HW1 Solution

MECHANICS OF MATERIALS

ThirdEdition

Beer • Johnston • DeWolf

Normal Strain

stress==APσ

AP

APσ ==

22

APσ =

© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 2 - 1

strain normal==Lδε

Lδε =

LLδδε ==

22


ThirdEdition


Sample Problem 2.1

SOLUTION:

• Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC.

• Evaluate the deformation of links ABThe rigid bar BDE is supported by two links AB and CD.

Link AB is made of aluminum (E = 70

and DC or the displacements of Band D.

• Work out the geometry to find theLink AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm2. Link CD is made of steel (E = 200 GPa) and has a cross-sectional area of (600

Work out the geometry to find the deflection at E given the deflections at B and D.

GPa) and has a cross sectional area of (600 mm2).

For the 30-kN force shown, determine the d fl ti ) f B b) f D d ) f E


deflection a) of B, b) of D, and c) of E.


ThirdEdition


Sample Problem 2.1Displacement of B:

=AEPL

BδFree body: Bar BDE

SOLUTION:

( )( )( )( )

1014

Pa1070m10500m3.0N1060

6

926-

3

−

××

×−=

m10514 6−×−=

↑= mm 514.0BδDisplacement of D:

( ) F

M B

2060kN300

0=∑

( )( )m4.0N1090 3×

=AEPL

Dδ( )

tensionF

F

CD

CD

0M

kN90

m2.0m6.0kN300

D =

+=

×+×−=

∑ ( )( )( )( )

m10300

Pa10200m10600m4.0N1090

6

926-

−×=

××=

( )ncompressioF

F

AB

AB

kN60

m2.0m4.0kN300

−=

×−×−=


↓= mm 300.0Dδ


ThirdEdition


Sample Problem 2.1

Displacement of D:

′ BHBB

( )mm 200mm 0.300mm 514.0 −

=

=′

xx

HDDD

mm 7.73=x

′ HEEE

( )mm 7.73

mm7.73400mm 300.0

+=

=′

E

HDDDδ

↓9281δ

mm 928.1=Eδ


↓= mm928.1Eδ


ThirdEdition


Static Indeterminacy• Structures for which internal forces and reactions

cannot be determined from statics alone are said to be statically indeterminate.

• A structure will be statically indeterminate whenever it is held by more supports than are

i d t i t i it ilib i

• Redundant reactions are replaced with unknown loads which along with the other

required to maintain its equilibrium.

• Deformations due to actual loads and redundant

unknown loads which along with the other loads must produce compatible deformations.

0=+= RL δδδ

reactions are determined separately and then added or superposed.



ThirdEdition


Example 2.04Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are applied.

SOLUTION:

• Consider the reaction at B as redundant, release

• Solve for the displacement at B due to the

the bar from that support, and solve for the displacement at B due to the applied loads.

• Require that the displacements due to the loads

• Solve for the displacement at B due to the redundant reaction at B.

q pand due to the redundant reaction be compatible, i.e., require that their sum be zero.


• Solve for the reaction at A due to applied loads and the reaction found at B.


ThirdEdition


SOLUTION:Example 2.04

SOLUTION:• Solve for the displacement at B due to the applied

loads with the redundant constraint released, 33 10900106000

LLLL

AAAA

PPPP

4321

2643

2621

34

3321

m1500

m10250m10400

N10900N106000

====

×==×==

×=×===

−−

EEALP

LLLL

i ii

ii9

L

4321

10125.1

m 150.0

×=∑=δ

• Solve for the displacement at B due to the redundant constraint,

−== BRPP 21

( )==

×=×= −−

LL

AA

21

262

261

m 300.0

m10250m10400


( )∑

×−==

iB

ii

iiR E

REALPδ

31095.1


ThirdEdition


Example 2.04

• Require that the displacements due to the loads and due to the redundant reaction be compatible,

0=+= δδδ

( ) 01095.110125.1

0

39=

×−

×=

=+=

B

RL

ER

Eδ

δδδ

kN 577N10577 3 =×=BR

• Find the reaction at A due to the loads and the reaction at B

kN323

kN577kN600kN 3000

=

∑ +−−==

A

Ay

R

RF

kN323AR

kN577

kN323

=

=A

R

R


kN577=BR


ThirdEdition


Net Torque Due to Internal Stresses

• Net of the internal shearing stresses is an internal torque, equal and opposite to the

( )∫ ∫== dAdFT τρρ

applied torque,

• Although the net torque due to the shearing stresses is known, the distribution of the stresses is not

• Distribution of shearing stresses is statically indeterminate – must consider shaft deformations

• Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional l d t b d if


loads can not be assumed uniform.


ThirdEdition


Axial Shear Components

• Torque applied to shaft produces shearing stresses on the faces perpendicular to the

iaxis.

• Conditions of equilibrium require the existence of equal stresses on the faces of the

• The existence of the axial shear components is

existence of equal stresses on the faces of the two planes containing the axis of the shaft

The existence of the axial shear components is demonstrated by considering a shaft made up of axial slats.

The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.



ThirdEdition


Shaft Deformations

• From observation, the angle of twist of the shaft is proportional to the applied torque andshaft is proportional to the applied torque and to the shaft length.

L

T∝

φ

φ

L∝φ

• When subjected to torsion, every cross-section of a circular shaft remains plane and

di t t dundistorted.

• Cross-sections for hollow and solid circular shafts remain plain and undistorted because a

• Cross-sections of noncircular (non-axisymmetric) shafts are distorted when

circular shaft is axisymmetric.


subjected to torsion.


ThirdEdition


Shearing Strain

• Consider an interior section of the shaft. As a torsional load is applied, an element on the i t i li d d f i t h binterior cylinder deforms into a rhombus.

• Since the ends of the element remain planar, the shear strain is equal to angle of twist

ρφ• It follows that

the shear strain is equal to angle of twist.

• Shear strain is proportional to twist and radius

LL ρφγρφγ == or

maxmax and γργφγcL

c==



ThirdEdition


Stresses in Elastic Range• Multiplying the previous equation by the

shear modulus,maxγργ G

cG =

c

maxτρτc

=

From Hooke’s Law, γτ G= , so

• Recall that the sum of the moments from

421 cJ π= The shearing stress varies linearly with the

radial position in the section.

JdAdAT max2max ττ∫∫

• Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section,

Jc

dAc

dAT max2max ρρτ ∫ =∫ ==

( )41

42

1 ccJ −= π

• The results are known as the elastic torsion formulas,


( )122 ccJ −= π and max J

TJ

Tc ρττ ==


ThirdEdition


Sample Problem 3.1

SOLUTION:

• Cut sections through shafts ABd BC d f iand BC and perform static

equilibrium analysis to find torque loadings

Shaft BC is hollow with inner and outer

• Apply elastic torsion formulas to find minimum and maximum stress on shaft BC

Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm, respectively. Shafts AB and CD are solid of diameter d. For the loading shown,

• Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the g ,

determine (a) the minimum and maximum shearing stress in shaft BC, (b) the required diameter d of shafts AB and CD

required diameter


if the allowable shearing stress in these shafts is 65 MPa.


ThirdEdition


SOLUTION:Sample Problem 3.1• Cut sections through shafts AB and BC

and perform static equilibrium analysis to find torque loadings

( )

CDAB

ABx

TT

TM

=⋅=

−⋅==∑mkN6

mkN60 ( ) ( )mkN20

mkN14mkN60

⋅=

−⋅+⋅==∑

BC

BCx

T

TM



ThirdEdition


A l l ti t i f l t Gi ll bl h i t dSample Problem 3.1• Apply elastic torsion formulas to

find minimum and maximum stress on shaft BC

• Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter

( ) ( ) ( )[ ]4441

42 045.0060.0

22−=−=

ππ ccJ mkN665 34max⋅

=== MPaTcJ

Tcππ

τ

46m1092.13

22−×=

( )( )109213

m060.0mkN2046

22max

⋅=== −J

cTBCττm109.38 3

32

42

−×=c

ccJ ππ

mm8772 == cd

MPa2.86m1092.13 46

=

× −J

mm45min1min ==ττ c

mm8.772 == cd


MPa7.64

mm60MPa2.86

min

2max

=τ

τ c

MPa7.64

MPa2.86

min

max

=

=

τ

τ


ThirdEdition


Angle of Twist in Elastic Range• Recall that the angle of twist and maximum

shearing strain are related,

Lcφγ =max L

γmax

• In the elastic range, the shearing strain and shear are related by Hooke’s Law,

TcτJGTc

G== max

maxτγ

• Equating the expressions for shearing strain and solving for the angle of twistsolving for the angle of twist,

JGTL

=φ

• If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotations

∑= iiGJLTφ


i iiGJφ


ThirdEdition


Symmetric Member in Pure Bending• Internal forces in any cross section are equivalent

to a couple. The moment of the couple is the section bending moment.

• From statics, a couple M consists of two equal and opposite forces.

• The sum of the components of the forces in anyThe sum of the components of the forces in any direction is zero.

• The moment is the same about any axis perpendicular to the plane of the couple and

• These requirements may be applied to the sums of the components and moments of the statically

perpendicular to the plane of the couple and zero about any axis contained in the plane.

∫∫ ==

dAMdAF xx σ

00

of the components and moments of the statically indeterminate elementary internal forces.


∫ =−=

∫ ==

MdAyM

dAzM

xz

xy

σ

σ 0


ThirdEdition


Bending DeformationsBeam with a plane of symmetry in pure

bending:• member remains symmetricmember remains symmetric

• bends uniformly to form a circular arc

• cross-sectional plane passes through arc center• cross-sectional plane passes through arc centerand remains planar

• length of top decreases and length of bottom increases

• a neutral surface must exist that is parallel to the upper and lower surfaces and for which the lengthupper and lower surfaces and for which the length does not change

• stresses and strains are negative (compressive) b th t l l d iti (t i )


above the neutral plane and positive (tension) below it


ThirdEdition


Strain Due to Bending

Consider a beam segment of length L.

After deformation, the length of the neutral surface remains L. At other sections,

( )( ) yyLL

yL

θρθθρδ

θρ

′

−=′

( )

xyy

L

yyLL

ρρθθδε

θρθθρδ

−=−==

−=−−=−=

linearly) ries(strain va

mm

y

cρc

εε

ερε

−=

== or

mx cεε =



ThirdEdition


Stress Due to Bending• For a linearly elastic material,

mxx EcyE εεσ −==

linearly) varies(stressmcyσ−=

• For static equilibriumFor static equilibrium,

∫

∫∫ −=== dAcydAF mxx

σ

σσ0• For static equilibrium,

dAcyydAyM mx ⎟

⎠⎞

⎜⎝⎛−−=−= ∫∫ σσ

∫−= dAycmσ0

First moment with respect to neutral plane is zero Therefore the neutral

MMcc

IdAyc

M mm

==

== ∫

σ

σσ 2

plane is zero. Therefore, the neutral surface must pass through the section centroid. c

ySI

mx

m

−=

==

σσ

σ

ngSubstituti


IMy

x −=σ


ThirdEdition


Deformations in a Transverse Cross Section• Deformation due to bending moment M is

quantified by the curvature of the neutral surfaceMcmm ===

11 σε

EIM

IEcEcc

=

ρ

• Although cross sectional planes remain planar when subjected to bending moments, in-plane deformations are nonzero,

ρννεε

ρννεε yy

xzxy =−==−=

• Expansion above the neutral surface andExpansion above the neutral surface and contraction below it cause an in-plane curvature,

curvature canticlasti 1==

′ ρν

ρ


ρρ


ThirdEdition


Sample Problem 4.2SOLUTION:

• Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.

( )∑ +=∑∑= ′

2dAIIAAyY x

• Apply the elastic flexural formula to find the maximum tensile and compressive stresses

A cast-iron machine part is acted upon

compressive stresses.

IMc

m =σ

p pby a 3 kN-m couple. Knowing E = 165 GPa and neglecting the effects of fillets, determine (a) the maximum

• Calculate the curvature

EIM

=ρ1


tensile and compressive stresses, (b) the radius of curvature.


ThirdEdition


Sample Problem 4.2SOLUTIONSOLUTION:

Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.

×=× 3

32

109050180090201mm ,mm ,mm Area, Ayy

∑ ×==∑

×=××=×

3

3

101143000104220120030402109050180090201

AyA

mm 383000

10114 3=

×=

∑∑=

AAyY

( ) ( )( ) ( )23

12123

121

231212

18120040301218002090 ×+×+×+×=

∑ +=∑ +=′ dAbhdAIIx


49-31212

m10868 mm10868 ×=×=I


ThirdEdition


Sample Problem 4.2

• Apply the elastic flexural formula to find the maximum tensile and compressive stresses.

Mc

49mm10868m022.0mkN 3

−×

×⋅==

=

IcM

IA

A

m

σ

σ

MPa0.76+=Aσ

49mm10868m038.0mkN 3

−×

×⋅−=−=

IcM B

Bσ MPa3.131−=Bσ

• Calculate the curvature1=

EIM

ρ

( )( )49- m10868GPa 165mkN 3×

⋅=

m 7.47

m1095.201 1-3

=

×= −

ρρ



ThirdEdition


Eccentric Axial Loading in a Plane of Symmetry• Stress due to eccentric loading found by

superposing the uniform stress due to a centric load and linear stress distribution due a pure bending moment

( ) ( )MyP

xxx += bendingcentric σσσ

Iy

A−=

• Eccentric loading

PF =• Validity requires stresses below proportional

limit deformations have negligible effect onPdMPF

== limit, deformations have negligible effect on

geometry, and stresses not evaluated near points of load application.



ThirdEdition


Example 4.07SOLUTION:

• Find the equivalent centric load and bending momentbending moment

• Superpose the uniform stress due to the centric load and the linear stress due to the bending moment.

• Evaluate the maximum tensile and

An open-link chain is obtained by

compressive stresses at the inner and outer edges, respectively, of the superposed stress distribution.

bending low-carbon steel rods into the shape shown. For 160 lb load, determine (a) maximum tensile and compressive

(b) di b i

• Find the neutral axis by determining the location where the normal stress is zero


stresses, (b) distance between section centroid and neutral axis

is zero.


ThirdEdition


Example 4.07

( )in25.0 22 == cA ππ

• Normal stress due to a centric load

in1963.0lb160

in1963.0

20

2

==

=

APσ

• Equivalent centric load

psi815=

• Normal stress due toqand bending moment

( )( )in60lb160lb160==

=PdM

P ( )25.043

4414

41 == cI ππ

Normal stress due to bending moment

( )( )inlb104

in6.0lb160⋅=

PdM

( )( )in10068.

in25.0inlb104in10068.3

43

43

×

⋅==

×=

−

−

IMc

mσ


psi8475=


ThirdEdition


Example 4.07

• Maximum tensile and compressive stresses

• Neutral axis locationstresses

84758150

=+=

+= mt

σσσ

σσσpsi9260=tσ

( )ilb105

in10068.3psi815

0

430

0

×==

−=

−

MI

APy

IMy

AP

84758150−=−= mc σσσ

psi7660−=cσ

( )inlb105

p0 ⋅MAy

in0240.00 =y



ThirdEdition


Introduction

• Beams - structural members supporting loads at various points along the member

• Objective - Analysis and design of beams

various points along the member

• Transverse loadings of beams are classified as concentrated loads or distributed loads

• Applied loads result in internal forces consisting of a shear force (from the shear stress di ib i ) d b di l (f hdistribution) and a bending couple (from the normal stress distribution)

• Normal stress is often the critical design criteriaNormal stress is often the critical design criteria

SM

IcM

IMy

mx ==−= σσ

Requires determination of the location and


Requires determination of the location and magnitude of largest bending moment


ThirdEdition


Shear and Bending Moment Diagrams

• Determination of maximum normal and shearing stresses requires identification of maximum internal shear force and bending gcouple.

• Shear force and bending couple at a point are d i d b i i h h hdetermined by passing a section through the beam and applying an equilibrium analysis on the beam portions on either side of the sectionsection.

• Sign conventions for shear forces V and V’and bending couples M and M’g p



ThirdEdition



• Treating the entire beam as a rigid body, determine the reaction forcesbody, determine the reaction forces

• Section the beam at points near supports and load application points.

For the timber beam and loading

Apply equilibrium analyses on resulting free-bodies to determine internal shear forces and bending

lshown, draw the shear and bend-moment diagrams and determine the maximum normal stress due to b di

• Identify the maximum shear and bending-moment from plots of their

couples

bending. bending moment from plots of their distributions.

• Apply the elastic flexure formulas to


determine the corresponding maximum normal stress.


ThirdEdition



• Treating the entire beam as a rigid body, determine the reaction forcesthe reaction forces

∑ ∑ ==== kN14kN40:0 from DBBy RRMF

• Section the beam and apply equilibrium analyses on resulting free-bodies

( )( ) 00m0kN200

kN200kN200

111

11

==+∑ =

−==−−∑ =

MMM

VVFy

( )( ) mkN500m5.2kN200

kN200kN200

222

22

⋅−==+∑ =

−==−−∑ =

MMM

VVFy

mkN28kN14

mkN28kN26

mkN50kN26

55

44

33

⋅+=−=

⋅+=+=

⋅−=+=

MV

MV

MV


0kN14

N8N

66

55

=−= MV

V


ThirdEdition


Sample Problem 5.1• Identify the maximum shear and bending-

moment from plots of their distributions.

mkN50kN26 === MMV mkN50kN26 ⋅=== Bmm MMV

• Apply the elastic flexure formulas to determine the correspondingdetermine the corresponding maximum normal stress.

( )( )2612

61 m250.0m080.0== hbS

3

36

mN1050

m1033.833 −

⋅×

×=

M B36 m1033.833

mN1050−×

×==

SM B

mσ

Pa100.60 6×=mσ


m


ThirdEdition


Deformation of a Beam Under Transverse Loading• Relationship between bending moment and

curvature for pure bending remains valid for general transverse loadings.

EIxM )(1

=ρ

• Cantilever beam subjected to concentrated• Cantilever beam subjected to concentrated load at the free end,

EIPx

−=ρ1

EIρ

• Curvature varies linearly with x

1• At the free end A, ∞== A

A ρ

ρ,01

• At the support BEI

B =≠ ρ01


• At the support B, PLBB

=≠ ρρ

,0


ThirdEdition


Deformation of a Beam Under Transverse Loadingh i b• Overhanging beam

• Reactions at A and C

B di t di• Bending moment diagram

• Curvature is zero at points where the bending moment is zero, i.e., at each end and at E.

EIxM )(1

=ρ

• Beam is concave upwards where the bendingBeam is concave upwards where the bending moment is positive and concave downwards where it is negative.

M i t h th t• Maximum curvature occurs where the moment magnitude is a maximum.

• An equation for the beam shape or elastic curve


is required to determine maximum deflection and slope.


ThirdEdition


Equation of the Elastic Curve

• From elementary calculus, simplified for beam parameters,

2d

2

2

232

2

2

1

1dx

yd

dy

dxyd

≈

⎥⎤

⎢⎡

⎟⎞

⎜⎛+

=ρ

1dx ⎥

⎥⎦⎢

⎢⎣

⎟⎠

⎜⎝

+

• Substituting and integrating,

( )2

21

d

xMdx

ydEIEI

x

==ρ

( )

( )

10

CdxxMdxdyEIEI

xx

+=≈

∫∫

∫θ


( ) 2100

CxCdxxMdxyEI ++= ∫∫


ThirdEdition


Equation of the Elastic Curve

( ) CCdMdEIxx

∫∫

• Constants are determined from boundary conditions

( ) 2100

CxCdxxMdxyEI ++= ∫∫

• Three cases for statically determinant beams,

– Simply supported beam0,0 == BA yy

– Overhanging beam0,0 == BA yy

– Cantilever beam0,0 == AAy θ

• More complicated loadings require multiple i t l d li ti f i t f


integrals and application of requirement for continuity of displacement and slope.


ThirdEdition


Direct Determination of the Elastic Curve From the Load DistributionLoad Distribution

• For a beam subjected to a distributed load,dVMddM 2

( ) ( )xwdxdV

dxMdxV

dxdM

−=== 2

• Equation for beam displacement becomes

( )xwdx

ydEIdx

Md−== 4

4

2

2

( ) ( )2131

dxxwdxdxdxxyEI −= ∫∫∫∫• Integrating four times yields

432

2213

161 CxCxCxC ++++

• Constants are determined from boundary conditions


conditions.


ThirdEdition


Statically Indeterminate BeamsC id b i h fi d A d ll• Consider beam with fixed support at A and roller support at B.

• From free-body diagram, note that there are four k ti tunknown reaction components.

• Conditions for static equilibrium yield000 =∑=∑=∑ Ayx MFF y

The beam is statically indeterminate.

xx• Also have the beam deflection equation,

( ) 2100

CxCdxxMdxyEIxx

++= ∫∫which introduces two unknowns but provides pthree additional equations from the boundary conditions:

0At000At ===== yLxyx θ


0,At 00,0At ===== yLxyx θ


ThirdEdition


Sample Problem 9.1

SOLUTION:

• Develop an expression for M(x) and derive differential equation for elastic curve.

ft 4ft15kips50

psi1029in7236814 64

===

×==×

aLP

EIW • Integrate differential equation twice and apply boundary conditions to obtain elastic curve

For portion AB of the overhanging beam, (a) derive the equation for the elastic curve, (b) determine the maximum deflection

obtain elastic curve.

• Locate point of zero slope or point of maximum deflection.

(b) determine the maximum deflection, (c) evaluate ymax. • Evaluate corresponding maximum

deflection.



ThirdEdition



• Develop an expression for M(x) and derive differential equation for elastic curve.q

- Reactions:

↑⎟⎞

⎜⎛ +=↓=

aPRPaR 1 ↑⎟⎠

⎜⎝+=↓=

LPR

LR BA 1

- From the free-body diagram for section AD,y g ,

( )LxxLaPM <<−= 0

ayd 2

- The differential equation for the elastic curve,


xLaP

dxydEI −=2


ThirdEdition


Sample Problem 9.1• Integrate differential equation twice and apply

boundary conditions to obtain elastic curve.

121 CxaPdyEI +=

213

1

61

2

CxCxLaPyEI

CxL

Pdx

EI

++−=

+−=

PaLCLCLLaPyLx

Cyx

61

610:0,at

0:0,0at

113

2

=+−===

===x

LaP

dxydEI −=2

2

L 66

xPaLdyady 11 22 ⎥

⎤⎢⎡

⎟⎞

⎜⎛

Substituting,

⎤⎡ ⎞⎛32PaLxxaPyEI

Lx

EIPaL

dxdyPaLx

LaP

dxdyEI

11

3166

121

3

2

+−=

⎥⎥⎦⎢

⎢⎣

⎟⎠⎞

⎜⎝⎛−=+−=


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

32

6 Lx

Lx

EIPaLyL

y66


ThirdEdition


Sample Problem 9.1• Locate point of zero slope or point

of maximum deflection.

LP Ld 2⎤⎡ ⎞⎛

⎤⎡ 32

LLxL

xEI

PaLdxdy

mm 577.0

331

60 ==

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−==

• Evaluate corresponding maximum

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

32

6 Lx

Lx

EIPaLy

• Evaluate corresponding maximum deflection.

( )[ ]32

57705770 −=PaLy ( )[ ]max 577.0577.06

=EI

y

EIPaLy6

0642.02

max =

( )( )( )( )( )46

2max

in723psi10296in180in48kips500642.0

×=y


in238.0max =y


ThirdEdition


Sample Problem 9.3

SOLUTION:

• Develop the differential equation for the elastic curve (will be functionally dependent on the reaction at A).

For the uniform beam, determine the reaction at A, derive the equation for th l ti d d t i th

• Integrate twice and apply boundary conditions to solve for reaction at Aand to obtain the elastic curve.

the elastic curve, and determine the slope at A. (Note that the beam is statically indeterminate to the first degree)

• Evaluate the slope at A.

degree)



ThirdEdition


Sample Problem 9.3• Consider moment acting at section D,

M D 0

⎞⎛

=∑

MxLxwxRA 0

321

3

20 =−⎟

⎟⎠

⎞⎜⎜⎝

⎛−

LxwxRM A 6

30−=

• The differential equation for the elastic

LxwxRM

dxydEI A 6

30

2

2−==

The differential equation for the elastic curve,

Ldx 6



ThirdEdition


Sample Problem 9.3• Integrate twice

14

02242

1 CLxwxREI

dxdyEI A +−== θ

215

031206

1 CxCL

xwxRyEI A ++−=

LxwxRM

dxydEI A 6

30

2

2−==

• Apply boundary conditions:

1

0:0,0at 3

2 ===

L

Cyx

01:0,at

0242

1:0,at

214

03

13

02

=++−==

=+−==

CLCLwLRyLx

CLwLRLx

A

Aθ

01206

:0,at 21 ++ CLCLRyLx A

• Solve for reaction at A

011 40

3 =− LwLRA ↑= LwRA 01


0303 0 =− LwLRA ↑= LwRA 010


ThirdEdition


Sample Problem 9.3

⎞⎛⎞⎛ 5 111

• Substitute for C1, C2, and RA in the elastic curve equation,

xLwL

xwxLwyEI ⎟⎠⎞

⎜⎝⎛−−⎟

⎠⎞

⎜⎝⎛= 3

05

030 120

112010

161

( )xLxLxwy 43250 2+= ( )xLxLxEIL

y 2120

−+−=

Diff ti t t fi d th l

( )42240 65120

LxLxEIL

wdxdy

−+−==θ

• Differentiate once to find the slope,

EILw

A 120

30=θat x = 0,


Documents

HW1 Solution