TIMBER DESIGN

PROPERTIES OF WOOD

a.) EDGE GRAIN – growth rings are not approximately at right angles with the surface lumber.

b.) FLAT GRAIN – face is approximately tangent to the growth ring.

c.) CROSS GRAIN – is the deviation of the direction of the fibers of the wood from a line parallel to the edges of the piece.

MODULUS OF ELASTICITY (E)- is the measure of stiffness and durability of materials of a beam, it is

the measurement of the resistance to deflection.

COMPRESSION- ability of wood to resist compressive stresses, depends upon the

direction of the load with respect to the grain of wood then in compression perpendicular to grain.

TENSION PARALLEL TO GRAIN- same as unit fiber stress in bending.

SHEARING STRESS- ability of timber to resist slippage of one part another along the grain.

In beam, it is known as the horizontal shear.

NOMINAL SIZE- is the same as undressed size. Dressed size is the actual dimension of

a finished product.

WOODEN BEAM

BEAM AND BENDING

BEAM – is a structural member subjected to bending flexure induced by traversed loads.

BENDING MOMENT – is the summation of moment taken to left or right of the section about the neutral axis.

TWO TYPES OF BENDINGS

a.) SYMMETRICAL BENDING – is that in which the plane of application of the load is perpendicular to any principal area.

b.) UNSYMMETRICAL BENDING – the application of the load is not at any principal axes but causes both the major and minor axis.

CONDITIONS OF EQUILIBRIUM

W

W

1.) The summation of tensile stress is equal to the summation of compressive stress.

∑FH = 0

2.) The summation of external shear, VE is equal to summation of resisting shear, VR

∑FV = 0; ∑FE = ∑VR

3.) The summation of external moment ME is equal to summation of resisting moment.

FOUR MAJOR CONSIDERATION OF BENDING DESIGN

1.) The beam must be safe from flexural bending.2.) The beam must be safe from allowable shear stress.3.) The beam must be safe from deflection.4.) The beam must be safe from end bearing and end connection.

RESISTING MOMENT OF RECTANGULAR BEAM

The resisting moment must be equal or > the external moment.

ME

VE

COMPRESSION

TENSIONVR

MR

OTHER FORMULA FOR FLEXURAL STRESS:

Rectangular: S= bh 2 ; fb = 6M 6 bh2

S= πr 3 ; fb = 4M 4 πr3

Triangle: S= bh 2 ; fb = 6M 24 bh2

Tube: S= π (R4 - r4) 4R

FLOOR SYSTEM

S

DERIVATION OF FLEXURAL FORMULA:

FLOORING

TRIBUTARY AREA

JOIST

GIRDER (Main Beam)

COLUMN / POST

W

ELASTIC CURVE

C

N.A.

Xdx

d

By Ratio and Proportion Stress Diagram

f = fb

c x

f = c fb

x

f = x f c

therefore: fb = f

Summing of Moment from N.A.

M = fb dAx fb = f = f ∫ dAx fb = Mc

c I

I = ∫ dAx2

FOR RECTANGULAR SECTION:

f = Mc

I fb = 6M

bd2

where:I = bd 3 ; c = d 12 2

ILLUSTRATIVE EXAMPLE:

fb = 12Md/2 bd2

1. A wooden beam 150 mm x 250 mm is to carry the loads shown. Determine the maximum flexural stress of the beam.

∑MA =0

RB = 15 (2) + 6 (3) (1.5)

3 RB = 19 kN

∑MB =0

RA = 15 (2) + 6 (3) (1.5)3

RA = 19 kN

Mmax = 16 KN-m

fb = 6M bd2

= 6 (16) (1000) 2 150 (250) 2

fb = 10.24 MPa

2. A timber beam 100 mm x 300 mm x 8 m carries he loading as shown. If the max. flexural stress is 9 MPa. For what max. value of w will the shear be zero under P. What is the value of P?

150mm250m

m

2 m 1 m

14

2

-13

-19

16

15 Kn

6 Kn / m

P

W

6m 2m

∑MA =0

RB = 6P + w(8) (4) 8

RB = 4w + 0.75P ∑MB =0

RA = 2P + 8w (4) 8

RA = 0.25P + 4w

V & P = 0

Then: Substitute P = 8w

0 = (4w + 0.25P) – 6w 4.5 = 4w + 2.05 (8w) 0 = -2w +0.25P2w = 0.25P w = 0.75 KN ≈ 750 NP = 8w

Mmax = 3 (4w + 0.25P)

fb = 6M Solve for P:

bd2

9MPa = 6(3)(4w + 0.25P)(1000) 2 P = 8(750) (100)(300)2

4.5 = 4w + 0.25P P = 6000N

3. A floor joist 50 mm x 200 mm simply supported on a 4 m span, carry a floor joist load at 5 KN/m2. Compute the centerline spacing between the joists to develop a bending stress of 8 MPa. What safe floor load could be carried on a centerline spacing of 0.40 m?

Sol’n:

a.) b.) s = 0.40m ; w = ?

Mmax = wL 2 Mmax = wL 2 8 8

= 5s(4) 2 = 0.40(P)(4) 2 8 8

Mmax = 10s KN-m Mmax = 0.80P

fb = 6M fb = 6M

bd2 bd2

8 = 6(10s)(1000) 2 8 = 6(0.8P)(1000) 2

(50)(200)2 (50)(200)2

s = 0.27m P = 3.33 KN-m

HORIZONTAL SHEARING STRESS

General Equation

fv = VQ Ib

where:fv = shear stress; MPa

W

6m

50mm

200 mm

V = shear force or Vmax; KN or NQ = statistical moment

I = moment of inertia

FOR RECTANGULAR SECTION (DERIVATION)

where: I = bh 3 12 Q = h · bh

4 2 Q = bh 2 8

Substitute Q in the gen. eq’n FOR ANY SECTION

fv = VQ fv = k(V/A) Ib k = 3/2 for rectangular

V {bh 2 } k = 4/3 for circular

= 8 k = 2 for circular thinning

bh 3 · b k = bet. 3/2 ≥ 4/3 12 for trapezoidal section

fv = 3V = 3V 2bh 2A

DESIGN OF BEAMS

PROCEDURE:

1.) Load Analysis – compute the loads, the beam will be required to support and make a dimension sketch (beam); V and M diagram to shown the loads their location.

2.) Determine the max. bending moment and max. shear. Compute the required section modulus from the flexural or solve for bh², select adequate section.

b/h = ¼ or R/h ≤ 203.) Investigate/ analyze the beam selected for bending and horizontal

shear if it falls, revise the section. Investigate the beam for deflection.

h/4

h/2

N.A.h

b

Problem: The second floor of an apartment building is a constructed out of 1 in. thick T&G flooring on floor joist are supported by girders spaced @ 2.5 m o.c. Design the floor joist using 63% stress grade guijo . LL = 4.8 KPa; Gs = 0.65.

Data: flooring: T & G = 1” Joist = 63% Stress grade fb = 17.1 Mpa fv = 1.89 Mpa

Soln.

[1] Load Analysis

w = D.L + L.L L.L = 4800(0.4) D.L. = 0.65(9810)(0.0254)(0.4) L.L = 1920

D.L = 64.79 N/m w = D.L + L.L

= 64.79 + 1920

w = 1984.79

[2.] Max. Moment and Max Shear

Vmax = wL 2

= 1984.79 (2.50)/2

Vmax = 2480.99 N

TRIBUTARY AREA

FL

OO

RN

G

0.4 m

2.5 m

W

L / 2 L / 2

Mmax = wL 2 8 = 1984.79(2.5) 2

8Mmax = 1550.62 N-m

FLEXURE:

fb = MC = 6M b = 1 = b= h I bd2 h 4 417.1 = 6(1550.62)(1000)

bh2 therefore:bh2 = 544,077.19 mm3 bh2 = h · h2

4therefore: h 3 = 544,077.19 mm3

4b= 544,007.19 therefore: h = 129.59 mm ≈ 150 mm

(129.59) 2 b = 32.4 mm ≈ 50 mm

[3.] ANALYSIS

w = DL + LL from flexureDL = 64.79 + (0.65)(9810)(0.050)(0.15)

= 112.61 N/m fb = 6MLL = 1920 N/m bd2

therefore: = 6(1587.98)(1000)w = DL + LL 50(150) 2

= 112.61 +1920 fb = 8.47 MPa < 17.1 MPaw = 2032.61 N/m SAFEMmax = wL 2

8 from shear= 2032.61(2.5) 2 8 fv = 3 · V

Mmax = 1587.98 N-m 2 AVmax = wL 3 · 2540.76

2 2 50(150) = 2032.61(2.5) fv = 0.51 MPa < 1.89 MPa

2 SAFEVmax = 2540.76 N Adopt: 50 mm x 150 mm floor joist

PROBLEM:Timber 200 mm x 300 mm and 5 m long, supported at top and bottom,

back up a dam restraining water deep (3 m). Compute the centerline spacing of the timber to cause a flexural stress of 7 MPa.

Mmax

P = h = 9.81 (3)P = 29.43 KN-m

Solving the reactions due to fluid pressure

∑MR1 = 0R2 = 29.43s(3)[2/3(3) + 2]

2(5)R2 = 35.32s

∑MR2 = 0R1 = 29.43s(3)[(1/3)(3)]

(2)5R1 =8.83s

f

P

2 m5 m

Solving for x and Mmax: in terms of s:

by ratio & proportion_ _y/ x = 29.43(s)

3 _

y = 9.81(s)(x)

∑ fv = 0VA = 0 = 8.83s – 0.5(9.81s)x2

0 = 8.83s – 4.905s(x2)x = 1.34 m∑ MA = Mmax

Mmax = 8.83s(3.34) - 0.5(9.81)s(1.34)[(1/3) 1.34]Mmax = 26.56s

Therefore: fb = 6M bd2

7 = 6(26.56s)(1000) 2 200(300) 2

s = 0.79 mm

y

2 m X = 1.34m

VA

MA

R1

NOTCHED BEAMS

THE REQUIRED ARE GIVEN BY:

fv = 3 · V or 3 · V 2 A 2 bdfv = 3 · V ; d = bd1

2 b d1 OR

d1 = 3 · V · d 2 fv b

LENGTH OF A NOTCH

R = Vmax = ½(ZQ)b therefore:

L = 2Vmax ZQb

d1 d

b

ZQ

V

where: L = length of the notch ZQ = allowable stress in compression to the given

DEFLECTION OF BEAMS

Allowable Limits:

L/360 = for beams carrying plastered ceilingL/700 = for beams carrying a line of shaftingL/200 = for railroad stringerL/480 = for beams supporting concrete forms of R.C. slabs or beamsL/360 = for all other unless specified

1. Use full value of Young’s Modulus of Elasticity, if the deflection is due to transmit loads like live loads.

2. Use a portion of E ranging from ½ to ¾ if the deflection is due to a constant load like DL.

= kwL2

EI

where:k = numerical coeff. depending on the load of the beamw = total load; KN/mL = span in meterE = Modulus of ElasticityI = Moment of Inertia

METHODS IN CALCULATING DEFLECTIONS

1. Unit load method2. Method of superposition3. Conjugate method4. Double integration5. 3 – Moment Equation

Example: Design the beam shown using 80% stress grade guijo.

Data: For 80% guijo

fb = 21.8 Mpa fv = 2.4 MpaW

Za = 4.26 Mpa

Reactions: ∑Ma =0 Rb = 2(4.5)(.75)+10(3)+1.67 6 R b = 6.4 KN

∑Ma =0 Ra = 10(3)+2(4.5)(5.25)-1.67 6 R a = 12.6 KN

Mmax = 17.55 KN.MVmax = 9.60KN

2.) TRIAL SECTION LOAD:

fb = 21.8 Mpa b = h/4 fv = 2.4 Mpa h/4· h² = 4.83 x106 h = 268.33 mm say 300mm for flexure: b = 67.08 mm say 100mm

fb = 6M try: 100X 300 mm BEAM. bh² 21.8 N/mm² = 6(17.55 KN.m)(1000²) bh² bh² = 4.83 x106 CHECK FROM fv LENGTH OF THE NOTCH

fv = 3/2 Vmax L = 2V ZQ = 4.26 Mpa A ZQb 2.4 = 3/2·9.6(1000) L= 2(9.6)(1000) 100(300) 4.26(100)= .48 L = 45.07 mm

fvall > fvact safe! Adopt:100X300mm (BEAM)

3.) DEPTH OF THE NOTCH

d = 3/2 · V/fv · d or h b d = 3/2 (9.6 x 1000) · 300 2.4 100

1.5 3.0 1.5 1.5

d1 = 134.16

d = 134.16 mm

ANALYSIS OF IRREGULAR SECTION:

Problem: The T – section is the cross – section of the beam formed by the joining two regular pieces of wood together the beam is subjected to a max. shearing force of 60 KN. DET.

a.) Shearing stress of the N.A.b.) Shearing stress @ the junction bet. two pieces of wood.

FIGURE:

SOLUTION:

45.07

200 mm

20 mm

40 mm

100 mm

CONECTION

a.) Solve for N.A. (V. Theorem) A1 = 200 (40) = 8,000A2 = 100 (20) = 2,000

AT = 10,000 mm² _

10,000 y = 8000(20) + 2000(90) _

therefore: y = 34 mm from top

for: b = 200; Q = 200(34)(17) Q = 115,600 mm3

Solve for I (by transfer formula)

INA = I + Ad²

= [200(40) + 8000(14)² + 20(100) 3 + 200(56)²] 12 12

INA = 10.57x106 mm4

fv = VQ = 60(1000)(115,600) = 3.28 MPa Ib 10.57x106 (200)

b.) Shear stress from the joint

For: b = 20 mm; Q = 20(100)(56) Q = 112,000 mm3

fv = VQ = 60(1000)(112,000) = 3.28 MPa Ib 10.57x106 (200)

fv = 31.79 MPa

For: b = 200 mm; Q = 200(40)(14) Q = 112,000 mm3

fv = VQ = 60(1000)(112,000) = 3.28 MPa Ib 10.57x106 (200)

fv = 3.18 MPaPROBLEM:

Determine the safe concentrated load P @ the center of the trapezoidal section having a simple span of ? m; if fbALL = 10.34 MPa, neglect beam weight.

APPLIED LOAD

Solution:

Mmax = PL/4 = 6P/4Mmax = 1.5P eq’n 1

From: fb = MC I

Locate N.A.:

A1 = 75(125) = 9375 mm² A2 = 2[1/2 (115) (125)] = 14375 mm²

AT = 23750 mm²

By V. Theorem:

23750 yb = 9375 (62.5) + 14375 (14.67)yb = 49.89 mm

FOR INA

INA = I + Ad²

= [75 (125) 3 + 9375 (12.61)2 ] + 2 [115 (125) 3 + 115 (125 ) (8.22) 2 ] 12 36 2

INA = 27.15x106 mm4

fb = MC I

10.34 = 1.5P (49.89) (1000)² INA

P = 3.75 KNBEAMS LATERALLY UNSUPPORTED

Allowable Extreme Fiber Stress:

12.6162.5

8.2241.67

REF (ў)

N.A

75 mm

115 mm

3 m 3 m

P

f = fb [1 – L ] 100b

- when the span with no lateral supports exceeds 20 times the width of the member or:

L > 20B

where: f = allowable extreme fiber stress for a beam laterally supported fb or fp = allowable extreme finer stress for the timber when beam is laterally supported L = unsupported span of the beam b = width of the beam

PROBLEM:A beam having a span length of 4.5 m has two concentrated load of 13.5

KN at the third points of the span, the beam is laterally unsupported. Design the approximate size of the beam to carry these loads. If the allowable bending stress is 9.7 MPa for a beam laterally supported and an allowable shearing stress of 0.83 MPa. Assume weight of wood is 0.3 KN/m3.

Given: fb = 9.7 MPa fv = 0.83 MPa g = 6.3 KN/m

Sol’n:from flexure:fb = 6M bd2

9.7 = 6 (20.25) (1000) 2 bd2

therefore: bd2 = 12.53x106

but: b = d/4 ; d/2try: b = d/2

d/2 (d) = 12.53x106

therefore: d = 292.64 mm ≈300 mm R1 = 13.5 KN b = 146.31 mm ≈ 150 mm R2 = 13.5 KN

try: 150 mm x 300 mm Vmax = 13.5 KNMmax =20.25 KN-m

From Shearingfv = 3/2 (V/A) = 3 (13.5) (1000) 2 (150) (300)fv = 0.45 MPa < 0.83 MPa ok!

CHECK:L/b > 20

1.5 1.5 1.5

13.5 13.5

20.25

L/b = 4500 = 30 > 20 ok! 150

Considering the weight of the beam:

Beam wt. = 6.3 (0.15) (0.30) = 0.28 KN/m

Vmax = 14.13 KNMmax = 20.96 KN-m

From Shearing: From Bending:

fv = 3 (14.13) (1000) f = 9.7 [1 – 4500 ] 150 (300) 100(150)

fv = 0.47 < 0.83 ok! f = 6.79 MPa

Then:fb = 6M bd2

= 6 (20.96) (1000)2

100 (300)2

fb = 9.32 MPa > 6.79 MPa not ok!

Revise the Dimension:bd2 = 12.53x106

d = 368.66 mm ≈ 375 mmb = 92.17 mm ≈ 200 mm

try: 200mm x 375 mm

f = 9.7 [1 – 4500 ] 100 (200)

f = 7.52 MPa

fb = 6M bd2

= 6 (20.96) (1000)2

200 (375)2

fb = 4.97 MPa < 7.52 MPa ok!

1.5 1.5 1.5

13.5 13.5

14.1313.71

0.21

-0.21

-13.71 -14.13

20.8820.88

20.96

BEAM IN UNSYMMETRICAL BENDING

METHOD 1: MOMENT RESOLUTION METHOD

Where:Wn = Wcos∅WT = Wsin∅

a. Due to normal load

fbN = 6MN

bd2

b. Due to tangential load

fbT = 6MT

bd2

fbWT = fbW + fbf ≤ fbALL

(Bi-axial Bending)

METHOD 2: JACOBY’S METHOD

tanß= d2/b2 tan£ = Ix/Iy tan£

where: y = ½ (dcosß + bsinß) I = bd/2[(dcos ß) 2 + (bsinß)2](dcosß + bsin£)

= b 2IX 2IY

WN

WT

W

WN

d

b

WT

bd

WT

WN

W

Y

PROBLEM:Check the adequacy of 50 mm x 70 mm, purlin spaced at 0.6 m O.C. if it

has a single span of 1.5 m and a 63% stress grade tanguile is used.

Sol’n:WDL = Wt roof + Wt purlins WDL = 0.77 (0.6) cos20° = 0.10 (0.6) + 1 (0.6) = 0.434 KN/mWDL = 0.66 KN/m WTOTAL = 1.09 KN/m

fbALL ≥ fbACT

fbALL = 10.9 MPa

fbACT = fbN + fbT

PROBLEM:In the figure shown, the support @ A is 12 mm below the level of B. If the

beam is 75 mm x 150 mm, E = 13.8 GPa, determine the flexural stress of the beam.PROBLEM:

A 100 mm x 200 mm beam 6 m long is supported @ the ends and at midspan. It carries uniform load of 7.5 KN/m exceeding its own weight. Determine the max. flexural stress of the beam if the allowable deflection is limited to 10 mm; E = 13.8 GPa. g = 5.6 KN/m3

SPACING OF RIVETS/BOLTS IN BUILT-UP BEAMS:

PROBLEM:

0.6

0.6

1.5 m

A distributed load w (KN/m) is applied over the entire length of the simply supported beam 4 m long. The beam section is a box beam built-up as shown, and secured by screw spaced 50 mm apart. Determine the maximum value of w if fb = 10 MPa; fv = 0.80 MPa and the screws have a shearing strength of 800 N each.

PROBLEM:Three planks 75 mm x 200 mm are bolted together to form a built-up

beam with 100 mm bolt in a single row-spaced 125 mm apart. If the bolts can∅ develop 90 MPa shear, what is the safe uniform load a cantilever beam with 3 m span could carry neglecting beam weight.