ch. 8 deflectionsdue tobending 8... · 2018. 4. 17. · m2794.001000(solid mechanics) professor...
TRANSCRIPT
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 1 / 28
CH. 8
DEFLECTIONS DuE TO BENDING
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 2 / 28
8.1 Introduction
i) We consider the deflections of slender members which transmit
bending moments.
ii) We shall treat statically indeterminate beams which require
simultaneous consideration of all three of the steps (2.1).
iii) We study mechanisms of plastic collapse for statically
indeterminate beams.
iv) The calculation of the deflections is very important way to analyze
statically indeterminate beams and confirm whether the deflections
exceed the maximum allowance or not.
8.2 The Moment – Curvature Relation
▶ From Ch.7
àWhen a symmetrical, linearly elastic beam element is subjected to
pure bending, as shown in Fig. 8.1, the curvature of the neutral axis
is related to the applied bending moment by the equation.
�
�= ���∆�→�
∆�
∆�=
��
��=
��
����=
��
��(8.1)
For simplification, ��� → �
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 3 / 28
▶ Simplification
i) Assume that the shear forces which necessarily accompany a
varying bending moment do not contribute significantly to the
overall deformation.
ii) Accordingly, although M� is not a constant, the expressions defined
from pure bending can be applied.
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 4 / 28
▶ Differential equations between the curvature and the deflection
1 ▷ The case of the large deflection
The slope of the neutral axis in Fig. 8.2 (a) is ��
��= ��� �
Next, differentiation with respect to arc length s gives�
�����
��� =
�
��(����)
∴ ���
�����
��= �����
��
��
→��
��=
���
�����
������� (a)
From Fig. 8.2 (b)
(��)� = (��)� + (��)�
→ ���
����= 1+ �
��
����
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 5 / 28
→ ���
����=
�
��(��/��)�(b)
&���� =��
��=
�
[��(��/��)�]�/�(c)
If substituting (b) and (c) into the (a), ��
��=
���/���
[��(��/��)�]�/�=
���
[��(��)�]�/�= � (8.2)
∴ � =���
[��(��)�]�/�=
��
��
When the slope angle � shown in Fig. 8.2 is small, then ��/�� is
small compared to unity. If we neglect (��/��)� in the
denominator of the right-hand term of (8.2), we obtain a simple
approximation for the curvature
��
��≈
���
���≈
��
��(8.3), (8.4)
à There is less than a 1 percent error involved in the approximation
(8.3) to the exact curvature expression (8.2) when � < 4.7°.
2 ▷ The case of the small deflection
The slope of the neutral axis in Fig. 8.2 (a) is ��
��= ����
When the deflection is very small,
�� ≈ ��, tan� ≈ �
→ � =��
��≈
��
��≈
�
�����
��� =
���
���(8.3)
∴ � =�
�=
��
��=
���
���=
��
��(8.4)
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 6 / 28
▶ Comment on Eq.(8.4)
� =��
��≈
���
���≈
��
��(8.4)
i) This is essentially a linear “force-deformation” or “stress-strain”
relation. �� = �����
���
ii) EI is called the flexural rigidity or the bending modulus.
iii)The sign convention of the Mb and the curvature are same as what
we discussed in chapter 3.
▶The solution of the deflection-curvature
i) Integration of the moment-curvature relation
ii) Method of the singularity functions
iii) Moment-area method
iv) Superposition technique
v) Load-deflection differential equation
vi) Elastic energy method
8.3 Integration of the moment-curvature relation
▶Differential equation of deflection-curvature in case of the linear elastic materials and very small deflection.
�����
���= ��,��
���
���= −�,��
���
���= �
��
��= �� → ����� = ��,���
��� = −�,������� = �
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 7 / 28
▶ Example 8.1
Determine the deflection curvature of the deformed neutral axis in simple
beam shown in Fig. 8.3
Sol)
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 8 / 28
à Using the singularity functions and bracket notation introduced in Sec
3.6, we can write a single expression for the bending moment M,
directly from the free body of Fig. 8 4b.
����� = �� =��
�� −�〈� − �〉� (a)
���� =��
�
��
�−�
〈���〉�
�+ �� (b)
��� =��
�
��
�−�
〈���〉�
�+ ��� + �� (c)
B.C.) �(0) = �(�) = 0
�(0) = 0; 0 = ��
�(�) = 0; 0 =��
���� −
���
�+ ���
→ �� =��
��(�� − ��),�� = 0
∴ From eq.(c)
� = −�
������
�(�� − �� − ��) + 〈� − �〉��
(�)��� = (�′)��� = −���
����
To give some idea of order of magnitudes, let us consider the following
particular case:
� = 3.70m,� = � = 1.85m
� = 1.8kN,� = 11GN/m�,� = 3.33(10)�mm�
Then, the maximum deflection and slope is
(�)��� = (�)���/� = −���
����= −
����(�.��)�
��(�����)(�.������)
= −5.19(10)��� = −5.19mm
���� = ��(0) = ��(�) = −
���
����= −0.00420rad = −0.2409°
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 9 / 28
▶ Example 8.2
Determine the deflection � and the slope angle � of the beam in Fig.8.5
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 10 / 28
In order to obtain the bending moment in the interior of the beam, we
isolate the segment of length L - x shown m Fig. 8.5 (b). From this free
body we obtain the bending moment
�� = −�(� − �) − � (a)
Inserting (a) into the moment-curvature relation (8.4), we find the
differential equation for the beam displacement �(�).
∴ �����
���= −�� + �� −� (b)
���� = −��� + ���
�−�� + �� (d)
��� = −����
�+ �
��
�−�
��
�+ ��� + �� (e)
B.C.) ��(0) = 0,�(0) = 0
∴ �� = �� = 0
∴ � = −�
����
��
�(3� − �) +�
��
�� (f)
∴ � = −�(�) =���
���+
���
���
� = −��(�) =���
���+
��
��
Figure 8.5 (c) shows the case where the moment � = 0.
Figure 8.5 (d) shows the case where the moment � = 0.
▶ Example 8.3
Determine the deflection δ of the beam in Fig.8.6
Sol)
����� = �� = ��� − ���� +�
�� −
�〈���〉�
�(a)
���� = ����
�−�� �� +
�
�� � −
�〈���〉�
�+ �� (b)
��� = ����
�− ���� +
�
����
�−
�〈���〉�
��+ ��� + �� (c)
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 11 / 28
B.C.) �(0)� = 0,�(0) = 0
∴ �� = �� = 0
∴ � = −�(� + �) =��
�����
�+
����
�+
���
�+
��
��
When � = 0;
� =���
���
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 12 / 28
▶ Example 8.4
Draw the B.M.D in Fig.8.7, statically indeterminate beam
The conditions of equilibrium applied to Fig. 8.7c yield
����� = �� = ��� − �〈� − �〉� (a)
���� = ����
�− �
〈���〉�
�+ �� (b)
��� = ����
�− �
〈���〉�
�+ ��� + �� (c)
B.C.) �(0) = �(�) = ��(�) = 0
�(0) = 0;
0 = ����
�− �
(���)�
�+ ��
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 13 / 28
→ �� =���
�−
����
�
�(�) = 0;
0 = ��
∴ 0 = ����
�− �
��
�+ �
���
�−
����
�
∴ �� =���
���(3� − �)
By equating the magnitudes given in Fig. 8.8, we find that the bending
moments at B and at C have equal magnitude when � = �√2 − 1�� =
0.414�.
�� < 0.414� → (��)������
� > 0.414� → (��)�������
▶ Example 8.5
A long uniform rod of length L, weight w per unit length, and bending
modulus EI is placed on a rigid horizontal table. Determine the length b in
Fig.8.9 which lifts up from the table.
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 14 / 28
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 15 / 28
▶Verbal analysis
i) The curvature of the beam is 0 in the region AB. à ∴ Between A and
B, �� = 0
ii) In section AB, � = 0. à ∴ Net load intensity is 0
iii) From Fig (d), �� is positive. ∵ The positive curvature is needed
in order that the beam is detached from the table.
iv) From Fig (e), the reaction force R� should exist which offsets the
bending moment in order to satisfy the equilibrium.
v) The deflection and the slope angle are 0 at the point B since the
deformation should be continuous.
vi) The bending moment at the point B is 0 but the shear force appears
suddenly.
▶Formulated analysis
i) it is convenient to deal only with the segment of the beam between B
and D.
ii) We obtain �� from the free body of Fig. 8.9g and insert in (8.4).
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 16 / 28
����� = �� = ��������
��� + �
(���)�
��〈� − �〉� −�
��
�(a)
���� =��������
���� +
�(���)�
��〈� − �〉� −
���
�+ �� (b)
��� =��������
����� +
�(���)�
���〈� − �〉� −
���
��+ ��� + �� (c)
B.C.) �(0) = �(�) = ��(0) = 0
∴ �� = �� = 0 → �(�) = 0
∴ 0 =��������
����� + 0 −
���
��
→ � = √2�
8.4 Superposition
à The method is based on linear relation between the load and the
deflection.
▶ The linearity is based on the followings.
i) Linearity between the bending moment and the curvature.��
��=
��
��
ii) Linearity between the curvature and the deflection.��
��≈
���
���
à This expression can be applied only when the load-deflection is
linear and the deformation is infinitesimal.
▶ Using superposition to obtain the solution to a beam-deflection
problem we again are using the three steps in Eq. (2.1) and nothing
additional; Equilibrium, geometric compatibility and the force-
deformation relation.
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 17 / 28
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 18 / 28
▶ Example 8.7
Draw the B.M.D.
.
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 19 / 28
▶ Assumptions
i) There is no stress in the beam when � = 0.
ii) Although the shear force P is applied on the beam, the deflection is
infinitesimal sufficiently. Thus the effect of the axial stress on the
bending can be ignored.
This beam is statically indeterminate since there are four
unknowns,���� ���� , but two equilibrium equations, ∑�� =
0,∑� = 0.
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 20 / 28
▶ Analysis
Superposition of the three loadings of Fig. 8.12(b), (c), and (d) gives
the loading of Fig. 8.12(a). Zero slope and zero displacement
conditions at C lead to
��� − �� + �� = 0�� − �� + �� = 0
� (a)
From the Table 8.1, we find
⎩⎪⎨
⎪⎧�� =
���(����)
���,�� =
���
���
�� =���
�
���,�� =
����
���
�� =���
�
���,�� =
���
��
�
�� =���(�����)
��
�� =���(���)
��
�� =����
��
▶ Example 8.8
Determine the ����� at the point D.
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 21 / 28
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 22 / 28
Sol)
▶ Idealization
i) The bolt-joint is completely effective in clamping the beam at C.
ii) The axial compressive force of the beam doesn’t affect the bending.
In case of ii), we next consider the case that the axial compressive
force affect the bending.
▶ Analysis
From Fig. 8.15(d),
�� = ��� + √2��� (a)
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 23 / 28
From Fig. 8.15(b),
���� =
√��
����√2� =
���
����
��� =��
����
� (b), (c)
From Table 8.1 – Case 1,
�� =(���)��
���(d)
Insert (b), (c), and (d) into the compatibility relation (a) :
� =�
����/(�����)��√��/(�����)(e)
=��
���.������.����= 19.60kN (f)
∴ ��� = 0.090mm�� = 1.760mm
� (g)
▶ Effect of the compressive force on the bending.
��(��������������������)
��(�������������������)= � ∙
��
(���)�
= 19.60�.�����
�.�(�)= 0.02875
≒ 2.9%
∴ The bending from compressive load can be ignored.
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 24 / 28
▶ Comparison of peak stress between the pin joint and clamping.
i) σ�� at the pin joint
→ ��� =��.��
�.�(��)��= 6.25MN/m�
ii) σ�� at the clamping
→ ��� =��.��
�.�(��)��+
(�.��)(�.���)
��(��)��= 15.12MN/m�
∴ The peak stress can be main factor of local deformation in the
slender member.
▶ Truss
à If the joints are pinned, the structure is called a truss
▶ Frame
à If the joints are rigid, the structure is called a frame.
The structure in Fig. 8.15 is the one example of the mixed structure.
8.5 Load-Deflection Differential Equation
à As an alternative to using the moment-curvature equation (8.4) to solve
beam-deflection problems, we can make use of an equation which
directly relates the external loading to the beam deflection.
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 25 / 28
▶Load-deflection differential equation
From chapter 3,
��
��+ � = 0 (3.11)
���
��+ � = 0 (3.12)
∴ ����
���= � (8.5)
→��
������
���
���� = ������� = � (8.6)
&�
�����
���
���� = ������ = −� (8.7)
▶Summary
����� = �� (8.4)
������ = −� (8.7)
������� = � (8.6)
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 26 / 28
The Fig. 8.19, 8.20, 8.21, and 8.22 represent the support condition at
the support point.
▶Problem solving process
i) Set up the loading intensity equation �(�). It’s efficient to use
singularity functions.
ii) Integrate the governing equation and find the four constants of
integration.
iii)This procedure is available regardless of whether it is determinate or
not.
According to ‘Timoshenko & Gere’, the constant of integration is
always zero if the governing equation �(�) contains all reaction forces.
However, in ‘Crandall’ it can have non zero value, and actually it is true.
See the example 3.9.
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 27 / 28
▶ Example 8.9
Determine the deflection at the point B in offset arm. Ignore the weight.
Sol)
The load intensity function q for 0 < � < � is
� =��
�〈� − �/3〉�� −�〈� − �/3〉�� (a)
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 28 / 28
B.C.)
��(0) = �(�) = 0
��(0) = ��(�) = 0
� (b)
Insertion of (a) into the load-deflection differential equation (8.6) yields
�����
���= � �
�
�< � − �/3 >�� −< � − �/3 >��� (c)
Expressions for ��/�� and � are obtained by integrating (c).��
��=
�
����
�〈� − �/3〉� −
〈���/�〉�
�+ ��
��
�+ ��� + ��� (d)
� =�
����
�〈� − �/3〉� −
〈���/�〉�
�+ ��
��
�+ ��
��
�+ ��� + ��� (e)
Substitution of (d) and (c) into the boundary conditions (b) gives four
simultaneous equations for the constants of integration. Their solution is
�� =�
�� , �� = −
�
���,�� = 0,�� = 0 (f)
Inserting these in (e) we find
� =�
������
��〈� − �/3〉� −
�
�〈� − �/3〉� +
�
��� − 2���� (g)
We obtain the desired deflection by setting � = �/3
∴ �� = −(�)���/� =�����
�,�����(h)
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 1 / 5
8.6 Energy Method
à In this section the formulas for strain energy in torsion and bending,
developed in Section 6.8 and 7.8, are used to illustrate the
application of Castigliano’s theorem to more complicated elastic
systems.
▶ Strain energy of bending
à In case of linear elastic beam, small deflection and small slope angle.
�� =�
���, � = ∫
��
����� = ∫
��
�����
�����
��
Shear strain energy can be neglected in case of common beam
which has much bigger length than width. For example �/� > 6.
▶ If a slender elastic shaft oriented along the axis of x carries a tensile
force �(�), a twisting moment ��(�), and a bending moment ��(�),
then according to (2.11), (6.13), (7.31) the total strain energy in the
member is
� = ∫��
�����
�+ ∫
���
�����
�+ ∫
���
�����
�(8.8)
▶Castigliano’s theorem
�� =��
���(8.9)
1▷If a deflection δ is desired at a point where there is no load (or in a
direction which is not in line with a load),
� = ���
������
(8.10)
2▷If the elastic energy is expressed in terms of the external load �� and
the reactions ��, a set of equations for determining the �� can be obtained from the conditions that there be no in-line deflection at each of the statically indeterminate reaction points.
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 2 / 5
��
���= 0 (8.11)
▶ Example 8.10
Solve the example 8.2 only using energy method.
�� = −�(� − �) − �
∴ � = ∫[�(���)��]�
�����
�
�
∴ � =��
��= ∫
�(���)���(���)
����
�
�=
���
���+
���
���
� =��
��= �
�(� − �) + �
����
�
�
=���
2��+��
��
▶ Example 8.11
Solve the example 8.8 only using energy method.
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 3 / 5
The total strain energy in this case is due to longitudinal and bending loading in CD and to longitudinal loading in BD,
� = ∫��
�������
�
�+ ∫
(�����)�(���)�
�����
�
�+ ∫
�√�(���)��
�������
√��
�(a)
The vertical and horizontal deflections at point D are
�� = ���
������
= ∫(���)(���)�
����
�
�=
(���)��
���(b)
�� = ���
������
= ∫(���)(���)�
����
�
�− ∫
��
������
√��
�
=(���)��
���−
�√���
����
(c)
in terms of the statically indeterminate reaction X. To determine X we usethe fact that the deflection at point C in Fig. 8.24a is zero; i.e., according to(8.11) we have
0 = ���
������
= ∫�
������
�
�− ∫
(���)(���)�
����
�
�+ ∫
��
������
√��
�
=��
����−
(���)��
���+
�√���
����(d)
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 4 / 5
fromwhichwesolveforX,
∴ � =�
����/(�����)��√��/(����
�)(e)
Note that this result agrees with (e) in Example 8.8 and also that (b) aboveagrees with (d) in Example 8.8. The horizontal deflection (c) above can berewritten in the following form by using (d),
�� =��
����(f)
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M2794.001000 (Solid Mechanics) Professor Youn, Byeng Dong
Ch. 8 Deflections due to bending 5 / 5
▶ Example 8.12
Determine the deflection of the n times coiled spring.
At each section of the spring, the wire carries a transverse shear force P, a twisting moment ��, and a bending moment ��, as indicated in Fig 8.25b.By applying the equilibrium requirements to this free body, we find�� = ��(1 − cos�),�� = �� sin � (a)
The total strain energy (8.8) in the wire (of uncoiled length 2���) due to the twisting and bending contributions is
� = ∫����(������)�
������
���
�+ ∫
���������
������
���
�
=����
���3�� +
����
����� (b)
(where, � =���
�, � =
���
�)
∴ � =��
��= ������
�
��+
�
��� =
�����
�����
�+
�
��
=�����
��������
����� (c)
Note that under the same load this spring deflects nearly twice as much as the spring with centered ends in Example 6.4