When forces are applied to structures or machine parts, not only are stresses generated but strains and deflections are also produced. This concept has already been illustrated for the case of a straight uniform elastic bar loaded by a direct axial force in Figures 2.1 and 2.2.
The force-deflection equation (2-1) for this simple axial loading case is repeated here for convenience, as
(4-46) where
(4-47) is axial spring rate of the loaded bar, Fis applied force, and is force-induced elastic de- formation.
Torsional moment loading also produces torsional shearing strain and consequent an- gular deflection of the torsionally loaded member. The angular deflection for torsional loading of elastic members, given by (4-44), is repeated here, as
(4-48) where is angular deflection in radians, Tis applied torque, Lis bar length, Gis shear mod- ulus of elasticity, and parameter Kis a function of the cross-sectional shape, tabulated for several cases in Table 4.5. If the member is cylindrical, Kis the polar moment of inertia, J.
Bending loads produce transverse beam deflections that may be found (assuming elastic behavior) by twice integrating the differential equation of the deflection curve and using the particular boundary conditions of interest. The governing differential equation is
(4-49) whereyis transverse beam deflection at any location xalong the length of the beam, Mis applied bending moment, Eis modulus of elasticity, and Iis area moment of inertia about
d2y dx2 = -M
EI
15
u
u = TL KG
df
k = AE L F = ky = kdf
14
14See 4.6 for a more complete discussion of multiaxial Hertz contact stress distributions.
15See, for example, ref. 1, p. 139.
the neutral axis of bending. The deflection equations for several common cases are given in Table 4.1.
In all of these cases, when an elastic member is loaded slowly by a force, a torsional mo- ment, or a bending moment, to produce corresponding axial, torsional, or bending deflections or displacements, the external forces or moments do workon the body (average force or mo- ment times corresponding displacement), which is stored in the strained body as potential en- ergy of strain, usually just called strain energy. If the strain does not exceed the elastic limit, the stored strain energy can be recovered during a gradual unloading of the body.
Stored Strain Energy
If a material follows Hooke’s Law, and if deformations are small, the displacements of an elastic structure are linear functions of the external loads. Thus, if the loads increase in pro- portion to each other, the displacements increase in like proportion. To illustrate, if a ma- chine part or a structure is simultaneously loaded by external forces , the corresponding displacements may be found from linear force-displacement diagrams of the type shown in Figure 2.2, and the work done on the machine part, which is equal to the stored strain energy, is
(4-50) Forces and displacements in this context are generalized terms which include moments and corresponding angular displacements.
For example, if the cubic element of material shown in Figure 4.13 is subjected to three mutually perpendicular tensile forces , , and , the stored strain energy is
(4-51) or, since
(4-52)
(4-53) (4-54) and
(4-55) (4-56) (4-57) the strain energy expression (4-51) may be rewritten as
(4-58) U = csxex
2 + syey
2 + szez
2 ddxdydz ez = dz
dz ey = dy
dy ex = dx
dx sz = Pz dxdy sy = Py
dxdz sx = Px
dydz U = Pxdx
2 + Pydy
2 + Pzdz
2 Pz Py Px U = P1d1
2 + P2d2
2 + Á + Pidi
2 d1,d2, Á , di
P1,P2, Á , Pi
Pz
Py
Px
z
y
x
dz
dx dy
Figure 4.13
Unit cube of material subjected to mutually perpendicular forces Px, Py, and Pz.
Deflection Analysis; Common Types of Loading 163
Noting that is the volume of the cube, the stored strain energy per unit volume,u, may be written as
(4-59) Based on the concept of (4-50) and the details of (4-46), (4-48), and (4-49), strain energy expressions for various common loading conditions may be written. For exam- ple, tension or compression loading gives, for the case of constant loading and constant geometry16
(4-60) For torsion
(4-61) For pure bending
(4-62) and for direct shear
(4-63) For the case of transverse shear in a beam, expressions for strain energy due to the trans- verse shear can be developed but are complicated functions of the beam cross section.
Except for very short beams, stored strain energy due to the transverse shear is negligible compared to the stored strain energy due to the bending, given in (4-62).
Acurved beam can be treated as a straight beam, from an energy standpoint, as long as the radius of curvature of the beam is more than twice it depth. In addition the energy due to axial load and shear can be neglected if the length to depth ratio is greater than 1017. Consider a typical curved beam as shown in Figure 4.14.
The internal forces acting on an arbitary of the beam will consist of a normal (N) and shear (V) forces as well as a bending moment (M). These internal loads are functions of the applied load P, and the angle As a result, the strain energy for a curved beam is ex- pressed in a somewhat different manner from that of a straight beam. for a curved meme- ber we have the following expressions for the strain energy
Axial (4-64)
Shear U = (4-65)
3
u
0
kV2 2GAR du U =
3
u
0
N2 2EAR du u
Udir-sh = P 2 aPL
AGb = P2L 2AG Ubend = M
2 aML
EIb = M2L 2EI Utor = T
2 aTL
KGb = T2L 2KG Utens = F
2 aFL
AEb = F2L 2AE u = U
dxdydz = sxex
2 + syey
2 + szez
2 dxdydz
16If load, cross-sectional shape or size, or modulus of elasticity vary along the length of the member, the appro- priate integral expression from Table 4.7 must be written instead.
17See, for example, ref. 8. p. 165
Figure 4.14 Curved beam model showing (a) the original beam, and (b) the internal forces acting on an arbitrary section of the beam.
(a)
P R
(b)
P R V
N M
Bending (4-66)
wherekis a shear correction factor that depends on the beam cross-section (see Table 4.4) U =
3
u
0
M2 2EIR du
Example 4.9 Stored Strain Energy in a Loaded Beam
A beam is simply supported at each end and loaded by a vertical midspan load Pand a pure momentMapplied at the left support. Calculate the strain energy stored in the beam if it has an area moment of inertia for the cross section equal to Iand modulus of elasticity E.
Solution
For this case, the loading is constant and the cross section all along the beam is constant.
Therefore, from Table 4.1, superposing cases 1 and 4, the midspan deflection at (under load P) is
The end slope at the left end (where the moment is applied) is
Based on (4-50), the stored strain energy in the beam with both PandMacting is
or, substituting
where Uis the total stored strain energy in the beam.
U = P 2 aPL3
48EI + ML2 16EIb + M
2 aPL2 16EI + ML
3EIb = 1 EI aP2L3
96 + M2L
6 + PML2 16 b U = Py
2 + MuA
2 uA = PL2
16EI + ML 3EI y = PL3
48EI + M 6EI aL3
8L + 2L2 2 - 3L2
4 b = PL3
48EI + ML2 16EI
x = L>2
Castigliano’s Theorem
By writing an expression for stored strain energy in any loaded elastic structure or machine part, a simple energy methodfor calculating the displacements of points in the elastic body may be utilized. This energy method, called Castigliano’s theorem, may be stated as follows:
When any combination of external forces and/or moments act on an elastic member, the small displacements induced at any point and in any direction may be determined by computing the partial derivative of total stored strain energy in the loaded member with respect to any selected force or moment to obtain the displacement at the corresponding point and in the corresponding direction of the selected force or moment.
For example, in the case of simple tension in a uniform prismatic member, the stored strain energy is given by (4-60) as
(4-64) Utens = F2L
2AE
Deflection Analysis; Common Types of Loading 165
Differentiating partially with respect to force ,
(4-65) This expression for force-induced elastic deformation in the direction of is confirmed by (2-6).
Castigliano’s theorem may also be used to find the deflection at a point in a loaded member even if no force or moment acts there. This is done by placing a dummyforce or dummymoment, , at the point and in the direction of the deflection desired, writing the expression for total stored strain energy including the energy due to , and partially dif- ferentiating the energy expression with respect to . This procedure yields the deflection at the desired point. Since is a dummy force (or moment), it is set equal to zero in the expression to obtain the final result. Table 4.7 provides strain energy expressions and de- flection equations for common types of loading.
di
Qi
di
Qi
Qi
Qi
df F 0U
0F = FL AE = df
F
TABLE 4.7 Summary of Strain Energy Equations and Deflection Equations for Use with Castigliano’s Method Under Several Common Loading Conditions
Type of Beam Type of Load Strain Energy Equation Deflection Equation
Straight Axial
Bending
Torsion
Direct shear1
Curved2 Normal
Bending
Bending
1Transverse shear associated with bending gives a similar but more complicated function of the particular cross-sectional shape (see Figure 4.8 and associated discussion) and is usually negligible compared to the strain energy of bending.
2Reference Figure 4.14 for description of N,M, and V.
d = 3
u
0
kV10M/0P2 GA R du U =
3
u
0
kV2 2GAR du
d = 3
u
0
M10M/0P2 EI R du U =
3
u
0
M2 2EIR du
d = 3
u
0
N10N/0P2 EA R du U =
3
u
0
N2 2EAR du
d = 3
L
0
P10P>0Q2
AG dx
U = 3
L
0
P2 2AGdx
d = 3
L
0
T10T>0Q2
KG dx
U = 3
L
0
T2 2KGdx
d = 3
L
0
M10M>0Q2
EI dx
U = 3
L
0
M2 2EIdx
d = 3
L
0
P10P>0Q2
EA dx
U = 3
L
0
P2 2EAdx
A P A
A
A
A
B
B
B
B P
C I, E
C
C x
x
x
x L
2 L
L
2 L
M
M
M
MAB
MBC RL
RL
RL
QB
RR
RR
RR yB
(a)
(b)
(c)
(d)
L – x
Figure E4.10
Simply supported beam with midspan load Pand momentMat the left support.
18See, for example, ref. 1. pp. 340 ff.
Redundant reactionsin statically indeterminate structures or machine parts may also be found by applying Castigliano’s theorem.18This is accomplished by partially differentiat- ing the total strain energy stored in the loaded member with respect to the chosen redundant reaction force or moment, setting the partial derivative equal to zero (since immovable re- actions do no work and store no energy), and solving for the redundant reaction.
Example 4.10 Beam Deflection and Slope Using Castigliano’s Theorem
The simply supported beam shown in Figure E4.10(a) is subjected to a load Pat midspan and a moment Mat the left support. Using Castigliano’s theorem, find the following deflections:
a. Midspan beam deflection
b. Angular deflection of the beam at the left support c. Angular deflection of the beam at midspan Solution
a. To find the midpsan deflection , where Pis acting, the procedure will be to write an expression for total strain energy in the beam under load, U, then differentiate Upar- tially with respect to P. From Table 4.7, the deflection equation for the case of bend- ing may be seen to be
Since the beam of Figure E4.10(a) is not symmetrically loaded, the integration must be separately executed for beam section and beam section , then added. That is,
d = 1 EI c
L
L>2 0
MAB0MAB 0Q dx +
L
L
L>2
MBC0MBC 0Q dxd
BC AB
d = L
L
0
M10M>0Q2
EI dx
yB uB
uA
yB
Deflection Analysis; Common Types of Loading 167
Using the sketch of Figure E4.10(a), and taking counterclockwise moments as pos- itive, the reactions and may be found from static equilibrium requirements as
and
Next, based on the free-body diagrams of Figures E4.10(b) and (c), the moment equa- tions for sections and of the beam may be found to be
and
These may be partially differentiated with respect to midspan load to give
and
Substituting gives
which yields
b. To find the angular deflection at the left support, the approach is similar, except that MABandMBCmust be differentiated with respect to the moment Mapplied at the left support, to give
and
0MBC 0M = x
L - 1 0MAB
0M = x L - 1 uA
yB = PL3 48EI + ML2
16EI +
L
L
L>2
aPx 2 + Mx
L - PL
2 - Mb ax 2 - L
2b dxd yB = 1
EIc L
L>2 0
a- Px 2 + Mx
L - Mb a-x 2b dx 0MBC
0P = x 2 - L
2 0MAB
0P = - x 2
P MBC = Px
2 + Mx L - PL
2 - M L
2 … x … L MAB = - Px
2 + Mx
L - M 0 … x … L 2 BC
AB
RR = P 2 + M
L RL = P
2 - M L RR
RL
Substituting gives
which yields
c. To find , a dummy moment, , must be applied at midspan, in addition to the actual loads on the beam, as shown in Figure E4.10(d). Referring to Figure E4-10(d), the reactions and may be found on the basis of static equilibrium to be
and
With these reactions, the free-body diagrams of Figure E4.10(b) and (c) may again be used to write the moment equations for sections and of the beam as
and
From Table 4.7, the total stored strain energy expression, , for the loaded beam may be written as
SubstitutingMABandMBCinto this gives
+ L
L
L>2aPx 2 + Mx
L + QBx L - PL
2 - M- QBb2dxd U = 1
2EIc L
L>2 0
a- Px 2 + Mx
L + QBx
L - Mb2dx U = 1
2EIc L
L>2 0
MAB2 dx + L
L
L>2MBC2 dxd U MBC = Px
2 + Mx L + QBx
L - PL
2 - M - QB L
2 … x … L MAB = - Px
2 + Mx L + QBx
L - M 0 … x … L 2 BC AB RR = P
2 + M L + QB
L RL = P
2 - M L - QB
L RR
RL QB uB
uA = PL2 16EI + ML
3 +
L
L
L>2aPx 2 + Mx
L - PL
2 - Mb ax
L - 1b dxd uA = 1
EIc L
L>2 0
a- Px 2 + Mx
L - Mb ax
L - 1b dx Example 4.10
Continues
Deflection Analysis; Common Types of Loading 169
which yields
Differentiating partially with respect to ,
Finally, setting dummy load equal to zero, gives
This may be verified by superposing cases 1 and 4 of Table 4.1.
uB = - ML 24EI QB
uB = 0U
0QB = - ML
24EI + QBL 12EI QB
U = P2L3
96EI + PML2 16EI + M2L
6EI - MQBL 24EI + Q2BL
24EI
Example 4.11 Curved-Beam Deflection Using Castigliano’s Theorem
The curved beam in Figure E4.11A has a radius of curvature Rand a circular cross-section of radius r. It is subjected to a vertical load Pat point B.
a. Determine the horizontal displacement at Bin terms of P, R, Iand modulus of elasticity E. Consider the bending only.
b. Determine the horizontal displacement at Bas in (a) above including the normal force N.
Solution
A section cut from the curved beam is shown in Figure E4.11B.
Summing forces in the x-direction and moments about point C gives
a. Applying Castigliano’s theorem yields dHorz = 0U
0Q`
Q=0 = 0U 0M
0M 0Q`
Q=0
aMC= 0: M= -QR11 - sinu2 - PRcosu aFx = 0: N = Qsinu- Pcosu
Figure E4.11A Curved beam with vertical load at end B.
P B
A
R
P
B Q
R
R–Rsin V
N C Rsin
x
Rcos M
Figure E4.11B Free-body diagram of a segment of the curved beam.
The energy due to bending is
and we have
Thus, the horizontal deflection is
b. If the axial load is included, we have
We have
Thus
The horizontal deflection at Bdue to bending and axial load is given as dHorz = dHorzM + dHorzN= PR3
2EI - PR 2EA dHorzN = 0UN
0Q `Q = 0 = -PR 2EA 0UN
0Q = R EA3
p>2
0
1Q sin2u - Psinucosu2R du = R EAap
4Q - 1 2Pb 0N
0Q = sinu and 0UN
0Q = 3
x>2
0
N
EAsinuR du 0UN
0Q = 0UN 0N
0N 0Q UN =
3
x>2
0
N2 2EAR du dHorzM = PR3
2EI 0U
0Q`
Q=0
= PR3 EI 3
p>2
0
[cosu - sinu]du 0U
0Q = 3
p>2
0
-[QR11 - sinu2 + PRcosu][-R11 - sinu2]R du 0U
0M = 3
p>2
0
M
EIR du and 0M
0Q = -R11 - sinu2 U = Ubend =
3
p>2
0
M2 2EIR du Example 4.11
Continues
Deflection Analysis; Common Types of Loading 171
Note, that for a circular section
Therefore, we can write
We see that the term
Hence,UNcan be neglected.
aR
rb2>>1 which implies PR3
2EI >> PR 2EA PR3
2EI = 2PR Epr2aR
rb2 I = p
4r4, A = pr2
Example 4.12 Statically Indeterminate Beam–Method of Superposition
The 150 mm ⫻150 mm steel beam in Figure E4.12A has been placed over a creek bed that has a span of 6 m. The construction is such that the beam is fixed at one end and sim- ply supported at the other. If the supported end settles 20 mm, determine the reaction at the supported end due to the beam carrying a uniform load of 7 kN/m.
Solution
The problem is statically indeterminate and can be solved using the method of superposi- tion. The problem can be expressed as follows, as shown in Figure E4.12B:
The deflection due to the load R is found from Table 4.1 (case 10) and is
and for the deflection due to the uniform load w,we have from Table 4.1 (case 11)
Thus we have for the end deflection,
d = dR + dw = -RL3 3EI + wL4
8EI dw = wL4
8EI dR = RL3 3EI
Figure E4.12A
Cantilevered beam simply supported at one end.
20 mm 7 kN/m
L = 6/m
R
7 kN/m 7 kN/m
20 mm = +
L = 6/m R
R
w
Figure E4.12B Superposition model
Solving for the reaction R gives
The moment of inertia of the cross section is
andRis given as R = 3170002162
8 - 31207 * 1092142.1875 * 10-62
63 10.0202 = 13.32 kN
I = bh3 12 = h4
12 = 1504
12 = 42.1875 * 106 mm4 R = 3wL
8 - 3EI L3 d Example 4.12
Continues
Example 4.13 Statically Indeterminate Beam–Castigliano’s Method
The beam C-Bin Figure E4.13A is fixed at Cand simply supported on the cantilever A-B atB.
Both beams are steel with E⫽207 GPa. Beam A-BhasI⫽25⫻106mm4and beam C-BhasI⫽15⫻106mm4. Determine the deflection of the intermediate support B.
Solution
The free-body diagrams of the beams are shown in Figure E4.13B.
For beam I,the moment equation is
M1 = Rx - wx2 2
Figure E4.13A Statically indeterminate beams.
w = 8 kN/m
B
A C
5 m 5 m
w = 8 kN/m
LBC
LAC B
B A
C VC
VA
MC
MA
x
x R
R Beam I
BeamII Figure E4.13B
Free-body diagrams of upper (Beam I) and lower (Beam II) beams.
Deflection Analysis; Common Types of Loading 173
and the strain energy is
The deflection at Bis given as
In the same manner, we have for beam II
and the deflection is
Let’s assume that deflection upward is positive; thus, we have at Bthat
or
and
Substituting the given values gives
R =
8000(5)4 8A15 * 10-6B
53 3A25 * 10-6B +
53 3A15 * 10-6B
= 9.375 kN R =
wL4CB 8EII aL3CB
3EII
+ L3AB 3EIII
b RaL3CB
3EII + L3AB
3EIIIb = wL4CB
8EII RL3CB
3EII
- wL4CB 8EII
= - RL3AB 3EIII
dBI = dBII
= 1 EIIIaRx¿3
3 bLAB
0
= RL3AB 3EIII dBII = 0UII
0R = 1 EIII3
LAB
0
Rx¿x¿dx¿ = 1 EIII3
LAB
0
Rx¿2dx¿ UII =
3
L
0
M2
2EIIIdx = 1 2EIII3
LAB
0
(Rx¿)2dx MII = Rx¿
= 1 EIIaRx3
3 - wx4 8 bLCB
0
= RL3CB
3EII - wL4CB
8EII dBI = 0UI
0R = 1 EII3
LCB
0
aRx - wx2
2 bx dx = 1 EII3
LCB
0
aRx2 - wx3 2 bdx UI =
3
L
0
M2
2EIIdx = 1 2EII3
LCB
0
aRx - wx2 2 bdx
z
y F
F Sphere 1
Sphere 2 d1
d2
E1,v1
E2,v2 2a pmax
x Circular
contact
footprint Semielliptic
pressure distribution Figure 4.15
Two spheres in contact, loaded by force F.
19See, for example, ref. 8, pp. 581 ff., or ref. 4, p. 652.