff v* dA
for V*:
i
PRINCIPLE OF VIRTUAL DISPLACEMENTS
SEC. 14-4.
CHAP. 14
PLANAR DEFORMATION OF A PLANAR MEMBER
(14-24)
where
A
e = A
o
-I
-1
J
=7 1
(°1
Y2
dA
(1
-j 'sfr(1
dA
-
y3
--
i
_ Y2)dS
F1
e, = e + -+
0
F2
-
GA*
du 2
'
du,
M
AEi= =
U
+-1
dS
R
F,1
M
-+
k = k +
EI*
AER
u2
- (1
(14-25)
J
U2
R
(14-28)
co
Thick Member
f{(e
AI, 1\+
F,
AE
/.^
Note that the axial force and moment are coupled, due to the curvature.
Inverting (14-25) leads to expressions for the forces in terms ofthe deformations:
AR2 -- AR 2 [1
dul
To complete the treatment of the linear elastic case, we list the expanded
forms of the principle of virtual forces for thick and thin members. Note that
these expressions are based on a linear variation in normal stress over the cross
section.
do
IdS
A
Ei*
kO)
F 1 = .EA.(e
- e) -(ki - 6 1
R(1 - )
EI*
EI*
(e) - e) + --- 6 (k - k° )
_
M = - R((1
1- )
F2 = GAie 2
Fl-
d1
+ Ael
AE
dS
du2+ ul
F2
e2 =
GA2 -- dS
M
do
- d
k =k 0 +
El
dS
e =
dA)
If the section is symmetrical with respect to the Y3 axis, I* = I and A = A2 .
The deformation-force relations corresponding to this choice for V* are
e2 =
Assuming a curved member to be thin is equivalent to using the expression
for V* developed for a prismatic member. The approximate form of (14-25)
for a thin member is
+ k`
2
AF
2
AER
GA2
)
F,
M
AM dS = L di APi
* a)
+ AER +
(14-29)
Thin Member
(14-26)
1 1±
-- AF, + GA AF2
2
EI··MAM} dS =
YdA]
di AP
(14-30)
We observe that
AR2 = (
=
(a)
where p is the radius of gyration and d is the depth of the cross section. For
example,
d2 ,
I
(b)
2
AR
12R2
2
for a rectangular cross section. Then, 6 is of the order of (d/R ) and can be
2
neglected when (d/R) << 1.
2
A curved member is said to be thin when O(d/R) << 1 and thick when O(d/R) <<
1. We set 6 = 0 for a thick member. The thinness assumption is introduced
14-4.
FORCE-DISPLACEMENT RELATIONS-DISPLACEMENT
EXPANSION APPROACH; PRINCIPLE OF VIRTUAL
DISPLACEMENTS
In the variational procedure for establishing one-dimensional force-displacement relations, it is not necessary to analyze the deformation, i.e., to determine
the strains at a point. One has only to introduce suitable expansions for the
stress components in terms of the one-dimensional force parameters. Now, we
can also establish force-displacement relations by starting with expansions for
the displacement components in terms of one-dimensional displacement pa­
rameters and determining the corresponding strain distribution. We express the
436
SEC. 14-4.
CHAP. 14
PLANAR DEFORMATION OF A PLANAR MEMBER
= dy
2
Q--v2 = -dy
=y2
12)i
12
=
-
E_
a
We use a prime superscript to denote quantities associated with the deformed
position of the member, which is shown in Fig. 14-10; for example:
F'= '(y) = position vector to point P(y) in the deformed position (point P').
t1 = tangent vector to the deformed centroidal axis.
- = position vector to Q(y, y2, y3) in the deformed position (point Q').
2
(+4-CV12
(14-32)
4
= t2 +
)dy
V,/ 2
--
The analysis of strain consists of determining the extensions and change in
angle between the line elements. We denote the extensional strains by
(j = 1, 2) and the shearing strain by y1 2. The general expressions are
(14-31)
t2
1
dvy=
QQ =
dy = ac'dyt
+ d
dy'
dv = 2 d
QQ 2 = ( 2 dy 2
437
From Fig. 14-10, and noting (14-31):
stresses in terms ofthe displacement parameters using the stress-strain relations,
and then substitute the stress expansions in the definition equations for F 1, F2 ,
and M. The effect of transverse shear deformation is usually neglected in this
approach. To determine the strain distribution, we must first analyze the
deformation at a point. This step is described in detail below.
Figure 14-9 shows the initial position of two orthogonal line elements, QQ1
and QQ2 , at a point (y, v2, y3 ). The vectors defining these elements are
QQ
PRINCIPLE OF VIRTUAL DISPLACEMENTS
i P
_-1
sin y
-
j=(,
i
2
~~inF~~
__(14-33)
-,
Q
ie ',l Iq
VW
Now, we restrict this discussion to small strain. Substituting for the deformed
vectors and neglecting strains with respect to unity, (14-33) expands to
I
-
(I
(2
C2
2
e
l
1
2(0( 2)2
Y
(3
2t
+
Y
0(
a ll2
(7u
a
2
,
I
_ ,
21
2
Y12
+
' 2
+ iltey7t2
'
2-
x
2
CYOV
(14-34)
1
(2
+
t2....
2
i2
2 a .
0(2
( 2
y Y2
The nonlinear terms are associated with the rotation of the tangent vector.
Neglecting these terms corresponds to neglecting the difference between the
deformed and undeformed geometry, i.e., to assuming linear geometry.
The next step involves introducing an expansion for 2 in terms of Y2. We
express fl2 as a linear function of Y2.
u2 = U -
(ov
2t
(14-35)
where co = co(y) and
t = uIt +
·-,Y,I
Fig. 14-9. Initial geometry for orthogonal curvilinear line elements.
2 t2
=i'(y)
(14-36)
is the displacement vector for a point on the centroidal axis. Equation (14-35)
implies that a normal cross section remains a plane after deformation. One can
interpret co as the rotation of the cross section in the direction from tt toward
t2. This notation is illustrated in Fig. 14-11.
In what follows, we consider only linear geometry. Substituting for 12, taking
y = S, and evaluating the derivatives lead to the following strain expansions:
PLANAR DEFORMATION OF A PLANAR MEMBER
438
el
1
- y 2 /R (el ­
C2 =
- y
=
el = d dS
R
y 2k)
1
d
dS2
+
tl
R
y2=
- 0) = 7t2jY2=°
(14-37)
SEC. 14-4.
- y 2 /Re2
i - yyIR
expressions for q1 and y12. Note that the assumption that a normal cross
section remains plane does not lead to a linear variation in extensional strain
over the depth when the member is curved.
We introduce the assumption of negligible transverse deformation by setting
e2 = 0. The resulting expressions for c) and k in terms of u1 and u2 are
e2 =0
du 2 + 1
R
dS
One could include an addiThe vanishing of e2 is due to our choice for 12-.
and additional terms in the
=
give
e2
would
t
.
This
/iy
term,
linear
tional
2 2
, .
2'
(14-38)
2
d ul\
dS
d u2
dS 2
kc
d
dS
A,
X2
439
PRINCIPLE OF VIRTUAL DISPLACEMENTS
do
k=dS
1
712
du2
e2
CHAP. 14
When transverse shear deformation is neglected, one must determine F2 using
the moment-equilibrium equation.
The next step involves expressing F1 , F2, and M in terms of the one-dimensional deformation parameters ee 2 and k. Illn what follows, we consider the
material to be linearly elastic and take the stress-strain relations for al1 , 0r12
as: ­
a, 1 = E(. 1 - eo)
(a)
1t2 = Gy 1 2
12 ._
Substituting for el, 712, using (14-37),
(e l
al1 = 1
2/R
1-
u
'
12
G
-y/R
=
-
2 k)-
E
EI
(14-39)
e2
and then evaluating Fl, F2 , and M, we obtain
F = Eel
Fig. 14-10. Deformed geometry for orthogonal curvilinear line elements.
U
2 t2
-
Ek
Y2
I - y2/R
F
IJ f
Centroidal axis
i e°I cldA
-E
dA
(14-40)
I - Y21R
(ul -- y 2 )tl
l
- Y2/R
M = -Eel JiY2 d
+
Iff
Y1
y/R
1 - dA
Ek
+
y2
dA
The various integrals can be expressed in terms of only one integral by using
the identity
_ Y2 /R
1
(a)
1 - y /R
1 - y2 /R
2
and noting that Y3 is a centroidal axis:
ilt­
Fig. 14-11. Displacement expansion.
fSY2 dA =O
t The relation for al is exact only when c22
member.
=
33 = O. We generally neglect
(b)
22, 33 for a
PLANAR DEFORMATION OF A PLANAR MEMBER
440
CHAP. 14
One can easily show that
PRINCIPLE OF VIRTUAL DISPLACEMENTS
441
xi < 1. Then
This series converges for
dA
ff1
SEC. 14-4.
I'
- y 2 /R
R
Id
- + ff2R3
R
2
1 -_fi
2R/
(14-41)
+
+
5
+
(c)
+2R
and
rf Y2 dA
1 =
i - Y2/R
1
3
!)2
1
)
(d)
For completeness, we list the inverted form of (14-40),
eF = e + F
F1
+
EA
Fig. E14-1
M
EAR
F2
e2 =GA,
k = k° +
F1
M
+ M
EAR
El"
d
where
I"=A' A
Y3
+ -
(14-42)
'-
A'O1 ± AR2)
=-l-' J82
AR
dot
The expressions for e1 are identical with the result (see (14-25)) obtained with
the variational approach. However, the result for k differs in the coefficient
for M. This difference (I' or I") is due to the nonlinear expansion used for ao 1
c
I~~
Wdm,,e
n-!A
The relations listed above involve exact integrals. Now, when the member
is thick, we neglect (y 2/R1)2 with respect to unity. This assumption is introduced
by taking
"
for
the
r
co
scio
show
i'_- F.
1 -the
foY2R
expansions
and I':
1
We determine I' for the rectangular cross section shown in Fig. E4-1.
ff. dA
_
1-
J-di2 1
2/R
22
in the expansions for
i}.d
Y2/R
(a)
a
, o2 and I
11 ~ eell
E
2
3
+
R
|-
k
'2
+
+
+---- +
(
Y2
Yk
_
1
R
2
To obtain a more tractable form, we expand the log terms, using
=
+
d
G
-
1
2R
= -R bd + R b In
In
(a)
FI {
(b)
dAY il
1 - 2 Y2/R
= I + 1
(
3 14
(14-43)
442
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
To be consistent, we must also neglect I'/AR 2 with respect to unity in the
expression for A2 and I". When the member is thin, we neglect y 2/R with
respect to unity.
1
1 - y2 /R
Ut11
°
y2k - e
el-
E
G2
(14-44)
e2
U12
G
It is of interest to establish the one-dimensional form of the principle of virtual
displacements corresponding to the linear displacement expansion used in this
development. The general three-dimensional form for an orthogonal coordinate
system is (see Sec. 10-6):
jJ(ol11 b& +
+
Pi Adi
'.)d(vol.)=
+
1'2 (5' 1 2
(a)
where Pi represents an external force quantity and di is the displacement quantity
corresponding to Pi. We consider only i and y,, to be finite, and express the
differential volume in terms of the cross-sectional coordinates Y2, Y3 and arc
length along the centroidal axes (see Fig. 14-9):
-
d(vol.) = dS2 dy 2 dv3 =
) dS
dy 2 dy 3
SEC. 14-4.
we obtain
fs[F, e, + F2 e 2 + M Ak]dS = P Adi
Example 14-2
The assumption of negligible transverse shear deformation is introduced by setting e2
equal to zero. This leads to an expression for the rotation, co, in terms of the translation
components,
o =-
C
=
EPi Adi
duz
u2
dS
R
1
Ao =
Aut
R
+
d
St12,
d1
I1
R
Id
bet =/:Au,
dS
- v 2 k)
k =
(c)
1
du
=
s
U2
dS
R
du2
ut1
ez = dS
R
1 d
Au2 + - -- Au
R dS
[Fl be + M (bk]dS
SRA
(d)
dco
7t2
dS2
­
k = dS
Substituting for e,
d
(d)
- ---Au 2
and integrating by parts, the left- and right-hand sides of (b)expand to
Y12 = 1-V2/R e2
el
(c)
Substituting for Aco and the strain variations,
2
1-/R(e
(b)
ul
d du2
dco
= = I _ d+
R
dS dS
dS
k
We take (14-45) as the form of the principle of virtual displacements for planar
deformation.
The strains corresponding to a linear expansion for displacements and linear
geometry are defined by (14-37), which are listed below for convenience:
q =
(a)
R
dS
fs[Fl, e1 + M 5k]dS = EPj Ad,
(b)
(14- -45)
- Y2) dAj dS
Y12)
Il
(at1 2
and the relations for negligible transverse shear deformation reduce to
el =
(ll1 b3 +±12
(14-46)
This result depends only on the strain expansions, i.e., (c). One can apply it
for the geometrically nonlinear case, provided that (c) are taken as defining
the strain distribution over the cross section.
We use the principle of virtual displacements to establish consistent force­
equilibrium equations. One starts with one-dimensional deformation-displacement relations, substitutes in (14-46), and integrates the left-hand side by parts.
Equating coefficients of the displacement parameters leads to a set of force
equilibrium equations and boundary conditions that are consistent with the geo­
metrical assumptions introduced in establishing the deformation-displacement
relations. The following example illustrates this application.
Then (a) reduces to
ILT
443
PRINCIPLE OF VIRTUAL DISPLACEMENTS
and using the definition equations for F1, F2, and M,
=
I(F1 +dS
F1 +
·
fSB·[
JAs
S=S
d
dAM
Au1
--lA
dFI
f
l
±JAuL~~!
2
-
M -
I dM] +
-
dS
dM]
(e)
A
2
S-SA
F,
dZM
)
+
dS' l
CHAP. 14
PLANAR DEFORMATION OF A PLANAR MEMBER
444
di=
{
irs7
+ (
A /
(
R) +
~~~
.[A"
b
+
A
2
(
U
+ R) AUBI + (FB2
(F,
R )- A
d l
d
+nB) AUB 2
+
(FA2
MA
s"")
_1}j
2
-
+ MB A ,dSB
lA) AlA2±
MAA
14-5.
CARTESIAN FORMULATION
CS
We consider the case where the equation defining the centroidal axis has the
form x 2 = f(xj). The geometrical relations for this parametric representation
are obtained by taking y = x in (14-7). They are summarized belowt for
convenience and the notation is shown in Fig. 14-12:
dS =
SA < S < S,
m
dM
dF1
+
+ b + -- = 0
R
RdS
dS
Ft
+ --R
d2 M
S2
dS2
+ b2
dx,
tan 0 =- 1
dxd
dm
- dS-= 0
dS
t
1
prescribed or
u2
prescribed or
dt
2
dS
F1 = - FA
dM
­
= FA2 -
dx
(g)
R =-I II2
prescribed or
_Cos 0
+"(dx I
il·[-
S = SA
UI
445
(f)
The consistent equilibrium equations and boundary conditions for negligible transverse
shear deformation follow by equating corresponding coefficients of the displacement
variations in (e) and (f):
+-
FORMULATION
The other equilibrium equation and the boundary conditions are not changed. Using
(h) instead of (a) eliminates the shear term, F2 /R, in the tangential force-equilibrium
equation.
and
zPi
CAR-TESIAN
SEC. 14-5.
12
++ (d)
(14-47)
M = - IA
+ d-2-
S = S
utl
u2
du 2
dS
prescribed or
Ft = FBI
cdM
prescribed or
.
dS
d71
.
-
F
2
- m
d
M =
B
One can obtain (g) by solving the last equation in (14-14) for F2 and substituting in the
first two equations.
Suppose we neglect ut/R in the expression for co:
dS
2
ke-
t
(h)
d u2
~vdS 2
dF 1
- = b
dS
R
d2
1
dS
R
In the previous formulation, we worked with displacement components and
external force components referred to the local frame. An alternate approach,
originally suggested by Marguerre,. involves working with components referred to the basic frame rather than the local frame. The resulting expressions
differ, and it is therefore of interest to describe this approach in detail. We
start with the determination of the force-equilibrium equations.
Consider the differential element shown in Fig. 14-13. The vector equilibrium
equations are
dF
ax
This assumptiont is generally referred to as Mushtari's approximation. The equilibrium
equations for the tangential direction reduce to
t See Ref. 5.
-2
dS
prescribed or
dT 2
I
+ p 0
dM+
dp
­
dx- + dx x F.
dxl
(i)
f See Prob. 14-1.
+See Ref. 6.
dr,
(14-48)
l
+ h=
PLANAR DEFORMATION OF A PLANAR MEMBER
446
CHAP. 14
SEC. 14-5.
CARTESIAN FORMULATION
447
where , fl are the external applied force and moment vectors per unit projected
length, i.e., per unit x1. They are related to b and i (see Fig. 14-4) by
fp dx = b dS = (ab)dx1
-hdx1 =771 dS (¼)d(14-49)
X2
hdx
II
Y2
r
''
= N dS = (N4-z)dxI
r
Substituting for the orce and moment vectors,
I
F+ = Ftl + F2t 2 = N 171 + N 272
.Yi
fi = h3
-M+ = M 3
P = Pi11 + P272
N 1 = F1 cos 0 aU
X2 =f(xtd)
12
I
I
df dxI
(14-50)
F2 sin 0
N 2 = F1 sin 0 + F2 cos 0
dx
the equilibrium equations expand to
XI
il
dN
d
- -- (F1 cos 0 - F 2 sin 0) = -p,
Fig. 14-12. Notation for Cartesian formulation.
dN 2 = d
dx(
-1 (
ac
X2
dx l
dx I
(F, sin 0 + F 2 cos 0) = - P2
+ h = F2 = -N,
(14-51)
sin 0 -t+N2 cos
We restrict this treatment to an elastic material and establish the force­
displacement relations, using the principle of virtual forces,
,,,
t*
.,
dx =
[e AF 1 + e2 AF 2 + k AM]c dx =
di AP
(a)
where V' = * (F,1 F 2 M) is the complementary energy per unit arc length.
Consider the differential element shown in Fig. 14-14. The virtual-force sys­
tem is statically permissible, i.e., it satisfies the force-equilibrium equations
identically:
-- AF+ =
dx,
0
(b)
-AM + at x AF+ = 0
Expanding C di APi,
12
X1
d APi=
AF+
r_.
-+c t
I
x Fc + AM+
-
I
lx
dx
(c)
-I
and then substituting for the displacement and rotation vectors,
Fig. 14-13. Differential element for equilibrium analysis.
U =
UVll + V212
(0 = (013 = (Ct3
C(1A
'
448
CHAP. 14
PLANAR DEFORMATION OF A PLANAR MEMBER
we obtain
SEC. 14-6.
DISPLACEMENT METHOD OF SOLUTION
Marguerre starts with
dx, + AN dx,
N1 dx
1
2 d- -AF
dx
E di zPi =
(d)
2 co + AM d-) dx,
Finally, substituting for N,, N 2 in terms of F,, F 2 and equating coefficients of
the force increments result in
el
OV*
- F=
dv,
dV
cos 2 0--- + sin 0 cos 0 -- 2
V
dx,
dx,
e2 =
?F=
- sin 0 cos 0
k
F,
e=FU
2
dx,
+ cs2 0
N,
F,
N2
F 2 + 4-)
dF,
co
dx,
(14-53)
sil 0
1
1
d.x,-+
) + P2
d.F+ d (F
dM
F2 -
aV*
dco
=-cos 0
aM
dx,
cos0
(a)
(14-54)
dx,
e2
=
in (14-50), which relate the cartesianand localforces.
= 0O
m
dxl
tall 0 =
F
and the resulting equations are
The member is said to be shallow when 02 << 1. One introduces this assumption by setting
OtV*
F,
dvl
d.x,
c/f dv2
dx, dx,
V*
(?F2
dv2
dx,
O)
V*
d(o
OM
(14-55)
d.x,
One step remains, namely, to establish the boundary conditions. The general
conditions are
X2
I
VI
AMl.
-AM
449
+x
dx
V2
(2
M
AM+(
-A,'+
I
"dxlk
2?
t
1-Yul2(~
ZI
I
I
I I
I
I
dxtI
I
N,]
N 2 ' prescribed at each end
cJ
(14-56)
We obtain the appropriate boundary conditions for the various cases considered
above by substituting for N,, N 2 and o. For example, the boundary conditions
for the Marguerre formulation are
dx- ( 21i1
V2
or
or
or
14-6.
x1
Fig. 14-14. Virtual force system.
Marguerre's equations are obtained by assuming the member is shallow
and, in addition, neglecting the contribution of F2 in the expression for N,.
vI
or
F,
v2
or
F2 + d
co
or
M
worM~~
F,
prescribed at each end
(14-57)
DISPLACEMENT METHOD OF SOLUTION-CIRCULAR MEMBER
The displacement method involves solving the system of governing dif­
ferential equations which, for the planar case, consist of three force-equilibrium
equations and three force-displacement equations. If the applied loads are
independent of the displacements, we can first solve the force equilibrium
equations and then integrate the force-displacement relations. This method
is quite straightforward for the prismatic case since stretching and flexure are
uncoupled. However, it is usually quite difficult to apply when the member is
PLANAR DEFORMATION OF A PLANAR MEMBER
450
CHAP. 14
curved (except when it is circular) or the cross section varies. In what follows,
we illustrate the application of the displacement method to a circular member
having a constant cross section, starting with1. the exact equations (based on stress expansions) for a thick member
2. Marguerre's equations for a thin member
d¥
Eq. (14-19)
b
- dM =-2( R2
-1
-- RF1 =
2
--R (C
2
(-
F2 =
sin 0 + M,)
d
ioi~i
l~ins
Il - R ib,(b1 +± R)
R dO
COS 0 + C 3
C 2 sin 0 + C 3 COs 0 +
S=R
F2
Eq. (14-25) =>
F
R
dS = RdO
dmin
(14-58)
b2 -
R =
M
AER
1 (duZ
RK dO
-
+AER
AER
(14-59)
M
+
E
-R
EP"
R
d)
dO
Fig. 14-15. Notation for circular member.
1
Solution of the Force-EquilibriumEquations
Integration of the Force-DisplacementRelations
We consider the external forces to be independent of the displacements.
Integrating the first equilibrium equation, we have
We start with (14-59) written in a slightly rearranged form:
du,
RF, = -M-R2
(b
+
-
R
d + C1
2
+ M of
= (b)R
C +
dOsolution
The
general
1dm
2F
2
-
dOis-
U2 =
(a)
du 2
where Cl is an integration constant. Substituting for F1 in the second equation
results in a second-order differential equation for M:
d M
F,
112
1 (dU
Ot 2
F1
k
2
dOm
-n
F1
AF
(14-61)
dMP
F2
1 dM
F2 =
-n
+-)
dO
d2 M2
451
where Mp denotes the particular solution due to the external distributed loading
and C 2, C 3 are constants. Once M is known, we find F1 using (a) and F2 from
the moment equilibrium equation. The resulting expressions are
F1
The results obtained for this simple geometry provide us with some insight as to
the relative importance of transverse shear deformation and stretching deformation versus bending deformation.
When the centroidal axis is a circular segment, R = const, and the equations
simplify somewhat. It is convenient to take the polar angle 0 as the independent
variable in this case. We list the governing equations below for convenience
and summarize the notation in Fig. 14-15:
dF 1
R--
DISPLACEMENT METHOD OF SOLUTION
SEC. 14-6.
f(b + )+d]
-+ ui
dO
dco
dO
(b)
R e°
RF 2
=---
GA*
do= Rkio
I
+
(M + KF,)
(a)
+ Ro
R
M + AI
EJ* L
*
R
(RF)
i
To determine ul and u2, we transform the first two equations to
The general solution of(b) is
M = C
+ C cos 0 + C3 sin 0 + Mp
(14-60)
(d=I
t 2
+ Re
+-
(M + RF 1 )
(14-62)
452
PLANAR DEFORMATION OF A PLANAR MEMBER
SEC. 14-6.
CHAP. 14
DISPLACEMENT METHOD OF SOLUTION
453
Note that 6, is associated with transverse shear deforlation. Substituting for ¢ in (14-63)
and integrating, we obtain
and
d2 U2
d22
R dF2
dco
1
R
2=+GR
-Re o (M + RF1)
GA* dO
(10
AE
R dF 2 + 2 0
aiR 2
R k - Re ° +
M
GA * dOI
e
U
2
- 0 sin(O - 0)
(e)
The solution for u, follows from (14-62):
(14-63)
I
U
1 =
C4 sin
0
- C
,
Os 0 + C6
+ ' 2 [0 cos(
+ -R
E-V
= AR2 = 0 ()
al = 1 -
= C4 cos 0 + C5 sin 0 +
-0) + sin(Os - 0)]
(f)
Next, we determine co using (14-64),
2
We have previously shownt that 6, is of the order of (d/R) . It is reasonable
to neglect 6, with respect to I but we will retain it in order to keep track of the
influence of extensional deformation. We solve (14-63) for u2, determine u,
from (14-62), and co from the second equation in (a),
F2
GA*
-
[ (dtI + tll
dO
(14-64)
This leads to three additional integration constants. The six constants are
determined by enforcing the three boundary conditions at each end. Various
loading conditions are treated in the following examples.
C6
r0
-R
FB1R2
+ - E i­
{( + al sin(OB
-
0)}
(g)
Finally, the constants are found by enforcing the displacement boundary conditions
at 0 = 0:
3
C4 = -at FPiR
El*
C5 =
(2I ~-2
a)
- sin OB
El*
(h)
3
C6 = -a,
FBI R
--sin 0B
Example 14-3
Consider a member (Fig. E14-3) fixed at the negative end (A) and subjected only
to FBI at the right end (B). The boundary conditions for this case are
Fl = FRo;
u
F2
= u2 = 0) = 0
=M = 0
atO = 0B
at 0 = 0
To determine the relative importance of stretching and shear deformation versus bending
deformation, we evaluate the displacements at 0 = O0and write the resulting expressions
(a)
1ig. E14-3
Specializing the force solution for no external distributed loading and enforcing the
boundary conditions at B, we obtain
F1 = F
cos(Os - 0)
Sin(0B - 0)
M = RFB(I -- cos (B -
(b)
F2 = F
))
FBI
To simplify the analysis, we suppose there is no initial deformation. Using (b), ¢/ takes
the form
=
EI*
[al - a2 cos(0B - 0)]
(c)
where
5 GA=R2
a2 = a
t See Sec. 14-3, Eq. (14-26).
+
(d) 2
=
I -
(d)
e + 5s
Constant cross section
PLANAR DEFORMATION OF A PLANAR MEMBER
454
AS2
(sB - sin 0B)(1 + 1) (Se)
OB =
* ( 0s - 2 sin 0B -½ sin 0? cos 0B)(
El
d2
1
I*
FBlR2
UB2 =
455
For example,
in the following form:
UB1 = -f-
DISPLACEMENT METHOD OF SOLUTION
SEC. 14-6,
CHAP. 14
2
(1 - cos 0B - 2 sin 0B)(
1
+ b2
De
l*S
,
2
I
sin 0B
expression for uB is
(i)
UBI = T
O= - Sill 0
OB
1 - coS B -
b4
1 + cos
Sin 2 0ll
OB
1- cos 0
cos 0 = 1 - -- +
2
3
0
sin2 =
2224
(1__ +
(j)
*'mI.
A
-A
*U-L·
The internal force distributions due to F,, acting on the cantilever member shown
in Fig. E14-4 are given by
F1 = --F;B2 sin(O - 0)
04)
duI
dO-
...
5u1m
UB2
i-sr
F" S2
I*
FI1S3 0 [l
EI*
El ] +
3 GASS
P5 1S3
O
4
EP'_k 1.2 L-±--
<-
(k)
1ESl
2
2122U
dO2 + u2 = R2 k 0 -Re,
du,
dO = 2 + Re °
j
Now,
d R
EI
0 (d)2
GAS 2 = 0-)
(b)
Eliminating n from the first two equations, we obtain
*
4A
CFE5 2 11
(;A
°
2 = Re
du 2
+ u, = Rct)
dO
RM
do)
°
._ = Rk +
El*
dO
and neglect 02 with respect to unity. The resulting expressions are
)B --
(a)
We suppose the member is not shallow and neglect stretching and shear deformation.
The force-displacement relations reduce to (we set A - A* =-, in (14-59))
)
0-
15
2
(n)
, 2
M = F R sin(0B - 0)
24
2+
-
F2 = Fi 2 cos(OB - 0)
04
02
sin 0 cos 0= (l
-IA
'�"�······P
i2 (
(1 -
[
2
(d/R) .
The coefficients (b1 , ·., b,) are of order unity or less when 0~ is not small with respect
to unity, i.e., when the segment is not shallow. Also, (5,and 6 are of order (d/R)2. It 2follows
that the displacements due to stretching and shear deformation are of order (dIR) times
the displacement due to bending deformation for a nonshallow member.
To investigate the shallow case, we replace the trigometric terms in (i) by their Taylor
series expansions,
sin 0 = 0
/2 o
In sum, we have shown that the percentage of error due to neglecting stretching and
2
transverse shear deformation is of the order of (d/R) for a onshallow circular member.
15'), we cannot neglect stretching deformation. Actually,
If the member is shallow (0s <
the stretching term dominates when the member is quite shallow. The error due to ne­
glecting transverse shear deformation for the shallow case is still only of the order of
½(01 - sin 0B cos On)
B - 2 sin OB+ Sil 0B3COS UOs
-l-sin
+
EI* _20
-- 0B + 2 sin 0O -- I sin O0 COS ,O
2
0 - 2 sin 0s + - sin , cos OB
b=
(m)
0.26 ()
S
=
GA½S - 10G
2
for a rectangular section and v = 0.3. Since (d/S) << I for a member, we can neglect the
transverse shear terms in UBI, 11B2 and the stretching term in c)B. However, we must retain
the stretching term in uBtsince it is of the same order as the bending term. The appropriate
+ b3 56)
b4 5s)
- (e
12\S
R
-- M
(c)
dO
(1)
We determine u2, then ut, and finally . Note that (c) corresponds to (14-62), (14-63)
. The final expressions (for no initial deformation or support
= coA2
and (14-64) with A
456
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
mOVement) are
movement)
are
112
SEC 14-6
FB2 R3
-
{(O
0 cos 0n
COS(OB - 0) - sin
DISPLACEMENT METHOD OF SOLUTION
}
457
457
Fig. E14-4
Ff2R"
R2
FB2
EI*
{ COS(OB
- 0) - COS O4
Example 14-5
We analyze the shallow parabolic member shown in Fig. E14-5 using Marguerre's
equations. We consider the member to be thin and neglect transverse shear deformation.
Taking f = ax/2 and P = nm= 0, the governing equations (see (14-55) and (14-57))
reduce to
dF
-=01
dxj
I
d2M
-
dX 2
F
-P2 =
(a)
,A const
dM
F2 = -
dx
1 {el = e 4- =
AF
d,co l
i
...
-t
=el
+
4
1l-=
,
i
aq d'.I I
Fig. E14-5
X.
.
(b)
dxI
k =
-
2
M
d 2
El
djI
2
L
= 0
vI, v2, co prescribed at x
(h/L) 2 ((1
N = F
N2 =
= Ns1
dM
-- - + axlF, =
dx=
(c)
at x = L
M =0
Integrating (a) and using the boundary conditions at xt = L, we obtain
F.
_
AT
I(2
_
XI)2
-2)(d
P2(L
2
F2 = P2(L - x)
a
NBI
2
(d)
y)
(L2
X1
- aN,3
We suppose e ° = k ° = 0 to simplify the discussion. Integrating the moment-curvature
relation,
d2lV2
EI-M=-(L
WT =
P2 (LX a2 I
-x)
2
- )
-
I
2
2
(L2 - x)
(e)
458
PLANAR DEFORMATION OF A PLANAR MEMBER
where et, k0 , 42, and 7 are defined by (14-24) for the stress-expansion approach
and (14-42) for the displacement-expansion approach.
and noting that v2 = dv2 /dxl = 0 at x1 = 0 lead to the solution for v2 ,
P2
(I
L
2
4
1
3
L2X2 _ 1 X4
alB
(f)
The axial displacement is determined by integrating the extensional strain displacement
relation,
dV2
Fl
Avl
dxl
AE
dxl
(g)
ii~L~~~~~~~~~~~
ap2 (
23I1
1L5x~
Lx}I
+
R
+
+
I
o
Fl
e, = e + -
F2
k = k
(h'
2/L
[(=
I(¶6L)
(ia)
Now,
5
2h/L 2
a -
(i)
L.
.a2()A 2 ()2
(j)
3)
3p)
3
and we see that this term dominates when hzis larger with respect to the cross-sectional
depth.
FORCE METHOD OF SOLUTION
+ M--
EI
where A2 , e, k are the same as for a prismatic member.
When the member is not shallow, it is reasonable to neglect stretching and
transverse shear deformation. As shown in Example 14-3, this approximation
2
introduces a percentage error of 0(c/R) . Formally, one sets A = 42 = o.
If the member is shallow, we can still neglect transverse shear deformation
but we must include stretching deformation.
The basic steps involved in applying the force method to a curved member
are the same as for the prismatic case. H-Iowever, the algebra is usually more
complicated, due to the geometry. We will discuss first the determination of
the displacement at a point.
To determine the displacement at Q in the direction defined by to, we apply
7
an external virtual force, APQ Q, generate a statically determinate system of
to APQ,
corresponding
internal forces and reactions
AFj = Fj,Q APQ
AM = M QAPQ
ARk = Rk, Q APQ
Our starting point is the principle of virtual forces restricted to planar
deformation,
Js(el AF1 + e2 AF 2 + k AM)dS -
Jk ARk = E cli APi
(14-65)
where the virtual-force system is statically permissible, lk,represents a support
movement, and ARk is the corresponding reaction increment. The relations
between the deformation measures (el, e2, k) and the internal forces (F,, F2, M)
depend on the material properties and on whether one employs stress or displacement expansions. This discussion is limited to a linearly elastic material
but one should note that (14--65) is valid for arbitrary material. For convenience, we list the force-deformation relations below. The notation for
internal force quantities is shown in Fig. 14-3.
F=
M
F2
G2
GA 2
=
1,2)
(14-68)
and substitute in (14-65):
(14-69)
(IQ= s(elF,Q + e 2 F2 , C2 + kM Q)dS - YR,, Qk,,
This expression is valid for an arbitrary material. We set e2 = 0 if transverse
shear deformation is negligible and el = e if stretching deformation is
negligible.
Example 14-6
suppose the member is not shallow and neglect stretchingand transerseshear deformation.
A+ ER
+ A_
AE
el =
(i
We consider the thin linearly elastic circular segment shown in Fig. E14-6A. We
Arbitrary Linearly Elastic fember
e2
(14-67)
ii
°
A- {I +
14-7.
Thin Linearly Elastic Member
X1
We express the last term in (g)as
Then
459
FORCE METHOD OF SOLUTION
SEC. 14-7.
CHAP. 14
M
M
F,
0
+ 71±R +f7
k = k<
A EI
The reactions are the end forces at A for this example, and (14-69) expands to
(14-66)
dQ
=(e°t,,
Q + (k
+ -)
M Q dS +
IA1FA1.Q + tIA2FA2.Q
In what follows, we illustrate the application of (a)
+
AMA,Q
(a)
460
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
SEC. 14-7.
FORCE METHOD OF SOLUTION
461
Fig. E14-6A
Fi.
E14-6B
17_
I
I
i
i
i
F
r
o
U2
F2
,WB
t
611tl
Expressionsfr Displacemnents at B
To determine UB, we take APQ = AFB,.
The internal virtual-force system corresponds
to FBI = + 1. ,It is convenient to work with q -=0 B - 0 as the independent variable
rather than 0.
The force-influence coefficients (F1.Q, F2 Q.,M Q)follow directly from Fig. E14-6B:
F1,Q = FllAP =+ = F, I,=+ = cos 11
F , = sin n
M,Q = R(1 - cos )
(b)
Substituting (b) in (a) results in the following general expression for u,,:
UB = R
I'"I(
e ° cos
+ R k° +
JO '
+UAI
COS OB +
\
tiA2
M
Fig. E14-6C
) (1 - cos )di;
/
sin OB + )OAR( - cOS OB)
(C)
Once the loading is specified, we can evaluate the integral. Terms involving the support
displacements define the rigid body displacement at B.
Taking APQ = AFB2, AMB leads to expression for UB2 and con. We list them below
for future reference:
UB2 = R
{-e'°sin
B = CA + R
(k
+R(k
+)
+
'
sln'}d
l-7AlsinOB+T7A2cos0
OlARsinOB
(d)
Solution for a ConcentratedLoading at an Arbitrary Interior Point
We consider an arbitrary force vector, Pc, and moment, M, applied at point C as shown
in Fig. E14-6C.
A
~~j_______m___~~~~~·
~ ~~
·
TF111
PLANAR DEFORMATION OF A PLANAR MEMBER
462
Pc = Pcitl
CHAP. 14
+ PC2t2
The expressions for the displacements at B due to an external loading are obtained by
specializing (c) and (d) for no initial deformation or support movement and noting that
FORCE METHOD OF SOLUTION
463
terms of the applied loads and the force redundants:
(e)
MC = Mct3
M = RPcl[1 - cos(/
SEC. 14-7.
F1 = F,
.
+ E Fl,kZk
i
I
F2 = F2 , 0 +
F2.kZ,
k=1
M = 0
r/ < Ic
- tic)] + RPc 2 sin(7 - r/c) + Mc
,
qi > i'c
t)
(14-70)
I
I
f
M = M,
+
M,kZ
k=1
i
The solution for constant I is
Ri
Ri, o +
+
R.
kZk
k=l
PUB1
CR3
0c - sin
EI
(
-
2
-cos Oc
+
in 0B + sin r1 + 5- cos tic +
s
ill CCOS
0B
Substituting the virtual force system corresponding to AZj (which is statically
permissible and self-equilibrating) in (14-65) and letting j range from 1 to r
lead to the compatibility equations relating the actual deformations:
2
+ 2- sill c - - sin 0c sin
OH)
Js(eFl
+ _CR- (Oc + sin '1c -- sin OB)
EI
PclR3
uLB2=
-- I -cos
j + e 2 F2 j + kM,i)dS -
= 0
(a)
i =
sinc csin-
OB + cos
TiRi
sin Oc sin
(g)
?,..,
7
When the material is linearly elastic, the compatibility equations take the
form
ElR3 (2
PC
(2
1 2S
2
R2Mc
+
(cos /c - cos OB)
COB=
R2pc_
EI
2
(
ik =fkj
RMc
R P
-_ sin Oc) + ---- 2(1 - cos c) +--EI
El
If we take point C to coincide with B, qic = 0 and Oc = 0B The resulting equations
relate the displacement at B due to forces applied at B in the directions of the local frame
at B and can, be interpreted as member force-deformation relations. It is convenient to
express these relations in matrix form:
qt
=
,%B
(h)
UBI
R
EI
k=1
where
I
M4
We call f the member flexibility matrix.
FJFk
+
Aj =
.(jkZ
+
( = ,...,
A- (FjM
r)
(14--71)
k + F, kM j)
MM k)IS
2 ,F 2 k + ,--F
hiRi, j + E
= Aj
(Fi e + Mio
+
(Fl,jM, o + F,OM.j) +
E Fl. jF, o
-F2.F2.i
-
M,
)dS
We set 7 =,
2 = A2 , and 1/AR =
for a thin member.
Note that Ijk is the displacement of the primary structure in the direction of
Zj due to a unit value of Zk. Also, Aj is the actual displacement of the point of
application of Z
inius the displacement of the primary structure in the direc­
tion of Zj due to support movement, initial deformation, and the prescribed
external forces.
Example 14-7
We describe next the application of the principle of virtual forces in the
analysis of a statically indeterminate planar member. Let the member be in­
determinate to the rth degree and let Z1 , . . ., Z,. represent the force redundants.
Using the equilibrium equations, we express the internal forces and reactions in
Consider the symmetrical closed ring shown in Fig. E14-7. From symmetry,
F,=
F;2 ~
at = 0
(a)
--�--- X--�av;r�
I
PLANAR DEFORMATION OF A PLANAR MEMBER
464
CHAP. 14
SEC. 14-7
We take the moment at 0 = 0 as the force redundant. To simplify the algebra, we suppose
the member is thin and neglect stretching and shear deformation. The compatibility equations reduces to
FORCE METHOD OF SOLUTION
M = M, of
filZ = A1
fl = Ei M2 dS
A =
465
Because of symmetry, we need to integrate over only a quarter of the ring. Finally, the
total moment is
Z1, I
=PR (' '
cos0a
2
(e)
The axial and shear force variations are given by
(b)
F1
-- M,oM, dS
f,.fEl
P
-Y cos 0
(f)
P
F2 = ---2 sin 0
Note that fiI is the relative rotation (C() due to a unit value of ZI and A, is the relative
rotation () due to the applied load. Equation (b) states that the net relative rotation
must vanish.
When the equation defining the centroidal axis is expressed in the form
f (xl), it is more convenient to work with force and displacement quantities referred to the basic frame rather than to the local frame, i.e., to use the
Cartesian formulation developed in Sec. (14--5). The cartesian notation is sum­
marized in Fig. 14-16.
X2
Fig. E14-7
SK-4!
"IX
I
1,
ii
P
1)}
i
1, 1
.S
I
il
Now,
MI1 = M'z,=l = +1
P
M,0 =-R(1 - cos 0)
2
I
I
I
I
2
We consider I to be constant. Then, (b) reduces to
/2
7
J
-
-fMo, dS
-
M
M,
P
1
I
dS
-
(l 7z/2
2
__
Fig. 14-16. Notation for Cartesian formulation.
(c)
72
no in
cos tjdlt f(PR\
tan 0 = d
dx
­
i
- I-
2
The geometrical quantities and relations between the internal force components are
dS
(d)
dx
cos 0
F = N 1 cos 0 + N 2 sin 0
F2 = -N 1 sin 0 + N 2 cos 0
(14-72)
466
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
SEC. 14-7.
We first find N, N 2 and then determine F1 , F2. To obtain the equations for
the cartesian case, we just have to replace dS by dx,/cos 0 in the general ex­
pressions ((14-69) and (14-71)). In what follows, we suppose the member is
thin and linearly elastic.
When the member is not shallow, we can neglect the stretching and transverse
shear deformation terms. The equations for this case reduce to:
FORCE METHOD OF SOLUTION
467
CompatibilityEquation
YfJkZk = j
k
fjk =
A =
Ni. jNl,k + f
|
E IR,j
M,
x1
(14-78)
EI-l')
A
e° +
NkI/ j + k° + EI°) M Uj ax,
-
Displacement at PointQ
Example 14-8
+
(eFI,Q + (ko
dQ =
M )
M)
-os /
E Ri.Q
(14-73)
Consider the two-hinged arch shown in Fig. E14-8A. We work with reaction com­
ponents referred to the basic frame and take the horizontal reaction at B as the force
redundant.
CompatibilityEquations
fjkZ
=
Fig. E14-8A
X2,1;2
k
) dxI
;M k cos 0
(14­74)
F ,
Aj = j Ri,jai - Je
+
(ko +
Mj cos
b_--
We can express the terms involving F1 () in terms of N.1 () and N 2 () since
F1() = cos 0 N1 () + sil 0 N 2,()
B
i
(a)
Then,
I
A
I, )]O
0I,•
0
cos 11<[l
1
|
[N,
1
el
+f
N2x() ,1X1
(14-75)
One must generally resort to numerical integration in order to evaluate the
integrals, due to the presence of the term 1/cos 0.
When the member is shallow, 02 << 1, and we can approximate (14-72) with
cos 0
1
sin 0
tan 0 =
dl
F,1
N 1 + f'N
F2
-f'N
(14-76)
1
_
_.
-L,.
Primary Structure
We must carry out two force analyses on the primary structure (Fig. E14-8B), one for
the external forces (condition Z1 = 0) and the other for Z = 1. The results are displayed
in Figs. E14-8C and D, respectively.
If'
ds
L
X2,
X
v2
2, v2
Fig. E14-8B
2
+ N2
We cannot neglect the stretching deformation term in this case. However, it is
reasonable to take F1,
N1 . We also introduced this assumption in the devel­
opment of Marguerre's equations. The equations for the shallow case with
negligible transverse shear deformation and F, , N1 have the forms listed
below:
Displacement at PointQ
dQ= J[ye
°
E
NY,
,
X 01
+
El
M, Q dxu-
Rdi
(14-77)
--
_
cB
r;--�iC�L
-I"
p
I
CHAP. 14
PLANAR DEFORMATION OF A PLANAR MEMBER
468
Fig. E14-8C
SEC. 14-7.
FORCE METHOD OF SOLUTION
Using the results listed above, the various terms in (a) expand to
N2
-)M
-D
469
di =
ZRi,
Nj
-VA,
'
LX
-
I[
.fl=fl
+ BI + -B U2Z
VA2
L
cos
L
1 + f'N2. 1) + k Cos 0] dxl
:[e?(Ni.
f",-
]Ldx,
k (f
h+1Pk° f
±
Icos
E1 (0 +-L-Jx)
,
JXoEIcos
-I~';,:)
Om,
(b)
cjs
(--I- e.,:) ax,
(
ELcos
(+
L-
x-
Once the integrals are evaluated, we can determine Z, from
Fig. E14-8D
1
.
(c)
Finally, the total forces are obtained by superposition of the two loadings:
It.ib
Ih
L
Nj = Nj,
+ ZINj,1
j = 1,2
Mo
M
o
Ri Ri,
+ ZtM 1
+ Z1 Ri,
i = 1, 2, 3
R4
(d)
Z
To evaluate the vertical displacement at point Q, we apply a unit vertical load at Q on
the primary structure and determine the required internal forces and reactions plotted in
Fig. E14-8E.
s
Nl
A1
=
Z
1
Fig. E14-8E
/2
(+)
xQ
IVxQ
tQtXQII
N2 ,1
I
LQI
L
1
LQ2
i
I
i
I N2, Q
(+)
I ---xo 1IL
(-)
Compatibility Equation
We suppose the member is not shallow. The compatibility equations for Z1 follow
from (14-74):
filZ,-Al
I
2 dxt
(a)
E =R, cos 0
A1 = ZRi, 1'-
fe'(N,.,
MQ
\I~1
f'N2,I)(k
-
M-
+H-
P
M
=I1
I
-PQ
_=
7n
~i
OCftIATII0hl
AlNIAD
,
-
4IU
rL/AI~n5
(r
I IIUI
VLSr~nlVI
A
DI A^IAD
/IiNAADrD
~
ur n rilur\e aenriviur
IA
r'UAD
SEC. 14-7.
EC.
, 14-7.
Applying (14-73), we obtain
Vr2 =
471
A2 + (B2
XA2)
Q
1
-
+ fIe°f' Idx 1
xl
-
Jxl +
-'V
~"t ,'~
J
d)
E)l cos
Q
(ko + M dx
E-l cos 0
JX
x
L J
A numerical procedure for evaluating these integrals is described in the next section.
l
FEamnl
Th
14-9
Or mc,.
· l.oyJt.tlllx.,I.
uIr
n
~r]
nc,
n
l;
ItxlltVItVVV
VVu-llll~3~
.VCtJIc
?rh
-1
.ll ~
ch~crt;
........ in~r T,, C71A-Q
,
CttIXt
ot1./VY t L11I
It?
a
;r -A-,
-r
.
l
Al.-r
o
­
jected to a uniform load per unit horizointal length, that is, per unit x t . The equation for
the centroidal axis is
x2
4h /
x2
-L X=
:
(a)
where h is the elevation at mid-span (x1 = L/2). We take the horizontal reaction at the
right end as the force redundant and consider only bending deformation. Figures E14-9B
and C carry through an analysis parallel to that of the preceding example.
tructure
Determzination ofZ and Total Forces
=
The equation for Z. follows from (14-74):
M
A.
A
Z
o
cdx
,
1
.
FTI
m
d3
q
rnq0
f
L--
dx=
)
-
(b)
ELcosFig.
Nnt,,
tl,.t 4h,; T-lt
;C
-,V1.. .tl~[ tllo
...-/- ,tW;?.t
-1;A
vctt
Gt,,
- ar"r
cll
Ct.Vlt.,v
t-1
,,,;,;
vc.lJ
N, = N 1o,0 + ZNI,
N 2 = N2
M =M
CO.i
L
0111U, 1
f
+
_
=
,1
t L .lA.....A
, Lilt; UCIUIIIICU
0
,r
+ ZiN2
+ZM,
1
pL
1 illc1av
1.,1.
t
itntl G",-p
I-
tut
-l~
,,,
C,,.
2
2
(c)
(c)
=0
.+1,th t+..
W.{
VILII Lll
ih-ltii
IilLldtl
h+dp1. w..t...
b11ilcI
Wll
;o(('
El )
X1
.. 1
U(.LIt
deformation is neglected. It follows that (c) also apply for the fixed lnonshallowv case.
When the arch is shallow, the effect of axial deformation cannot be neglected. The
expression for Z, follows from (14-78):
Z-
I
N2 1
= p (x1
" Call,, fl-
=
;
.I l-h ^
,aL., I tl
I.
Otl111 LUIIIUIIlU5
bil:
E14-9B
FI
i
N1, 0 =or
I
N1 oN,,
=
_
-JoN
+ -
+ 1
M oM,
iV2.0 =P(xs -t
M
p(
6dx
Force System Due to p
(d)
i
2)
PL2
'!
~~~~~~~~~~~~~~~.
C II%-V
LeJf dxl
x (k° +KElI cos 0
+
I~;,~
'
r
_L
-o
FORCE METHOD OF SOLUTION
472
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
SEC. 14-8.
NLIMFI(,Ai (11\1T1=rALI
If E is constant, (d) reduces to
2
pL2
8h
LIAnr1P,·11
tvikUttUUFlS
473
Ii,un
.........
..
._....
fjIpLI
It is of interest to determine the rotation at B. The "Q" loading consists ofa unit moment
applied at B to the primary structure (see Fig. E14--9D). Applying (14-77) (note that
dx1
iIf2
Id
- d.x +
pL2
1
8h 1+6
1fl2 d
I f x
Fig. E14-9D
$1M
(e)
JL
fA- dx
fo 1'
,Q = I
dx
The parameter 8 is a measure of the influence of axial deformation. As an illustration, we
consider A and I to be constant and evaluate (5for this geometry. The result is
1
5=15
8 Ah 2
15t p\2
8
.
I,COB
Xl
B
L
L:
(f)
L
where p is the radius of gyration for the cross section.
PQ =+1
Fig. E14-9C
the stretching terms vanish since N 1 ,Q = ), we obtain
,(:
=
:
P'
iLM.
o EI L
22
When EI is constant, (i) reduces to
1
1
N2 ,1 = 0
J
=
14-8.
One should note that (e) applies only for the shallow case, i.e., for (f') 2 <<1. Now,
2Ž)1
(g)
For the assumption of shallowness to be valid, 16(h/L)2 must be small with respect
to
unity. The total forces for the shallow case are
N, =
pL 2
l
N2 = P
h
(i)
El
1
NUMERICAL INTEGRATION PROCEDURES
One of the steps in the force method involves evaluating certain integrals
which depend on the member geometry and the cross-sectional properties.
Closed-form solutions can be obtained for only simple geometries, and one
usually must resort to a numerical integration procedure. In what follows, we
describe two procedurest which can be conveniently automated and illustrate
their application in deflection computations.
We consider the problem of evaluating
1+ 6
(h)
x--
J
f(x) dx
I
X'-a
8h
\(1+8-
(j)
f
Force System Due to Z1 = +1
I'9=-(1
+
Since cow¢ 0, the results for the fixed end shallow case will differ slightly
from (h).
0
N1, = +1
I
dX,
¢B = 5 _24_r
Xl
0
)_
= 2 (
L
\,1 + *)
t See Ref. 8 for a more detailed treatment of numerical integration schemes.
(14-79)
PLANAR DEFORMATION OF A PLANAR MEMBER
474
CHAP. 14
where f(x) is a reasonably smooth function in the interval xA < x < xB. We
divide the total interval into n equal segments, of length h:
- XB
ho=
-
XA
SEC. 14-8.
NUMERICAL INTEGRATION PROCEDURES
f(x)
' -
15fk
J,, =33.fo + f, + 4(,
f,
AJk, k+l
(14-80)
Iff(x) is discontinuous, we work with subintervals and use a different spacing
for each subinterval. For convenience, we let x0, x , ... , x, represent the
coordinates of the equally spaced points on the x axis, and fo, J,... J the
corresponding values of the function. This notation is shown in Figure 14-17.
475
be determined using
ft d
-
+
8
k+
Jk+2]
(14-84)
Finally, one can express J as
+.f
+
+ ;, -
(14-85)
+ 2(f2 +f + .. + . -2)j
Equation (14-85) is called Simpson's rule.
f
fl
A
A
I
xo
--
f2
B
h
h
XI
I;?
X2
XnI
- 1
x
X?
Fig. 14-17. Coordinate discretization for numerical integration.
The simplest approach consists in approximating the actual curve by a set
of straight lines connecting (Jo, f), (i.f ,), etc., as shown in Fig. 14-18. With this
approximation,
AJk-I,
k
=
Jk =
k-
|xf
xo
f(x)dx
dx
=
h
2 (fk
- +.1
)
-k
.s.
Xk+ l
Fig. 14-18. Linear approximation.
(14-81)
Vi1
k-1 + AJk- , k
2)
If only the total integral is desired, we use,
J, =
(14-82)
f(x)dx = AJ,-1, i It (Jo + f 1) +
fi
o
i=t
i=l
which is called the trapezoidalrule.
A more accurate formula is obtained by approximating the curve connecting
three consecutive points with a second-degree polynomial, as shown in Fig.
14-19. This leads to
rk+
AJk k+2
=
2
fdx
[J +
4
fk+ + fk+2]
(14-83)
Xk
Jk+2 = Jk + AJk. k+ 2
To apply (14-83), we must take an even number of segments, that is, n must
be an even integer. If the values of J at odd points are also desired, they can
Ak
'
I--
Xk
l
Yk +2
1n
Fig. 14-19. Parabolic approximation.
--- I-IY"-·-UUll"l�i-----I�
PLANAR DEFORMATION OF A PLANAR MEMBER
476
CHAP. 14
SEC. 14-8.
NUMERICAL INTEGRATION PROCEDURES
477
To evaluate (b), we divide the total length into n equal segments of length h, number
the points from 0 to n, and let
Example 14-10
Consider the problem of determining the vertical displacement at Q for the straight
member of Fig. E14-10. We suppose shear deformation is negligible. The deflection
due
Hk ==
Fig. E14-10
Hn, + k Xk.Jk
(d)
If, in determining Jy, Hn, we also evaluate the integrals at the interior points, then we can
readily determine the displacement distribution using (d).
7=XQ
.f
- L
dk = Xk
I Q
.XQ
(c)
x -E dx
With this notation, (b) takes the form
dQ
..-
E
Example 14-11
,
(fl
El
Consider the simply supported nonshallow arch shown. We suppose there is some
distribution of M and we want to determine the vertical deflection at Q. Considering
M
Fig. E14-11
dQ
P2 = +1
lI
1
a
{lL
t_-
t
A'
L
Q(M,-
_ _
_
_
_ _
)
...
__/
...
'.L
M,Q
t
) m
Al
to bending deformation (we consider the material to be linear elastic) is given by
dQ =
- MQ dx
(a)
where M is the actual moment and MQ is due to the "" loading. Substituting for M Q,
(a) expands to
deX
":
LM
I- dx
de,ET =
E(
J-
M
f
d+
o E
"a\Jo
,QEl
)
t
M
xdxE
Q
TL(f
J
L
M
x
x7[
E
Ed
(b)
only bending deformation, dCis given by
de=Q
M, Qds =
,--o--
)
M, Q dx
(a)
478
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
PROBLEMS
Now, the distribution of M, Qis the same as for the straight member. Then, the procedure
followed in Example 14-10 is also applicable here. We just have to replace M/EI with
M/EI cos 0 in Equation (c) of Example 14-10.
479
Prob. 14-2
2t
REFERENCES
T
1. TIMOSHENKO, S. J.: Advanced Strength of Materials,Van Nostrand, New York, 1941.
2. BORG. S. F., and J. J. GENNARO: Advanced Sructural Analysis, Van Nostrand,
New York, 1959.
3. REISSNER, E.: "Variational Considerations for Elastic Beams and Shells," J. Eng.
Mech. Div., A.S.C.E, Vol. 88, No. EM 1, February 1962.
4.
b - 0.75d
t = d/20
d
_L_
MARTIN, H. C.: Introduction to MatrixMethods of StructuralAnalbsis, McGraw-Hill,
New York, 1966.
5.
V
MUSHTARI, K. M., and K. Z. GALIMOV: "Nonlinear Theory of Thin Elastic Shells,"
Israel Program for Scientific Translations, Jerusalem, 1962.
6. MARGUERRE, K.: "Zur Theorie der gekriimmten Platte grosser Formanderung,"
Proc. 5th Int. CongressAppl. Mech. pp. 93-101. 1938.
7. ODEN, J. T.: Mechanics of Elastic Structures, McGraw-Hill, New York, 1967.
8. HILDEBRAND, F. J.: Inlroduction to Numerical Anallsis, McGraw-Hill, New York,
1956.
14-6.
Evaluate I' and I" for the symmetrical section shown.
Prob. 14-6
t
__L
PROBLEMS
14-1. Specialize (14-7) for the case where 'v, = x. Let x 2 ,=f(x1) and
let 0 be the angle from X to Y, as shown below. Evaluate the various terms
for a parabola
= 0.75d
d
-
_L
f = Clxl + C2 x1
Finally, specialize the relations for a shallow curve, i.e., where 02 << 1.
Prob. 14-1
X2
}2
y1
d/20
I'
6---b
b
Tt
14-7. Consider a circular sandwich member comprised of three layers,
as shown. The core layer is soft (E << 0), and the face thickness is small in
comparison to the depth (,
d). Establish force-deformation relations based
on strain expansions (see (14-37)).
Prob. 14-7
f(xl )
t
1' I
- I
14-2. Evaluate I* and (see Equation 14---24) for the section defined by
the sketch.
14-3. Verify (14-34).
14-4. Verify (14-41) and (14-42).
14-5. Discuss the difference between the deformation-force relations based
on stress and displacement expansions (Equations (14-25) and (14-42)).
Illustrate for the rectangular section treated in Example 14-1. Which set of
relations would you employ?
Core
-f­
d,
T
14-8. Starting with (14-34) and (14-35), derive a set of nonlinear strain
displacement relations for a thin member. Assume small finite rotation, and
PLANAR DEFORMATION OF A PLANAR MEMBER
480
(d)
linearize the expressions with respect to Y2, i.e., take
y2 k
-
e
E1 =
Derive the nonlinear deformation-displacement and equilibrium equa­
tions for the cartesian formulation. Refer the translations and loading
to the basic frame, i.e., take
U2= Vt11 +
e2
}'12 ;
Determine the corresponding force-equilibrium equations with the principle of
virtual displacements.
14-9. Refer to Fig. 14-10 and Equation (14-31). If we neglect transverse
shear deformation, t2 is orthogonal to t'1 and we can write
-
' dy
Prob. 14-10
(a)
[l3T2
+
A
1.+ eIl ,
R'
dt2
dS-
1+ e ,
RF' 2
h2 = const
= Pfltl + fl2t2
t 2 = -2tl
dt'l
d-S-
22
+ P2 1 2
p
P P=
Specialize the equations for the case of a shallow member.
14-10. The accompanying sketch applies to both phases of this problem.
1 (1 + gl)a 2 = ( I _/)OC
t'
481
PROBLEMS
CHAP. 14
M
- 0
+_i -l- = (1 -t- el)a
ct
' =a
st
.
(a)
'L
Verify that &ccan be expressed as
E=1 1- y2/Ri
Il
R'
- Y2
(b)
(a)
1
1- y
re,
Also determine el and R' for small strainr.
initial tangent vectors,
=
and take y - S (i.e.,
(b)
Ult
+
Express ii in terms of the
-
t2 t 2
= I).
Derive the force-equilibrium equations, starting with the vector equa­
tions (see (14-12) and Fig. 14-4),
dF++
dS
dM+ +
Determine the complete solution for the circular member shown.
Utilize symmetry at point A (ul = o = F2 = 0) and work with (14-58),
(14-59). Discuss the effect of neglecting extensional and shear de­
formation, i.e., setting (1/A) = (l/A 2 ) - 0.
(b) Repeat (a), using Mushtari's equations for a thin member with no
transverse shear deformation, which are developed in Example 14-2.
Show that Mushtari's approximation ( << d2dO) is valid when the
segment is shallow.
14-11. The sketch presents the information relevant to the problem:
Prob. 14-11
P2 = const
=0
I
I
+ t' x F+ =
XI
_
-
Ih
2=
i_
and expanding the force vectors in terms of components referred to
the deformed frame:
h= bt't + b2t2
F+ = Fit'l + F2 t 2
M+
(c)
Mt3
Assume small strain.
Derive the force-equilibrium equations with the principle of virtual
displacements. Take the strain distribution according to Equation (b).
I
A
ax
B
I
14
r
/,-_
I
= 21h
e
F .
:B
X2
_
L
L
MEMBER
PLANAR DEFORMATION OF A PLANAR
AOQ
40h
483
PROBLEMS
CHAP. 14
14-15.
parabolic arch
(a) Apply the cartesian formulation to the symmetrical
shear
transverse
neglect
and
shown. Consider the member to be thin
deformation.
(set 1/A = 0).
(b) Specialize (a) for negligible extensional deformation
the validity of
investigate
and
case
(c) Specialize (a) for the shallow
Marquerre's approximation.
due to a uniform distributed
14-12. Refer to Example 14-6. Determine ulB2
loading, b2 = constant.
(see sketch). Consider
14-13. Determine the displacement measures at B
to integrate the
convenient
more
be
may
It
Note:
deformation.
only bending
(14--69).
governing equations rather than apply
(a)
Refer to Example 14-7,
in Fig. E14-7.
Determine the radial displacement at B defined
shown.
(b) Determine the force solution for the loading
Prob. 14-15
P
Prob. 14-a3
Thin ci
P
14-16.
The sketch defines a thin parabolic two-hinged arch.
b
Prob. 14-16
X
14-14.
information sketched:
Solve two problems with the
Prob. 14-14
A
A
F1 -*)
f =I
I - I,
I
1~1
/cosa
Determine the horizontal reaction at B due to the concentrated load.
Consider the arch to be nonshallow.
for a distributed loading
(b) Utilize the results of(a) to obtain the solution
p2 (x) per unit x 1 .
temperature increase,
(c) Determine the reactions resulting from a uniform
AT.
by a prismatic member
(d) Suppose the horizontal support at B is replaced are frictionless hinges.
extending from A to B. Assume the connections
Repeat parts a and c.
(a)
displacement at point B
(a) Determine the fixed end forces and radial
deformation and utilize
bending
only
with the force method. Consider
B.
symmetry at
force.
(b) Generalize for an arbitrarily located radial
Icos
484
14-17.
PLANAR DEFORMATION OF A PLANAR MEMBER
CHAP. 14
Consider the arbitrary two-hinged arch shown. Discuss how you
Prob. 14-17
Al
H -~~~~~~~~'~~
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
en1gineering theory. Examples illustrating the application of the displacement
and force methods are presented. Next, we outline a restrained warping for­
mulation 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
l
= t
(a)
+ sin /)b
t3 = - sin Oh + cos ib/
t 2 = COS q0fi
where
= (s). Differentiating (a) and using the Frenet equations (4-20),
we obtain
dt
dS
-K sin
K cos
(iS
-=K cos
Note
that
ais
skew-sym
0
K sin
T + dS
dSale
Note that a is skew-symmetric.
485
) ot t 34
t1
t2
(15-1)