Mechanics of Materials Chap 10-01

8/9/2019 Mechanics of Materials Chap 10-01

1/10

Differential Equations of the Deflection Curve

The problems for Section 10.3 are to be solved by integrating the

differential equations of the deflection curve. All beams have constant

flexural rigidity EI. When drawing shear-force and bending-moment

diagrams, be sure to label all critical ordinates, including maximum

and minimum values.

Problem 10.3-1 A propped cantilever beamAB of lengthL is loaded

by a counterclockwise momentM0

acting at supportB (see figure).

Beginning with the second-order differential equation of the

deflection curve (the bending-moment equation), obtain the reactions,

shear forces, bending moments, slopes, and deflections of the beam.

Construct the shear-force and bending-moment diagrams, labeling all

critical ordinates.

Solution 10.3-1 Propped cantilever beam

10Statically Indeterminate Beams

A

L

y

xB

M0

MA

RA

RB

M0 applied load

SelectMA as the redundant reaction.

REACTIONS (FROM EQUILIBRIUM)

(1) RB R

A(2)

BENDING MOMENT (FROM EQUILIBRIUM)

(3)

DIFFERENTIAL EQUATIONS

(4)EIvMA

L

x2

2LxM0x

2

2LC1

EIvMMA

L(xL)

M0x

L

MRAxMAMA

L

(xL) M0x

L

RAMA

L

M0

L

B.C. 1 C1 0

(5)

B.C. 2 v(0) 0 C2 0

B.C. 3 v(L) 0 MA

REACTIONS (SEE EQS. 1 AND 2)

SHEAR FORCE (FROM EQUILIBRIUM)

BENDING MOMENT (FROM EQ. 3)

M2M0

2L(3xL)

VRA3M0

2L

RB3M0

2LRA

3M0

2LMA

M0

2

M0

2

EIvMA

Lx3

6

Lx2

2M0x3

6LC2

v(0) 0


2/10

Problem 10.3-2 A fixed-end beamAB of lengthL supports a uniform

load of intensity q (see figure).Beginning with the second-order differential equation of the

deflection curve (the bending-moment equation), obtain the reactions,

shear forces, bending moments, slopes, and deflections of the beam.

Construct the shear-force and bending-moment diagrams, labeling all

critical ordinates.

Solution 10.3-2 Fixed-end beam (uniform load)

634 CHAPTER 10 Statically Indeterminate Beams

SLOPE (FROM EQ. 4)

DEFLECTION (FROM EQ. 5)

v

M0x2

4LEI

(L

x)

vM0x

4LEI(2L 3x)

SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS

3M0 2L

V

O

Mo2

Mo

MO

L

3

A

L

y

xB

MBMA

RA RB

q

SelectMA

as the redundant reaction.

REACTIONS (FROM SYMMETRY AND EQUILIBRIUM)

MB MA


(1)


(2)

B.C. 1 C1 0

(3)

B.C. 2 v(0) 0 C2 0

B.C. 3 v(L) 0 MA

qL2

12

EIvMAx

2

2

q

2

Lx3

6

x4

12C2

v(0) 0

EIvMAx

q

2 Lx2

2

x3

3 C1

EIvMMAq

2

(Lxx2)

MRAxMAqx2

2MA

q

2

(Lxx2)

RARB

qL

2

REACTIONS

SHEAR FORCE (FROM EQUILIBRIUM)

BENDING MOMENT (FROM EQ. 1)

SLOPE (FROM EQ. 2)

DEFLECTION (FROM EQ. 3)

maxvL2 qL

4

384EI

vqx2

24EI

(Lx)2

v qx12

EI(L2 3Lx 2x2)

Mq

12

(L2 6Lx 6x2)

VRA qxq

2

(L 2x)

MAMBqL2

12RARB

qL

2


3/10

Problem 10.3-3 A cantilever beamAB of lengthL has a fixed support

atA and a roller support at B (see figure). The support atB is moved

downward through a distance B

.

Using the fourth-order differential equation of the deflection curve

(the load equation), determine the reactions of the beam and the equation

of the deflection curve. (Note: Express all results in terms of the imposed

displacement B

.)

Solution 10.3-3 Cantilever beam with imposed displacement B

SECTION 10.3 Differential Equations of the Deflection Curve 635


qL2

12

qL2

24

M O

qL2

12

x

qL

2

qL

2

V

O

B

L

A

B

x

y

RA

MA

RB


RAR

B(1) M

AR

BL (2)


(3)

(4)

(5)

(6)

EIv C1x36 C

2x22 C

3x+ C

4(7)

B.C. 1 v(0) 0 C4 0

B.C. 2 C3 0

B.C. 3 C1L C

2 0 (8)

B.C. 4 v(L) B

C1L 3C

26EI

BL2 (9)

SOLVE EQUATIONS (8) AND (9):

C2

3EIB

L2C

1

3EIB

L3

v(L) 0

v(0) 0

EIv C1x22C2xC3

EIv MC1xC2

EIv VC1

EIv q 0

SHEAR FORCE (EQ. 4)

REACTIONS (EQS. 1 AND 2)

DEFLECTION (FROM EQ. 7):

SLOPE (FROM EQ. 6):

v3Bx

2L3

(2Lx)

vBx

2

2L3

(3Lx)

MARBL3

EIB

L2RARB

3

EIB

L3

RAV(0) 3

EIB

L3V

3EIB

L3


4/10

Problem 10.3-4 A cantilever beamAB of lengthL has a fixed support atA and a spring support atB (see figure). The spring behaves in a linearly

elastic manner with stiffness k.

If a uniform load of intensity q acts on the beam, what is the

downward displacement B

of endB of the beam? (Use the second-order

differential equation of the deflection curve, that is, the bending-moment

equation.)

Solution 10.3-4 Beam with spring support


y

xA B

L

k

q

MA

RA

RB

q intensity of uniform load

EQUILIBRIUM RA qL R

B(1)

(2)

SPRING RB k

B(3)

B downward displacement of pointB.



EIvMRAxMAqx2

2

MRAxMAqx2

2

MAqL2

2

RBL B.C. 1 C1 0

B.C. 2 v(0) 0 C2 0

B.C. 3 v(L) B

SubstituteRA

andMA

from Eqs. (1) and (2):

Substitute forRB

from Eq. (3) and solve:

B3

qL4

24EI 8kL3

EIBRBL

3

3

qL4

8

EIBRAL

3

6

MAL2

2

qL4

24

v(0) 0

EIvRAx3

6MA

x2

2

qx4

24C1xC2

EIvRAx2

2MAx

qx3

6C1

Problem 10.3-5 A propped cantilever beamAB of lengthL supportsa triangularly distributed load of maximum intensity q

0(see figure).

Beginning with the fourth-order differential equation of the deflection

curve (the load equation), obtain the reactions of the beam and the

equation of the deflection curve.


A

L

y

q0

xB

MA

RA

RB

Triangular load q q0(L x)L


(1)

(2)EIvVq0xq0x

2

2LC1

EIv qq0

L

(Lx)

(3)

(4)

(5)EIv q0x

4

24

q0x5

120LC1

x3

6C2

x2

2C3xC4

EIvq0x

3

6

q0x4

24

LC1

x2

2C2xC3

EIvMq0x

2

2

q0x3

6LC1xC2


5/10

Problem 10.3-6 The load on a propped cantilever beam AB of lengthLis parabolically distributed according to the equation q q

0(1 x2/L2),

as shown in the figure.






B.C. 1 (6)

B.C. 2 C3 0

B.C. 3 v(0) 0 C4 0

B.C. 4 v(L) 0 (7)

Solve Eqs. (6) and (7):

SHEAR FORCE (EQ. 2)

Vq0

10L

(4L2 10Lx 5x2)

C2 q0L

2

15C1

2q0L

5

C1L 3C2q0L

2

5

v(0) 0

C1LC2q0L

2

3v(L) 0 REACTIONS

From equilibrium:

DEFLECTION CURVE (FROM EQ. 5)

or

vq0x

2

120LEI

(4L3 8L2x 5Lx2 x3)

EIvq0x

4

24

q0x5

120L

2

q0L

5

x3

6 q0L

2

15

x2

2

MAq0L

2

6RBL

q0L2

15

RBV(L) q0L

10

RAV(0) 2

q0L

5

A

L

y

x2

L2q =q0 1

xB

MA

RA

RB

q0 ( )

Parabolic load q q0(1 x2L2)


(1)

(2)

(3)

(4)

(5)

B.C. 1 C1L C

2 5q

0L212 (6)

B.C. 2 C3 0

B.C. 3 v(0) 0 C4 0

B.C. 4 v(L) 0 C1L 3C

2 7q

0L230 (7)

v(0) 0

v(L) 0

EIvq0x4

24

x6

360L2C1x

3

6C2

x2

2C3xC4

EIvq0x3

6

x5

60L2C1x

2

2C2xC3

EIvMq0x2

2

x4

12L2C1xC2

EIvVq0(xx33L2) C1

EIv qq0(1 x2L2)

Solve Eqs. (6) and (7):

C1 61q

0L120 C

211q

0L2120

SHEAR FORCE (EQ. 2)

REACTIONS RA V(0) 61q

0L120

RBV(L) 19q

0L120

From equilibrium:


q0x

2(Lx)

720L2EI

(33L3 28L2x 2Lx2 2x3)

vq0x

2

720L2EI

(33L4 61L3x 30L2x2 2x4)

MA

2

3(q

0)(L)

3L

8 R

BL

11q0L2

120

Vq0

120L2

(61L3 120L2x 40x3)


6/10

Problem 10.3-7 The load on a fixed-end beamAB of lengthL isdistributed in the form of a sine curve (see figure). The intensity

of the distributed load is given by the equation q q0

sin x/L.




Solution 10.3-7 Fixed-end beam (sine load)


A

L

yxL

q =q0 sin

xB

MA MB

RA RB

q q0

sin xL

FROM SYMMETRY: RAR

B M

AM

B


(1)

(2)

(3)

(4)

(5)

B.C. 1 From symmetry, C1 0

B.C. 2 C3 q

0L33

B.C. 3 C2 2 q

0L23

B.C. 4 v(0) 0 C4 0

v(L) 0

v(0) 0

VL2 0

EIvq0L

4

4

sinx

LC1

x3

6C2

x2

2C3xC4

EIvq0L

3

3

cosx

LC1

x2

2C2xC3

EIvMq0L

2

2

sinx

LC1xC2

EIv V

q0L

cos

x

L C1

EIv qq0sin xL

SHEAR FORCE (EQ. 2)

BENDING MOMENT (EQ. 3)


or

vq0L

2

4EI

L2sin xL

x2 Lx

EIvq0L

4

4

sinx

L

q0L2x2

3

q0L3x

3

MBMA2q0L

2

3

MAM(0) 2q0L

2

3

Mq0L2

3

sin xL

2

RBRAq0L

RAV(0) q0L

V

q0L

cos

x

L

Problem 10.3-8 A fixed-end beamAB of lengthL supportsa triangularly distributed load of maximum intensity q

0(see figure).




Solution 10.3-8 Fixed-end beam (triangular load)

A

L

y

q0

xB

MA MB

RA RB

q q0(1 xL)


(1)

(2)EIvVq0x x2

2LC1

EIv qq01xL

(3)

(4)

(5)EIvq0x4

24

x5

120LC1x

3

6C2

x2

2C3xC4

EIvq0x3

6

x4

24LC1x

2

2C2xC3

EIvMq0x2

2

x3

6LC1xC2


7/10

Problem 10.3-9 A counterclockwise momentM0

acts at the midpoint

of a fixed-end beamACB of lengthL (see figure).


deflection curve (the bending-moment equation), determine all reactions

of the beam and obtain the equation of the deflection curve for the

left-hand half of the beam.

Then construct the shear-force and bending-moment diagrams for the

entire beam, labeling all critical ordinates. Also, draw the deflection curve

for the entire beam.

Solution 10.3-9 Fixed-end beam (M0

= applied load)


B.C. 1 C3 0

B.C. 2 (6)

B.C. 3 v(0) 0 C4 0

B.C. 4 v(L) 0 (7)

Solve eqs. (6) and (7):

SHEAR FORCE (EQ. 2)

REACTIONS

RBV(L)

3q0L

20

RAV(0) 7q0L

20

Vq0

20L

(7L2 20Lx 10x2)

C2 q0L

2

20C1

7q0L

20

C1L 3C2 q0L

2

5

C1L 2C2 q0L

2

4v(L) 0

v(0) 0 BENDING MOMENT (EQ. 3)

REACTIONS

DEFLECTION CURVE (EQ. 5)

or

vq0x

2

120LEI

(Lx)2(3Lx)

vq0x

2

120LEI

(3L3 7L2x 5Lx2 x3)

MBM(L) q0L

2

30

MAM(0) q0L

2

20

Mq0

60L

(3L3 21

L2x 30

Lx2 10x3)

A

L

2

L

2

y

xBC

MB

M0

MA

RA RB

Beam is symmetric; load is antisymmetric.

Therefore, RA R

B M

A M

B

C 0

DIFFERENTIAL EQUATIONS (0 xL2)

(1)

(2)

(3)

B.C. 1 C1 0

B.C. 2 v(0) 0 C2 0

B.C. 3 Also,MBRAL

6MA

RAL

6vL

2 0

v(0) 0

EIvRAx3

6 MAx2

2 C1xC2

EIvRAx2

2MAxC1

EIvMRAxMA

EQUILIBRIUM (OF ENTIRE BEAM)

MB 0 M

AM

0M

BR

AL 0

or,

DEFLECTION CURVE (EQ. 3)

0 x L2vM0x

2

8LEI

(L 2x)

MAMBM0

4MA

RAL

6

RARB3M0

2L

RAL

6M0

RAL

6RAL 0

a


8/10

Problem 10.3-10 A propped cantilever beamAB supportsa concentrated load P acting at the midpoint C(see figure).


deflection curve (the bending-moment equation), determine all reactions

of the beam and draw the shear-force and bending-moment diagrams for

the entire beam.

Also, obtain the equations of the deflection curves for both halves

of the beam, and draw the deflection curve for the entire beam.



DIAGRAMS

At point of inflection: max2

maxM0L

2

216EI

3M02L

V

O

M04

M04

MO

M0

2

M02

L

6

L

3

L

6

L

6max

A

L

2

L

2

y

xBC

MA

RARB

P

P applied load atxL2

SelectRB

as redundant reaction.


RA P R

B(1) (2)

BENDING MOMENTS (FROM EQUILIBRIUM)

MRB

(L x) L2xL

0 x L2

MRAxMA (PRB)x PL2RBL

MAPL

2RBL

DIFFERENTIAL EQUATIONS (0 xL2)

(3)

(4)

(5)

B.C. 1 C1 0

B.C. 2 v(0) 0 C2 0

DIFFERENTIAL EQUATIONS (L2 xL)

(6)

(7)

(8)EIvRBLx2

2RB

x3

6C3xC4

EIvRBLxRBx2

2C3

EIvMRB(Lx)

v(0) 0

EIv (PRB)x3

6 PL

2RBLx

2

2C1xC2

EIv (PRB)x2

2 PL

2RBLxC1

EIvM (PRB) x PL2RBL


9/10


B.C. 3 v(L) 0 (9)

B.C. 4 continuity condition at pointC

at :

or (10)

From eq. (9): (11)

B.C. 5 continuity condition at pointC.

at :

or

From eq. (1):

From eq. (2):

SHEAR-FORCE AND BENDING MOMENT DIAGRAMS

MAPL

2RBL

3PL

16

RAPRB11P

16

RB5P

16

RBLL2

8RBL

3

48 PL

2

8L

2RBL

3

3

PL3

8

(PRB)L3

48 PL

2RBLL

2

8

(v)Left (v)RightxL

2

C4 RBL

3

3

PL3

8

C3 PL2

8

RBL2RBL

2

8C3

(PRB)L2

8 PL2 RBLL2

(v)Left (v)RightxL

2

C3LC4 RBL

3

3DEFLECTION CURVE FOR 0 xL/2 (FROM EQ. 5)

(0 xL/2)

DEFLECTION CURVE FORL/2 xL (FROM EQ. 8)

(L/2 xL)

SLOPE IN RIGHT-HAND PART OF THE BEAM

From eq. (7):

Point of zero slope:

0.5528L

MAXIMUM DEFLECTION

DEFLECTION CURVE

max(v)xx1 0.009317PL3

EI

x1L5 555x

21 10Lx1 4L

2 0

vP

32EI

(4L2 10Lx 5x2)

P

96EI

(Lx) (2L2 10Lx 5x2)

vP

96EI(2L3 12L2x 15Lx2 5x3)

vPx2

96EI(9L 11x)

11P

16

5P

16

V

O

3PL16

3L11

M

O

5PL 32

x1

3L

11 max

CA B


10/10

Method of Superposition

The problems for Section 10.4 are to be solved by the method of

superposition. All beams have constant flexural rigidity EI unless

otherwise stated. When drawing shear-force and bending-moment

diagrams, be sure to label all critical ordinates, including maximum

and minimum values.

Problem 10.4-1 A propped cantilever beamAB of lengthL carries

a concentrated load P acting at the position shown in the figure.

Determine the reactionsRA

,RB

, andMA

for this beam. Also, draw the

shear-force and bending-moment diagrams, labeling all critical ordinates.



A

a b

B

L

MA

RARB

P

SelectRB

as redundant.

EQUILIBRIUM

RA P R

B M

A Pa R

BL

RELEASED STRUCTURE AND FORCE-DISPLACEMENT

RELATIONS

COMPATIBILITY

B (

B)1 (

B)2 0

OTHER REACTIONS (FROM EQUILIBRIUM)


MAPab

2L2(L b)RA

Pb

2L3(3L2 b2)

RBPa2

2L3(3L a)

B Pa

2

6EI(3L a)

RBL3

3EI 0

A Ba

P

b

(B)1Pa2

(3La) 6EI

(B)2

PBL

3

(3L

a) 3EIABL

RB

RA

RB

V

O

MA

M1

M1RBbM

O

Mechanics of Materials Chap 10-01

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