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.
W
b.) UNSYMMETRICAL BENDING – the application of the load is not at any principal axes but causes both the major and minor axis. W
CONDITIONS OF EQUILIBRIUM
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.
ME VE
COMPRESSION
MR
VR
TENSION
FOUR MAJOR CONSIDERATION OF BENDING DESIGN 1.) 2.) 3.) 4.)
The beam must be safe from flexural bending. The beam must be safe from allowable shear stress. The beam must be safe from deflection. 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. OTHER FORMULA FOR FLEXURAL STRESS:
Rectangular:
Triangle: Tube:
FLOOR SYSTEM
S= bh2 ; fb = 6M 6 bh2 S= πr3 ; fb = 4M 4 πr3 S= bh2 ; fb = 6M 24 bh2 S= π (R4 - r4) 4R
S TRIBUTARY AREA FLOORING
JOIST GIRDER (Main Beam)
COLUMN / POST
DERIVATION OF FLEXURAL FORMULA:
W
ELASTIC CURVE
N.A. d dx
X C
By Ratio and Proportion Stress Diagram
f
=
c
fb x
f = c fb x f=x f c therefore: fb = f Summing of Moment from N.A.
M = fb dAx = f ∫ dAx c
fb = f fb = Mc I
I = ∫ dAx2 FOR RECTANGULAR SECTION:
f = Mc I
fb = 12Md/2 bd2 fb = 6M bd2
where: I = bd3 ; c = d 12 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. 150mm
15 Kn 250mm
6 Kn / m
2m
1m
∑MA =0
14
RB = 15 (2) + 6 (3) (1.5) 3 RB = 19 kN
2 -13 -19
∑MB =0
16
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? 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 Substitute P = 8w
Then:
0 = (4w + 0.25P) – 6w (8w) 0 = -2w +0.25P 2w = 0.25P P = 8w
4.5 = 4w + 2.05 w = 0.75 KN ≈ 750 N
Mmax = 3 (4w + 0.25P) fb = 6M bd2 9MPa = 6(3)(4w + 0.25P)(1000)2 (100)(300)2 4.5 = 4w + 0.25P
Solve for P:
P = 8(750) 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?
W
6m 50mm
200 mm
Sol’n:
a.)
b.) s = 0.40m ; w = ?
Mmax = wL2
Mmax = wL2 8 2
= 5s(4) 8 Mmax = 10s KN-m
8 = 0.40(P)(4) 2 8 Mmax = 0.80P
fb = 6M
fb = 6M
bd2
bd2
8 = 6(10s)(1000) 2 (50)(200)2 s = 0.27m
HORIZONTAL SHEARING STRESS General Equation
fv = VQ
8 = 6(0.8P)(1000) 2 (50)(200)2 P = 3.33 KN-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: h/4 h/2 N.A. h
I = bh3 12 Q = h · bh 4 2 2 Q = bh 8
b Substitute Q in the gen. eq’n
fv = VQ Ib V {bh2 } = 8 3 bh · b 12
FOR ANY SECTION
fv = k(V/A) k = 3/2 for rectangular k = 4/3 for circular k = 2 for circular thinning k = bet. 3/2 ≥ 4/3 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 ≤ 20
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. TRIBUTARY AREA FLOORNG
2.5 m
0.4 m
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
W L/2
L/2
Vmax = wL
2 = 1984.79 (2.50)/2 Vmax = 2480.99 N
Mmax
Mmax = wL2 8 = 1984.79(2.5) 2 8 Mmax = 1550.62 N-m
FLEXURE: fb = MC = 6M I bd2 17.1 = 6(1550.62)(1000) bh2 2 bh = 544,077.19 mm3 therefore: b= 544,007.19 (129.59) 2
b = 1 = b= h h 4 4 therefore: bh2 = h · h2 4 h3 = 544,077.19 mm3 4 therefore: h = 129.59 mm ≈ 150 mm 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 = wL2 8 from shear 2 = 2032.61(2.5) 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
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
P =f
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 2m
5m
∑MR2 = 0 R1 = 29.43s(3)[(1/3)(3)] (2)5 R1 =8.83s
Solving for x and Mmax: in terms of s:
y
VA 2m
X = 1.34m
R1
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
MA
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
NOTCHED BEAMS b d1
d
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 V ZQ
R = Vmax = ½(ZQ)b therefore: L = 2Vmax ZQb where:
L = length of the notch ZQ = allowable stress in compression
to the given
DEFLECTION OF BEAMS Allowable Limits: L/360 L/700 L/200 L/480 L/360
= for beams carrying plastered ceiling = for beams carrying a line of shafting = for railroad stringer = for beams supporting concrete forms of R.C. slabs or beams = 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. 2. 3. 4. 5.
Unit load method Method of superposition Conjugate method Double integration 3 – Moment Equation
Example: Design the beam shown using 80% stress grade guijo. Data: For 80% guijo
W
fb = 21.8 Mpa fv = 2.4 Mpa Za = 4.26 Mpa
1.5
3.0
1.5
1.5
Reactions: ∑Ma =0 Rb = 2(4.5)(.75)+10(3)+1.67 6 Rb = 6.4 KN ∑Ma =0 Ra = 10(3)+2(4.5)(5.25)-1.67 6 Ra = 12.6 KN
Mmax = 17.55 KN.M Vmax = 9.60KN
2.) TRIAL SECTION
LOAD:
fb = 21.8 Mpa fv = 2.4 Mpa
b = h/4 h/4· h² = 4.83 x106 h = 268.33 mm say 300mm b = 67.08 mm say 100mm
for flexure:
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
= .48
L = 2V ZQb
ZQ = 4.26 Mpa
A 2.4 = 3/2·9.6(1000) L= 2(9.6)(1000) 100(300) 4.26(100) 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
d = 134.16 mm
d1 = 134.16
45.07
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: 200 mm
40 mm CONECTION
100 mm
20 mm
SOLUTION: a.) Solve for N.A. (V. Theorem)
A1 = 200 (40) = 8,000 A2 = 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)
12 INA = 10.57x106 mm4
3
+ 200(56)²
]
12
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.
75 mm 62.5
12.61
N.A
8.22
41.67
REF (ў)
115 mm
APPLIED LOAD P
3m
Solution: Mmax = PL/4 = 6P/4 Mmax = 1.5P
3m
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)
12 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:
36
3
+ 115 (125) (8.22) 2 2
]
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 = 12.53x106 but: b = d/4 ; d/2 try: b = d/2 d/2 (d) = 12.53x106
13.5 1.5
13.5 1.5
1.5
20.25
therefore: d = 292.64 mm ≈300 mm b = 146.31 mm ≈ 150 mm try: 150 mm x 300 mm 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!
R1 = 13.5 KN R2 = 13.5 KN Vmax = 13.5 KN Mmax =20.25 KN-m
150
Considering the weight of the beam: 13.5 13.5 1.5
14.13
1.5
Beam wt. = 6.3 (0.15) (0.30) = 0.28 KN/m Vmax = 14.13 KN Mmax = 20.96 KN-m
1.5
13.71 0.21 -0.21 -13.71
-14.13
20.96
20.88
20.88
From Shearing:
From Bending:
[
fv = 3 (14.13) (1000) 150 (300) fv = 0.47 < 0.83 ok! 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 mm b = 92.17 mm ≈ 200 mm try: 200mm x 375 mm
[
]
f = 9.7 1 – 4500 100(150) f = 6.79 MPa
]
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!
BEAM IN UNSYMMETRICAL BENDING METHOD 1: MOMENT RESOLUTION METHOD W WN Where: Wn = Wcos∅ WT WT = Wsin∅
a. Due to normal load fbN = 6MN bd2 WN d b
b. Due to tangential load WT d
fbT = 6MT bd2 b
fbWT = fbW + fbf ≤ fbALL (Bi-axial Bending) METHOD 2: JACOBY’S METHOD tanß= d2/b2 tan£ = Ix/Iy tan£ WN
W WT
Y
=b
where: y = ½ (dcosß + bsinß) I = bd/2[(dcos ß) 2 + (bsinß)2] (dcosß + bsin£) 2IX 2IY
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.
0.6 0.6
Sol’n: WDL = Wt roof + Wt purlins = 0.10 (0.6) + 1 (0.6) WDL = 0.66 KN/m
WDL = 0.77 (0.6) cos20° = 0.434 KN/m WTOTAL = 1.09 KN/m fbALL ≥ fbACT fbALL = 10.9 MPa
1.5 m 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: 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.
