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ftH\f*sffi& HM*Y*SM

Timber Design cPert CPM oEstimates olRR of PD 1ig4 oSeismic Analysis . H ig hway Engi neeri ng . Theory of Structu res o Geotech nical Engineering .concrete Mixture Problems olnfluence Diagram for Beams

Coptright 1987

Venancio (BS.CE,

(hilPlace) Engineer ineer - CIT (2rilPlace) Engineer - CIT (7th place)

-

I.' besavilla, Jr.

MSME, AS, E (PTCE)

- August, August, l: 1969 Iuli 1966

lnstructor: C*bu Institute ol Teihnolog

trnstructor: Unfuusity of theVkayas Chairman: C,E, Deot. Univusitt of theVisanas ',ollege of Engineeing Engineeriig anil College anil Archiie&ure, Untuersity Ihiierc of theVkayas As an Ouktanililg Bifucator from the Phil.Veterans Legion on May 1984 AsOutstaniling Avarilee: As Outstanilins Afumnw in the thc Fielil Field of Bdrcatioa Eilucation ttom CIT //lumni Associationr ltc., Mirch 1990 As_ Ouktaniling F ngineering Eilucator from the -December CIT Higlr School Alumni Associatiory 199 I Cebu Instituk of Technologyr Alumni Association (ZOO3-up to the present) )resiilort: Philippine Philippine Institute Institu* oJ (HCE) Cebu Chaliter oJCivil'Engineerc Civil Engineerc (ptCB) Chapter 2009 )rrsidant: Philippine Institute oJ Civil Engineen (PICE) National-Boanl 2009 Mcrrbet: Geo-Institute of the American Sociery of Cioil Fngineerc (dSCn) Motber: StruchralEngineeitglnstitute of theAnqrcan-society olCitilEngineers (eSCn) tulember American Society of Civil Engineirc (NCE) (Memberihili No. 3469-60) (Membershio No. l?4SSi) Member: American Concieti Memben Concreti hrstitute-(Aq) hstitute(ACl) (Memberihip ru155,3) Member: PICE Delegation to the Anerican Society of Ciil Engineers (eSCn) Convantion and Co$erence, Minneapolk, Minnaota, USA (Oct. 1997) ril PICB Delegation to the 77th Executive Comiittee Meeting oJ the Asian C\il Engineefiag - Coorilinating Council (lCnCC), Hanoi, Vieinim (April 2009) PICE Delegation to the Japan Society of Civil lngineerc (JSCB) ConJerence, Fukuoka, Japan (Septembr 2009) PICE Dclegation to the Americin Society of Ctuil Engitecrs (,ESCB) lnaul Conference, Kansas City, pSA (Octobtr 2009) PICE Intenational Committee Delegation to the 78th Executiye Commiltee Meeting oJthe Asiott Citil Eaginenin! Coorilinating Council (,*C&CC) Busan,

Korea (F ebruary

20

1

0

)

Intenational Committee Delegation to the Sth Cieil Engineering ConJerence in,kianRe$on (CACen), {yilney,Arstralia (,+ugttsi2OlO) " PICE Inturational Committee Delegatlon to the Japan Society ol Cfuil Engineers jSCn) PICE

Annual Conference, Hokkaido, Japan (Septimbo 2010)

PlCElnteraational CommitteeDelegatiinto thiAnedcan Society oJCivilEtgineus (.*SCB) Convention

atd

Con;ference, Las Vegas, tuSA (October 201 0) ISBN y,1- E510.14.1 Available at:

BISlUlllt

Enginetiring Review Genter

DAVAO Floor, Bldg. St., City

4th Ponas Magallanes Davao Tel. No. (082) Chalet Bldg. s Grotto tinican Road 68 San Roque St. Baguio City No. (074) 44$5918

222-3305

DEORO Floor, Bldg. St. City

CAGAYAN 3rd Ecology Bank Tiano Bros. Cagayan De oro Tel. (08822) 723-167(Samsung)

TACLOBAN Door A-303, F. Mendoza Commercial Complex 141 Sta Nifio Skeet Tadoban City Tel. No. (053) 32t3706

MANII-A 2nd Floor, Concepcion Villaroman Bldg., P. Campa St., Sampaloc Metro Manila Tel. (02) 736-0966

GENERAL SA}ITOS R.D. Rivera Bldg. Constar Lodge, PioneerAve. General Santos City Tel. No. (083) 301-0987

l l

i

TIMBER DESIGN

.

Copyright 1987 by Venancio I. Besavilla, Jr. All Righs Reserved. No part of this publication mav be reproduced, storeil in a retri.val system, or trarsmitte( in any form or by any rneans, llecronic,

o

. .

mechanical, photocopying, recording, or otherwise, without the prior wriuen permission of tlre publisher.

o o

. r r . . . . r

ISBN 971- t510-14-l

o

B.

C.

Prrnta

E2 5rrnr Johr Streer. Don Bo$, Vrllegr Prinucrr. fchu ( ir) fct 272 28 l-l

I

I

Other

. . r . . .

96

97

Composite Beams

Florural and Axid Tension

166

150 165 168

Flexural and Axid Compression Wooden Colums Tapered Columns Axial toadwith Bending Composite Column Deflection of Beams Notching on Trusses

TIIEORYOfSTRUCTURBS o Determinacy of Beams and Trusses o Deflection byDouble Integration Mettrod

o

74 80 88

r3g 15r

Furlins

.

. .

73 79 87 138

SBISMIC ANAI,YSIS Earttrquake Forces Acting on Structures o Portal Method o Cantilever Mettrod

o

VIB Puhlirhcr

Stresses Beams Stress

Bending and Shearing Notching on Allowable Bending Curvature of Beams Placed on Top of ttre Bolted Connections

1

Deflection byArea Moment Method Coniugate Beam Mettrod Moment Distribution Three Moment Equation Slope Deflection Method Analysis of Determinate Beams Moment Distribution Applied to Beams Three Moment Equation Applied to Frames Slope Defleciion Method for Beams VirtualVork Method Virtual Work Method Applied to Beams Virtual Work Method Applied to Frames Virtud Work Method Applied to Trusses

t6g

t7r

172 201 203

200

202

2t0

ztt

214

213 240

241

25,6

s-1

s-18

s-19

s-26

S-27

s-35

TH.1

TH.6

TH.7

TH.l1

TH-t?

-

TH-41 TH-46

TH-42 TH-47 TH-58 TH-72 TH-78 -

TH-7r

.-

TH-94

TH-83

Ttl-57 TH-77 TH-82

TH-95 - TH-99 TH-100 -TH104 TH-105 - TH-108 TH-109 - TH-tt? TH-113 - TH-l15 TH-l15 - TH-l19

,:||

,i

D.

INEI,UENCE DHGRAM

I-l

I-10

E.

ESTIMAf,ES

E-1

E-7

PERT CPM

P-l

P-8

MC

I 6M

G.

GEOIBCHNICAI, ENGINEERING

o Atterberg's Limits o USDA Mettrod of Soil Classification

. . .

and AASHTO Method of Soil Classification Soil Compaction Permeability USCS

e Effective Stress ofSoil e Shesses in Saturated Soil o Flow Nets

. . . r .

o o

. . . . . . . . . . .

o H.

Compressibility of Soils Teragtu's Bearing Capacity Tri-Axial Test (Non Cohessive Soil) Tri-Axial Test (Cohessive Soil) Tri-Axial Test (Unconsolidated Undrained Test) Tri-Axial Test (Unconftned Compression Test) Laterul Earth Pressure Anchored Sheet Pile Slope Stability Braced Sheetingsin Braced Sheetings in Clay Piles on Clay Piles on Sand Drilled Piles on Clay Drilled Piles on Sand Settlement of Individual Piles Capacity of Group of Piles Settlement of Group of Piles Pile Driving Problems

Sand

HIGTNYAY ENGINEERING

l

c-l c-6 c-7

m c-5 c-6

a

G-g

bmding

- G-12 G-r3 G-18 G-rg - G-22 G,23 - G-23 c-24 - c-24 G-25 - G-52 G-31 - c-37 c-38 - c-39 c-40 - c-41 c-42 - c-42 c-43 - G-43 G-44 - G50 G-51 - c-55 G-56 c-58 G-59 - G-60

s

G-r0

c-61,

c-64

c-65 ,G-70 G-68

G-69 G-71 G-73

- G-72 - G-74

G-75

'G-83 G-78

c-77

G-87

G-82 G-86 G-89

H-l -

H-12

b&

1-

stress

gedion modulus)

monent of inertia at the neutral axis.

bd3 t2

d rt

"H nax. aaticalsbear

fulicqlrnanenrof mea

I

M4 24 b&

E_

8

nnnenl of insti4 ot ncutral o*is

bd3 t2 a,

ritimlsbear

-d

2

Tim[erlleslgn A

llesign

For rectangulat sectlon:

,_3V ru 2hrj

mm x 300 mm rectangular bearn carries a uniformly distributed Ioad of "tr, over its entire span. Ihe beam is freely supported at its ends. If the max. hle bending stress is 8.27 NIPa and simultaneously the max. allowable shearing

@ For trlangalar sectlan:

bd3 I__

'l/

I

i

is 0.70 MPa.

36

of the following gives the value of max. vertical shear. ich of the following $ves the value of the span of the beam. ich of the following gives the value of "IZ' in klVm.

G)@$ -24 b& 2

' h=u -2 =

=

-

ation:

,

ru -W- Ih

bil V-

uertical shear:

-

f tu-

1lt -,

6436

362

!?=

zbd

o'70=,(l,{ffioo) I/=

,,, _3V ,b4 "2

Y= 74 kN

, -3! Jt'zA

Span of bean:

@ For clrcalar ctoss-sectlon:

, -YQ Jalb n14

I Q =A! xrz 4,

v- z 3n

i

a=11 b=2r

a

Y:13

14000N

l: ;;i":1

I;r.r;'31 _l

-6M

8.27=ffiF It =

12.405xt06 tt.mm = 12.405liN.m wLZ

M=

@ Value

oJ"W"

W =28

,rr- *

"-

3.54

a

12.405

=Y

VLz=W.24

_. wL n=T W t[=T

VL=B

, --1! 3n

I

lu=m

,-4v _.-Ju -

3n'2

t-"..:..i-t

[i;:i:-i] 300

W=?3

4

100

tv'41

Di:IiJ

") ,tu= nf,2r

t'u

i

91,2=9.24

. 99.24 ,=-il{ L = 3.54 m.

W= 7.91H{/m

l

4

':

lleslgn

Iimterlleslgn

4 m. carries ttatal load including its own mm and a depth of 250 mm, used dressed 200 of l0 litrUm. tlnas a wiOtf of mm. lhe wooden secfion is made up of 10 by its dimensions reducing by Apitong. Use table on ftgureltl-6.

r

Aflgor is-supported by 75 mm x 200 rryn wooden ioists spaced at 400 nm on centerr $0, -* effective !p-an of 3 p. Ile total flogr load transmitted to the ioisrs is 5 kpn. $wood of 6.i kMm3. E11t = tax tO3 Upa _WgiSht

O Which o.f the following gives the maximum bending stress. @-Vhich of the foliowing gtves the maximuni shearing stress. @ trhich of the following gves the maximum deflectiJn of the ioist

beam havlng a simple'span of

of the followlng gives the max. flexural stress of the beam. of the followtng gves the max. sheadng stresi of the beam. of the fottoving gives the max. deflecdon of the beam.

Solutlon: _A Ma*lmam betdlng stress: wt. ofwood = 6.3(0.075X0.2X1000) M. ofumod=94.5N/m

:

flentral

Total wt. of wood

,=

=

m

W=2W4.5tVm

"t" ,m

1v1,2

h=

stress of tbe bean:

16.5 MPa 7310 MPa

l.75MPa,

S

,=ryy M =2356.3N.m

i*=tu *=ulix

. M =2356.1.x103.N.mm &V

Ib=irt 6(zlse$ rct

M = 20 kN.m

16=_7@_ Io=4.71Wa @ Mw. shearingstress:

, _3V t'zbd 2M4.5$) ,, y =1

V

@ lla$.de{lectlon: 5W4

bna*=ffi 75(2w)3 ,' -t2

I =50xtO6

=3141.75N

, _3V - zbd

ru

" 3G141.75) ro=iil)M Ia=O.31'

Wa

t'n$

=

-_-J'"i

'

mm4

3l4(tzttoij5o) lor bna*=3,6Emm. i

" ,b-

6M

.

6 (20)

b d,

lS

'o= r96'E5ox

fr=10.105

Wa


1l r(x) =

r

By slnar:

,,)

t

t,l(x)

^ o(,/11@,r,''

d =244 Say 250 mm,

, J,

l()()

rJr),650N >99,326.5N (safe)

t u, 100 x 150 struts

= J,"3V zbd

\&96f1\

0.85

=766 r rlrlhrr flume for carrying water is constructed as shown. The water pressure

d =296

Sayd =3ffi Ilse 3M x 300 wales

@

Deslgn

of strwts;

fiz = 10125 kB Rz= 99326.5N

Arial txad P Icngth of strut = l, = 1400 mm

W326.5N for eath strut 2000 - 2(300)

rug hy

; t .

I onrpute the diameter of the steel

rlmn.r of wood. l,r,orpute the size of the wooden post so as not to exceed the allowable shearing

tltts.s.

rMn,

,\olrttion: t)

r' 45 kN )J/r 0

99326.5

q6

t = ti346j

ttl(r) mmz

= fi346.5

d = $1.1 mm

rod.

Lnrpule the size of the wooden post so as not to exceed the allowable bending

,)

Jc

at the

r rtcel md extending through the opposite post

i0(

A=;P

dz

,, lll

ltlaneter of steel rod:

Trial area:

.

en a triangular hydrogtafc load against each vertical post, reaching a maximum LN/m at the bottorn. Each post is supported at the b6ttom as shoil and

lrerl*r

, /

=P(1) 4s(U

l6 t25kN

--.,

60

61

lnlm

IlmtorDoslm T = AsIs

=f,oz rrul

r25oo

d=ll.ffisay12mm [Jsed=12nmC

@

ttrcsses ofwood:

Pst tuu to baadtng:

Slzc of

-8.27 MIa

R+12.5=45

ctr€ss = 0.76

R = 32.5IN

t

_30

the

x5

l=lOx

the max.

45 kN

Vc=0

uP=o r = 1.58 M = rz.5(r.58.o.A

//=

-\ (p lalform lod:

1.58(lo) (1.t8)(1.58) 6

20.68 kN.m

-

&t'I

rb- bdz

614

6(20.6s)

ld

'fo6oof

10.2=--ibd" Tty b

"" ,L--

=

It,6tx

150

6(20.58)

10b

Slze of lmst dae to shear: Ily b =200

N.mm

8

150 x 3(M mm

l_*

@

106

g

to.z(t5o)

d =285sal3N Use

of concrete holloy block watt that the.beam could

many layers of 1i0 x 200 x 400 mm hollow blocks that the beam courd assuming 25 mmmortar in between layers of hollow blocks.

n.s-!=o

M =12.5Q.18) -

load that the beam coutd carry lncluding lts own

u 6)2

;-8 r

{,t4 kxrm

t

ru -3v - zbd

. 0.82=m d= Ilse

297 sa! 300 mm. 200 x 3N)mn.

j

I 63

62

Ilmtol0oclu

0oslgn

bean having a dimenslon shown has a span of 6 n. lte beam catrles e load of 50 kI{ at its 4idspan. Negtecdng the wetght of tte beam.

Bysbwr#rzrri:

,

J0

_3V

-zM

thsrnalmum flcxural &emax. sheadng

n=!=y,

o*=ffi

of the beam. of the beam.

;tl#)

ra = 7600N/m ar = 7.6 kl'{/m

.@ [@o*ztroor I 3 L 600+100 J

lJ&u=4.14hN/f,,

E ll4.29mm

l*r,. @

14O3.6

Iletgbt of corctete uall: 4.14 = o.l5(0.3)(5)

x

r00(300x35

106

7r),

.

mml

w

*

!ff.Ot.z,Dr)2

flennal stres of the beon:

+ 0.05(0.150) (23.5) + //(0.15X23.5)

50

tN

ug

I

H = 7.06m,

PL ----:50(5) l?

44=

@

No.

r 75 kN.m r 75x lo6N.mm 75x td (185.71) .-+---..r------

oJlayen of ffiB:

Height olone CHB = 0.20 m.

1403.6

Tonl hQghtpr CuBuitlt morbr

=0.225m. No.

oftayrsof

nU

100

strass.' V(2b +B)z (?,8 + b'tz

=# U

No. of hyers of CHB

= 4.7 say 5

x

9,92 MPa

loz

*

4u6

*

sz)ftz

+ ab +

9=114.29

ilf

laycn

15O

mm

_ r

150000

[2(100) + 6oo]2 [z(Oool

3(105000) [(too)z + 4(600)(100) +

1,96Wa

*

roo]2

tfool2] ltrool2 + 600(100) + (500)?]

64

65 l

Dorlun

IlmDfi ll0slen

l

Trro movtng lmds onststs of 65 kN md 45 klt is moving at e constant dtstance of I m. alongaspan of 7 m. Compute the locadon of tte 65 kI{ load from the end support of the beam to

l j

O @ @

o

Locatton of tbc 65

lNlron

the

n '--n

eadsapprt of tbe beam:

obiln max. momtnt, ptace tbe IMA in srcb a uay that the certter with th center of tbe bigg$laqdandtbemultant lM, Ibe 65 kNnu,stuPlMat 2.8E85 m.funA.

Mar. momett fure to tbls Ywnz=o Rr = 45.341d'{

Max, noment = 45.j4(2.8il Mdu. noment = 730,EI hl{,m

Max. bendlng stress of.thg bean:

ainckkt

r= u =ffQ..x)

,=*9Y@

//

= 8.01 kN.m Total moment = 130.81 + 32.70 + 8.01 171.J2 kN.m

^ 6u 6(tzt.sz)

-

|t)

65 kN

too

lb=iA, = -3oo{doolZfb=9.r3wo

1--

--1zoO:-1

(L)

.vI" n @\L_ 7+@i r 98l(3)(1.5) + P(1.5)

45 kN

ei$ 65 kN

coacentrated bdat tbe fup to bendJng:

l*lan

t4414,.5

+ l.SPl{/m

:44145N + 1500PN.mm

r

150

il! I

(44145N

---.

45 kN

+ -n3*tT*' 1500D(150)

l8335tl = 1E.335

= 0.25(130.81)

Inpacl momm, = 32.70 kN.m l,lonent duc to fudlmd: 0.3(0.6)(7.5) = 1.35 kN/m

rnonent =

of ttn beam

lod:

7fi1 = l16(2.335;

Total

axis of the beam. that the beam could carry at its mld shess of 12ltIP:* that the beam could carr! atits mtdspan stress of 0.82 lIPa

:!l$s tN

To

Impd momml

l

at the

oJtnertta at the nedral 200(1003 50Q00)3 .

ll0r=65(0) +45(3)

o

inerta

the addltional concentrated go as not to exdred tte the addidonal concentrated not to exceed the iallowable

l'

Solution: x = 1.23m.

@

the moment of

produce mulntummomcnt Comlute tte mu moment due to thls movlng loads. If a 300 mm x 6(X) mm beam ls used to carry thls load on a span of 7 m.' compul! the maxlmum bendlng stress of the beam. Ldd 25% hpact stress due to lho moving loads. Gonsider the weight of beam whtch is 7.50 klVmr. Alowablt bending strtss ls 10.5 MPa.

b{

corccn roted

lodat

tbe

fue tosbean

.W : r

2fr0(50) (tzil

+ loo(looX5o)

1750000

r--2oo--_-1

-# Y(t75cf[f.l

G_-.--.-7*

. 323x l@ (100)

V:179ll!5N

:981$)

+P

=2W+P

15003N= 15.U3

At

l

66

67

IlmDorDoslgn

Four 50 mm x 2(X) mm dressed plank are used to cary a super lmposed load of

'S

antform

lod lf lt k plued

on q hollou box:

kIVm.

O

5)3 If the planks arc placed stde by stde wtth the 200 mm side ln

a verdcd positlon,

@

determtnb the safe unlform load lt could lt carry lf lt has an allowable bendln; stress of E.30 MPaand aspan of 6 m. If the planks are placed in a form of a hollow box, determtne the safe uniforfl

@

load lt could carry. IIow should the four planks be framed to glve the maxlmum sdfrness as a beam?

O

Safc ualform

Solution:

lod lt otld utty tJplued ddc by sldc:

Note For Drcsed dimensions ddud ttre follovdng'ttricknes. For msnbers les than 150 mm deduct 9.5 mm (!13) For membes greater than 150 mm deduct 12.5

b = 50 -9.5

b =40.5mm d =200 -

12,.5

On)

t2

2{4xto6

wl

106.50(rs7.5)3'

t2

'ilc I 15.085 kN.m

VL2 8

w6)2 8

,.35 kN/ril

d =187.5mm

.--162(1s7,il3 t2 /i 8!x tdmma ,

funngocttuxlmun

retbcftwnnx bx.

- tlc lo=T

w=ffffil ldN.ilm

,,/=

7.88x

//=

7.88h'lm

vL2 M=t

v6)2

7.88=--

V=7.79 kVtn

l l

68

69

Ilm[erllesign

@ncentrated load tt uould carry: 125) + 10060)60)(2)

Four 50 mm x 200 mm section is to be framed to carry maximum shear on a span of 4 m. Neglecting the weight of the beam Allowable shear stress ts 0.70 Mpa.

F--200----r tt 100(2oo)3

O

t2 x 106

Compute the safe concentrated load that Ore beam could ranl al a distance of 1.5 m. from tfie flxed support if the

12

beam is arranged as shown. N KN

hN

ponoentrated load

3

roo(zoo)3

t2 x

@

lt could carry:

12

106

)(125) + 100(50X50)(2)

Compute the safe concentrated load that the beam could r:rtry al distance of 1.5 m. from the ftxed if the beam ls

1750000

arranged as shown.

tb

w r/ (r750000)

*&-3;106(too) N

15.32

N

concentrated load corry:

@

Compute the safe concenttzled load thattte beam could czrra! e.aitiCtanrr. oII'.5 m. from tte fixed sulrport if it is arranged as shown.

E--

'3v

2(200)(200) 18667 N

1s,667

W

tt

r_zo0_____*-l

7t

70

Tim[ullesign

A 100 mm x 250 mm wooden beam 3.60 m. long was designed to carty a uniform of "F' kN/m at the teft end and increasing to 3W kNrrr at the right end of the si supported beam. Neglecting the weight of the beam.

6tr

'imE5oP 12,92

O @ O

Compute the location of max. moment from the left end of the beam. Compute the value of tV' lN/m if the altowable bending stress'is 12.4 MPu Compute the max. shearing stress of the beam.

x

106 N.mm

12,92 kN.m

:'R1r-w*q

Te)

,9X 1,80

Solution:

O

'e

Logattoi of max, mommt:

l,l8,3u

h=w(3'6) tD^t_-2W

0.95) 1.80

vw(tu*y

(3.6) Z

ry(1ffi3w)

3,262u)

Pz=3.6W

= 3.25h1

3.96

Wm

blpr=o 36R1=36w

e) .*w()

ltet

'tfil

R1= JIY

.'36.96) xy/2

r-_fi.

t/

2W

+R2=Jg^.ryy

WX

t7,2w -3a

1.80

t 4,Zw

4={+w 3* =[,

;11.88,

ffi.,w

xZ+3.6x-10.8=0 x = 1.95 m.

3 4.2(3.96)

RF3W

16,632 kN 16.632 kN

l6632N

v_

'm 306632)

'ilrobi(2jo) -0,998 Wa

Rr=11.88 I 1.88

Rz=16.632

72

IIm[u0eslgn

A simply supported wooden beam carrying a uniform load has a span of 9 m. beam has an adequate lateral supports. Allowable stresses: Bending =

l2.40MIa

Deflection =

I

ffi

(tooo)3

ofsnan

3ooo)t

f, = 13fi)0 MPa

i$o+ar

r

544350

O

tlat when the allowable stress of l2A0 ofthe beam is fr ofspan.

Compute the depth of the beam so reached, the deflection

@

If the width of the beam is 300 mm, how much total uniform load can it

Mpa

lt

t

m.

650mm

load:

salcly

support?

O

Compute the shearing stress of the beam due to this uniform load.

6M

=fr-r(esof 26t.9s t 0.60 Fy = 150 f" = 150 MPa 20.67

f*=150

f- =7.26MPa

< 7.3 MPa (safe)

momentof steel:

rO Determine the total resisting moment capacity of the flitch beam.

@ If 800 kN equals the total uluformly distributed load supported by the beam, what

25(300F

portion of the load will be supported by the steel?

@ Determine the shear capacity of the steel. Allowable shear stre-ss of steel

is 0.40Fr.

+SS.ZSkN.m

momentof wood:

Solution: O Total resisting moment capacity of the flitch beam.

5M j----lg,

bd2

-6M

,r_w

2(200X300)'

=43.56kN.m

resisting mement:

n=

:'M!

E

M*

56.25 + 43.56

----!-

E

200000 9697

n=20.67

+

:

.

,99.8L kN.m li l

163

162-H

Iimter llesign

@

Deslon

Porton of load that will be supported by steel:

i tO reinforred ,1"d1 = !5'25 %of momentcarriedby-

,,.g1

%

11gg1 .

,

of moment carried bY steel = ffi .%%

Therefore 0.5636(800) = 450.SS is

the steel bearn, a wooden planli is placed between the flanges as

$owable flexural stress of steel is 176 Upa aod for wood is 21 Mpa f" = NIPu, Eu = 11400 MPa Moment of inertia of steel Ix = 8.5 x 106'mm4 and its arca is 5680 mm2.

of the following givcs the max. bending moment trat the beam can support the wooden plank. of the following $ves the moment of inerda about the,neutal axis ryhen the

kN

pla+kis afrrched'.

cadeilby steel

of the.follorring $ves the max. bending moment traf the bem can support thcrooden plank is anached. Determine the shear capacity of the steel. Allowable shear stress of steel is 0.40F,

bdfng

.\-2bd -3V" o.4oFy

I

=#

c.

t4 (to5)

3V

o.+o(2so)=7offi V

matmnt uttboat the

t*plank.

'B:5*m x 106N,mm

76lN.m

= 500000N

v" = 5001t't

oftfrortla. o, N.A. uood to steel sectlon:

mm mm2 r7.10)

164

165

Ilm[erlleslgn A =A1 +AZ |

Ay=Atyt+Azlz

rlte beam is constructed from wood and reinforced with steel strap. The ,rected to a moment of 90 kl.in and a max. vertical shear of i0 lN. E; =

73Wy =5680(0) + l7l0(55)

I

=

12.73

Ir,tA

= 8.5

xL05

+ 5680(12.7rr,

^8; = 15000 MPa

.

!#@

+

fit0(42.27)z

Int= 13:9r.16mm4

pf the following gves the max. fleniiat stress oJwood. of the following gives the max. flexural stress of steel strap. Of the followinggives th'e max. shearing stress in the beam.

@

tlar. betdhg

momcnt uben the uoofun plank ls

ortobet:

stress of wood steel to equloalent woobn

Forileel:

.MC fi=t

-

200000 15000

M (n7.73) rrr=1$iffi

y

M =20-l9Xto6tt.mm

x

//

.

7333.3:3(10)(15,il1)2

106 mm4

= 20.19 kN.m

ForrM:

- MC nlu= t stress

t7.54ar)='tU# u =55.49x16 M =55.491rt{.m

ofsteel:

x 106 (l5o) '1515.30 x I

r7rfi

MPa

We M = 2O.79

k[.m (snel otttols)

stress

ln the bean:

10)(15, + 150(175X75) t6.67

x 106 mm4

166

167

trmter0esl0n

Dcslgn

lpen member is 3 m. in length is made up of Apitong 150 mm x 300 mm loctlon, with an allowable stress based on 80% stress grade as shown on the Members subiected to hoft flexure and axtal tension shall be propordoned such 11 F1

.'*,

1.0 and

'ff

=

r.o

grain in bending

and tension parallel to

=

of elasticity

= 7310 Mpa

grain perpendicular to the grain prrallel to the grain parallel to the

w klVm

16.5 Mpa

= 9.56 Mpa

= Z.Z0 MIa = 1.73 Mpa

lcarries a uniform load of 18 kMm besides its own weight t. the beam carries an axial tensile load of 180 kN.

Vei$t

of wood =

of the following gives the actual tensile strcss if ody tensile forrce is acting.

where: -ft= actual arial

of the foltowing gives th6 interaction value of both bending and teniile stress

of the following gives the ratio of the difference between its actual bendlng stress to the adiusted bending stress for slenderness.

^T J,= I

hnrlle

-fb = actual bending shess

.

t|c

Jb=T

Fr = allowable

axial

stress

Fb= alhuabb bmding s'tras

T

I

Fb' = albuable unit strasfor

ertremefibu in

b mding

adj w te d fw s len dtrn as

_lEuoo It0(300)

I.ttPa

18 kIVm

168

169

Ilmtar0oslgn

@ Iateroctlon oqlae:

ft, .ft Ft Fo w=18+0.15(0.3X7., W= 18.3375k0il/m 'g4z

M=,

to

btb flacarp and drhl amprwbn slnll b poportilnd ilcb tfut

< 1.0

u'18'33ts$)z

wkMm

8

/1=

20.63 kN.m

6l'l

rrobil 5(20.63) rc6

" rb=1fr(Nri4 fo = g.t7 tlPa

rt .fo

4

F,* ro= r6J.

s.n

iG

4 rt **tb = o.rg, @

Ratio=W Le

= 1.g2 Lu = 1.92(30N)

1P

=

Le

5/@y

,r={4 Cs

=8.76

= zstcs

d = 1586 Try 200 x 200

L1 l --u\

40oo tnn

Lvv

--'

Fall

(5.28) 260000

(ok)

x 25Amm sectlon.

t;

/ll

--

? = 1lP = 35 > 245t oongcolumn)

rc between

=

capacttt ofselected secilon to the reqalred load:

264000 - 250000

= 40OA

N

183

t82

Ilm[orlleslgn

,E .ffi

An old Apitong post 200 mm x 300 mm x 4.25 m. long has been previously des wtth an atowatie compressive stress of 9.56 MPa and a modulus of elasdcity of MPa It is desigred to srbstitute the old post vith a Yakal post of the-same lengtlt the,old poqt Altgyble compressive stress for Ydd is 15.8 MPa with a modulur elasticity of 9780 MPa

CI n/hat is the capactty of Apttong? @ What stue of Yakal post is required to replace Apitong? @ What ts the percentage increase in the capacity of the nev post to the old post?

longmlumn

'0'30(9780)

(2s3il2

O

Solution: Capacttl ofAPttong:

3,66

f,=ffi="'s

D

(3.66)

(fait)

1r.,,

K=0.67t

\ .,

x 200 (3.66)

K=0.6714ffi

N

x 200 18.55

bng column

Q/ff 0.30(7310) -?ott=-dffi

Fail = 4,86MP?^ P =A Fall P = 200(2oo)(4.86) P=

1944(MN

Slze ofYakal Post: Trial area dz

=

=21.25

0.30 Eu

=

>

16,69

0.30(9780)

'erffi MPa

Foa 50)

N>l)22N

x2il)mm

(safe)

YakalPost

194400

(15S)

12304

d = 110.9

Try 150x150

t==@=2a.33

lnctease ln capQtjt:

*

(2500@. lg44oo) loo tg4400

=33.74%

t84

185

tlmDff Ds$lgn

A steel beam is used as a girder with 7.3 m. spar carrying a total uniform load ol

kMm includlng its own weight

Pnoperdes of steet

'itodutus of elasdcity = 20O000 Mpa Moment of Inertta ir = 723 x 106 mm4 Properties ofwood: Allowable comprrssive stress parallel to the graln = Modulus of elasdcity = 13790i/Pa

O

ffi fO.fl

13&

i|k

10.35

,LO

Which of the following glver the deflecdon at the mtdspan.

@ vhich of the fotlovtng gves the reactlon at rhe midspan suptr,ort if tt is planned

underpin at the midspan by ustng a'wooden post so as to reduce the deflection

'ol,

12.7'm,[n. @ vhich of the folloving gves the size of the square wooden post if it has a helght

3m.

ro.Es

Solution:

f*o**A*

column)

i (g-l

[r-

;(#;'l

10.15 MPa

@ DeJlectlott at mldslnn 5WL4

11=

of 250 x 250 rnm mlumn seclion:

fi4EI 5(l45ooo) (7.3)4 (looo)3

/1=J84(2mxzjtio[

) (10.15)

?l=37.(Rmm

N >434985N (safe)

@ Rerctlon at mtdspan to redrce tbe

deflectlon to 12.7 mm.

! =lt-!2

.

,r.Z=iZ.*-nftffiffi R = 4j4,985

N

@ Slze of uoofunlmst:

i(fu)-] /,,,[, rura

B.a6

P

t.nal area = Fc A _434985 -t0.35

A

=

) (9.86)

394400 K

wooden post to the actual loadt r30 kN

0"671

Solution: Posltlon to obtalmd tbe nnst strassed past:

- ItlC J- I a

=ZA0(7) 1400 kN.m

./=

colunm)

0.N Eu

I =2(2.n)z I =8.99m4

. "

Qang

=wffi

=2.12m.

f =-

15.80

t5.69 :K

iI= C

ruq

0,571

Yarttcal reutlon due to moment.

o.30(9780)

$7.,2

t4N(2.12) 8.99

= 9.58MPa

330.14 kN

Vutical reaction due to utt. of tanh:

Falt

tr=ff=3z.su,N Total

(9.58) N >36W20 (safe)

urticalrwtion =

330.14

+

32.5

=

mm x 2(N) mm

362.64 kN

Actilal laod:

t,of

caputty of selutedpost

lod:

=M

3832A0

p

= 7.(M

*5.t0.

352.64P

5

= 369.82 kN

PNillffi

to obtdlncd tbc

srffi,,sl

mt

191

190

Ilm[fi

llo$l0n + 2(t.15)z

A water ank having a total weight of 140 kI{ ts supported by three (3) wooden It ts $biected to a selsmlc force of 90 kN actlng at 9 m. above tte eoncrpte rvhtch is arranged forming an equtlateral triangle having e length of one side equrl 4 m. wooden posts ts battered t horizontal to 4 vertical. Assume shess parallel to the grain ts length of the post to be 4 m. MPa Modulus olelasdcity of wood is 13800 MPa Neglect wind cffect on pracingp.

lte

Witkn

Allorvable

doleto

nranmt:

(e)

)kN.m

140t(N

hrd

'

d*

to

lM

*iybt

oI

*k =+

={6.67 loi

r$.57 28l.l4 kN =234.47 + =

c.c.l

Pr=281.1{

Posttton offootfng fn otht to obtaln MLr. SffestorA

-,

4.123 4.0

= 289.79

o @ @

o

tle max. reaction of the most stressed wooden post Determine the min. reaction of the wooden post Determine the size of the wooden post to carry such load.

Hl

(rnaa. reac$on of

nwt

Compute

l,

Solution: remtlon of tbe most i=4Sin60'

Ma.x. h

=3.46m, ,)

* ="; $.46) J

x =2.31m.

y=3.46-2.3t ?=

1.15

rcuuon of the uoofualpst:

stressed colamn:

7,98 73 kN

aertiml

lM = tt6.n +ff lM -763.40 hll

s.tresvl mlumn)

192

193

nm[fiDrslgn Min.r@ion: water taak with 4 m. inside diarneter is supported by four wooden posts diameter of 250 mm. Ite four points of suppotls A, B, C and D under a square with 4.6 m. sides and 6 m. below the tank, tte coresponding of supports form a btgger square of 7.6 m. sides. When empty the tank LI{. Each post wtll carry one fourth of the total load equally and trat the with the axis of the post Allowable comprcsslve stress parallel to MPa }todirlus of elasticity is 135(X) MPa 10.50 =

Dral6t.1O

' Pt* _t63.40 4t23-

P66=

4

16t8.42

Et

O $tnofsoob;pst: Ilhl aw: P A=E

, _w790

" - (10.5) & =27599

d= l66mm

Ifi 2A0imx

2A0mm

fi=ffi=zorn the allowable compressive stress of the wooden posl the capacity of the wooden pos-t.the maximum height of water in the tank that the posts could safely

,(=o.67raH K=0.671 K=24.33>20

IntMialealunn:

Fc,=Fclr;@)^1

compressloe stress

fun bost: (1.5)2

'Fc'=10.5

[' iQtJ'*]

F'c'= 8.S MPa P =AFc' P = 200(200)(8.90) P = 3560000 N > 2W7g0N (wfe) Use

2(M mm x 2(N mm awdcn post

+ (t.S)z

4,5

(6)2

+x2

!6 +

4.5

6.36rr..

n

(250)2 4

221.56

of

194

195

Ilm[orlloslxn L_ 6ffi d- 22t.56

i

'z=u,o

x 300 mm rectangular beam is supported ln a horizontal position shown. At

It ls being held by a pin.and at'B'i by a cable BD inclined 3 verticd to 4

K=0.571

Assume all forces are applied to the beam along its central axis. Allowable ve stress pamllel to the grain = 10.50 MPa Neglecting the weight of the cable. Modulus of elasticity of wood = 13800llPa

K=0.671\m K=24.M