Timber Design

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

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

∑F

H

= 0

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

shear, VR

∑F

V

= 0; ∑F

E

= ∑V

R

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

ME VE COMPRESSION TENSION VR MR

 The resisting moment must be equal or > the external moment. OTHER FORMULA FOR FLEXURAL STRESS:

Rectangular:

S= bh

2

_{; f}

b

= 6M

6 bh

2

S= πr

3

_{; f}

b

= 4M

4 πr

3

Triangle: S= bh

2

_{; f}

b

= 6M

24 bh

2

Tube: S= π (R

4

_-

_r

4

₎

4R

FLOOR SYSTEM

DERIVATION OF FLEXURAL FORMULA:

FLOORING TRIBUTARY AREA S

JOIST

GIRDER (Main Beam)

COLUMN / POST W ELASTIC CURVE C N.A. X dx d

By Ratio and Proportion Stress Diagram

f = f

b

c x

f = c f

b

x

f = x f

c

therefore:

f

b

= f

Summing of Moment from N.A.

M = f

b

dAx

f

b

= f

= f ∫ dAx

f

b

= M

c

c

I

I = ∫ dAx

2

FOR RECTANGULAR SECTION:

f = M

c

I

f

b

= 6M

bd

2

where:

I = bd

3

_{; c = d}

12 2

f

b

= 12Md/2

bd

2

ILLUSTRATIVE EXAMPLE:

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

∑M

A

=0

R

B

= 15 (2) + 6

(3) (1.5)

3

R

B

= 19 kN

∑M

B

=0

R

A

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

3

R

A

= 19 kN

Mmax = 16 KN-m

f

b

= 6M

bd

2

= 6 (16) (1000)

2

150 (250)

2

f

b

= 10.24 MPa

150mm 25 0m m 2 m 1 m 14 2 -13 -19 16 15 Kn 6 Kn / m

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?

W

P

∑M

A

=0

R

B

= 6P + w(8) (4)

8

R

B

= 4w + 0.75P

∑M

B

=0

R

A

= 2P + 8w (4)

8

R

A

= 0.25P + 4w

V & P = 0

Then:

Substitute P = 8w

0 = (4w + 0.25P) – 6w

4.5 = 4w + 2.05

(8w) 0 = -2w +0.25P

2w = 0.25P

w = 0.75 KN ≈ 750 N

P = 8w

Mmax = 3 (4w + 0.25P)

f

b

= 6M

Solve for P:

bd

2

_{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

f

b

= 6M

f

b

= 6M

bd

2

_bd

2

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

f

v

= VQ

W

6m 50mm 20 0 m m

Ib

where:

fv = shear stress; MPa

V = shear force or Vmax; KN or N Q = 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

f

v

= VQ

f

v

=

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

f

v

= 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 ≤ 20

h/4 h/2 N.A. h b

3.) Investigate/ analyze the beam selected for bending and horizontal shear if it falls, revise the section. Investigate the beam for deflection.

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 TRIBUTARY AREA FL O O R N G 0.4 m 2.5 m

W

L / 2 L / 2

2 = 1984.79 (2.50)/2 Vmax = 2480.99 N Mmax = wL 2 8 = 1984.79(2.5) 2 8 Mmax = 1550.62 N-m FLEXURE: fb = MC = 6M b = 1 = b= h I bd2 _{h 4 4} 17.1 = 6(1550.62)(1000) bh2 _therefore: bh2 _{= 544,077.19 mm}3 _bh2 _{= h · h}2 4 therefore: h 3 _{= 544,077.19 mm}3 4 b= 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 flexure DL = 64.79 + (0.65)(9810)(0.050)(0.15) = 112.61 N/m fb = 6M LL = 1920 N/m bd2 therefore: = 6(1587.98)(1000) w = DL + LL 50(150) 2 = 112.61 +1920 fb = 8.47 MPa < 17.1 MPa w = 2032.61 N/m SAFE Mmax = wL 2 8 from shear = 2032.61(2.5) 2 8 fv = 3 · V Mmax = 1587.98 N-m 2 A Vmax = wL 3 · 2540.76 2 2 50(150) = 2032.61(2.5) fv = 0.51 MPa < 1.89 MPa 2 SAFE

Vmax = 2540.76 N Adopt: 50 mm x 150 mm floor joist Mmax

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.

P = h

= 9.81 (3) P = 29.43 KN-m

Solving the reactions due to fluid pressure

∑MR1 = 0 R2 = 29.43s(3)[2/3(3) + 2] 2(5) R2 = 35.32s ∑MR2 = 0 R1 = 29.43s(3)[(1/3)(3)] (2)5 R1 =8.83s f P 2 m 5 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 = 0 VA = 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 bd fv = 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 where: d1 d b ZQ V

L = length of the notch

ZQ = allowable stress in compression to the given

DEFLECTION OF BEAMS

Allowable Limits:

L/360 = for beams carrying plastered ceiling L/700 = for beams carrying a line of shafting L/200 = for railroad stringer

L/480 = for beams supporting concrete forms of R.C. slabs or beams L/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 beam w = total load; KN/m

L = span in meter

E = Modulus of Elasticity I = Moment of Inertia

METHODS IN CALCULATING DEFLECTIONS 1. Unit load method

2. Method of superposition 3. Conjugate method 4. Double integration 5. 3 – Moment Equation

Example:

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

For 80% guijo

fb

= 21.8 Mpa

f

v = 2.4 Mpa

Za = 4.26 Mpa

W

Reactions:

∑

M

a =0

R

b = 2(4.5)(.75)+10(3)+1.67 6

R

b = 6.4 KN ∑

M

a =0

R

a = 10(3)+2(4.5)(5.25)-1.67 6

R

a = 12.6 KN

M

max = 17.55 KN.M

V

max = 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 Z

Q

= 4.26

Mpa A

Z

Q

b

2.4 = 3/2·9.6(1000)

L=

2(9.6)(1000)

100(300) 4.26(100) = .48

L

= 45.07 mm

fv

all

> fv

act 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

d = 134.16 mm

1.5 3.0 1.5 1.5 d1 = 134.16

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:

a.) Solve for N.A. (V. Theorem)

45.07 200 mm 20 mm 40 mm 100 mm CONECTION

A

1

=

200 (40) = 8,000

A

2 = 100 (20) = 2,000

A

T =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 MPa PROBLEM:

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/4 Mmax = 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) + 115 (125) (8.22)} 3 2

]

12 36 2 INA = 27.15x106 mm4 _{fb = MC} I 10.34 = 1.5P (49.89) (1000)² INA P = 3.75 KN

BEAMS LATERALLY UNSUPPORTED

Allowable Extreme Fiber Stress:

12.61 62.5 8.22 41.67 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 > 20

B

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 _{= 12.53x10}6 but: b = d/4 ; d/2 try: 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 KN Mmax =20.25 KN-m From Shearing fv = 3/2 (V/A) = 3 (13.5) (1000) 2 (150) (300)

fv = 0.45 MPa < 0.83 MPa ok! CHECK: L/b > 20 L/b = 4500 = 30 > 20 ok! 1.5 1.5 1.5 13.5 13.5 20.25

150

Considering the weight of the beam:

Beam wt. = 6.3 (0.15) (0.30) = 0.28 KN/m Vmax = 14.13 KN Mmax = 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 _{= 12.53x10}6 d = 368.66 mm ≈ 375 mm b = 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 14.13 13.71 0.21 -0.21 -13.71 -14.13 20.88 20.88 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£ = I} x/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 b d 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/m WDL = 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:

0.6

0.6

PROBLEM:

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.