1
VBCOA – Region 5
May 15, 2014
Structural Analysis by Hand
2
Presenter
Brian Foley, P.E.Fairfax County
Deputy Building Official
703‐324‐1842
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Logistics
Exits Restrooms
Cell Phones No Smoking
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4
Questions?
SURE– Go ahead and ask your question!
This class is interactive!
There are no stupid questions!
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Attendees
■ Inspector
■ Plan reviewer
■ Contractor
■ Engineer
■ Architect
■ Designer
■ Suppliers
■ Attorney
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Objectives
Determine loading requirements for joists and beams
Apply loads using free body diagrams
Calculate moment and deflection
Analyze compliance for flexure, shear and deflection
Analyze simple steel beam
Analyze spread footings
3
7
Assumptions
No structural theory
Knowledge of arithmetic and basic algebra
Simple loading
Uniform loads
Point loads at midspan
Simple spans
Wood, LVL, steel
Residential construction
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Math 101
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Operation Sequence
Please Excuse My Dear Aunt Sally
Parenthesis
Exponent
Multiplication
Division
Addition
Subtraction
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10
P.E.M.D.A.S.
52 + (3 – 1) / 2 – 4 x 3
Step 1: Parenthesis 52 + 2 / 2 – 4 x 3
Step 2: Exponents 25 + 2 / 2 – 4 x 3
Step 3: Multiply 25 + 2 / 2 – 12
Step 4: Divide 25 + 1 – 12
Step 5: Add 26 – 12
Step 6: Subtract 14
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Mathematical Formulas
Volume of a box
Length x Width x Height
Variables
L = Length
W = Width
H = Height
V= Volume
V = L x W x H
V = LWH
Height
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Mathematical Formulas
V = LWH
Values
L = 5’
W = 2’
H = 6’
V = 5 x 2 x 6
V = (5)x(2)x(6)
V = (5)(2)(6)
V = 60 feet3
Hei
ght
5
13
You Try It, Find V
Values
2’ w = 3’
h = 6”
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You Try It, Find V
Values: 2’,w 3’,h 6” Convert h value from inches to
feet: 6” = 0.5’
Insert values into formula:
2 3 0.5 1 ft3
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A Word About Units
Always include units with every calculation
Ensure all calculations are completed in the same units (inches, feet)
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16
Vertical Load Path
Vertical load path transfers gravity load (snow):
■ to roof sheathing
■ to rafters
■ to top plate
■ to studs
■ to bottom plate
■ to foundation & footings
■ to ground
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Design Methodologies
Load & Resistance Factor Design (LRFD)
Applied loads adjusted up
Resistance capacity of structural member adjusted down
Compare values:capacity > loads
Allowable Stress Design (ASD)
Actual stress calculated using applied loads
Structural member’s allowable stresses calculated
Compare values:allowable > actual
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Joist/Beam Analysis
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19
Sample First Floor Framing Plan
2x8 joists
16"
16"
(2)
1¾x9
½ M
icro
lam
Hem-Fir #2
4'-0" 10'-0"
12'-0
"
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Step 1: Determine uniform dead load
Units for uniform dead load
pounds per square foot
lbs/ft2
psf
Dead Load: weight of structure
Assume 10 PSF for floors
Assume 15 PSF for roofs
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Step 2: Determine uniform live load (psf)
Live Load: weight produced byuse and occupancy
People
Furniture
Vehicles
Units: pound per squarefoot (psf)
IBC Table 1607.1
IRC Table R301.5
8
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Step 2: Determine uniform live load (psf)
For floors: IRC Table R301.5
40
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Step 2: Determine uniform live load (psf)
For roofs, greatest of live load or snow load
Snow load: IRC Section R301.2.3
Northern Virginia counties 20 – 30 psf
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Step 2: Determine uniform live load (psf)
Roof live load: IRC Table R301.6
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Step 3: Calculate tributary width (feet)
Load influence distance from each side of a framing member
Joists: half the distance to next adjacent joists on each side
Beams: half of the joists’ span that bear on each side of the beam
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Step 3: Calculate tributary width (feet)
16"
16"
'33.112
16
"162
16
2
16
feettoConvert
TW
EXAMPLE: JOIST
8" 8"
Tributary width
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Step 3: Calculate tributary width (feet)
EXAMPLE: BEAM
4'-0" 10'-0"
12'-0
"
Tributary width
10
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Step 3: Calculate tributary width (feet)
EXAMPLE: BEAM
2' 5'
'72
10
2
4TW
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Step 4: Calculate linear load (plf)
w = uniform load x TW Units: pounds per linear foot = lbs/ft = plf EXAMPLE:JOIST: wLL = (40)(1.33) = 53.33 plf
wDL = (10)(1.33) = 13.33 plfw = 66.67 plf
BEAM: wLL = (40)(7) = 280.0 plfwDL = (10)(7) = 70.0 plf
w = 350.0 plf
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Free Body Diagram
A two‐dimensional graphic symbolization of a structural member which models bearing locations and loading elements
Linear load, w:
Point load, P:
Bearing locations (reaction), R:
100 plf
500 lbs
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31
Free Body Diagram
Joist
66.67 plf
10’-0”
Beam
350 plf
12’-0”
w
lR R
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Example: Free Body Diagram
On plans provided, first floor joist adjacent fireplace hearth extension:
15’-4”
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You Try It
Draw the free body diagram of the sunroom beam
Show load and its value
Total uniform load = 50 psf
Tributary width = 5’
Total linear load = (50)(5) = 250 plf
Show span length
250 plf
10’-4”
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Step 5: Bending Analysis
Flexure, bending, moment, torque
Highest at midspan for uniform load
Pulling stress or tension on bottom face of member
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Step 5A: Determine F’b (psi)
Allowable bending stress, F’b The maximum bending stress permissible for a
specified structural member
Units for stress:
pounds per square inch
lbs/in2
psi
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Step 5A: Determine F’b (psi)
“Raw” value based on wood species: Fb Adjusted allowable bending stress,
F’b = Fb CM CF Cr CD, where:
CM = Service condition (wet or dry)
CD = Load duration (normal or snow)
Cr = Repetitive use (joists, 3+ply beams)
CF = Member size (2x?)
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Step 5A: Determine F’b (psi)
Use tables in chart based on:
Species
Service condition (wet or dry)
Load duration (normal or snow)
Single (2‐ply beam) or repetitive (joists)
Member size
EXAMPLE:
Joists: Repetitive, dry, 2x8, normal load duration, Hem‐Fir#2: F’b = 1,173 psi
Beam: Single, dry, (2)1¾x9½, normal load duration, Microlam: F’b = 2,684 psi
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Step 5B: Determine b (in), d (in), S (in3)
EXAMPLE:
2
6
1bdS
b
d
32 in6.52)5.9)(75.1(6
1)2(:
SBEAM
Section Modulus:
32 in1.13)25.7)(5.1(6
1: SJOIST
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EXAMPLE:
JOIST:
S=13.1 in3
BEAM:
S=(2)(26.3)=52.6 in3
Step 5B: Determine S (in3)
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Step 5C: Determine Span Length, l (ft) Calculate Moment, M (lbs‐ft)
l
w
8
2wlM where: l = span length, ft
w = total linear load, plfM = moment, lbs-ft
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Step 5C: Determine Span Length, l (ft) Calculate Moment, M (lbs‐ft)
where: l = span length, ftw = total linear load, plfM = moment, lbs-ft
JOIST: w = 66.67 plfl = 10'
BEAM: w = 350 plfl = 12'
ftlbs4.8338
)10)(67.66( 2
M
EXAMPLE:
8
2wlM
ftlbs300,68
)12)(350( 2
M
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Step 5D: Calculate fb (psi)
Actual bending stress, fb The bending stress a specified structural member
is experiencing under maximum applied load
where: S = section modulus, in3
M = moment, lbs-ft
fb = actual bending stress, psiS
Mfb
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15
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Step 5D: Calculate fb (psi)
where: S = section modulus, in3
M = moment, lbs-ft
fb = actual bending stress, psiS
Mfb
12
JOIST: M = 833.4 lbs-ftS = 13.1 in3
BEAM: M = 6,300 lbs-ftS = 52.6 in3
psi4.7631.13
)4.833)(12(bf
EXAMPLE:
psi3.437,16.52
)300,6)(12(bf
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Step 5E: Compare F’b with fb
If fb ≤ F’b, then member is good for bending
EXAMPLE:
JOIST: fb = 763.4 psi < F’b = 1,173 psi OK!
BEAM: fb = 1,437.3 psi < F’b = 2,684 psi OK!
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Step 6: Shear Analysis
Shear is similar to a cutting stress
Highest at ends = reaction
Wood shear analysis uses shear value at a distance from the end equal to member’s depth
Member experiencesa slicing action invertical plane
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Step 6A: Determine F’v (psi)
Allowable shear stress, F’v The maximum shear stress permissible for a
specified structural member
Units for stress:
pounds per square inch
lbs/in2
psi
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Step 6A: Determine F’v (psi)
Allowable shear stress
“Raw” stress based on wood species: Fv Adjusted allowable shear stress: F’v ,
F’v =Fv CM CD, where:
CM = Service condition (wet or dry)
CD = Load duration (normal or snow)
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Step 6A: Determine F’v (psi)
Use tables based on:
Species
Service condition (wet or dry)
Load duration (normal or snow)
EXAMPLE:
Joist: Dry, Hem‐Fir#2: F’v = 150 psi
Beam: Dry, Microlam: F’v = 285 psi
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Step 6B: Calculate Area, A (in2)
EXAMPLE:
b
d
2in9.10)25.7)(5.1(:JOIST A
2in2.33)5.9)(75.1)(2(:BEAM A
A=bd
Area:
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Step 6B: Calculate Area, A (in2)
EXAMPLE:
JOIST:
A=10.9 in2
BEAM:
S=(2)(16.6)=33.2 in2
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Step 6C: Calculate Shear, V (lbs)
where: l = span length, ftw = total linear load, plfd = member depth, in
122
dlwV
JOIST: w = 66.67 plfl = 10'd = 7.25"
BEAM: w = 350 plfl = 12'd = 9.5"
lbs0.29312
25.7
2
1067.66
V
EXAMPLE:
lbs9.822,112
5.9
2
12350
V
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Step 6D: Calculate fv (psi)
Actual shear stress, fv The shear stress a specified structural member is
experiencing under maximum applied load
where: fv = actual shear stress, psiA = area, in2
V = shear, lbsA
Vfv 2
3
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where: A = area, in2
V = shear, lbs
fv = actual shear stress, psiA
Vfv 2
3
JOIST: V = 293.0 lbsA = 10.9 in2
BEAM: V = 1,822.9 lbsA = 33.2 in2
psi4.82)2.33)(2(
)9.822,1)(3(vf
EXAMPLE:
psi4.40)9.10)(2(
)0.293)(3(vf
Step 6D: Calculate fv (psi)
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Step 6E: Compare F’v with fv
If fv ≤ F’v, then member is good for shear
EXAMPLE:
JOIST: fv = 40.4 psi < F’v = 150 psi OK!
BEAM: fv = 82.4 psi < F’v = 285 psi OK!
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Step 7: Deflection Analysis
"Sag" a member experiences
Analysis is based on live load only
Highest at midspan
Deflection, ∆
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Step 7A: Determine allowable deflection (in)
Use Table R301.7
where l = span length of member, in.(span length must be converted from feet to inches)
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Step 7B: Calculate allowable deflection (in)
EXAMPLE: Floor,
JOIST:
BEAM:
"33.0360
)10)(12(max
"40.0360
)12)(12(max
360
12max
l
20
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Step 7C: Calculate I (in4)
EXAMPLE:
3
12
1bdI
b
d
43 in250)5.9)(75.1(12
1)2(:
IBEAM
Moment of Inertia:
43 in6.47)25.7)(5.1(12
1: IJOIST
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Step 7C: Calculate I (in4)
EXAMPLE:JOIST: I= 47.6 in4
BEAM:I=(2)(125)
=250 in4
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Step 7D: Determine E (psi)
Modulus of elasticity, E
The mathematical description of a structural member’s elastic characteristics
Units for modulus of elasticity:
pounds per square inch
lbs/in2
psi
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Step 7D: Determine E (psi)
Modulus of elasticity
“Raw” value based on wood species: E
Adjusted modulus of elasticity: E’
E’ =E CM, where:
CM = Service condition (wet or dry)
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Step 7D: Determine E (psi)
Use tables based on:
Species
Service condition (wet or dry)
EXAMPLE:
Joist: Hem‐Fir#2, dry: E’ = 1.3 x 106 psi
Beam: Microlam, dry: E’ = 1.9 x 106 psi
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Step 7E: Calculate actual deflection (in)
where:
∆act = actual deflection
l = span length, ftwLL = linear live load, plfE = modulus of elasticity, psiI = moment of inertia, in4
EI
lwLLact
45.22
22
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Step 7E: Calculate actual deflection, Δact (in)
JOIST: wLL = 53.3 plfl = 10'E = 1.3x106 psiI = 47.6 in4
BEAM: wLL = 280 plfl = 12'E = 1.9x106 psiI = 250 in4
"194.0)6.47)(10x3.1(
)10)(3.53)(5.22(6
4
act
EXAMPLE:
"275.0)250)(10x9.1(
)12)(280)(5.22(6
4
act
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Step 7F: Compare ∆max to ∆act
If Δact ≤ Δmax, then member is good for deflection.
EXAMPLE:
Joist: Δact = 0.194" < Δmax = 0.333" OK!
Beam: Δact = 0.275" < Δmax = 0.400" OK!
JOIST PASSES!
BEAM PASSES!
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When good beams go bad…
When one component fails, the structural member fails
RULE OF THUMB: the bending stress is the gauge
In a uniformly loaded member, shear will pass if flexure passes
Deflection is critical when flexure barely passes
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Shortcuts…“Dirty numbers”
Span length: no need to be exact
More than one live load on a member?
Use value of live load member mostly sees, or
use highest value
Round values up
Drop decimals
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Examples – Go to the Plans
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Example: Garage Rafters
Go to Sheet S4 on the provided plans
Do rafters comply?
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70
Example: Garage Rafters
Step 1: Determine uniform dead load, DL = 15 psf
Step 2: Determine uniform live load, LL = 20 psf
Step 3: Determine tributary area, TW = 2’
Step 4: Calculate linear load
wDL = (2)(15) = 30 plf
wLL = (2)(20) = 40 plf
w = 70 plf
Step 5A: Determine F’b Repetitive, dry, snow load duration, SPF So. #2
F’b = 1,230 psi
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Example: Garage Rafters
Step 5B: Determine S For 2x8, from chart, S = 13.1 in3
Step 5C: Determine l, calculate M l = 11’
w = 70 plf
Step 5D: Calculate
ftlbs059,18
)11)(70( 2
M
psi9671.13
)059,1)(12(bf
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Example: Garage Rafters
Step 5E: Compare:
fb = 967 psi < F’b = 1,230 psi
Flexure analysis OK!
Step 6A: Determine F’v
Dry, SPF So. #2
F’v = 135 psi
Step 6B: Determine A (in2)
For 2x8, from chart, A = 10.9 in2
Step 6C: Calculate lbs34312
25.7
2
1170
V
25
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Example: Garage Rafters
Step 6D: Calculate
Step 6E: Compare:
fv = 47.2 psi < F’v = 135 psi
Shear analysis OK!
Step 7A: Determine ∆max
Roof, slope > 3:12, no ceiling finish
l/180
Step 7B: Calculate
psi2.47)9.10)(2(
)343)(3(vf
"733.0180
)11)(12(max
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Example: Garage Rafters
Step 7C: Determine I For 2x8, from chart, I = 47.6 in4
Step 7D: Determine E’
Dry, SPF So. #2
E’ = 1.1 x 106 psi
Step 7E: Calculate
Step 7F: Compare:
∆act = 0.25” < ∆max = 0.733”
Deflection analysis OK!
"25.0)6.47)(10x1.1(
)11)(40)(5.22(6
4
act
RAFTER PASSES!
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Example: Second Floor Header
Go to Sheet S3 on the provided plans
Does header comply?
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Example: Second Floor Header
Step 1: Determine uniform dead load, DL = 10 psf
Step 2: Determine uniform live load, LL = 30 psf
Step 3: Determine tributary area, TW = 14’
Step 4: Calculate linear load
wDL = (14)(10) = 140 plf
wLL = (14)(30) = 420 plf
w = 560 plf
Step 5A: Determine F’b Repetitive, dry, normal duration, SPF So. #2
F’b = 891 psi
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Example: Second Floor Header
Step 5B: Determine S For 2x12, from chart, S = (3)(31.6) = 94.8 in3
Step 5C: Determine l, calculate M l = 10’
w = 560 plf
Step 5D: Calculate
ftlbs000,78
)10)(560( 2
M
psi8868.94
)000,7)(12(bf
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Example: Second Floor Header
Step 5E: Compare:
fb = 886 psi < F’b = 891 psi
Flexure analysis OK!
Step 6A: Determine F’v
Dry, SPF So. #2
F’v = 135 psi
Step 6B: Determine A For 2x12, from chart, A = (3)(16.9) = 50.7 in2
Step 6C: Calculate lbs275,212
25.11
2
10560
V
27
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Example: Second Floor Header
Step 6D: Calculate
Step 6E: Compare:
fv = 67 psi < F’v = 135 psi
Shear analysis OK!
Step 7A: Determine ∆max
Floor
l/360
Step 7B: Calculate
psi67)7.50)(2(
)275,2)(3(vf
"333.0360
)10)(12(max
80
Example: Second Floor Header
Step 7C: Determine I For 2x12, from chart, I = (3)(178) = 534 in4
Step 7D: Determine E’
Dry, SPF So. #2
E’ = 1.1x106 psi
Step 7E: Calculate
Step 7F: Compare:
∆act = 0.16” < ∆max = 0.333”
Deflection analysis OK!
"16.0)534)(10x1.1(
)10)(420)(5.22(6
4
act
HEADER PASSES!
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You Try It
Using the house plans provided, do the sunroom joists and beam comply?
28
82
JOISTS1. Uniform dead load:2. Uniform live load:
3. Tributary width:
4. Linear load:
5A. Allowable bending stress:5B. Section modulus:5C. Span length, moment:
DL = 10 psfLL = 40 psf
F’b = 1,070 psiS = 13.1 in3
ftlbs250,18
)10)(100( 2
M
wDL= 2 x 10 = 20 plfwLL = 2 x 40 = 80 plf
w = 100 plf
l = 10’
TW = 2’
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JOISTS
5D.Actual bending stress:
5E.Compare:
psifb 6.141,1
1.13
250,112
fb = 1,141.6 psi > F’b = 1070.0 psi NG
JOIST FAILS!
84
BEAM1. Uniform dead load:2. Uniform live load3. Tributary width:
4. Linear load:
5A. Allowable bending stress:5B. Section modulus:5C. Span length, moment:
DL = 10 psfLL = 40 psf
wDL= 5 x 10 = 50 plfwLL = 5 x 40 = 200 plf
w = 250 plfF’b = 1,070 psi
S = (3)(13.1) = 39.3 in3
ftlbs335,38
)33.10)(250( 2
M
l = 10.33’
TW = 5’
29
85
BEAM
5D.Actual bending stress:
5E.Compare:
6A.Allowable shear stress:6B.Area:6C.Shear:
psifb 016,1
3.39
335,312
fb = 1,016 psi < F’b = 1,070 psi OK
F’v = 135 psiA = (3)(10.9) = 32.7 in2
lbs 140,112
25.7
2
33.10250
V
86
BEAM
6D.Actual shear stress:
6E.Compare:
7A.Allowable deflection:
7B.Moment of inertia:7C.Modulus of elasticity:
psifv 5.52
7.322
140,13
fv = 52.5 psi < F’v = 135 psi OK
I = (3)(47.6) = 142.8 in4
"344.0
360
33.1012max
E’ = 1.1x106 psi
87
BEAM
7D.Actual deflection:
7E.Compare:
"326.0)8.142)(x101.1(
)33.10)(200)(5.22(6
4
max
Δact = 0.326" < Δmax = 0.344” OK
BEAM PASSES!
30
88
Steel Beams
Wide flange section
A992, high strength
A36, normal strength
Design check for uniformly loaded beam
Basement location
Simple spans
89
Step 1: Determine Span Length (ft)
From plans, choose longest span of a beam type, W8x18
For example, assume 17’
90
90
Step 2: Determine Allowable Load (kips)
31
91
Step 3: Determine Uniform Loads (psf)
First floor
Live load = 40 psf
Dead load = 10 psf
Second floor
Live load = 30 psf (if sleeping)
Dead load = 10 psf
92
Step 4: Determine Tributary Width (ft)
'132
75.12
2
25.13TW
93
Step 5: Calculate Actual Linear Load (plf)
First floor:
Second floor:
Total:w = 1,170 plf
klf 650)1040)(13(1 w
plf 520)1030)(13(2 w
32
94
Step 6: Calculate Actual Total Load (kips)
Actual Load = (17)(1,170) = 19,890 lbs = 19.9 kips
17’
1,170 plf
95
Step 7: CompareActual and Allowable Loads
Actual = 19.9 kips > Allowable = 14 kips NG!
Try W8x28, Lmax = 23 kips > 19.9 kips, OK!
BEAM FAILS!
96
Rules of Thumb
The deeper the beam the more it can span
A steel beam can usually span 2 feet for every inch of depth
W8, span≈16’
W10, span≈20’
33
97
You Try It
Go to plans
Check steel beam in basement
98
Solution W8x18, span length = 12’
Allowable load = 20 kips
Total uniform load: w1= 14 x (40 + 10) = 700 plfw2= 14 x (30 + 10) = 560 plf
1,260 plf
Actual = 12 x 1,260 = 15.1 kips < 20 kips OK!1,000
'142
33.15
2
75.12TW
99
99
Steel Beams
34
100
Spread Footings
Soil has maximum capacity to withstand pressure
Footing distributes point load to soil based on maximum capacity
101
Step 1: Determine Presumptive Soil Bearing Pressure, Qmax (psf)
Determine using Table R401.4.1
Maximum assumable often identified by AHJ
For example, Qmax = 1,500 psf
102
How to Calculate Reactions
l
w
2
wlR where: l = span length, ft
w = total linear load, plfR = reaction, lbs
R R
35
103
Step 3: Calculate Point Load
Point load = R1 + R2
For example:
P = 17,550 lbs.1,170 plf
17’ 10’13’
lbs 605,72
)13)(170,1(1 R
lbs 945,92
)17)(170,1(2 R
104
Step 2: Calculate Qact (psf) Calculate actual bearing
pressure:
P = column load, lbs
L = footing length, ft
W = footing width, ft
For example:
LW
PQact
lbs 097,1)4)(4(
550,17actQ
105
Step 4: Compare Qact to Qmax
Qact= 1,097 psf < Qmax = 1,500 psf, OK!
FOOTING PASSES!
36
106
You Try It
Go to the plans provided
Are the spread footings supporting the steel columns adequate?
107
Solution
Soil type = silty sand (SM), Qmax = 2,000 psf
L = W = 2.5'
Qact = 2,416 psf > Qmax = 2,000 psf, NG!
lbs 648,132
)12)(170,1(
2
)33.11)(170,1(P
psf 184,2)5.2)(5.2(
648,13actQ
The End