18 Masupp Presentation

. . . . . .

Review Moment-Area method Method of Superposition

......

Lecture : Bending (VI) — Methods of Moment-Areaand Superposition

Yubao Zhen

Lecture : Bending (VI) — Methods of Moment-Area and Superposition

. . . . . .

.. Review: Integration method for deflection curve

For constant EI, differential eqn. of deflection

EIdvdx=M(x) bending-moment equation

EIdvdx=V(x) shear-force equation

EIdvdx= − q(x) load equation

Problem solving procedure:... obtain either one ofM(x),V(x) and q(x)... integrate the corresponding equation... determine the integration constants by

boundary/continuity/symmetry conditions... substitute back for deflection equation, and further results

. . . . . .

.. Outline

... Moment-areamethod (弯矩 -面积法)deflection angle and deflection at a single location.

... first theorem: deflection angle

... second theorem: deflection

... Method of superposition for deflection (叠加法)application of tabulated formulas

. . . . . .

.. Outline

. . . . . .

.. Outline

. . . . . .

.. Outline

. . . . . .

.. Introducing two new methods for slope and deflection

method of integration:() standard but tedious procedure; () results as function of x

what if only those results at some specific locations are required?getting v(x)Ð→ v(x)——waste of time!

Moment-areamethod (弯矩 -面积法，面积 -力矩法)... Best for finding slope and deflection at discrete points... A semigraphical technique (easy to use)... Best suited for simplemoment diagrams

i.e., with only concentrated and couple moments applied

Method of superposition... Based on the linear relation between deflection and load... How it works: tabulated results for various beam loadings

appendix G

. . . . . .

appendix G

. . . . . .

appendix G

. . . . . .

appendix G

. . . . . .

appendix G

. . . . . .

appendix G

. . . . . .

appendix G

. Moment-areamethod

(弯矩 -面积法)

. . . . . .

Review Moment-Area method Method of Superposition First theorem Second theorem Examples

.. Moment-area method: overview

two theorems:

... st theorem: deals with deflection angle

... nd theorem: deals with deflection

features:

... theoretical background:

bending moment equation EIdvdx=M(x)

... information requiredM-diagram and a sketch of deflection curve

... operationsintegration(s) ofM-diagram: area and its first moment.advantage: no need to solve integration constants

. . . . . .

two theorems:

features:

. . . . . .

two theorems:

features:

. . . . . .

two theorems:

features:

. . . . . .

two theorems:

features:

. The first theoremofmoment-areamethod

—— relative rotation of tangents

弯矩 -面积点法第一定理：切线间的相对转角

. . . . . .

.. Deflection angle: relative rotation

θ = dvdx

←Ð small slope assumptionÐ→ κ = dθds≈ dθdx= MEI

relative rotation of tangents on a differential element

EIdvdx=M(x) = EI d

dx(dvdx)Ð→ d

dxθ = M

EIÐ→ dθ = M

. . . . . .

θ = dvdx

EIdvdx=M(x) = EI d

dx(dvdx)Ð→ d

dxθ = M

EIÐ→ dθ = M

. . . . . .

θ = dvdx

EIdvdx=M(x) = EI d

dx(dvdx)Ð→ d

dxθ = M

EIÐ→ dθ = M

. . . . . .

.. First moment-area theorem

9.6 MOMENT-AREA METHOD

In this section we will describe another method for finding deflectionsand angles of rotation of beams. Because the method is based upon twotheorems related to the area of the bending-moment diagram, it is calledthe moment-area method.

The assumptions used in deriving the two theorems are the same asthose used in deriving the differential equations of the deflection curve.Therefore, the moment-area method is valid only for linearly elasticbeams with small slopes.

First Moment-Area TheoremTo derive the first theorem, consider a segment AB of the deflectioncurve of a beam in a region where the curvature is positive (Fig. 9-22).Of course, the deflections and slopes shown in the figure are highlyexaggerated for clarity. At point A the tangent AA! to the deflectioncurve is at an angle uA to the x axis, and at point B the tangent BB! is atan angle uB. These two tangents meet at point C.

The angle between the tangents, denoted uB/A, is equal to the dif-ference between uB and uA:

uB/A " uB # uA (9-60)

Thus, the angle uB/A may be described as the angle to the tangent at Bmeasured relative to, or with respect to, the tangent at A. Note that theangles uA and uB, which are the angles of rotation of the beam axis at

626 CHAPTER 9 Deflections of Beams

M—EI

FIG. 9-22 Derivation of the first moment-area theorem

.First Moment-area theorem..

......

The angle θB/A between the tangentsto the deflection curve at two pointsAand B equals the area of theM/EIdiagram between these two points.

i.e. θB/A = ∫B

θB/A = ∫B

Adθ = ∫

M(x)EI

dx (use algebraic value forM)

. . . . . .

uB/A " uB # uA (9-60)

M—EI

......

i.e. θB/A = ∫B

θB/A = ∫B

Adθ = ∫

M(x)EI

. . . . . .

uB/A " uB # uA (9-60)

M—EI

......

i.e. θB/A = ∫B

θB/A = ∫B

Adθ = ∫

M(x)EI

. . . . . .

uB/A " uB # uA (9-60)

M—EI

......

i.e. θB/A = ∫B

θB/A = ∫B

Adθ = ∫

M(x)EI

. . . . . .

.. On the sign ofM and θB/A

counter-clockwise

clockwise

Obviously, θB/A = −θA/B. If the area underM/EI betweenA − B is

... positiveÐ→ θB/A > (tangent lines: atA→ at B: counter-clockwise)

... negativeÐ→ θB/A < (tangent lines: atA→ at B: clockwise)

. . . . . .

counter-clockwise

clockwise

. . . . . .

counter-clockwise

clockwise

. . . . . .

counter-clockwise

clockwise

. . . . . .

counter-clockwise

clockwise

. . . . . .

.. On application of First moment-area theorem

θB/A = θB − θA = ∫B

... one of θA and θB should be given (from B.C.s)

... integral on the r.h.s in general can be obtained by reading areas(with signs)

... don’t forget the EI , it isMEI

, notM(x) diagram

. . . . . .

, notM(x) diagram

. . . . . .

, notM(x) diagram

. The second theoremofmoment-areamethod

——deviation of a point from a tangent

弯矩 -面积点法第二定理：一点与另点切线的偏距

. . . . . .

.. Deviation of a point from a tangent line

= xdθ

tA/Bdθdt

dt ≈ ds′ = xdθ ,along with

dθ = MEI

Ð→ dt = xMEI

Ð→ tA/B = ∫B

Adt = ∫

dx = x∫B

dxtA/B: relative deviation of tangents

. . . . . .

= xdθ

tA/Bdθdt

dθ = MEI

Ð→ dt = xMEI

Ð→ tA/B = ∫B

Adt = ∫

dx = x∫B

. . . . . .

= xdθ

tA/Bdθdt

dθ = MEI

Ð→ dt = xMEI

Ð→ tA/B = ∫B

Adt = ∫

dx = x∫B

. . . . . .

.. Second moment-area theorem

tA/B = ∫B

dx = x∫B

dx = QMEI

.Secondmoment-area theorem..

......

The tangential deviation tA/B of pointA from tangent at point B is equal tothe first moment of the area of theM/EI diagram betweenA and B,

evaluated with respect toA. i.e. tA/B = x∫B

. . . . . .

.. On the sign of tA/B

... tA/B: tangent at B as reference;

... x: measured fromAÐ→ x ≥ ;

... x = ∣AC∣ > (C: centroid ofM/EI betweenA and B).

... the area∫B

dx can be positive/negative

positiveÐ→A being above the tangent extended from BnegativeÐ→A being below the tangent extended from B

. . . . . .

... the area∫B

. . . . . .

... the area∫B

. . . . . .

... the area∫B

. . . . . .

... the area∫B

. . . . . .

... the area∫B

. . . . . .

.. The mirror term tB/A

tB/A =∫A

Bx′MEI

=x′ ∫A

x′ = ∣BC∣ > , but the area∫A

dx′can be positive/negative

... positiveÐ→ B being above the tangent extended fromA

... negativeÐ→ B being below the tangent extended fromA

. . . . . .

tB/A =∫A

Bx′MEI

=x′ ∫A

. . . . . .

tB/A =∫A

Bx′MEI

=x′ ∫A

. . . . . .

tB/A =∫A

Bx′MEI

=x′ ∫A

. . . . . .

tB/A =∫A

Bx′MEI

=x′ ∫A

. . . . . .

.. tA/B and tB/A

x x′

dθdθ

tA/B tB/A

Caution: tA/B ≠ tB/A (∣AC∣ ≠ ∣BC∣)

to memorize tX/Y:the first sub-indexX() on the elastic curve; () the base for x = ∣XC∣;the second sub-index Y:reference point to send out a tangent

. . . . . .

.. tA/B and tB/A

x x′

dθdθ

tA/B tB/A

Caution: tA/B ≠ tB/A (∣AC∣ ≠ ∣BC∣)

to memorize tX/Y:the first sub-indexX() on the elastic curve; () the base for x = ∣XC∣;the second sub-index Y:reference point to send out a tangent

. . . . . .

.. On the application of moment-area method

Applicable areas:

... st theorem—— angle between two tangents/deflection angle

nd theorem—— tangential deviation/deflection

... best for simpleM/EI diagrams (straight line segments)concentrated loads, some simple distributed loads

... trick: may split the area belowM/EI to pieces for area/first momentcalculation

... less convenient for higher order loadings

. . . . . .

Applicable areas:

. . . . . .

Applicable areas:

. . . . . .

Applicable areas:

. . . . . .

.. Centroids of typical areas

parabolic(spandrel)

parabolic

(semisegment)

A = lh

rectangle

parabola (semi)segment:抛物线 (半)弓形面parabola spandrel:抛物线半余面

. Applications of

moment-areamethod

. . . . . .

.. Example

Given: cantilevered beam, EI =const, loaded at C a force PSolve: . θB and θC; . δB and δC

Solution: θB and θC;... constructM/EI diagram... sketch the deflection curve... only slopes to determineÐ→

apply the st theorem... θA = Ð→ θB = θB/A, θC =θC/A

... By the first moment-areatheorem

θB = θB/A =L/(− PL

EI− PL

EI) = −PL

EIθC = θC/A =

(−PL

EI)L = −PL

EIBoth slopes are clockwise.Lecture : Bending (VI) — Methods of Moment-Area and Superposition

. . . . . .

.. Example

EI− PL

EI) = −PL

EIθC = θC/A =

(−PL

EI)L = −PL

. . . . . .

.. Example

EI− PL

EI) = −PL

EIθC = θC/A =

(−PL

EI)L = −PL

. . . . . .

.. Example

EI− PL

EI) = −PL

EIθC = θC/A =

(−PL

EI)L = −PL

. . . . . .

.. Example

EI− PL

EI) = −PL

EIθC = θC/A =

(−PL

EI)L = −PL

. . . . . .

.. Example

EI− PL

EI) = −PL

EIθC = θC/A =

(−PL

EI)L = −PL

. . . . . .

.. Example

EI− PL

EI) = −PL

EIθC = θC/A =

(−PL

EI)L = −PL

. . . . . .

.. Example

EI− PL

EI) = −PL

EIθC = θC/A =

(−PL

EI)L = −PL

. . . . . .

.. Example (cont.)

deflections: δB, δC... Observe: tangents to A, B, and C, note also δA = tB/A = δB − δA = δB = Qtrapezoid,B = Qrectangle,B +Qtriangle,B =

(× L) × (L

× (− PL

EI)) + (

L) × (

L× (− PL

EI)) = −PL

tC/A = δC − δA = δC =L × (

L × (−PL

EI)) = −PL

. . . . . .

.. Example (cont.)

(× L) × (L

× (− PL

EI)) + (

L) × (

L× (− PL

EI)) = −PL

tC/A = δC − δA = δC =L × (

L × (−PL

EI)) = −PL

. . . . . .

.. Example (cont.)

(× L) × (L

× (− PL

EI)) + (

L) × (

L× (− PL

EI)) = −PL

tC/A = δC − δA = δC =L × (

L × (−PL

EI)) = −PL

. . . . . .

.. Example (cont.)

(× L) × (L

× (− PL

EI)) + (

L) × (

L× (− PL

EI)) = −PL

tC/A = δC − δA = δC =L × (

L × (−PL

EI)) = −PL

. . . . . .

.. Example : example -, page

Find the angle of rotation θB and deflection δB at the free end B of acantilever beamACB supporting a uniform load of intensity q acting overthe right-hand half of the beam. (Note: The beam has length L andconstant flexural rigidity EI.)

Find the angle of rotation uB and deflection dB at the free end B of a cantileverbeam ACB supporting a uniform load of intensity q acting over the right-handhalf of the beam (Fig. 9-25). (Note: The beam has length L and constant flexuralrigidity EI.)

SolutionThe deflection and angle of rotation at end B of the beam have the direc-

tions shown in Fig. 9-25. Since we know these directions in advance, we canwrite the moment-area expressions using only absolute values.

M/EI diagram. The bending-moment diagram consists of a parabolic curvein the region of the uniform load and a straight line in the left-hand half of thebeam. Since EI is constant, the M/EI diagram has the same shape (see the lastpart of Fig. 9-25). The values of M/EI at points A and C are !3qL2/8EI and!qL2/8EI, respectively.

Angle of rotation. For the purpose of evaluating the area of the M/EI dia-gram, it is convenient to divide the diagram into three parts: (1) a parabolicspandrel of area A1, (2) a rectangle of area A2, and (3) a triangle of area A3.These areas are

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

According to the first moment-area theorem, the angle between the tangents atpoints A and B is equal to the area of the M/EI diagram between those points.Since the angle at A is zero, it follows that the angle of rotation uB is equal tothe area of the diagram; thus,

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

Deflection. The deflection dB is the tangential deviation of point B withrespect to a tangent at point A (Fig. 9-25). Therefore, from the second moment-area theorem, dB is equal to the first moment of the M/EI diagram, evaluatedwith respect to point B:

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

in which x#1, x#2, and x#3, are the distances from point B to the centroids of therespective areas. These distances are

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

FIG. 9-25 Example 9-11. Cantilever beamsupporting a uniform load on the right-hand half of the beam

. . . . . .

.. Example (cont.)

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

Solution:

... construction ofMEI

diagram

... θB — by the first moment-areatheorematA: θA = θB/A = θB−θA = θB = A+A+A

A =L(−qL

EI) = − qL

A =L(−qL

EI) = − qL

L(−qL

EI+ qL

EI) = − qL

EIÐ→ θB = A +A +A = −

EI(clockwise)

. . . . . .

.. Example (cont.)

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

Solution:

diagram

A =L(−qL

EI) = − qL

A =L(−qL

EI) = − qL

L(−qL

EI+ qL

EI) = − qL

EIÐ→ θB = A +A +A = −

EI(clockwise)

. . . . . .

.. Example (cont.)

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

Solution:

diagram

A =L(−qL

EI) = − qL

A =L(−qL

EI) = − qL

L(−qL

EI+ qL

EI) = − qL

EIÐ→ θB = A +A +A = −

EI(clockwise)

. . . . . .

.. Example (cont.)

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

Solution:

diagram

A =L(−qL

EI) = − qL

A =L(−qL

EI) = − qL

L(−qL

EI+ qL

EI) = − qL

EIÐ→ θB = A +A +A = −

EI(clockwise)

. . . . . .

.. Example (cont.)

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

Solution:

diagram

A =L(−qL

EI) = − qL

A =L(−qL

EI) = − qL

L(−qL

EI+ qL

EI) = − qL

EIÐ→ θB = A +A +A = −

EI(clockwise)

. . . . . .

.. Example (cont.)

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

Solution: (cont.)... δB — by the second moment-area

theoremδB/A = δB − δA = δB =Ax̄ +Ax̄ +Ax̄x̄ =

, x̄ =

x̄ =L+ L= L

recall:

A = −qL

EI, A = −

A = −qL

EIÐ→ δB = −

EI(downward)

. . . . . .

.. Example (cont.)

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

, x̄ =

x̄ =L+ L= L

recall:

A = −qL

EI, A = −

A = −qL

EIÐ→ δB = −

EI(downward)

. . . . . .

.. Example (cont.)

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

, x̄ =

x̄ =L+ L= L

recall:

A = −qL

EI, A = −

A = −qL

EIÐ→ δB = −

EI(downward)

. . . . . .

.. Example (cont.)

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

A1 " #13

#"!#8qEL2

I#" " #

I# A2 " #

#!#8qEL2

I#" " #

A3 " #12

#"!#38qELI

# ! #8qEL2

I#" " #

uB " A1 $ A2 $ A3 " #478qEL3

I# (9-66)

dB " A1x#1 $ A2x#2 $ A3x#3 (g)

x#1 " #34

# !#L2

#" " #38L# x#2 " #

# $ #L4

# " #34L# x#3 " #

# $ #23

#" " #56L#

Example 9-11

8EI—!

L2— L

, x̄ =

x̄ =L+ L= L

recall:

A = −qL

EI, A = −

A = −qL

EIÐ→ δB = −

EI(downward)

. . . . . .

A simple beamADB supports a concentrated load P acting at the positionshown. Determine the angle of rotation θA at supportA and thedeflection δD under the load P. (Note: The beam has length L andconstant flexural rigidity EI.)634 CHAPTER 9 Deflections of Beams

FIG. 9-26 Example 9-12. Simple beamwith a concentrated load

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

. . . . . .

.. Example (cont.)634 CHAPTER 9 Deflections of Beams

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

Solution:

... theM/EI diagram

... the sketch of the deflected curve

... θAno perfect reference with θ = turn to the deviation between tangents

θA =tB/AL

tB/A =L + b×(

LPabLEI) = Pab

EI(L+b)

Ð→ θA =PabLEI

(L + b)

. . . . . .

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

Solution:

... theM/EI diagram

θA =tB/AL

tB/A =L + b×(

LPabLEI) = Pab

EI(L+b)

Ð→ θA =PabLEI

(L + b)

. . . . . .

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

Solution:

... theM/EI diagram

θA =tB/AL

tB/A =L + b×(

LPabLEI) = Pab

EI(L+b)

Ð→ θA =PabLEI

(L + b)

. . . . . .

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

Solution:

... theM/EI diagram

θA =tB/AL

tB/A =L + b×(

LPabLEI) = Pab

EI(L+b)

Ð→ θA =PabLEI

(L + b)

. . . . . .

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

Solution:

... theM/EI diagram

θA =tB/AL

tB/A =L + b×(

LPabLEI) = Pab

EI(L+b)

Ð→ θA =PabLEI

(L + b)

. . . . . .

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

Solution:

... theM/EI diagram

θA =tB/AL

tB/A =L + b×(

LPabLEI) = Pab

EI(L+b)

Ð→ θA =PabLEI

(L + b)

. . . . . .

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

... δDgeometric analysis:δD = DD −DD = DD − tD/A

the two components:

DD = aθA =PabLEI

(L + b)

tD/A =a× (

aPabLEI) = Pab

Ð→ δD =Pab

. . . . . .

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

the two components:

DD = aθA =PabLEI

(L + b)

tD/A =a× (

aPabLEI) = Pab

Ð→ δD =Pab

. . . . . .

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

the two components:

DD = aθA =PabLEI

(L + b)

tD/A =a× (

aPabLEI) = Pab

Ð→ δD =Pab

. . . . . .

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

tB/AtD/A

PabLEI—

L ! b3—x1 =

PabLEI—

3a—x2 =

the two components:

DD = aθA =PabLEI

(L + b)

tD/A =a× (

aPabLEI) = Pab

Ð→ δD =Pab

. . . . . .

.. Example

Given: a cantilever beam, EI =const, loaded at C aMSolve: ∆B and ∆C

Solution:

... constructM/EI diagram(concentrated load only)

... sketch the deflection curve

... only deflections to determineÐ→ apply the nd theorem

... θA = , ∆A = Ð→∆B = tB/A, ∆C = tC/A

... apply the theorem

∆B = tB/A = (L)[(−

EI)L] = −

∆C = tC/A = (L) [(−

EI)L] = −

EINote: Both B and C are below the tangent atA.

. . . . . .

.. Example

Solution:

... θA = , ∆A = Ð→∆B = tB/A, ∆C = tC/A

∆B = tB/A = (L)[(−

EI)L] = −

∆C = tC/A = (L) [(−

EI)L] = −

. . . . . .

.. Example

Solution:

... θA = , ∆A = Ð→∆B = tB/A, ∆C = tC/A

∆B = tB/A = (L)[(−

EI)L] = −

∆C = tC/A = (L) [(−

EI)L] = −

. . . . . .

.. Example

Solution:

... θA = , ∆A = Ð→∆B = tB/A, ∆C = tC/A

∆B = tB/A = (L)[(−

EI)L] = −

∆C = tC/A = (L) [(−

EI)L] = −

. . . . . .

.. Example

Solution:

... θA = , ∆A = Ð→∆B = tB/A, ∆C = tC/A

∆B = tB/A = (L)[(−

EI)L] = −

∆C = tC/A = (L) [(−

EI)L] = −

. . . . . .

.. Example

Solution:

... θA = , ∆A = Ð→∆B = tB/A, ∆C = tC/A

∆B = tB/A = (L)[(−

EI)L] = −

∆C = tC/A = (L) [(−

EI)L] = −

. . . . . .

.. Example

Solution:

... θA = , ∆A = Ð→∆B = tB/A, ∆C = tC/A

∆B = tB/A = (L)[(−

EI)L] = −

∆C = tC/A = (L) [(−

EI)L] = −

. . . . . .

.. Example

Solution:

... θA = , ∆A = Ð→∆B = tB/A, ∆C = tC/A

∆B = tB/A = (L)[(−

EI)L] = −

∆C = tC/A = (L) [(−

EI)L] = −

. . . . . .

.. Example

Given: a simple beam, EI =const, loaded atA a concentratedMSolve: deflection at C

. . . . . .

.. Example (cont.)

Solution:

... only deflection todetermineÐ→ apply thend theorem

... draw tangents at the twosupports. (note theconditions ∆A = ∆B = )

. . . . . .

.. Example (cont.)

Solution:

. . . . . .

.. Example (cont.)

Solution:

. . . . . .

.. Example (cont.)

Solution:

. . . . . .

.. Example (cont.)

Solution:

. . . . . .

.. Example (cont.)

Solution:

. . . . . .

.. Example (cont.)

Solution:

. . . . . .

.. Example (cont.)

Geometric analysis: ∆C = ∆′− tC/B =

tA/B− tC/B

(or take the tangent toA as the reference, ∆C =tB/A− tC/A)

... apply the nd theorem

tA/B = (L) [(M

EI) L] = ML

tC/B = (L) [(M

EI) L] = ML

∆C =tA/B− tC/B =

EI(downward)

... double check: ∆C =tB/A− tC/A

tB/A = (L) [(M

EI) L] = ML

EI, tC/A = (

L) [(M

EI) L] = ML

∆C =tB/A− tC/A = (

− )ML

EI= ML

EI(downward)

. . . . . .

.. Example (cont.)

tA/B− tC/B

tA/B = (L) [(M

EI) L] = ML

tC/B = (L) [(M

EI) L] = ML

EI(downward)

tB/A = (L) [(M

EI) L] = ML

EI, tC/A = (

L) [(M

EI) L] = ML

∆C =tB/A− tC/A = (

− )ML

EI= ML

EI(downward)

. . . . . .

.. Example (cont.)

tA/B− tC/B

tA/B = (L) [(M

EI) L] = ML

tC/B = (L) [(M

EI) L] = ML

EI(downward)

tB/A = (L) [(M

EI) L] = ML

EI, tC/A = (

L) [(M

EI) L] = ML

∆C =tB/A− tC/A = (

− )ML

EI= ML

EI(downward)

. . . . . .

.. Example (cont.)

tA/B− tC/B

tA/B = (L) [(M

EI) L] = ML

tC/B = (L) [(M

EI) L] = ML

EI(downward)

tB/A = (L) [(M

EI) L] = ML

EI, tC/A = (

L) [(M

EI) L] = ML

∆C =tB/A− tC/A = (

− )ML

EI= ML

EI(downward)

. . . . . .

.. Example (cont.)

tA/B− tC/B

tA/B = (L) [(M

EI) L] = ML

tC/B = (L) [(M

EI) L] = ML

EI(downward)

tB/A = (L) [(M

EI) L] = ML

EI, tC/A = (

L) [(M

EI) L] = ML

∆C =tB/A− tC/A = (

− )ML

EI= ML

EI(downward)

. . . . . .

.. Example

Given: overhanging steel beam, EI = × ksi× in, loaded atC aforce kip.Solve: deflection at C

. . . . . .

.. Example (cont.)

Solution:

... constructM/EI diagram (concentrated load only)

... only deflection to determineÐ→ apply the ndtheorem

... draw tangents at the two supports. (note theconditions ∆A = ∆B = )Geometric analysis:∆C = ∣tC/A∣ − ∆

′= ∣tC/A∣ − ∣tB/A∣ (mag.)

. . . . . .

.. Example (cont.)

Solution:

′= ∣tC/A∣ − ∣tB/A∣ (mag.)

. . . . . .

.. Example (cont.)

Solution:

′= ∣tC/A∣ − ∣tB/A∣ (mag.)

. . . . . .

.. Example (cont.)

Solution:

′= ∣tC/A∣ − ∣tB/A∣ (mag.)

. . . . . .

.. Example (cont.)

Solution:

′= ∣tC/A∣ − ∣tB/A∣ (mag.)

. . . . . .

.. Example (cont.)

Solution:

′= ∣tC/A∣ − ∣tB/A∣ (mag.)

. . . . . .

.. Example (cont.)

Solution:

′= ∣tC/A∣ − ∣tB/A∣ (mag.)

. . . . . .

.. Example (cont.)

Solution (cont.):... apply the nd theorem:

tC/A = ∣CB∣S△M =

− × × /EI = − kip ⋅ ft

tB/A =(

× × (−/EI)) =

− kip ⋅ ft

∆C = ∣tC/A∣ − ∣tB/A∣ = kip ⋅ ft

. in (downward)

. . . . . .

.. Example (cont.)

− × × /EI = − kip ⋅ ft

tB/A =(

× × (−/EI)) =

− kip ⋅ ft

∆C = ∣tC/A∣ − ∣tB/A∣ = kip ⋅ ft

. in (downward)

. . . . . .

.. Example (cont.)

− × × /EI = − kip ⋅ ft

tB/A =(

× × (−/EI)) =

− kip ⋅ ft

∆C = ∣tC/A∣ − ∣tB/A∣ = kip ⋅ ft

. in (downward)

. . . . . .

.. Example (cont.)

− × × /EI = − kip ⋅ ft

tB/A =(

× × (−/EI)) =

− kip ⋅ ft

∆C = ∣tC/A∣ − ∣tB/A∣ = kip ⋅ ft

. in (downward)

. . . . . .

.. Application procedure of the moment-area method

... construct theM/EI diagram

... indicate the unknowns (slope or deflection) on the sketch

... choose a proper reference tangent, conduct a geometric analysison the unknowns←Ð themost important step

... apply corresponding theorem and check results against thesketched deflection

. . . . . .

.. Sub-summary on the moment-area method

When to use... simple loads (concentrated force and moment, constant q)... only slope or deflections at specified points are needed

(if deflection curve is required, use integration method instead)

How to use... Justify the theorems to be used:

. first theorem: relative rotation/deflection angle. second theorem: tangential deviation/deflection. combination

... tangents at boundaries are quite useful

... a geometric analysis is required to relate the unknowns

... apply theorems for numerics

. . . . . .

. Method of superposition

(叠加法)

. . . . . .

Review Moment-Area method Method of Superposition Overview Examples

.. Method of Superposition —— Overview

With small deformations, deflection equation gives... linear structural response between load and deflection v(x)

external loadÐ→ linear q(x),V(x) andM(x))... no geometric nonlinearity

Ð→ problem is qualified for superposition

Operation:... determine deflection for each component load step by step... sum for overall effect

Supportingmaterials:tabulated results for various loadings on beams.refer Appendix G

. . . . . .

.. About the table in Appendix G

slopes and deflections(θ and v at characteristic points, deflection curve v(x))

beam types:... simply supported beam... cantilevered beam

in combination with... concentrated force... concentratedmoment... distributed force (const/linear)

Tricks:

notation substitution

directions and signs adjustment

. . . . . .

Tricks:

. . . . . .

Tricks:

. . . . . .

Tricks:

. . . . . .

Tricks:

. . . . . .

Tricks:

. . . . . .

Tricks:

. . . . . .

Tricks:

. . . . . .

Tricks:

. . . . . .

Tricks:

. . . . . .

A compound beamABC has a roller support atA, an internal hinge at B,and a fixed support at C. SegmentAB has length a and segment BC haslength b. A concentrated load P acts at distance a/ from supportA anda uniform load of intensity q acts between points B and C. Determine thedeflection δB at the hinge and the angle of rotation θA at supportA.(Note: The beam has constant flexural rigidity EI.)

A compound beam ABC has a roller support at A, an internal hinge at B, and afixed support at C (Fig. 9-20a). Segment AB has length a and segment BC haslength b. A concentrated load P acts at distance 2a/3 from support A and a uni-form load of intensity q acts between points B and C.

Determine the deflection dB at the hinge and the angle of rotation uA atsupport A (Fig. 9-20d). (Note: The beam has constant flexural rigidity EI.)

SolutionFor purposes of analysis, we will consider the compound beam to consist

of two individual beams: (1) a simple beam AB of length a, and (2) a cantileverbeam BC of length b. The two beams are linked together by a pin connection at B.

If we separate beam AB from the rest of the structure (Fig. 9-20b), we seethat there is a vertical force F at end B equal to 2P/3. This same force actsdownward at end B of the cantilever (Fig. 9-20c). Consequently, the cantileverbeam BC is subjected to two loads: a uniform load and a concentrated load. Thedeflection at the end of this cantilever (which is the same as the deflection dB ofthe hinge) is readily found from Cases 1 and 4 of Table G-1, Appendix G:

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

The angle of rotation uA at support A (Fig. 9-20d) consists of two parts: (1) an angle BAB$ produced by the downward displacement of the hinge, and(2) an additional angle of rotation produced by the bending of beam AB (orbeam AB$) as a simple beam. The angle BAB$ is

(uA)1 ! "d

aB" ! "

I" # "

The angle of rotation at the end of a simple beam with a concentrated load isobtained from Case 5 of Table G-2. The formula given there is

6(LLE#

in which L is the length of the simple beam, a is the distance from the left-handsupport to the load, and b is the distance from the right-hand support to the load.Thus, in the notation of our example (Fig. 9-20a), the angle of rotation is

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

FIG. 9-20 Example 9-8. Compound beamwith a hinge

. . . . . .

.. Example : example -, page (cont.)

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

Solution:

... load analysisinternal force by hinge B

... deformation analysisAB: . rigid rotation as B moves; .deflection by P

BC: . deflection by q; . by F = P

... superposition tables G-:(,), G-:

δB =qb

EI+ Fb

EI= qb

EI+ Pb

θA =δBa+Paa(a + a

aEI+ Pb

aEI+ Pa

. . . . . .

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

Solution:

δB =qb

EI+ Fb

EI= qb

EI+ Pb

θA =δBa+Paa(a + a

aEI+ Pb

aEI+ Pa

. . . . . .

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

Solution:

δB =qb

EI+ Fb

EI= qb

EI+ Pb

θA =δBa+Paa(a + a

aEI+ Pb

aEI+ Pa

. . . . . .

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

Solution:

δB =qb

EI+ Fb

EI= qb

EI+ Pb

θA =δBa+Paa(a + a

aEI+ Pb

aEI+ Pa

. . . . . .

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

Solution:

δB =qb

EI+ Fb

EI= qb

EI+ Pb

θA =δBa+Paa(a + a

aEI+ Pb

aEI+ Pa

. . . . . .

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

dB ! "8qEb4

I" # "

or, since F ! 2P/3,

dB ! "8qEb4

I" # "

29PEbI

3" (9-56)

(uA)1 ! "d

aB" ! "

I" # "

6(LLE#

(uA)2 ! ! "48P1E

P!"23a""!"

a3""!a # "

"""6aEI

Example 9-8

—2P3

Solution:

δB =qb

EI+ Fb

EI= qb

EI+ Pb

θA =δBa+Paa(a + a

aEI+ Pb

aEI+ Pa

. . . . . .

.. Example : example -, p.

Given: an overhanging beamABC, EI =const;Solve: δC

Combining the two angles, we obtain the total angle of rotation at support A:

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

This example illustrates how the method of superposition can be adapted tohandle a seemingly complex situation in a relatively simple manner.

SECTION 9.5 Method of Superposition 623

A simple beam AB of span length L has an overhang BC of length a (Fig. 9-21a). The beam supports a uniform load of intensity q throughout its length.

Obtain a formula for the deflection dC at the end of the overhang (Fig. 9-21c). (Note: The beam has constant flexural rigidity EI.)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

FIG. 9-21 Example 9-9. Simple beamwith an overhang

continued

. . . . . .

.. Example : (cont.)

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

... deformation analysistwo contributions of δC:. rigid rotation at B; . q onBC

... θB consists of two parts

θB =qL

EI− MBL

qL(L − a)EI

contribution from θB:

δ = aθB =qaL(L − a)

... q effect as in a cantilever beam to C: δ = −qa

EI... superpositionÐ→δC = δ + δ = −

qaEI(a + L)(a + aL − L)

. . . . . .

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

θB =qL

EI− MBL

qL(L − a)EI

. . . . . .

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

θB =qL

EI− MBL

qL(L − a)EI

. . . . . .

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

θB =qL

EI− MBL

qL(L − a)EI

. . . . . .

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

θB =qL

EI− MBL

qL(L − a)EI

. . . . . .

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

θB =qL

EI− MBL

qL(L − a)EI

. . . . . .

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

θB =qL

EI− MBL

qL(L − a)EI

. . . . . .

uA ! (uA)1 " (uA)2 ! #8qabE

I# " #

I# (9-57)

Example 9-9

qa2—2

P = qa

Point of inflection

uBD BA C x

continued

δC = δ + δ = −qaEI(a + L)(a + aL − L)

... further analysis

if δC = Ð→ a = ,−L,L(±√ − )

take the physical one a = .Lif a > .LÐ→ δC < (downward)if a < .LÐ→ δC > (upward)

. . . . . .

Given: a cantilever beam, EI =const;Solve: vB, θB

A cantilever beam AB with a uniform load of intensity q acting on the right-hand half of the beam is shown in Fig. 9-19a.

Obtain formulas for the deflection dB and angle of rotation uB at the free end (Fig. 9-19c). (Note: The beam has length L and constant flexuralrigidity EI.)

SolutionIn this example we will determine the deflection and angle of rotation by

treating an element of the uniform load as a concentrated load and then integrat-ing (see Fig. 9-19b). The element of load has magnitude q dx and is located atdistance x from the support. The resulting differential deflection ddB and differ-ential angle of rotation duB at the free end are found from the correspondingformulas in Case 5 of Table G-1, Appendix G, by replacing P with q dx and awith x; thus,

ddB ! duB ! "(q d

2xE)(Ix2)

By integrating over the loaded region, we get

dB ! !ddB ! "6qEI" !L

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

uB ! !duB ! "2qEI" !L

L /2x2dx ! "

478qEL3

I" (9-55)

Note: These same results can be obtained by using the formulas in Case 3of Table G-1 and substituting a ! b ! L /2.

(qdx)(x2)(3L # x)""

Example 9-7

FIG. 9-19 Example 9-7. Cantilever beamwith a uniform load acting on the right-hand half of the beam

. . . . . .

.. Example — Solution

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

refer to Appendix G, tableG- (,; )entries: ,. Note in entry , reverse q

vB =−qL

EI+ qL

EI= −qL

θB =−qL

EI+ qL

EI= −qL

entry : direct application with a = b = L, same result

. . . . . .

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

new idea that can be extended to moregeneral cases

strategy:... treat the distributed load as a

collection of differentialconcentrated ones

... integrate piece by piece

. . . . . .

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

. . . . . .

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

. . . . . .

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

. . . . . .

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

. . . . . .

.. Example — Solution (cont.)

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

for a differential concentrated load qdx (Table G-, entry )

dθB =−(qdx)x

EI, note the free segment (L − x)

dvB =−(qdx)x

EI+ (L − x)dθB = −

(qdx)x(L − x)EI

θB = ∫L

L/dθB = ∫

−(qdx)x

EI= −qL

EI(clockwise)

vB = ∫L

L/dvB = ∫

L/−(qdx)x

(L − x)EI

= −qL

EI(downward)

. . . . . .

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

dθB =−(qdx)x

dvB =−(qdx)x

EI+ (L − x)dθB = −

(qdx)x(L − x)EI

θB = ∫L

L/dθB = ∫

−(qdx)x

EI= −qL

EI(clockwise)

vB = ∫L

L/dvB = ∫

L/−(qdx)x

(L − x)EI

= −qL

EI(downward)

. . . . . .

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

ddB ! duB ! "(q d

2xE)(Ix2)

L /2x2(3L # x) dx ! "

34814qEL4

I" (9-54)

L /2x2dx ! "

478qEL3

I" (9-55)

(qdx)(x2)(3L # x)""

Example 9-7

dθB =−(qdx)x

dvB =−(qdx)x

EI+ (L − x)dθB = −

(qdx)x(L − x)EI

θB = ∫L

L/dθB = ∫

−(qdx)x

EI= −qL

EI(clockwise)

vB = ∫L

L/dvB = ∫

L/−(qdx)x

(L − x)EI

= −qL

EI(downward)

. . . . . .

.. Example

Given: L-shape bracket, load P at A. Fixed at C.Solve: θA

Solution:

... two contributions

θA by p. , entry Ð→ θA =PabEI

θA by p. , entry Ð→ θA =Pa

EI... final result:

θA = θA + θA =PaEI(b + a)

. . . . . .

.. Example

Solution:

EI... final result:

. . . . . .

.. Example

Solution:

EI... final result:

. . . . . .

.. Example

Solution:

EI... final result:

. . . . . .

.. Example

Solution:

EI... final result:

. . . . . .

.. Summary

moment areamethod

st theorem: θB/A = ∫B

dxrelative deflection angle

nd theorem: tA/B = x∫B

dx, tB/A = x′ ∫A

deviation of tangent/deflectionapplicable cases: simple loads; specified pointsapplication procedure: reference tangent + geometric analysis

method of superpositiontabulated results for various combinations of beams and loads

. . . . . .

.. Summary

moment areamethod

. . . . . .

.. Summary

moment areamethod

. . . . . .

.. Summary

moment areamethod

. . . . . .

.. Summary

moment areamethod

. . . . . .

.. Summary

moment areamethod

. . . . . .

.. Homework

Chap. — (superposition)Chap. — , (moment-area)

Due: .. (Tue.)

18 Masupp Presentation

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