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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 ...view middle of the document...

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...

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