Structural Analysis by Hand
VBCOA – Region 5
May 15, 2014
Presenter
Brian Foley, P.E.
Fairfax County
Deputy Building Official
[email protected]
703‐324‐1842
2
Logistics
Exits
Cell Phones
Restrooms
No Smoking
3
1
Questions?
SURE– Go ahead and ask your question!
This class is interactive!
There are no stupid
questions!
4
Attendees
■
■
■
■
■
■
■
■
Inspector
Plan reviewer
Contractor
Engineer
Architect
Designer
Suppliers
Attorney
5
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
6
2
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
7
Math 101
8
Operation Sequence
Please Excuse My Dear Aunt Sally

Parenthesis

Exponent

Multiplication

Division

Addition

Subtraction
9
3
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
Step 4: Divide
Step 5: Add
Step 6: Subtract
25 + 2 / 2 – 12
25 + 1 – 12
26 – 12 14
10
Mathematical Formulas


Volume of a box
Length x Width x Height
Variables






L = Length
W = Width
H = Height
V= Volume
Height

V = L x W x H
V = LWH
11
Mathematical Formulas

V = LWH
Values







L = 5’
W = 2’
H = 6’
Height

V = 5 x 2 x 6
V = (5)x(2)x(6)
V = (5)(2)(6)
V = 60 feet3
12
4
You Try It, Find V


Values



2’
w = 3’
h = 6”
13
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
14
A Word About Units


Always include units with every calculation
Ensure all calculations are completed in the same units (inches, feet)
15
5
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
16
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
17
Joist/Beam Analysis
18
6
16" 16"
12'-0"
(2) 1¾x9½ Microlam
Sample First Floor Framing Plan
2x8 joists
Hem-Fir #2
4'-0"
10'-0"
19
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
20
Step 2: Determine uniform live load (psf)

Live Load: weight produced by
use and occupancy






People
Furniture
Vehicles
Units: pound per square
foot (psf)
IBC Table 1607.1
IRC Table R301.5
21
7
Step 2: Determine uniform live load (psf)

For floors: IRC Table R301.5
40
22
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
23
Step 2: Determine uniform live load (psf)

Roof live load: IRC Table R301.6
24
8
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
25
Step 3: Calculate tributary width (feet)
EXAMPLE: JOIST
16" 16"
Tributary width
16 16
  16"
2 2
16
Convert to feet 
 1.33'
12
TW 
8"
8"
26
Step 3: Calculate tributary width (feet)
EXAMPLE: BEAM
12'-0"
Tributary width
4'-0"
10'-0"
27
9
Step 3: Calculate tributary width (feet)
EXAMPLE: BEAM
2'
5'
TW 
4 10
  7'
2 2
28
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 plf
w = 66.67 plf
BEAM:
wLL = (40)(7) = 280.0 plf
wDL = (10)(7) =
70.0 plf
w = 350.0 plf
29
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
30
10
Free Body Diagram
w
R
l
R
66.67 plf
Joist
10’-0”
350 plf
Beam
12’-0”
31
Example: Free Body Diagram

On plans provided, first floor joist adjacent fireplace hearth extension:
15’-4”
32
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”
33
11
Step 5: Bending Analysis


Flexure, bending, moment, torque
Highest at midspan for uniform load
Pulling stress or tension on
bottom face of member
34
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
35
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?)
36
12
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
37
Step 5B: Determine b (in), d (in), S (in3)
Section Modulus:
1
S  bd 2
6
d
b
EXAMPLE:
1
JOIST : S  (1.5)(7.25) 2  13.1 in 3
6
1
BEAM : S  (2) (1.75)(9.5) 2  52.6 in 3
6
38
Step 5B: Determine S (in3)
EXAMPLE:
JOIST: S=13.1 in3
BEAM:
S=(2)(26.3)
=52.6 in3
39
13
Step 5C: Determine Span Length, l (ft) Calculate Moment, M (lbs‐ft)
wl 2
M
8
where: l = span length, ft
w = total linear load, plf
M = moment, lbs-ft
w
l
40
Step 5C: Determine Span Length, l (ft) Calculate Moment, M (lbs‐ft)
M
wl 2
8
where: l = span length, ft
w = total linear load, plf
M = moment, lbs-ft
EXAMPLE:
JOIST:
w = 66.67 plf
l = 10'
M
(66.67)(10) 2
 833.4 lbs  ft
8
BEAM:
w = 350 plf
l = 12'
M
(350)(12) 2
 6,300 lbs  ft
8
41
Step 5D: Calculate fb (psi)


Actual bending stress, fb
The bending stress a specified structural member is experiencing under maximum applied load
fb 
12 M
S
where: S = section modulus, in3
M = moment, lbs-ft
fb = actual bending stress, psi
42
14
Step 5D: Calculate fb (psi)
12 M
fb 
S
where: S = section modulus, in3
M = moment, lbs-ft
fb = actual bending stress, psi
EXAMPLE:
JOIST:
M = 833.4 lbs-ft
S = 13.1 in3
BEAM:
M = 6,300 lbs-ft
S = 52.6 in3
fb 
fb 
(12)(833.4)
 763.4 psi
13.1
(12)(6,300)
 1,437.3 psi
52.6
43
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!
44
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 experiences
a slicing action in
vertical plane
45
15
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
46
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)
47
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
48
16
Step 6B: Calculate Area, A (in2)
Area:
A=bd
d
b
EXAMPLE:
JOIST : A  (1.5)(7.25)  10.9 in 2
BEAM : A  (2)(1.75)(9.5)  33.2 in 2
49
Step 6B: Calculate Area, A (in2)
EXAMPLE:
JOIST: A=10.9 in2
BEAM:
S=(2)(16.6)
=33.2 in2
50
Step 6C: Calculate Shear, V (lbs)
l d 
V  w  
 2 12 
EXAMPLE:
JOIST:
w = 66.67 plf
l = 10'
d = 7.25"
BEAM: w = 350 plf
l = 12'
d = 9.5"
where: l = span length, ft
w = total linear load, plf
d = member depth, in
 10 7.25 
V  66.67 
  293.0 lbs
12 
2
 12 9.5 
V  350 
  1,822.9 lbs
 2 12 
51
17
Step 6D: Calculate fv (psi)


Actual shear stress, fv
The shear stress a specified structural member is experiencing under maximum applied load
fv 
3V
2A
where:
fv = actual shear stress, psi
A = area, in2
V = shear, lbs
52
Step 6D: Calculate fv (psi)
3V
fv 
2A
where: A = area, in2
V = shear, lbs
fv = actual shear stress, psi
EXAMPLE:
JOIST:
V = 293.0 lbs
A = 10.9 in2
BEAM: V = 1,822.9 lbs
A = 33.2 in2
fv 
(3)(293.0)
 40.4 psi
(2)(10.9)
fv 
(3)(1,822.9)
 82.4 psi
(2)(33.2)
53
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!
54
18
Step 7: Deflection Analysis



"Sag" a member experiences
Analysis is based on live load only
Highest at midspan
Deflection, ∆
55
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)
56
Step 7B: Calculate allowable deflection (in)
EXAMPLE: Floor,
 max 
12l
360
JOIST:
 max 
(12)(10)
 0.33"
360
BEAM:  max 
(12)(12)
 0.40"
360
57
19
Step 7C: Calculate I (in4)
Moment of Inertia:
I
1
bd 3
12
d
b
EXAMPLE:
1
(1.5)(7.25) 3  47.6 in 4
12
1
BEAM : I  (2) (1.75)(9.5) 3  250 in 4
 12 
JOIST : I 
58
Step 7C: Calculate I (in4)
EXAMPLE:
JOIST: I= 47.6 in4
BEAM:
I=(2)(125)
=250 in4
59
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
60
20
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)
61
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
62
Step 7E: Calculate actual deflection (in)
 act 
22.5wLL l 4
EI
where:
∆act = actual deflection
l = span length, ft
wLL = linear live load, plf
E = modulus of elasticity, psi
I = moment of inertia, in4
63
21
Step 7E: Calculate actual deflection, Δact (in)
EXAMPLE:
JOIST:
wLL = 53.3 plf
l = 10'
E = 1.3x106 psi
I = 47.6 in4
BEAM: wLL = 280 plf
l = 12'
E = 1.9x106 psi
I = 250 in4
 act 
(22.5)(53.3)(10) 4
 0.194"
(1.3x106 )(47.6)
 act 
(22.5)(280)(12) 4
 0.275"
(1.9 x10 6 )(250)
64
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!
65
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
66
22
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
67
Examples – Go to the Plans
68
Example: Garage Rafters


Go to Sheet S4 on the provided plans
Do rafters comply?
69
23
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
70
Example: Garage Rafters

Step 5B: Determine S


Step 5C: Determine l, calculate M




For 2x8, from chart, S = 13.1 in3
l = 11’
w = 70 plf
(70)(11) 2
M
 1,059 lbs  ft
8
Step 5D: Calculate f b 
(12)(1,059)
 967 psi
13.1
71
Example: Garage Rafters

Step 5E: Compare:



Step 6A: Determine F’v



fb = 967 psi < F’b = 1,230 psi
Flexure analysis OK!
Dry, SPF So. #2
F’v = 135 psi
Step 6B: Determine A (in2)
For 2x8, from chart, A = 10.9 in2
11 7.25 
Step 6C: Calculate V  70 
  343 lbs
12 
2


72
24
Example: Garage Rafters


(3)(343)
 47.2 psi
Step 6D: Calculate f v 
(2)(10.9)
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
(12)(11)
Step 7B: Calculate  max 
 0.733"
180



73
Example: Garage Rafters

Step 7C: Determine I


Step 7D: Determine E’




For 2x8, from chart, I = 47.6 in4
Dry, SPF So. #2
E’ = 1.1 x 106 psi
( 22.5)(40)(11) 4
Step 7E: Calculate  act 
 0.25"
(1.1x10 6 )(47.6)
Step 7F: Compare:


∆act = 0.25” < ∆max = 0.733”
Deflection analysis OK!
RAFTER PASSES!
74
Example: Second Floor Header


Go to Sheet S3 on the provided plans
Does header comply?
75
25
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
76
Example: Second Floor Header

Step 5B: Determine S


Step 5C: Determine l, calculate M




For 2x12, from chart, S = (3)(31.6) = 94.8 in3
l = 10’
w = 560 plf
(560)(10) 2
M
 7,000 lbs  ft
8
Step 5D: Calculate f b 
(12)(7,000)
 886 psi
94.8
77
Example: Second Floor Header

Step 5E: Compare:



Step 6A: Determine F’v



fb = 886 psi < F’b = 891 psi
Flexure analysis OK!
Dry, SPF So. #2
F’v = 135 psi
Step 6B: Determine A
For 2x12, from chart, A = (3)(16.9) = 50.7 in2
10 11.25 
Step 6C: Calculate V  560 
  2,275 lbs
12 
2


78
26
Example: Second Floor Header


(3)(2,275)
 67 psi
Step 6D: Calculate f v 
( 2)(50.7)
Step 6E: Compare:



Step 7A: Determine ∆max



fv = 67 psi < F’v = 135 psi
Shear analysis OK!
Floor l/360
Step 7B: Calculate  max 
(12)(10)
 0.333"
360
79
Example: Second Floor Header

Step 7C: Determine I


Step 7D: Determine E’




For 2x12, from chart, I = (3)(178) = 534 in4
Dry, SPF So. #2
E’ = 1.1x106 psi
(22.5)(420)(10) 4
Step 7E: Calculate  act 
 0.16"
(1.1x10 6 )(534)
Step 7F: Compare:


∆act = 0.16” < ∆max = 0.333”
Deflection analysis OK!
HEADER PASSES!
80
You Try It

Using the house plans provided, do the sunroom joists and beam comply?
81
27
JOISTS
1. Uniform dead load:
2. Uniform live load:
3. Tributary width:
DL = 10 psf
LL = 40 psf
TW = 2’
4. Linear load: wDL= 2 x 10 = 20 plf
wLL = 2 x 40 = 80 plf
w = 100 plf
5A. Allowable bending stress: F’b = 1,070 psi
5B. Section modulus: S = 13.1 in3
5C. Span length, moment: l = 10’
M
(100)(10) 2
 1,250 lbs  ft
8
82
JOISTS
5D.Actual bending stress:
fb 
5E. Compare:
121,250  1,141.6 psi
13.1
fb = 1,141.6 psi > F’b = 1070.0 psi NG
JOIST FAILS!
83
BEAM
1. Uniform dead load: DL = 10 psf
LL = 40 psf
2. Uniform live load
3. Tributary width: TW = 5’
4. Linear load: wDL= 5 x 10 = 50 plf
wLL = 5 x 40 = 200 plf
w = 250 plf
5A. Allowable bending stress: F’b = 1,070 psi
5B. Section modulus: S = (3)(13.1) = 39.3 in3
5C. Span length, moment: l = 10.33’
M
(250)(10.33) 2
 3,335 lbs  ft
8
84
28
BEAM
5D.Actual bending stress:
fb 
5E. Compare:
123,335  1,016 psi
39.3
fb = 1,016 psi < F’b = 1,070 psi OK
6A. Allowable shear stress: F’v = 135 psi
6B. Area: A = (3)(10.9) = 32.7 in2
6C.Shear:
10.33 7.25

V  250
 2


  1,140 lbs
12 
85
BEAM
6D.Actual shear stress:
fv 
31,140  52.5 psi
232.7 
6E. Compare: fv = 52.5 psi < F’v = 135 psi OK
7A. Allowable deflection:
 max 
1210.33  0.344"
360
7B. Moment of inertia: I = (3)(47.6) = 142.8 in4
7C.Modulus of elasticity: E’ = 1.1x106 psi
86
BEAM
7D.Actual deflection:
 max 
(22.5)(200)(10.33) 4
 0.326"
(1.1x10 6 )(142.8)
7E. Compare: Δact = 0.326" < Δmax = 0.344” OK
BEAM PASSES!
87
29
Steel Beams






Wide flange section
A992, high strength
A36, normal strength
Design check for uniformly loaded beam
Basement location
Simple spans
88
Step 1: Determine Span Length (ft)


From plans, choose longest span of a beam type, W8x18
For example, assume 17’
89
Step 2: Determine Allowable Load (kips)
90
90
30
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
91
Step 4: Determine Tributary Width (ft)
TW 
13.25 12.75

 13'
2
2
92
Step 5: Calculate Actual Linear Load (plf)

First floor:
w1  (13)(40  10)  650 klf

Second floor:
w2  (13)(30  10)  520 plf

Total:
w = 1,170 plf
93
31
Step 6: Calculate Actual Total Load (kips)
1,170 plf
17’
Actual Load = (17)(1,170) = 19,890 lbs = 19.9 kips

94
Step 7: Compare
Actual and Allowable Loads

Actual = 19.9 kips > Allowable = 14 kips NG!
BEAM FAILS!
Try W8x28, Lmax = 23 kips > 19.9 kips, OK!
95
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’
96
32
You Try It
Go to plans
Check steel beam in basement


97
Solution





W8x18, span length = 12’
Allowable load = 20 kips
TW 
12.75 15.33

 14'
2
2
Total uniform load: w1= 14 x (40 + 10) = w2= 14 x (30 + 10) =
700 plf
560 plf
1,260 plf
Actual = 12 x 1,260 = 15.1 kips < 20 kips OK!
1,000
98
Steel Beams
99
99
33
Spread Footings


Soil has maximum capacity to withstand pressure
Footing distributes point load to soil based on maximum capacity
100
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
101
How to Calculate Reactions
R
wl
2
where: l = span length, ft
w = total linear load, plf
R = reaction, lbs
w
l
R
R
102
34
Step 3: Calculate Point Load


Point load = R1 + R2
For example:
(1,170)(13)
 7,605 lbs
2
(1,170)(17)
R2 
 9,945 lbs
2
R1 

P = 17,550 lbs.
1,170 plf
13’
17’
10’
103
Step 2: Calculate Qact (psf)
Calculate actual bearing pressure:
P

Qact 



LW
P = column load, lbs
L = footing length, ft
W = footing width, ft
For example:

Qact 
17,550
 1,097 lbs
(4)(4)
104
Step 4: Compare Qact to Qmax

Qact = 1,097 psf < Qmax = 1,500 psf, OK!
FOOTING PASSES!
105
35
You Try It


Go to the plans provided
Are the spread footings supporting the steel columns adequate?
106
Solution


Soil type = silty sand (SM), Qmax = 2,000 psf
P
(1,170)(11.33) (1,170)(12)

 13,648 lbs
2
2

L = W = 2.5'

Qact 

Qact = 2,416 psf > Qmax = 2,000 psf, NG!
13,648
 2,184 psf
(2.5)(2.5)
107
The End
36