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mraChapter 9
9.1 Introduction

Deflections of Beams

in this chapter, we describe methods for determining the equation of the deflection curve of beams and finding deflection and slope at specific points along the axis of the beam

9.2 Differential Equations of the Deflection Curve consider a cantilever beam with a concentrated load acting upward at the free end the deflection in the y v is the displacement

direction of the axis

the angle of rotation

(also called slope) is the angle between the x axis and the tangent to the deflection

curve point m1 is located at distance x point m2 is located at distance x + dx slope at slope at m1 m2 is is +d

denote O'

the center of curvature and

the radius of curvature, then d = ds is

and the curvature

1

=

1 C =

d C ds

the sign convention is pictured in figure slope of the deflection curve dv C dx for = tan or ds j dx d C dx d C = dx = cos j 1 and d 2v CC dx2 = dv tan-1 C dx tan dv C dx j , then

small = 1 C = 1 C =

=

if the materials of the beam is linear elastic = 1 C = M C EI [chapter 5]

then the differential equation of the deflection curve is obtained d C dx d2v = CC dx2 = M C EI and v dV CC dx d 4v CC dx4 = -q q -C EI

it can be integrated to find ∵ dM CC dx d 3v CC dx3 = V

then

V = C EI

=

2

sign conventions for M,

V

and

q are shown

the above equations can be written in a simple form EIv" = M EIv"' = V EIv"" = -q

this equations are valid only when Hooke's law applies and when the slope and the deflection are very small for nonprismatic beam [I = I(x)], the equations are d 2v EIx CC dx2 = M dM CC dx =

d d 2v C (EIx CC) dx dx2 d 2v d2 CC (EIx CC) dx2 dx2

=

=

V

dV CC dx

=

-q

the exact expression for curvature can be derived = 1 C = v" CCCCC [1 + (v')2]3/2

9.3 Deflections by Integration of the Bending-Moment Equation substitute the expression of M(x) into

the deflection equation then integrating to satisfy (i) boundary conditions (ii) continuity conditions (iii) symmetry conditions to obtain the slope and the

3

deflection

v of the beam

this method is called method of successive integration

Example 9-1 determine the deflection of beam supporting a uniform load of intensity also determine max AB q
B

and

A,

flexural rigidity of the beam is bending moment in the beam is M = qLx CC 2 q x2 CC 2

EI

differential equation of the deflection curve EI v" = Then EI v' = qLx2 CC 4 q x3 CC 6 ∴ + C1 = v' = q (L/2) 3 CCCC 6 0 at x=L/2 qLx CC 2 q x2 CC 2

∵ the beam is symmetry, qL(L/2)2 CCCC 4

0

=

-

+

C1

4

then

C1

=

q L3 / 24

the equation of slope is v' = q - CCC (L3 24 EI 6 L x2 + 4 x 3)

integrating again, it is obtained v = q - CCC (L3 x 24 EI 2 L x 3 + x 4) + C2

boundary condition : v = thus we have C2 =

0 0

at

x

= 0

then the equation of deflection is v = q - CCC (L3 x 24 EI max -

2 L x 3 + x 4)

maximum deflection = L - v(C) 2

occurs at center 5 q L4 CCC 384 EI

(x =

L/2)

max

=

(↓)

the maximum angle of rotation occurs at the supports of the beam = v'(0) = q L3 - CCC 24 EI q L3 = CCC 24 EI ( )

A

and

B

=

v'(L)

( )

5

Example 9-2 determine the equation of deflection curve for a cantilever beam AB to a uniform load of intensity q also determine
B

subjected

and

B

at the free end EI

flexural rigidity of the beam is bending moment in the beam q L2 - CC 2 q L2 - CC 2

M

=

+

qLx

q x2 - CC 2 q x2 CC 2 q x3 CC 6 0 + C1 0

EIv" =

+

qLx

EIy' =

qL2x - CC + 2 v' = 0

qLx2 CC 2 =

boundary condition C1 v' = =

at x =

qx - CC 6EI

(3 L2 -

3Lx

+

x 2)

integrating again to obtain the deflection curve v = qx2 - CC 24EI (6 L2 4Lx + x 2) + C2 0

boundary condition C2 =

v = 0

0

at x =

6

then v = qx2 - CC 24EI =
B

(6 L2 -

4Lx

+

x 2)

max

=

v'(L)

=

q L3 - CC ( ) 6 EI q L4 CC (↓) 8 EI

max

=

-

B

=

- v(L) =

Example 9-4 determine the equation of deflection curve,
A, B, max

and

C

flexural rigidity of the beam is bending moments of the beam M = Pbx CC L

EI

(0 ≦ x ≦ a)

Pbx M = CC - P (x - a) (a ≦ x ≦ L) L differential equations of the deflection curve EIv" = Pbx CC (0 ≦ x ≦ a) L

Pbx EIv" = CC - P (x - a) (a ≦ x ≦ L) L integrating to obtain

7

Pbx2 EIv' = CC + 2L

C1 (0 ≦ x ≦ a)

Pbx2 P(x - a)2 EIv' = CC - CCCC 2L 2 2nd integration to obtain EIv Pbx3 = CC + 6L C1 x +

+

C2

(a ≦ x ≦ L)

C3

(0 ≦ x ≦ a)

EIv

Pbx3 P(x - a)3 = CC - CCCC + C2 x 6L 6

+ C4

(a ≦ x ≦ L)

boundary conditions (i) v(0) = 0 (ii) y(L) = 0

continuity conditions (iii) v'(a-) = v'(a+) (iv) v(a-) = v(a+)

(i) (ii)

v(0) v(L)

= =

0 0

=> =>

C3 =

0 = Pba2 CC 2L 0

PbL3 Pb3 CC - CC + C2 L + C4 6 6 => Pba2 CC 2L + C1 =

(iii)

v'(a ) C1
-

-

= =

v'(a ) C2 v(a ) C4
+

+

+

C2

(iv) v(a ) = C3 =

=>

Pba3 Pba3 CC + C1a + C3 = CC + C2a + 6L 6L

C4

8

then we have C1 C3 = = C2 C4 = = Pb (L2 - b2) - CCCCC 6L 0

thus the equations of slope and deflection are v' = Pb - CC (L2 - b2 - 3x2) 6LEI Pb - CC (L2 - b2 - 3x2) 6LEI Pbx - CC (L2 - b2 - x2) 6LEI Pbx - CC (L2 - b2 - x2) 6LEI (0 ≦ x ≦ a) P(x - a )2 CCCC 2EI (0 ≦ x ≦ a) P(x - a )3 CCCC 6EI

v' =

(a ≦ x ≦ L)

v =

v =

(a ≦ x ≦ L)

angles of rotation at supports
A

=

v'(0)

=

Pab(L + b) - CCCCC 6LEI

( )

B

=

v'(L)

Pab(L + a) = CCCCC 6LEI

( )

A

is function of a (or b), to find = Pb(L2 - b2) - CCCCC 6LEI 0 => L2 - 3b2 =

( A)max,

set

d

A

/ db = 0

A

d

A

/ db =

0

=> b

=

L/ 3

9

( A)max = for maximum dv C dx = 0

PL2 3 - CCCC 27 EI occurs at x1, => x1 = if a L2 - b2 CCC 3 > b, (a x1 < a

≧ b)

max

=

Pb(L2 - b2)3/2 - v(x1) = CCCCC 9 3 LEI = - v(L/2) =

(↓) Pb(3L2 - 4b2) CCCCCC 48 EI

at

x =

L/2

C

(↓)

∵ the maximum deflection always occurs near the midpoint, ∴ gives a good approximation of the max C

in most case, the error is less than 3% an important special case is v' = P CC (L2 - 4x2) 16EI a = b = L/2

(0 ≦ x ≦ L/2)

v =

P CC (3L2 - 4x2) (0 ≦ x ≦ L/2) 48EI are symmetric with respect to PL2 = CC 16EI = PL3 CC 48EI x = L/2

v'

and v =

A

B

max

=

C

10

9.4 Deflections by Integration of Shear-Force and Load Equations the procedure is similar to that for the bending moment equation except that more integrations are required if we begin from the load equation, which is of fourth order, four integrations are needed

Example 9-4 determine the equation of deflection curve for the cantilever beam AB supporting a

triangularly distributed load of maximum intensity q0 also determine
B

and

B

flexural rigidity of the beam is q0 (L - x) CCCC L = -q =

EI

q

=

EIv""

q0 (L - x) - CCCC L

the first integration gives EIv"' ∵ = q0 (L - x)2 - CCCC + 2L = V = 0 C1 = 0

v"'(L)

=> C1

thus

EIv"'

=

q0 (L - x)2 - CCCC 2L
11

2nd integration EIv" ∵ thus = q0 (L - x)3 - CCCC + 6L = M = = 0 C2 => C2 = 0

v"(L)

EIv"

q0 (L - x)3 - CCCC 6L

3rd and 4th integrations to obtain the slope and deflection q0 (L - x)4 EIv' = - CCCC + 24L EIv = q0 (L - x)5 - CCCC + 120L C3

C3 x +

C4 0

boundary conditions : v'(0) the constants C3 = C3 and

= v(0) =

C4 can be obtained C4 = q0L4 CC 120

q 0 L3 - CC 24

then the slope and deflection of the beam are v' = q 0x - CCC (4L3 24LEI 6L2x + 4Lx2 - x 3)

v =

q 0x 2 - CCC (10L3 120LEI

10L2x +

5Lx2 - x3)

B

=

q0L3 v'(L) = - CCC 24 EI
12

( )

B

=

q 0 L4 - v(L) = CCC 30 EI

(↓ )

Example 9-5 an overhanging beam concentrated load P ABC with a

applied at the end

determine the equation of deflection curve and the deflection
C

at the end EI

flexural rigidity of the beam is

the shear forces in parts AB and BC are V = V = P -C 2 P (0 < x < L) 3L (L < x < C) 2

the third order differential equations are EIv'" = P - C (0 < x < L) 2 P 3L (L < x < C) 2

EIv'"

=

bending moment in the beam can be obtained by integration M = EIv" = Px -C + 2 Px + C1 C2 (0 ≦ x ≦ L) 3L (L ≦ x ≦ C) 2

M

=

EIv" =

13

boundary conditions : v"(0) = v"(3L/2) we get C1 = 0 C2 = 3PL - CC 2

= 0

therefore the bending moment equations are M = EIv" = Px -C 2 P(3L - 2x) - CCCCC 2 (0 ≦ x ≦ L) 3L (L ≦ x ≦ C) 2

M

=

EIv"

=

2nd integration to obtain the slope of the beam Px2 EIv' = - CC 4 + C3 (0 ≦ x ≦ L) 3L C4 (L ≦ x ≦ C) 2 v'(L+) C4

Px(3L - x) EIv' = - CCCCC 2 continuity condition : PL2 - CC 4 then C4 + C3 =

+

v'(L-) = - PL2 + 3PL2 CC 4

=

C3 +

the 3rd integration gives EIv = Px3 - CC 12 + C3 x + C5 (0 ≦ x ≦ L) 3L (L ≦ x ≦ C) 2

EIv =

Px2(9L - 2x) - CCCCC + 12

C4 x

+

C6

14

boundary conditions : we obtain C5 = 0 C3

v(0) = v(L-) = 0

=

PL2 CC 12

and then

C4

=

5PL2 CC 6 = 0

the last boundary condition : v(L+) then C6 = PL3 - CC 4

the deflection equations are obtained v = Px CC (L2 12EI x 2) (0 ≦ x ≦ L) 3L (L ≦ x ≦ C) 2

v =

P - CC (3L3 - 10L2x + 9Lx2 12EI P = - CC (3L - x) (L - x) (L - 2x) 12EI C is 3L - v(C) = 2 PL3 CC 8EI (↓)

-2x3)

deflection at
C

=

9.5 Method of Superposition the slope and deflection of beam caused by several different loads acting simultaneously can be found by superimposing the slopes and deflections caused by the loads acting separately

15

consider a simply beam supports two loads : (1) uniform load of intensity and (2) a concentrated load P 1 q

the slope and deflection due to load are ( C)1 5qL4 = CCC 384EI = ( B)1 = qL3 CC 24EI

( A )1

the slope and deflection due to load are ( C)2 PL3 = CC 48EI ( A )2

2

= ( B)2

PL2 = CC 16EI

therefore the deflection and slope due to the combined loading are = ( C)1 + ( C)2 5qL4 = CCC 384EI + ( A)2 = + PL3 CC 48EI

C

A

=

B

=

( A)1

qL3 PL2 CC + CC 24EI 16EI

tables of beam deflection for simply and cantilever beams are given in Appendix G

superposition method may also be used for distributed loading consider a simple beam ACB with a

triangular load acting on the left-hand half

16

the deflection of midpoint due to a concentrated load is obtained [table G-2] = Pa CC (3L2 - 4a2) 48EI q dx for P and x for a

C

substitute d

C

(qdx) x = CCC (3L2 - 4x2) 48EI

the intensity of the distributed load is 2q0x CC L due to the concentrated load q dx q0 x 2 = CCC (3L2 - 4x2) dx 24LEI due to the entire triangular load is q0 x2 = ∫ CCC (3L2 - 4x2) dx 0 24LEI
L/2 A

q

=

then d

C

acting on

x is

C

thus

C

C

=

q 0 L4 CCC 240EI from left

similarly, the slope end is d
A

due to

P acting on a distance a

Pab(L + b) = CCCCC 6LEI P with 2q0x dx/L, a with x, and b with (L - x)

replacing

17

2q0x2(L - x)(L + L - x) q0 d A = CCCCCCCCC dx = CCC (L - x) (2L - x) x2 dx 6L2EI 3L2EI thus the slope at A = ∫
0 L/2

throughout the region of the load is 41q0L3 CCC 2880EI

A

q0 CCC (L - x) (2L - x) x2 dx = 3L2EI

the principle of superposition is valid under the following conditions (1) Hooke's law holds for the material (2) the deflections and rotations are small (3) the presence of the deflection does not alter the actions of applied loads these requirements ensure that the differential equations of the deflection curve are linear

Example 9-6 a cantilever beam uniform load as shown determine EI =
B

AB

supports a P

q

and a concentrated load

and

B

constant

from Appendix G : for uniform load ( B)1 q ( B )1 qa3 = CC 6EI

qa3 = CC (4L - a) 24EI

for the concentrated load P

18

( B)2

PL3 = CC 3EI

( B )2

PL2 = CC 2EI

then the deflection and slope due to combined loading are = ( B)1 + ( B)2 = qa3 PL3 CC (4L - a) + CC 24EI 3EI + PL2 CC 2EI

B

B

=

( B )1 +

( B)2

qa3 = CC 6EI

Example 9-7 a cantilever beam load q AB with uniform

acting on the right-half
B

determine EI =

and

B

constant

consider an element of load has magnitude q dx and is located at distance x from the

support from Appendix G, table G-1, case 5 by replacing d = P with q dx and d a with x (qdx)(x2) = CCCC 2EI

B

(qdx)(x2)(3L-x) CCCCCC 6EI

B

by integrating over the loaded region = ∫
L

B

L/2

qx2(3L-x) CCCC dx = 6EI

41qL4 CCC 384EI

19

B

=

L

L/2

qx2 CC dx 2EI

7qL3 = CC 48EI

Example 9-8 a compound beam concentrated load load q as shown
B

ABC

supports a

P

and an uniform

determine EI =

and

A

constant

we may consider the beam as composed of two parts : (1) simple beam AB, (2) cantilever beam BC the internal force F = 2P/3 is obtained and

for the cantilever beam BC qb4 = CC 8EI AB, Fb3 CC 3EI
A

B

+

=

qb4 CC + 8EI

2Pb3 CC 9EI

for the beam parts : (1) angle

consists of two
B,

BAB' produced by

and P

(2) the bending of beam AB ( A )1 = C a P
B

by the load +

qb4 = CCC 8aEI

2Pb3 CCC 9aEI table G-2, Appendix

the angle due to G

is obtained from Case 5 of by 2a/3 and b by a/3

with replacing a

20

( A )2

P(2a/3)(a/3)(a + a/3) = CCCCCCCCC 6aEI
B

=

4Pa2 CCC 81EI
B

note that in this problem, continuous, i.e. ( B)L g ( B )R

is continuous and

does not

Example 9-9 an overhanging beam ABC determine EI
C C

supports a uniform load

q as shown

=

constant

may be obtained due to two parts
B

(1) rotation of angle

(2) a cantilever beam subjected to uniform load q
B

firstly, we want to find = qL3 - CC 24EI +

B

MBL CC 3EI qa2L CC = 6EI = qL(4a2 - L2) CCCCC 24EI

=

qL3 - CC + 24EI = a

then

1

B

qaL(4a2 - L2) CCCCCC 24EI produces an additional deflection
2

bending of the overhang BC = qa4 CC 8EI

2

21

therefore, the total downward deflection of C = + = qaL(4a2 - L2) CCCCCC 24EI +

is qa4 CC 8EI

C

1

1

=

qa CC (a + L) (3a2 + aL - L2) 24EI
C

for for

a
C

large, = a 0 =

is downward; for 3a2 + aL - L2 =

a 0

small,

C

is upward

L( 13 - 1) CCCCC 6
C

=

0.4343 L

a > 0.4343 L, point D

is downward;

a < 0.4343 L,

C

is upward

is the point of inflection, the curvature is zero because the

bending moment is zero at this point at point D, d2y/dx2 = 0

9.6 Moment-Area Method the method is especially suitable when the deflection or angle of rotation at only one point of the beam is desired consider a segment denote
B B/A A

AB of the beam

the difference between

and

B/A

=

B

m1

A

consider points

and

m2

22

d

=

ds C =

Mdx CC EI

Mdx / EI strip of the

is the area of the shaded Mdx / EI diagram and B

integrating both side between A ∫ d
A B

=

B M ∫ C dx A EI

B/A

=

B

-

A

=

area of the M/EI diagram between A

and B

this is the First moment-area theorem next, let us consider the vertical offset tB/A between points B and B1 (on

the tangent of A) ∵ dt = x1 d = Mdx x1 CC EI and B

integrating between A ∫ dt
A B

B Mdx = ∫ x1 CC A EI

i.e. tB/A A and

=

1st moment of the area of the M/EI

diagram between

B, evaluated w. r. t. B

this if the Second moment-area theorem

Example 9-10 determine cantilever beam
B

and AB

B

of a

supporting a
23

concentrated load sketch the A1 =

P

at

B

M/EI diagram first 1 PL -CLC = 2 EI PL2 - CC 2EI = = PL2 - CC 2EI ( )

B/A

=

B

A

B

Q1

=

A1 x

=

2L A1 C = 3

PL3 - CC 6EI (↓)

B

=

- Q1

PL3 = CC 6EI

Example 9-11 determine cantilever beam
B

and AB

B

of a

supporting a q acting

uniform load of intensity over the right-half sketch the A1 =

M/EI diagram first 1 L qL2 C C (CC) 3 2 8EI L qL2 C (CC) 2 8EI = = qL3 CC 48EI

A2 =

qL3 CC 16EI qL3 = CC 16EI

A3 =

1 L qL2 C C (CC) 2 2 4EI

24

B/A

= = = =

B

=

A1 + A2 + A3

7qL3 = CC 16EI

( )

B

tB/A A 1 x1 + A 2 x2 + A 3 x3 = 41qL4 CCC 384EI (↓) qL3 1 3L 1 3L 1 5L CC (C C + C C + C C) EI 48 8 16 4 16 6

Example 9-12 a simple beam ADB supports a concentrated load determine A1 =
A

P and

as shown
D

L Pab C (CC) 2 LEI

=

Pab CC 2EI

L+b Pab (L + b) tB/A = A1 CCC = CCCCC 6EI 3
A

=

BB1 CC L

=

Pab (L + b) CCCCC 6EIL
D

to find the deflection
D

at

D

=

DD1 = a
A

- D 2D 1 = Pa2b (L + b) CCCCCC 6EIL

DD1 D 2D 1

= =

tD/A = A2 x2 A Pab a C CC C 2 EIL 3 = Pa3b CCC 6EIL

25

D

=

Pa2b2 CCC 3EIL max to find the maximum deflection A3 = = x1 Pbx1 C CC 2 EIL
E

at E,

we set

E

=

0

=

Pbx12 CCC 2EIL = = - A3 = Pbx12 CCC 2EIL - Pbx12 / 2EIL

E/A

-

A

A

Pab (L + b) = CCCCC 6EIL x1 = =

then and

[(L2 - b2) / 3]2 x1 x1 - A3 C = 3 A Pb 3/2 CCCC (L2 - b2) 9 3 EIL

max

A

or

max

= =

offset of point 2 x1 A3 CC 3 =

from tangent at E

max

Pb 3/2 CCCC (L2 - b2) 9 3 EIL

Conjugate Beam Method EIv" = EI d /dx = M

Integrating = M ∫C dx EI M ∫∫ C dx dx EI

v =

26

beam theory dM / dx V = = V -∫q dx dV / dx M = = -q -∫∫ q dx dx

suppose we have a beam, called conjugate beam, whose length equal to the actual beam, let this beam subjected to so-called "elastic load" of intensity M/EI, then the shear force and bending moment over a portion M, can be obtained

of the conjugate beam, denoted by V and V = M -∫C dx EI M =

M - ∫∫ C dx dx EI

then (1) the slope at the given section of the beam equals the minus shear force in the corresponding section of the conjugate beam (2) the deflection at the given section of the beam equals the minus bending moment in the corresponding section of the conjugate beam i.e. = = -V -M

the support conditions between the actual beam and conjugate beam can be found Actual Beam fixed end free end simple end = 0, v = 0 g 0, g 0, g 0, g 0, vg0 v=0 v=0 vg0 Conjugate Beam V = 0, M=0 free end fixed end simple end interior hinge interior support V g 0, M g 0 V g 0, M = 0 V g 0, M = 0 V g 0, M g 0

interior support interior hinge

27

Example 1 1 PL PL2 B = - VB = - C CC L = - CC 2 EI 2EI PL2 2L PL3 B = - MB = - CC C = - CC 2EI 3 3EI ( )

(↓)

Example 2 1 2L wL2 wL3 A = - VA = - C C CC = - CC 2 3 8EI 24EI wL3 L wL2 L L C = - MC = - CC C - CC C C 24EI 2 8EI 2 4 1 wL2 L 3L 1 1 1 wL4 + C CC C C = - (C - C + CC) CC 48 64 128 EI 3 8EI 2 8 = 5 wL4 CCC 384EI (↓) ( )

28

Example 3
A

= - VA = -RA

1 M 2L ML = - C C C = - CC ( ) 2 EI 3 3EI

1 M L ML B = - VB = RB = C C C = CC 2 EI 3 6EI

( )

ML L 1 L M L C = - MC = - (CC C - C C CC C) 6EI 2 2 2 2EI 6 1 1 ML2 = - (C - C) CC = 12 48 EI ML2 - CC 16EI (↓)

Example 4
B

=

- VB

ML = CC EI ML2 CC 2EI

(

)

ML L B = - MB = CC C = EI 2

(↑)

Example 5 qL3 L qL2 B = - VB = - C CC = - CC 3 2EI 6EI qL3 3L qL4 B = - MB = - CC C = - CC 6EI 4 8EI

( )

(↓)

29

Example 6 1 L PL PL2 A = - VA = - C C CC = - CC 2 2 4EI 16EI

( )

PL2 L 1 L PL L C = - MC = - (CC C - C C CC C) 16EI 2 2 2 4EI 6 PL3 1 1 PL3 = - CC (C - C) = - CC EI 32 96 48EI

(↓)

9.7 Nonprismatic Beam EI g constant

Example 9-13 a beam ABCDE is supported a concentrated load P at midspan as shown IBD = 2 IAB = 2 IDE = 2I
A

determine the deflection curve, M = Px C 2

and

C

L (0 ≦ x ≦ C) 2 (0 ≦ x ≦ L/4) (L/4 ≦ x ≦ L/2)

then

EIv" = Px / 2 E(2I)v" = Px / 2

30

thus

Px2 v' = CC + C1 4EI Px2 v' = CC + C2 8EI

(0 ≦ x ≦ L/4)

(L/4 ≦ x ≦ L/2)

∵ ∴

v' = C2 =

0

at x =

L/2 (symmetric)

PL2 - CC 32EI = v'(L/4)+

continuity condition v'(L/4)C1 = 5PL2 - CCC 128EI

therefore

v' =

P - CCC (5L2 128EI P - CC (L2 32EI
A

-

32x2)

(0 ≦ x ≦ L/4)

v' =

- 4x2)

(L/4 ≦ x ≦ L/2)

the angle of rotation = v'(0) =

is ( )

A

5PL2 - CCC 128EI

integrating the slope equation and obtained v = P - CCC (5L2x 128EI P - CC (L2x 32EI 32x3 - CC) 3 4x3 CC) 3
31

+

C3

(0 ≦ x ≦ L/4)

v =

+

C4

(L/4 ≦ x ≦ L/2)

boundary condition we get

v(0) = C3 = 0

0

continuity condition v(L/4)- = we get C4 = PL3 - CCC 768EI

v(L/4)+

therefore the deflection equations are v = Px - CCC (15L2 384EI P - CCC (L3 768EI + 32x2) (0 ≦ x ≦ L/4)

v =

24L2x -

32x3)

(L/4 ≦ x ≦ L/2)

the midpoint deflection is obtained = - v(L/2) = 3PL3 CCC 256EI (↓)

C

moment-area method and conjugate beam methods can also be used

Example 9-14 a cantilever beam concentrated load IBC = 2 IAB
A

ABC

supports a

P =

at the free end 2I

determine denote
1

the deflection of A due to C fixed P(L/2)3 = CCC 3EI = PL3 CCC 24EI
32

1

and

C

=

P(L/2)3 (PL/2)(L/2)2 5PL3 CCC + CCCCC = CC 3E(2I) 2E(2I) 96EI PL2 = CC 16EI
C

C

P(L/2)2 (PL/2)(L/2) = CCC + CCCCC 2E(2I) E(2I)
C

addition deflection at A due to = + L CC 2 = =

and

2

C

5PL3 CC 48EI

A

=

1

+

2

5PL3 CC 16EI

moment-area method and conjugate beam methods can also be used

9.8 Strain Energy of Bending consider a simple beam AB subjected

to pure bending under the action of two couples M the angle = is L C = L = ML CC EI and

if the material is linear elastic, M has linear relation, then W = U = M CC 2 =

M 2L EI 2 CC = CC 2EI 2L

33

for an element of the beam d = dx d 2y = CC dx dx2 Md CC 2 = M 2dx CCC 2EI = EI(d )2 CCC 2dx

dU

= dW =

by integrating throughout the length of the beam U = M 2dx ∫ CCC 0 2EI
L

=

EI d 2y 2 ∫ C (CC) dx 0 2 dx2
L

shear force in beam may produce energy, but for the beam with

L/d > 8,

the strain energy due to shear is relatively small and may be disregarded deflection caused by a single load U = W = 2U CC P P C 2 or U = W = 2U CC M0 M0 CC 2

=

=

Example 9-15 a simple beam AB of length L

supports a uniform load of intensity q evaluate the strain energy M = qLx CC 2 qx2 - C = 2 q C (Lx - x2) 2

34

U

=

M2dx ∫ CC = 0 2EI
L

1 q L 2 CC ∫ [C(Lx - x2)] dx 2EI 0 2 = q2L5 / 240EI

=

q2 L CC∫ (L2x2 - 2Lx3 + x4)dx 8EI 0

Example 9-16 a cantilever beam AB is subjected to

three different loading conditions (a) a concentrated load P at its free end (b) a moment (c) both determine determine (a) U M M0 at its free end M0 acting simultaneously

P and
A A

due to loading (a) due to loading - Px P 2 L3 CC 6EI (b)

=
L

2 M2dx L (-Px) dx = ∫ CC = ∫ CCC = 0 0 2EI 2EI

W=U

P A P2L3 CC = CC 2 6EI =
L

PL3 A = CC 3EI

(b) U

M

- M0 (-M0)2dx ∫ CCCC 0 2EI
L

M2dx = ∫ CC = 0 2EI

=

M 02L CC 2EI

W=U

M 02L M0 A CC = CC 2 2EI

A

M0L = CC EI

35

(c)

M
L

=

- Px

L

M0

M2dx U = ∫ CC 0 2EI W=U

(-Px - M0)2dx P 2 L3 PM0L2 M02L =∫ CCCCC = CC + CCC + CC 0 2EI 6EI 2EI 2EI + M0 A CC = 2
A

P A CC 2

P2L3 CC 6EI
A

+

PM0L2 CCC 2EI

+

M02L CC 2EI

1 equation for two unknowns

and

9.9 Castigliano's Theorem (Energy Method) dU = Pd dU CC d dC CC dP = P

dC =

dP

=

where

C

is complementary strain energy = U

for linear elastic materials C then we have dU CC dP dU CC dM and = M =

similarly

=

for both P CC
∂P

acting simultaneously, U
∂U

=

U(P, M)

∂U

CC
∂M

=

in example 9-16 (c)

36

U

=

P 2 L3 CC 6EI CC
∂P ∂U

+

PM0L2 CCC 2EI PL3 CC 6EI PL2 CC 2EI +

+

M 02L CC 2EI

=

=

M 0 L2 CC 2EI M 0L CC EI

=

CC
∂M

∂U

=

+

in general relationship i =

CC
∂P i
∂U

∂U

Castigliano's Theorem M2dx M ∂M CC∫CC = ∫(C)(CC)dx ∂P i 2EI EI ∂P i

i

=

CC
∂P i

=

this is the modified Castigliano's Theorem in example 9-16 (c) M
∂M ∂P

=

- Px = -x

-

M0 CC
∂M
0

CC

∂U

=

-1

=

1 C∫(-Px - M0)(-x)dx EI

=

PL3 CC + 6EI PL2 CC 2EI +

M 0 L2 CC 2EI M 0L CC EI

=

1 C ∫(-Px - M0)(-1)dx = EI

37

Example 9-17 a simple beam uniform load q = P AB supports a

20 kN/m, and a = 25 kN 210 GPa

concentrated load L I = = 2.5 m

E =

31.2 x 102 cm4
C

determine M =

Px CC 2

+

qLx CC 2

-

qx2 CC 2

method (1) U M2dx = ∫CC 2EI P2L3 = CC + 96EI
∂U
C

=

1 Px 2∫ CC(CC 0 2EI 2
L/2

qLx + CC 2

-

qx2 2 CC) dx 2

5PqL4 CCC 384EI PL3 = CC 48EI

+

q 2 L5 CCC 240EI 5qL4 CCC 384EI

=

CC
∂P

+

method (2)
∂M / ∂P =

x/2 1 Px qLx qx2 x 2∫ C(C + CC - CC)(C) dx 0 EI 2 2 2 2
L/2

C

M ∂M = ∫(C)(CC)dx = EI ∂P = PL3 CC 48EI + 5qL4 CCC 384EI

= 1.24 mm +

1.55 mm

=

2.79 mm

38

Example 9-18 a overhanging beam ABC supports a

uniform load and a concentrated load as shown determine
C

and

C

the reaction at A qL RA = C 2 MAB =

due to the loading is P C 2

qx12 RA x1 - CC 2 qLx1 CC 2 Px1 CC 2 qx12 CC (0 ≦ x1 ≦ L) 2

= MBC

= - Px2 (0 ≦ x1 ≦ L/2)

then the partial derivatives are
∂ MAB / ∂ P

= =

- x1/2 - x2

(0 ≦ x1 ≦ L) (0 ≦ x2 ≦ L/2)

∂ MBC / ∂ P
C

=

∫(M/EI)( ∂ M / ∂ P)dx
L

= ∫ (MAB /EI)( ∂ MAB / ∂ P)dx
0

+ ∫ (MBC/EI)( ∂ MBC / ∂ P)dx
0

L/2

1 L qLx1 Px1 qx12 x1 1 L/2 = C∫ (CC - CC - CC)(- C)dx1 + C∫ (-Px2)(-x2)dx2 EI 0 2 2 2 2 EI 0 = PL3 CC 8EI qL4 - CC 48EI

39

to determine the angle qL RA = C 2 MAB = P C 2

C,

we place a couple of moment

MC

at C

MC - CC L

qx12 RA x1 - CC 2 qLx1 CC 2 Px1 CC 2 MC x 1 - CC L qx12 CC 2 (0 ≦ x1 ≦ L)

=

MBC

= - Px2

- MC

(0 ≦ x1 ≦ L/2)

then the partial derivatives are
∂ MAB / ∂ MC

= =

- x1/L -1

(0 ≦ x1 ≦ L) (0 ≦ x2 ≦ L/2)

∂ MBC / ∂ MC

C

=

∫(M/EI)(M/ MC)dx
L 0

= ∫ (MAB /EI)(MAB / MC)dx

+

L/2 0

(MBC /EI)(MBC / MC)dx qx12 x1 CC)(- C)dx1 2 L

1 L qLx1 Px1 MC x1 = C∫ (CC - CC - CC EI 0 2 2 L 1 L/2 + C∫ (-Px2 - MC)(-1)dx2 EI 0 since obtained 7PL2 = CC 24EI qL4 CC 24EI MC is a virtual load, set MC = 0,

after integrating

C

is

C

40

9.10 Deflections Produced by Impact 9.11 Temperature Effects

41

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