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10.1 Bending of a Cantilever Loaded at its End
Consider a cantilever beam with rectangular cross-section of unit width. The beam
is loaded by a concentrated force P applied at its tip in the manner shown in Figure 10.1.
c
c
P
y
x
L
1
Figure 10.1: Cantilever Beam Loaded by Concentrated Load
• Elasticity Solution
The elasticity solution for this problem is again obtained using a Airy stress func-
tion [1]. In this case,
Φ = b2xy −
d4
6
xy3
(10.1)
where b2 and d4 are constants. Recalling the relation between stresses in the x−y plane and
Φ, it follows that
σ11 =
∂2
Φ
∂y2
= −d4xy , σ22 =
∂2
Φ
∂x2
= 0 , σ12 = −
∂2
Φ
∂x∂y
= −b2 +
d4
2
y2
(10.2)
The constants b2 and d4 are evaluated by first noting that the longitudinal sides
y = ±c are free from force, implying that σ12 = 0. It therefore follows that
−b2 +
d4
2
y2
= 0 ⇒ b2 =
d4
2
c2
(10.3)
1
Thus, equations (10.2) become
σ11 = −d4xy , σ22 = 0 , σ12 =
d4
2
(
y2
− c2
)
(10.4)
To evaluate d4 we note that at the loaded end the shearing forces must sum to P.
Thus, accounting for the proper sign for σ12, leads to
−
∫
A
σ12 dA = P (10.5)
Specializing this equation for the cantilever beam under consideration and integrating
the resulting expression gives
−
∫ c
−c
d4
2
(
y2
− c2
)
(1)dy = −d4
(
1
3
y3
− c2
y
)
c
0
=
2
3
d4c3
= P (10.6)
But 2c3
/3 is equal to the moment of inertia I of the cross-section, thus d4 = P/I.
The stress state in the beam is thus described by
σ11 = −
P
I
xy (10.7)
σ22 = 0 (10.8)
σ12 =
P
2I
(
y2
− c2
)
(10.9)
Remark: These expressions for stress are identical to those associated with the ele-
mentary solution given in strength of materials textbooks; i.e., with standard
Bernoulli-Euler beam theory.
Remark: The present solution is an exact one only if the shearing forces are distributed
according to the same parabolic law as assumed herein. If the distribution of
forces is different from this parabolic law but is equivalent statically, then the
above expressions for σ11 and σ12 do not represent a correct solution at the ends
2
of the beam. Away from the ends (say a distance on the order of the depth of the
beam), Saint-Venant’s principle1
assures that the solution will be satisfactory.
The strains associated with the problem are are related to the stresses through the
constitutive relations. Assuming a linear, isotropic elastic material gives
ε11 =
1
E
(σ11 − νσ22) = −
Pxy
EI
(10.10)
ε22 =
1
E
(σ22 − νσ11) =
νPxy
EI
(10.11)
γ12 =
σ12
G
=
P
2IG
(
y2
− c2
)
(10.12)
The strains are next related to the displacements through the kinematic relations.
Assuming that the displacements and displacement gradients are infinitesimal, it follows
that
ε11 =
∂u
∂x
= −
Pxy
EI
(10.13)
ε22 =
∂v
∂y
=
νPxy
EI
(10.14)
γ12 =
∂u
∂y
+
∂v
∂x
=
P
2GI
(
y2
− c2
)
(10.15)
1
Saint-Venant’s principle states that the stresses some distance from the point of appli-
cation of the load are not affected by the precise behavior of the body close to the point
of application of the load. This principle is named in recognition of the French engineer
and mechanician Adhémar Jean Claude Barré de Saint-Venant (1797-1886). The original
statement of this principle was published in French by Saint-Venant in 1855 [2]. Begin-
ning with the work of von Mises in 1945 [3], the mathematical literature gives a more
rigorous interpretation of Saint-Venant’s principle in the context of partial differential
equations.
3
The displacements u and v are obtained by suitably integrating equations (10.13) and
(10.14), subjected to appropriate boundary conditions. In particular,
u = −
Px2
y
2EI
+ f1(y) (10.16)
v =
νPxy2
2EI
+ f2(x) (10.17)
where f1(y) and f2(x) are as yet unknown functions of y and x, respectively.
Differentiating equations (10.16) and (10.17) with respect to y and x, respectively,
gives
∂u
∂y
= −
Px2
2EI
+
df1(y)
dy
(10.18)
∂v
∂x
=
νPy2
2EI
+
df2(x)
dx
(10.19)
Substituting equations (10.18) and (10.19) into equation (10.15) gives
−
Px2
2EI
+
df2(x)
dx
+
νPy2
2EI
+
df1(y)
dy
=
P
2GI
(
y2
− c2
)
(10.20)
Equation (10.20) contains terms that are functions of x only and of y only, and one
term that is independent of both x and y. It is thus re-written as
F(x) + G(y) = H (10.21)
where
F(x) = −
Px2
2EI
+
df2(x)
dx
(10.22)
G(y) =
νPy2
2EI
−
Py2
2GI
+
df1(y)
dy
(10.23)
H = −
Pc2
2GI
(10.24)
4
The form of equation (10.21) means that F(x) must be some constant C1 and G(y)
must be some constant C2. Equations (10.22) and (10.23) are thus re-written as
df1(y)
dy
=
Py2
2GI
−
νPy2
2EI
+ C2 (10.25)
df2(x)
dx
=
Px2
2EI
+ C1 (10.26)
Integrating equations (10.25) and (10.26) gives
f1(y) =
Py3
6GI
−
νPy3
6EI
+ C2y + C3 (10.27)
f2(x) =
Px3
6EI
+ C1x + C4 (10.28)
where C3 and C4 are constants of integration. Substituting equations (10.27) and (10.28)
into equations (10.16) and (10.17) gives
u = −
Px2
y
2EI
+
Py3
6GI
−
νPy3
6EI
+ C2y + C3 (10.29)
v =
νPxy2
2EI
+
Px3
6EI
+ C1x + C4 (10.30)
The constants C1 to C4 are determined from equation (10.21) and from three boundary
conditions that prevent the beam from moving as a rigid body in the x − y plane. Assuming
that the centroid of the cross-section is fixed, it follows that for all boundary conditions
u = v = 0 at x = L and y = 0. It follows from equations (10.29) and (10.30) that
C3 = 0 , C4 = −
PL3
6EI
− C1L (10.31)
Equations (10.29) and (10.30) thus reduce to
u = −
Py
6EI
(
3x2
+ νy2
)
+
Py3
6GI
+ C2y (10.32)
5
v =
P
6EI
(
x3
+ 3νxy2
− L3
)
− C1(L − x) (10.33)
The general expression for the elastic curve is obtained by setting y = 0 in equa-
tion (10.33). This gives
v|y=0 =
P
6EI
(
x3
− L3
)
− C1(L − x) (10.34)
For determining the constant C1, at least three sets of boundary condition are possible
at the built-in end [1]. In particular,
• Boundary Condition 1: The first possible constraint condition assumes that an element
of the longitudinal beam axis is fixed; i.e.,
∂v
∂x
= 0 at x = L and y = 0. Differentiating
equation (10.33) and evaluating it at x = L and y = 0 gives
∂v
∂x

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cantilever_elas_P.pdf

  • 1. 10.1 Bending of a Cantilever Loaded at its End Consider a cantilever beam with rectangular cross-section of unit width. The beam is loaded by a concentrated force P applied at its tip in the manner shown in Figure 10.1. c c P y x L 1 Figure 10.1: Cantilever Beam Loaded by Concentrated Load • Elasticity Solution The elasticity solution for this problem is again obtained using a Airy stress func- tion [1]. In this case, Φ = b2xy − d4 6 xy3 (10.1) where b2 and d4 are constants. Recalling the relation between stresses in the x−y plane and Φ, it follows that σ11 = ∂2 Φ ∂y2 = −d4xy , σ22 = ∂2 Φ ∂x2 = 0 , σ12 = − ∂2 Φ ∂x∂y = −b2 + d4 2 y2 (10.2) The constants b2 and d4 are evaluated by first noting that the longitudinal sides y = ±c are free from force, implying that σ12 = 0. It therefore follows that −b2 + d4 2 y2 = 0 ⇒ b2 = d4 2 c2 (10.3) 1
  • 2. Thus, equations (10.2) become σ11 = −d4xy , σ22 = 0 , σ12 = d4 2 ( y2 − c2 ) (10.4) To evaluate d4 we note that at the loaded end the shearing forces must sum to P. Thus, accounting for the proper sign for σ12, leads to − ∫ A σ12 dA = P (10.5) Specializing this equation for the cantilever beam under consideration and integrating the resulting expression gives − ∫ c −c d4 2 ( y2 − c2 ) (1)dy = −d4 ( 1 3 y3 − c2 y )
  • 3.
  • 4.
  • 5.
  • 6. c 0 = 2 3 d4c3 = P (10.6) But 2c3 /3 is equal to the moment of inertia I of the cross-section, thus d4 = P/I. The stress state in the beam is thus described by σ11 = − P I xy (10.7) σ22 = 0 (10.8) σ12 = P 2I ( y2 − c2 ) (10.9) Remark: These expressions for stress are identical to those associated with the ele- mentary solution given in strength of materials textbooks; i.e., with standard Bernoulli-Euler beam theory. Remark: The present solution is an exact one only if the shearing forces are distributed according to the same parabolic law as assumed herein. If the distribution of forces is different from this parabolic law but is equivalent statically, then the above expressions for σ11 and σ12 do not represent a correct solution at the ends 2
  • 7. of the beam. Away from the ends (say a distance on the order of the depth of the beam), Saint-Venant’s principle1 assures that the solution will be satisfactory. The strains associated with the problem are are related to the stresses through the constitutive relations. Assuming a linear, isotropic elastic material gives ε11 = 1 E (σ11 − νσ22) = − Pxy EI (10.10) ε22 = 1 E (σ22 − νσ11) = νPxy EI (10.11) γ12 = σ12 G = P 2IG ( y2 − c2 ) (10.12) The strains are next related to the displacements through the kinematic relations. Assuming that the displacements and displacement gradients are infinitesimal, it follows that ε11 = ∂u ∂x = − Pxy EI (10.13) ε22 = ∂v ∂y = νPxy EI (10.14) γ12 = ∂u ∂y + ∂v ∂x = P 2GI ( y2 − c2 ) (10.15) 1 Saint-Venant’s principle states that the stresses some distance from the point of appli- cation of the load are not affected by the precise behavior of the body close to the point of application of the load. This principle is named in recognition of the French engineer and mechanician Adhémar Jean Claude Barré de Saint-Venant (1797-1886). The original statement of this principle was published in French by Saint-Venant in 1855 [2]. Begin- ning with the work of von Mises in 1945 [3], the mathematical literature gives a more rigorous interpretation of Saint-Venant’s principle in the context of partial differential equations. 3
  • 8. The displacements u and v are obtained by suitably integrating equations (10.13) and (10.14), subjected to appropriate boundary conditions. In particular, u = − Px2 y 2EI + f1(y) (10.16) v = νPxy2 2EI + f2(x) (10.17) where f1(y) and f2(x) are as yet unknown functions of y and x, respectively. Differentiating equations (10.16) and (10.17) with respect to y and x, respectively, gives ∂u ∂y = − Px2 2EI + df1(y) dy (10.18) ∂v ∂x = νPy2 2EI + df2(x) dx (10.19) Substituting equations (10.18) and (10.19) into equation (10.15) gives − Px2 2EI + df2(x) dx + νPy2 2EI + df1(y) dy = P 2GI ( y2 − c2 ) (10.20) Equation (10.20) contains terms that are functions of x only and of y only, and one term that is independent of both x and y. It is thus re-written as F(x) + G(y) = H (10.21) where F(x) = − Px2 2EI + df2(x) dx (10.22) G(y) = νPy2 2EI − Py2 2GI + df1(y) dy (10.23) H = − Pc2 2GI (10.24) 4
  • 9. The form of equation (10.21) means that F(x) must be some constant C1 and G(y) must be some constant C2. Equations (10.22) and (10.23) are thus re-written as df1(y) dy = Py2 2GI − νPy2 2EI + C2 (10.25) df2(x) dx = Px2 2EI + C1 (10.26) Integrating equations (10.25) and (10.26) gives f1(y) = Py3 6GI − νPy3 6EI + C2y + C3 (10.27) f2(x) = Px3 6EI + C1x + C4 (10.28) where C3 and C4 are constants of integration. Substituting equations (10.27) and (10.28) into equations (10.16) and (10.17) gives u = − Px2 y 2EI + Py3 6GI − νPy3 6EI + C2y + C3 (10.29) v = νPxy2 2EI + Px3 6EI + C1x + C4 (10.30) The constants C1 to C4 are determined from equation (10.21) and from three boundary conditions that prevent the beam from moving as a rigid body in the x − y plane. Assuming that the centroid of the cross-section is fixed, it follows that for all boundary conditions u = v = 0 at x = L and y = 0. It follows from equations (10.29) and (10.30) that C3 = 0 , C4 = − PL3 6EI − C1L (10.31) Equations (10.29) and (10.30) thus reduce to u = − Py 6EI ( 3x2 + νy2 ) + Py3 6GI + C2y (10.32) 5
  • 10. v = P 6EI ( x3 + 3νxy2 − L3 ) − C1(L − x) (10.33) The general expression for the elastic curve is obtained by setting y = 0 in equa- tion (10.33). This gives v|y=0 = P 6EI ( x3 − L3 ) − C1(L − x) (10.34) For determining the constant C1, at least three sets of boundary condition are possible at the built-in end [1]. In particular, • Boundary Condition 1: The first possible constraint condition assumes that an element of the longitudinal beam axis is fixed; i.e., ∂v ∂x = 0 at x = L and y = 0. Differentiating equation (10.33) and evaluating it at x = L and y = 0 gives ∂v ∂x
  • 11.
  • 12.
  • 13.
  • 14. x=L, y=0 = PL2 2EI + C1 = 0 ⇒ C1 = − PL2 2EI (10.35) From equation (10.21) it follows that C2 = H − C1 = P 2I ( L2 E − c2 G ) (10.36) where equation (10.24) has been used. Equations (10.32) and (10.33) thus become u = − Px2 y 2EI + Py3 6I ( 1 G − ν E ) + Py 2I ( L2 E − c2 G ) (10.37) v = P 6EI ( 3νxy2 + x3 − 3L2 x + 2L3 ) (10.38) The general equation for the elastic curve (equation 10.34) becomes v|y=0 = P 6EI ( x3 − 3L2 x + 2L3 ) (10.39) which is identical to the Bernoulli-Euler beam theory solution. The corresponding tip de- flection is 6
  • 15. v|x=0, y=0 = PL3 3EI (10.40) • Boundary Condition 2: The second possible constraint condition at the fixed end assumes that a vertical element of the cross-section is fixed; i.e., ∂u ∂y = 0 at x = L and y = 0. Differentiating equation (10.32) and evaluating it at x = L and y = 0 gives ∂u ∂y
  • 16.
  • 17.
  • 18.
  • 19. x=L, y=0 = − PL2 2EI + C2 = 0 ⇒ C2 = PL2 2EI (10.41) From equation (10.21) it follows that C1 = H − C2 = − P 2I ( L2 E + c2 G ) (10.42) where equation (10.24) has again been used. Equations (10.32) and (10.33) thus become u = Py 2EI ( L2 − x2 ) + Py3 6I ( 1 G − ν E ) (10.43) v = P 6EI ( 3νxy2 + x3 − 3L2 x + 2L3 ) + Pc2 2IG (L − x) (10.44) The corresponding equation of the elastic curve (equation 10.34) and the tip deflection are thus v|y=0 = P 6EI ( x3 − 3L2 x + 2L3 ) + Pc2 2IG (L − x) (10.45) v|x=0, y=0 = PL3 3EI + Pc2 L 2GI (10.46) The latter is seen to differ from equation (10.40) because of the presence of the shear term Pc2 L/2IG. • Boundary Condition 3: The third possible constraint condition at the fixed end assumes that three points of the boundary lie on a vertical line. As such, in addition to the common condition of u = v = 0 at x = L and y = 0, u = v = 0 at x = L and y = ±c. 7
  • 20. Evaluating equation (10.32) at x = L and y = c gives u = − Pc 6EI ( 3L2 + νc2 ) + Pc3 6GI + C2c = 0 (10.47) which leads to C2 = Pc 6EI ( 3L2 + νc2 ) − Pc3 6GI (10.48) From equation (10.21) it follows that C1 = H − C2 = − Pc2 3EI − P 6EI ( 3L2 + νc2 ) (10.49) where equation (10.24) has again been used. Equations (10.32) and (10.33) thus become u = − Px2 y 2EI + Py3 6I ( 1 G − ν E ) + Py 6I ( 3L2 E + ν c2 E − c2 G ) (10.50) v = P 6EI ( 3νxy2 + x3 − 3L2 x + 2L3 ) + Pc2 6I (L − x) ( ν E + 2 G ) (10.51) The corresponding equation of the elastic curve is thus v|y=0 = P 6EI ( x3 − 3L2 x + 2L3 ) + Pc2 6I (L − x) ( ν E + 2 G ) (10.52) Finally, the tip deflection is given by v|x=0, y=0 = PL3 3EI + Pc2 L 6I ( ν E + 2 G ) (10.53) The tip deflection is seen to differ from equation (10.40) because of the presence of the term Pc2 L (ν/E + 2/G) /6I, which contains both bending and shear contributions. 10.1.1 Insight Into Shear Deformations To gain some insight into the role and importance of shear deformations, recall that for the cross-section in question I = 2c3 /3. Equation (10.46) is thus re-written as 8
  • 21. v|x=0, y=0 = PL3 2Ec3 + 3PL(1 + ν) 2Ec (10.54) where the relation G = E/2(1 + ν) has been used. Based on equation (10.54) we make the following observations: • For given values of P, L, E and ν, increases in c decrease the flexural contribution to v by c−3 ; the shear contribution is reduced only by c−1 . • For given values of P, c, E and ν, decreases in L decrease the flexural contribution to v by L3 ; the shear contribution is reduced only by L. The following conclusions can thus be drawn: 1. The effect of shear deformations tends to be more pronounced for short, deep beams. Transverse displacements in such beams will thus be greater than those predicted using standard Bernoulli-Euler beam theory. 2. For “usual” beams with span-to-depth ratios greater than approximately 10, shear deformations will be negligible. Use of standard Bernoulli-Euler beam theory is thus appropriate in such cases. Consider a specific example with E = 30 x 106 , ν = 0.30 and P = 1000.The following cases are next investigated: • For c = 1 and L = 40: The span-to-depth ratio is 20:1. The associated tip displacement is v|x=0, y=0 = 1.067 + 0.0026 = 1.069. The shear deformation thus comprises only approximately 0.24 % of the total tip displacement. • For c = 1 and L = 20: The span-to-depth ratio is now 10:1. The associated tip displacement is v|x=0, y=0 = 0.1333 + 0.0013 = 0.1346. The shear deformation thus comprises only approximately 1.0 % of the total tip displacement. 9
  • 22. • For c = 1 and L = 5: The span-to-depth ratio is 2.5:1. The associated tip displacement is v|x=0, y=0 = 0.00208 + 0.00033 = 0.00241. The shear deformation now comprises 13.7 % of the total tip displacement. • For c = 5 and L = 20: The span-to-depth ratio is 2:1. The associated tip displacement is v|x=0, y=0 = 0.00107 + 0.00026 = 0.00133. The shear deformation thus comprises 19.6 % of the total tip displacement. 10
  • 23. BIBLIOGRAPHY [1] Timoshenko, S. P. and J. N. Goodier, Theory of Elasticity, Third Edition. New York: McGraw-Hill Book Co (1970). [2] Timoshenko, S. P., History of Strength of Materials. New York: Dover Publications, Inc. (1983). Original publication by McGraw-Hill Book Co. (1953). 11