Chapter 6

Beam DeflectionsINTRODUCTION

Beam Deflections, in this chapter we consider the rigidity of beams. From the word itself, it is the amount of deformation of beams. The deformation of a beam is usually expressed in terms of its deflections from its original unloaded position. The deflection is measured from the original neutral surface of the beam to the neutral surface of the deformed beam. The configuration assumed by the deformed neutral surface is known as the elastic curve of the beam.

One of the most important applications of beam deflections is to obtain equations with which, in combination with the conditions of static equilibrium, statically indeterminate beams can be analysed.

Methods used in determining beam deflections

1. Double Integration2. Area-Moment Method

Double-Integration Method

Formula/Process:

EI d y2

d x2 =M

EI d y2

d x2 =M - 1st Integration

EI dydx

=Mdx+C1

EI dydx

=Mdx+C1 - 2nd Integration

EI y=Mdxdx+C1 x+C2

Where:

Y = value of deflection (downward movement) of beam from the deflected beam (mm)

EI = Flexural Rigidity of beam (N.mm2)

E = Modulus of Elasticity (N/mm2)

I = Moment of Inertia of beam (cross section @ neutral axis) (mm4)

M = Moment equation of beam

C1∧¿ C2 = Constants of Integration

Illustrative Problems

1. Determine the maximum deflection δ in a simply supported beam of length L carrying a concentrated load P at midspan.Figure:

Solution:

EI y ' '=M

EI y ' '=12Px−P( x−1

2L)

EI y '=14P x2−1

2P (x−1

2L)

2

+C1

EIy= 112

P x3−16P(x−1

2L)

3

+C1 x+C2

@x=0 ; y=0 , Therefore ,C2=0

@x=L; y=0

0= 112

P L3−16P (L− 1

2L)

3

+C1L+0

0= 112

P L3− 148

PL3+C1L

C1=−116

PL2

Thus,

EIy= 112

P x3−16P(x−1

2L)

3

− 116

P L2 x

Maximum deflection will occur @x=12L (midspan)

EI ymax=1

12P( 1

2L)

3

−16P( 1

2L− 1

2L)

3

− 116

P L2( 12L)

EI ymax=1

96P L3−0− 1

32PL3

ymax=−PL3

48 EI

The negative sign indicates that the deflection is below the undeformed neutral axis.

Therefore,

δmax=−P L3

48EI

2. Determine the maximum deflection δ in a simply supported beam of length L carrying a uniformly distributed load of intensity wo applied over its entire length.Figure:

Solution:

EI y ' '=M

EI y ' '=12woLx−wo x (1

2x )

EI y ' '=12woLx−

12wo x

2

EI y '=14w o Lx

2−16wo x

3+C1

EIy= 112

wo Lx3− 1

24wo x

4+C1x+C2

@x=0 ; y=0 , Therefore ,C2=0

@x=L; y=0

0= 112

wo L4− 1

24wo L

4+C1L+0

C1=−124

w oL3

Therefore,

EIy= 112

wo Lx3− 1

24wo x

4− 124

wo L3 x

Maximum deflection will occur @x=12L (midspan)

EI ymax=1

12woL( 1

2L)

3

− 124

wo( 12L)

4

− 124

wo L3(1

2L)

EI ymax=1

96wo L

4− 1384

woL4− 1

48woL

4

EI ymax=−5384

wo L4

δmax=5wo L

4

384 EI

Taking W=w oL

δmax=5¿¿¿

δmax=5W L3

384 EI

3. As shown in the Fig. P-609, a simply supported beam carries two symmetrically placed concentrated loads. Compute the maximum deflectionδ .Figure:

Solution:

By symmetry

R1=R2=P

EI y ' '=Px−P ( x−a )−P(x−L+a)

EI y '=12P x2−1

2P ( x−a )2−1

2P (x−L+a )2+C1

EIy=16P x3−1

6P ( x−a )3−1

6P ( x−L+a )3+C1 x+C2

@x=0 ; y=0 , Therefore ,C2=0

@x=L ; y=0

(0=16P L3−1

6P (L−a )3+C1 L)6

0=PL3−P (L3−3L2a+3 La2−a3 )−Pa3+6C1 L

0=PL3−P L3+3 PL2a−3PLa2+Pa3−P a3+6C1 L

0=3P L2a−3PLa2+6C1 L

C1=−12

Pa(L−a)

Therefore,

EIy=16P x3−1

6P ( x−a )3−1

6P ( x−L+a )3−1

2Pa(L−a) x

Maximum deflection will occur @x=12L (midspan)

EI ymax=16P( 1

2L)

3

−16P( 1

2L−a)

3

−12Pa (L−a )(1

2L)

EI ymax=1

48P L3−1

6P [ 1

2(L−2a )]

3

−14P L2a+ 1

4PLa2

EI ymax=148

P L3−16P [L3−3 L2 (2a )+3 L (2a )2− (2a )3 ]−1

4P L2a+ 1

4PLa2

EI ymax=1

48P L3− 1

48PL3+ 1

8P L2a−1

4PLa2+ 1

6Pa3−1

4P L2a+ 1

4PLa2

EI ymax=−18

PL2a+ 16P a3

EI ymax=−124

Pa(3 L2−4 a2)

ymax=−Pa24 EI

(3 L2−4 a2)

δmax=Pa

24 EI(3 L2−4 a2)

4. Compute the value of EIδ at midspan for the beam loaded as shown in Fig. P-611. If E=10GPa, what value of I is required to limit the midspan

deflection to 1360 of the span?

Figure:

Solution:

∑MR 2=0

R1(4)=300 (2 )(3)

R1=450 N

∑MR 1=0

R2(4)=300 (2 )(1)

R2=150 N

EI y ' '=450 x−12

(300 ) x2+ 12

(300 ) ( x−2 )2

EI y '=4502

x2−3006

x3+ 3006

( x−2 )3+C1

EIy=4506

x3−30024

x4+ 30024

( x−2 )4+C1 x+C 2

@x=0 ; y=0 , Therefore ,C2=0

@x=4 ; y=0

0=4506

(4)3−30024

(4 )4+ 30024

(4−2 )4+C1(4)

C1=−450 N .m2

Therefore,

EIy=4506

x3−30024

x4+ 30024

( x−2 )4−450 x

@x=2m (midspan)

EI ymidspan=450

6(2)3−300

24(2 )4+ 300

24(2−2 )4−450(2)

EI ymidspan=−500 N .m3

EI δmidspan=500 N .m3

Maximum midspan deflection

δmidspan=1

360L= 1

360( 4 )= 1

90m

δmidspan=100

9mm

Thus,

10000 I( 1009 )=500(10003)

I=4 500 000mm4

I=4.5x 106mm4

5. Compute the midspan value EIδ for the beam loaded as shown in Fig. P-612.Figure:

Solution:

∑MR 2=0

R1(6)=600 (3 )(3.5)

R1=1050N

∑MR 1=0

R2(6)=600 (3 )(2.5)

R2=750 N

EI y ' '=1050 x−12

(600 ) (x−1 )2+ 12

(600 ) ( x−4 )2

EI y '=12(1050) x2−1

6(600 ) ( x−1 )3+ 1

6(600 ) ( x−4 )3+C1

EIy=16

(1050 ) x3− 124

(600 ) ( x−1 )4+ 124

(600 ) ( x−4 )4+C1 x+C2

@x=0 ; y=0 , Therefore ,C2=0

@x=6 ; y=0

0=16

(1050 )(6)3− 124

(600 ) (6−1 )4+ 124

(600 ) (6−4 )4+C1(6)

C1=−3762.5 N .m2

Therefore,

EIy=16

(1050 ) x3− 124

(600 ) ( x−1 )4+ 124

(600 ) ( x−4 )4−3762.5 x

@x=3m (midspan)

EI ymidspan=16

(1050 )(3)3− 124

(600 ) (3−1 )4+ 124

(600 ) (3−4 )4+C1(3)

EI ymidspan=−6962.5 N .m3

Thus,

EI δmidspan=6962.5 N .m3

Exercise 6.2 Beam Deflections

1. For the beam loaded as shown in Figure, determine (a) the deflection and slope under the load P and (b) the maximum deflection between the supports.

2. Compute the value of EIy at the right end of the overhanging beam shown in Fig. P-615.

3. If E = 29 x 106 psi, what value of I is required to limit the midspan deflection to 1

360 of the span for the beam in Fig. P-613?

4. For The beam loaded as shown in Fig. P-614, calculate the slope of the elastic curve over the right support.

5. Determine the value of the couple M for the beam loaded as shown in the Figure so that the moment of area about A of the M diagram between A and B will be zero. What is the physical significance of this result?

Theorems of Area-Moment Method

- A useful and simple method of determining slopes and deflection in beams it involves.

a. Area of the moment diagramb. Moment of the area of the moment diagram

Theorem I

The change of slope between tangents drawn to the elastic curve at any two points A

and B is equal to the product 1EI multiplied by the area of the moment diagram between those

two points.

- positive change of slope, θAB

is counterclockwise from the left tangent

- negative change of slope, θAB

is clockwise from the left tangent

Formula:

θAB=1EI

(Area)AB

Where:

(Area)AB = Area of the moment diagram under AB

Theorem II

The deviation of any point B relative to the tangent drawn to the elastic curve at any point A, in a direction perpendicular to the original position of the beam is equal to the product

of 1EI multiplied by the moment of the area about B of that part of the moment diagram

between points A and B.

Formula:

tB /A=1EI ( AreaAB ) ∙ XB

Where:

( AreaAB ) ∙ X B = moment of the area of the moment diagram between A and B

- positive deviation B is located above the tangent

- negative deviation B is located below the tangent

Determining the Area of the Moment Diagram and the moment of the area of the moment diagram

Using the moment diagram by Parts

- Process to compute easily the area of the moment diagram and its moment

Couple or Moment LoadA=CLM=−C

X=12L

Degree: zero

Concentrated LoadA=1

2P L2

M=−Px

X=13L

Degree: first

Uniformly Distributed LoadA=1

6PL3

M=−12

wo x2

X=14L

Degree: second

Uniformly Varying LoadA= 1

24P L4

M=−16

wo x2

X=15L

Degree: third

Application of Theorems of Area-Moment Method

DEFLECTION OF CANTILEVER BEAMS

Note: Tangential Deviation

- Can be applied at any point in the elastic curve is the distance from any point on the elastic curve to a tangent drawn to the curve at some other point.

But: Deviation generally is not equal to deflection

Cantilever Beam

Illustrative Problems:

1. The cantilever beam shown in Fig. P-636 has a rectangular cross-section 50 mm wide by h mm high. Find the height h if the maximum deflection is not to exceed 10 mm. Use E = 10 GPa.Figure:

Solution:

−δ=−t CA

= 1EI

¿

−t CA

= 1EI [ (16 ) (4 )( 2

3x 4)1

2(4 ) (2 )( 2

3x 2+2)]

−δ=−98.67EI

KN ∙m3

−EIδ=−98.67 KN ∙m3

(10 x103 ) I (10 )=98.67 (103)(109)

I=986.7 x 106mm4

I=bh3

12

−986.7 x 106=50h3

12

h=618.67mm

2. For the beam loaded as shown in the Fig. P-637, determine the deflection 6 ft from the wall. Use E = 1.5 x106 psi and I = 40 in4.

Figure:

Solution:

δ = tB/C = 1EI

(Area)BC∙ xB

δ = tB/C = 1EI [1

2(3840 ) (6 )(1

3x6)−2560 (6 ) (3 )−1

3(1440 ) (6 )( 1

4x 6)]

δ = −27360

EI lb∙ft3

δ = −27360(123)(1.5 x106)(40)

δ = 0.788 in

3. For the cantilever beam shown in Fig. P-638, determine the value of EI∂ at the left end. IS this deflection upward or downward?Figure:

Solution:

δ = tA/C = 1EI

(Area)AC∙ x A

δ = tA/C = 1EI [1

2(1000 ) ( 4 )( 2

3x 4)−2000 (2 ) (3 )]

EIδ = 6.67 KN ∙ m3 ; upward

4. For the cantilever beam shown in Fig. P-641, what value of P will cause zero deflection at A?

Figure:

Solution:

δ = tA/C = 1EI

(Area)AC∙ x A

0 = 1EI

[ 12

( 4 P ) (4 )( 23x 4)−400 (2 ) (3 )]

P = 112.5 N

5. Find the maximum deflection for the cantilever beam loaded as shown in Fig. P-642 if the cross section is 50 mm wide by 150 mm high. Use E = 69 GPa.Figure:

Solution:

δ = tC/A = 1EI

(Area)AC∙ xC

δ = tC/A = 1EI [1

2(12 ) (3 )( 1

3x 3)−10 (3 ) (1.5 )−1

3(2 ) (1 )( 1

4x1)]

δ = 27.17 KN ∙ m3

δ = 27.17 (103 )(109)

(69 x103 )(14.06 x 106)

∂ = 28.01 mm

Exercise 6.5 Cantilever Beams

1. Find the maximum value of EIδ for the cantilever beam.

2. Determine the maximum deflection for the beam loaded as shown.

3. For the beam shown, determine the value of I that will limit the maximum deflection to 0.50 in. Assume that E=1.5×106 psi .

4. Find the maximum value of EIδ for the cantilever beam.

5. For the cantilever beam loaded as shown, determine the deflection at a distance x from the support.

DEFLECTIONS OF SIMPLY SUPPORTED BEAMS

The deflection δ at some point B of a simply supported beam can be obtained by the following steps:

By Similar Triangle

CEL

= BDx

tC /A

L=δB+ tB/ A

x

δB=xLtC /A−tB /A

Where:

tC/ A=1EI

(Area)AC ∙ xC

tB /A=1EI

(Area)AB ∙ xB

ILLUSTRATIVE PROBLEMS

1. Compute the midspan value of EIδ for the beam shown. (Hint: Draw the M diagram by parts, starting from midspan towards the ends. Also take advantage of symmetry to note that the tangent drawn to the elastic curve at midspan is horizontal.Figure:

Solution:

t AB

= 1EI

¿

δ= 1EI [( 1

2×3000×2.5)( 2

3×2.5)+( 1

3×75×0.5)( 3

4(0.5 )+2)−( 1

3×1875×2.5)( 3

4×2.5)]

EIδ=3350 N ∙m3

2. For the beam shown, find the value of EIδ at 2 ft. from R2. (Hint: Draw the reference tangent to the elastic curve at R2.)

Figure:

Solution:

t AC

= 1EI

¿

t AC

= 1EI [( 1

2× 2560

3×4 )( 2

3×4)+( 1

2× 640

3×2)( 1

3(2 )+4)−(1

3×640×4)( 3

4×4)]

t AC

= 1EI

2986.67lb ∙ ft3

t Bc

= 1EI

¿

t AC

= 1EI

142.22 lb ∙ ft3

t AC

6=δB+t B

C

2

δB=26t AC

−t BC

δB=26 (2986.67

EI )−142.22EI

∴EI δB=853.34 lb ∙ ft3

3. Find the value of EIδ under each concentrated load of the beam shown.Figure:

Solution:

By ∆ ' syB

3=1400

8∴ y B=525

yC1

7=1400

8∴ y c1=1225

yC 2

4=1000

5∴ yC 2=800

t DA

= 1EI

¿

t DA

= 1EI [( 1

2×1400×8)( 1

3×8)−(1

2×1000×5)( 1

3×5)−( 1

2×400×1)( 1

3×1)]

t DA

= 1EI

10700 lb∙ ft3

t CA

= 1EI

¿

t CA

= 1EI [( 1

2×1225×7)( 1

3×7)−( 1

2×800×4)( 1

3×4 )]

t CA

= 1EI

7870.83lb ∙ ft3

t BA

= 1EI

¿

t BA

= 1EI [( 1

2×525×3)( 1

3×3)]

t BA

= 1EI

787.5lb ∙ ft3

t DA

8=δB+t B

A

3

δB=38t DA

−t BA

δB=38

(10700 )−787.5

∴EI δB=3225 lb ∙ ft3

t DA

8=δC+t C

A

7

δC=78t DA

−t CA

δC=78

(10700 )−7870.83

∴EI δC=1491.67 lb ∙ ft3

4. Find the value of EIδ at the point of application of the 200 Nm couple.Figure:

Solution:

t DA

= 1EI

¿

¿ 1EI [( 1

2×1275×3)( 1

3(3 )+1)+(1

2×75×1)( 2

3×1)−( 1

2×1000×2)( 1

3(2 )+1)]

t DA

= 1EI

2183.33 N ∙m3

t CA

= 1EI

¿

¿ 1EI [( 1

2×1275×3)( 1

3×3)−( 1

2×1000×2)( 1

3×2)]= 1

EI1245.83N ∙m3

t DA

4=δC+t C

A

3δC=

34t DA−t C

AδC=

34

(2183.33 )−1245.83

∴EI δC=391.67N ∙m3

5. Determine the midspan value of EIδ for the beam shown.Figure:

Solution:

t AB

= 1EI

¿

t AB

= 1EI [( 1

2×1600×6)( 1

3×6)−( 1

4×1600×4)( 1

5×4 )]

t AB

= 1EI

8320N ∙m3

t MB

= 1EI

¿

t MB

= 1EI [( 1

2×800×3)( 1

3×3)−( 1

4×25×1)( 1

5×1) ]

t MB

= 1EI

1198.75N ∙m3

t AB

6=δM+t M

B

3

δM=36t AB

−t MB

δM=12

(8320 )−1198.75

∴EIδM=2961.25N ∙m3

Exercise 6.6 Simply Supported Beams

1. The beam shown in Fig. P-658, find the value of EIδ at the point of application of the couple.

2. A simple beam supports a concentrated load placed anywhere on the span, as shown in Fig. P-659. Measuring x from A, show that the maximum deflection occurs at x = √[(L2 - b2)/3].

3. The middle half of the beam shown in Fig. P-664 has a moment of inertia 1.5 times that of the rest of the beam. Find the midspan deflection. (Hint: Convert the M diagram into an M/EI diagram.

4. For the beam shown in Fig. P-668, compute the value of P that will cause the tangent to the elastic curve over support R2 to be horizontal. What will then be the value of EIδ under the 100-lb load?

5. Determine the value of EIδ at the left end of the overhanging beam shown in Fig. P-670. Overhang Beam with Triangle and Moment Loads.

SOLUTIONS TO EXERCISE 6.2

SOLUTION 1

a.) Find for R1∧R2.

EI d2 y

d x2 =M

EI∫ d2y

d x2 =∫[−P (b )a

( x )+PLa

(x−a)]∫ dy

dx=∫ [−Pb x2

2a+PL (x−a )2

2a ]EIy=−Pb x3

6a+PL ( x−a )3

6a+C1 x+C2

Boundary Conditions:

At x=0; y=0; C2=¿0

At x=L; y=0; C1=−Pab

6

To get C1:

At x=a; y=0; C1=¿0

0=−Pba3

6a+0+C1 (a )+0

−C1 (a )=−Pba2

6

∴C1=Pba

6

Part (a): Slope and deflection under load P

Slope under load P: (note x=a+b=L)

EI dydx

=−Pb x2

2a+PL ( x−a )2

2a+ Pab

6

EI dydx

=−b2a

P (a+b )2+ a+b2a

(Pb )2+ Pab6

EI dydx

=−b2a

P (a2+2ab+b2 )+ ab2+b3

2aP+Pab

6

EI dydx

=−ab2

P+b2P – b3

2aP+ b2

2P+ b3

2aP+ Pab

6

EI dydx

=−12

b2 P– 13abP

EI dydx

=−16

b (3b+2a ) P

EI dydx

=−16

b [2 (a+b )+b ]P

∴EI dydx

=−16

b (2 L+b ) P

Deflection under the load P: (note x = a+b = L)

EIy=−Pb x3

6a+PL ( x−a )3

6a+Pab

6(x )

EIy=−b6a

P (a+b )3+ a+b6a

P (b3 )+ Pab6

(a+b )

EIy=¿ −b6a

P (a3+3a2b+3ab2+b3 )+ ab3+b4

6aP+ Pab

6(a+b )

EIy=−a2b6

P−ab2

2P−b3P

2− b4

6aP+ b3P

6+ b4

6 aP+ a2b

6P+ ab2

6

EIy=−−13

ab2P−13b3 P

EIy=−−13

(a+b )b2 Pab

∴EIy=−13

Lb2P

Part (b): Maximum deflection between the supports

The maximum deflection between the supports will occur at the point where dydx

=0.

EI dydx

=−b2a

P x2+ L2a

P ( x−a )2+ ab6

P

At dydx

=0 , ( x−a ) donot exist , thus ;

0=−b2a

P x2+ ab6

P

x2=13a2

x= 1√3

aAtx= 1√3

a

EIy=−b6a

P( 1√3

a)3

+ ab6

P( 1√3

a)EIy=−a2b

6¿¿

EIy= a2b6√3

P(−13

+1)EIy= a2b

6√3P( 2

3 )∴EIymax=

a2b9√3

P

SOLUTION 2

∑M R2=0]

R1(10)=1000(4)−400(3)(1.5). :R1 = 220 lb

∑M R1=0]R2(10)=1000(6)−400(3)(11.5 ). :R2 = 1980 lb

M=220 x−1000 ( x−6 )+1980 ( x−10 )− 400 ( x−10 )2

2

EI d2 y

d x2 =M

EI∫ d2 yd x2 =∫ [220 x−1000 ( x−6 )+1980 (x−10 )−400 ( x−10 )2

2 ]EI∫ dy

dx=∫ [ 220 x2

2−1000 ( x−6 )2

2+ 1980 ( x−10 )2

2−400 ( x−10 )3

6+C1]

EIy= 220 x3

6−

1000 ( x−6 )3

6+

1980 ( x−10 )3

6−

400 ( x−10 )4

24+C1 x+C2

Boundary condition

At x= 0, y= 0, therefore C2= 0

At x= 10ft, y= 0,

0=220 (10 )3

6−1000 (10−6 )3

6+ 1980 (10−10 )3

6−400 (10−10 )4

24+10C1

0=220 (10 )3

6−

1000 (10−6 )3

6+0−0+10C1

. :C1=−2600lb ft2

EIy=220 x3

6−1000 ( x−6 )3

6+1980 ( x−10 )3

6−400 ( x−10 )4

24+C1 x+C2

Therefore,

EIy=220 x3

6−1000 ( x−6 )3

6+1980 ( x−10 )3

6−400 ( x−10 )4

24−2600 x+0

At the right end of the overhanging beam, x= 13ft

EIy=220 (13 )3

6−

1000 (13−6 )3

6+

1980 (13−10 )3

6−

400 (13−10 )4

24−2600 (13 )

∴EIy=−2850 lbft3

SOLUTION 3

ΣM R2=012R1=2400(6)(5)∴R1=6000lbΣM R1=012R2=2400(6)(7)∴R2=8400 lb

EIy ' '=6000 x−12

(2400 ) ⟨ x−4 ⟩2+ 12

(2400 ) ⟨ x−10 ⟩2

EIy ' '=6000 x−1200 ⟨ x−4 ⟩2+1200 ⟨ x−10 ⟩2

EIy '=3000 x2−400 ⟨ x−4 ⟩3+400 ⟨ x−10 ⟩3+C1

EIy=1000 x3−100 ⟨ x−4 ⟩4+100 ⟨ x−10 ⟩ 4+C1 x+C2

At x=0 , y=0 , therefore ,C2=0

Atx=12 ft , y=0

0=1000(123)−100 (12−4 )4+100 (12−10 )4+12C1

∴C1=−110000 lb ⋅ ftThereforeEIy=1000 x3−100 ⟨ x−4 ⟩4+100 ⟨ x−10 ⟩ 4−110000 x

E=29×106 psiL=12 ft

Atmidspan, x=6 ft y=−1/360 (12)=−1/30 ft=−2/5∈¿

Thus ,EIy=1000 x3−100 ⟨ x−4 ⟩4+100 ⟨ x−10 ⟩ 4−110000 x

(29×106 ) I (−25 )=[1000(63)+100(24)−110000(6)](123)

∴ I=66.38 i n4

SOLUTION 4

∑MR 2=0

R1 (8 )+240(2)=100 ( 4 )(6)

R1=240 lb

∑MR 1=0

R2 (8 )=240 (10 )+100 (4)(2)

R2=400 lb

EI y ' '=240 x−12

(100 ) x2+ 12

(100 ) (x−4 )2+400(x−8)

EI y '=2402

x2−16

(100 ) x3+ 16

(100 ) ( x−4 )3+ 4002

(x−8)2+C1

EIy=2406

x3− 124

(100 ) x4+ 124

(100 ) ( x−4 )4+ 4006

( x−8 )3+C1 x+C2

@x=0 ; y=0 , Therefore ,C2=0

@x=8 ; y=0

0=2406

83− 124

(100 ) 84+ 124

(100 ) (8−4 ) 4+ 4002

(8−8 )2+C1(8)

C1=−560 lb . ft2

Therefore,

EI y '=2402

x2−16

(100 ) x3+ 16

(100 ) ( x−4 )3+ 4002

(x−8)2+C1

At the right support, x=8 ft

EI y '=2402

82−16

(100 ) 83+ 16

(100 ) (8−4 )3−560

EI y '=−10403

lb . ft2

y '=−10403 EI

lb . ft2

SOLUTION 5

ΣM A=04 R2+M=100(4)(2)∴R2=200−14 MΣM B=04 R1=100(4)(2)+M∴R1=200+14 M(AreaAB) . X A=012

(4 ) (800−M )( 43 )−1

3(4)(800)(1)=0

83(800−M )=3200

3

∴M=400 lb⋅ ft

The uniform load over span AB will cause segment AB to deflect downward. The moment load equal to 400 lb·ft applied at the free end will cause the slope through B to be horizontal making the deviation of A from the tangent through B equal to zero. The downward deflection therefore due to uniform load will be countered by the moment load.

SOLUTIONS TO EXERCISE 6.5

SOLUTION 1

EI tB /A=(AreaAB)Xb

EI tB /A=12L (PL )( 1

3L)−PaL( 1

2L)−1

2¿)P (L−a )[ 1

3(L−a )]

EI tB /A=16P L3−1

2PL2a−1

6P¿

EI tB /A=16PL3−1

2PL2a−1

6P (L3−3 L2a+3 La2−a3)

EI tB /A=16PL3−1

2PL2a−1

6P L3+ 1

2PL2a−1

2PLa2+ 1

6Pa3EI tB /A=

−12

PLa2+ 16Pa3

EI tB /A=−16

Pa2 (3L−a )

Therefore,

EI δmax=16Pa2 (3L−a )

SOLUTION 2

t A /B=1EI

(AreaAB)X a

t A /B=1EI [1

2(L )( 1

2wo L

2)( 13L)−3

8woL

2 (L )(12L)−1

3 ( 18wo L

2)( 12L)( 1

8L)]

t A /B=1EI [ 1

12wo L

4− 316

woL4− 1

384wo L

4]t A /B=

1EI [−41

384woL

4]t A /B=

−41woL4

384 EI

Therefore,

δmax=41wo L

4

384 EI

SOLUTION 3

M=550 lb ∙ ft

R=150lb

t A /B=1EI

(AreaAB)X a

−5= 1EI [ 1

2(300 ) (2 )( 26

3 )−550 (2 ) (9 )−14

(5 ) (250 ) (7 ) ] (123 )

−5= 1(1.5×106 ) I

(−16394400 )

I=2.18592¿4

SOLUTION 4

R=14wo L

M= 524

woL2

EI t A /B=( AreaAB ) X a

EI t A /B=124

wo L4− 5

48wo L

4− 11920

wo L4

EI t A /B=−1211920

woL4

Therefore,

EI δmax=121

1920woL

4

SOLUTION 5

yx=wo

L

y=w o

Lx

M=16wo L

2

R=12wo L

Moments about B:

Triangular force to the left of B:

M 1=−12

(L−x ) (wo− y )(13 )(L−x )

M 1=−16

¿

M 1=−wo ¿¿

Triangular upward force:

M 2=12

( xy )( 13x)=1

2x2 wo x

L=wo x

3

LRectangle(woby x ):

M 3=−wo x (12x)=−1

2wo x

2

Reactions R and M:

M 4=Rx=12wo Lx

M 5=−M=−16

wo L2

Deviation at B with the tangent line through C

EI tB /C=(AreaBC)X b

EI t BC

=14x (wo x

3

6 L )(15x)+ 1

2x ( 1

2woLx )( 1

3x)−( 1

6woL

2) x ( 12x)−1

3x( 1

2wo x

2)EI t B

C

=wo

120Lx5+

w oL12

x3−woL

2

12x2−

wo

24x4

EI t BC

=wo x

2

120L( x3+10 L2 x−10 L3−5 Lx2 )

Therefore,

EIδ=−wo x

2

120L(x3+10L2 x−10 L3−5 Lx2 )

EIδ=−wo x

2

120L(10 L3−10 L2 x+5 Lx2−x3 )

SOLUTIONS TO EXERCISE 6.6

SOLUTION 1

ya=M

L

y=MaL

EI tB /A=( AreaAB ) X B

EI tB /A=12(ay)( 1

3a)

EI tB /A=16a2( Ma

L )EI tB /A=

−M a3

6 L

EI tC/ A=(AreaAB ) X C

EI tC/ A=12

(LM )( 13L)−M (L−a )[1

2(L−a )]

EI tC/ A=16M L2−1

2M (L−a)2

tC /A

L=δB+ tB/ A

a

EIδB=aLEItC/ A−EI tB /A

EIδB=aL

¿

EIδB=aL [ 1

6M L2−1

2M (L−a )2−1

6Ma2]

EIδB=Ma6 L [L2−3 (L−a )2−a2 ]

EIδB=Ma6 L [L2−3 (L2−2 La+a2 )−a2 ]

EIδB=Ma6 L

[L2−3 L2+6 La−3a2−a2 ]

EIδB=Ma6 L

[−3 L2+6 La−−4 a2 ]

SOLUTION 2

ΣM R1=0LR2=Pa

∴R2=PaL

yx=Pb

L

y= PbL

x

t A /D=1EI (AreaAD ) X A

t A /D=1EI [ 1

2xy ( 2

3x)]

t A /D=1EI [ 1

3x2 y ]

t A /D=1EI [ 1

3x2(PbL x )]

t A /D=1EI [ PbL x3]

tC/D=1EI ( AreaCD) XC

tC/D=1EI [ 1

6(L−x )2 (Pb− y )+ 1

2(L−x )2 y−1

6Pb3]

tC/D=1EI [ 1

6(L−x )2(Pb− Pb

Lx)+ 1

2(L−x )2( PbL x)−1

6Pb3]

tC/D=1EI [ 1

6Pb (L− x )2(1− x

L )+ 12Pb (L−x )2( xL )−1

6Pb3]

tC/D=1EI [ Pb6 L

(L−x )3+ Pb2L

(L−x )2

x− Pb3

6 ]

From the figure:

t A /D=tC /D

1EI

Pb3 L

x3=1EI [ Pb6 L

(L−x )3

+Pb2L

(L−x )2

x− Pb3

6 ]Pb3L

x3= Pb6 L

(L−x )3

+ Pb2L

(L−x )2

x− Pb3

62x3

L=

(L−x )3

L+

3 (L−x )2 xL

−b2

2 x3=(L−x )3+3 (L−x )2 x−Lb2

2 x3=L3−3 L2x+3 Lx2−x2+3 L2 x−6Lx2+3 x2−Lb2

0=L3−3 Lx2−Lb2

0=L2−3 x2−b2

3 x2=L2−b2

x=√(L¿¿2−b2)/3¿

SOLUTION 3

t A /C=1EI ( AreaAC ) X A

t A /C=Pa3

6 EI+ Pa3

2EI+ 5 Pa3

18 EI

t A /C=17 Pa3

18 EI

Therefore,

δmidspan=17 Pa3

18 EI

SOLUTION 4

SOLUTION 5