WOOD, SOILS, AND STEEL INTRO
KNOWLEDGE BASE REQUIRED:
STRENGTH OF MATERIALS
TIMBER DESIGN
SOIL MECHANICS
STEEL DESIGN
REVIEW OF TIMBER DESIGN
•BENDING MEMBERS
•DEFLECTION MEMBERS
•SHEAR MEMBERS
•COLUMN MEMBER
•BEARING PROBLEM
REVIEW OF TIMBER:
BEARING PERPENDICULAR TO THE GRAIN- fc(perp)
P
lb
P
f 'cper   F 'cper
A
lb+3/8
where lb= bearing length
Note: When the bearing length is less than 6 in. and
when the distance from the end of the beam to the
contact area is more than 3 in., the allowable
bearing stress may be increased by Cb.
lb  0.375
Cb 
lb
The deformation limit of .04 inch. is provided by
ASTM D143 provides adequate service in typical
wood-frame construction.
Special Cases
In some designs where the deformation is critical,
a reduced value can be applied. ( WWPA P.9 Table F)
Deflection can be designed for a reduce limit of
.02 in. (also refer to P.251 in text)
Fc (perp .02) = 0.73 Fc (perp .04) + 5.60
Sample Problem:
Given a Hem-Fir Select Structural with 11,000#s on
supports:
a) check for the bearing of a cantilever support.
b) Assume critical deflection for heavy impact loads
at end of cantilever.
4x8
> 3”
3.5”
2 - 2x12
1.5” 1.5”
Fc(perp) = 405 psi
lb= 3”
therefore we can increase bearing stress, but lets
be conservative and use lb as recommended
11,000#
f cper 
 1047 psi  405 psi N.G.
2 x1.5 x3.5
We have to increase bearing
V
11,000
Req’d Area= ----- = ---------- = 27 sq in.
Fc(perp) 405psi
add
2-2X12 X 12
A= 6 X 3.5 = 21 sq in < 27 sq in NG
2-3X12 X 12
A=[(2X1.5)+(2X2.5)](3.5)= 28 sq in > 27 sq in OK
b) 4x8 bearing problem is O.K., now solve for critical
deflection with limit of .02 inch
F’c(perp .02) = 0.73 (405) + 5.60
F’c(perp .02)= 301.25 psi
Req’d Area = 11000/301.25 =36.5 sq in
add 2- 4x12x12
A=[(2X3.5)+(2X1.5)](3.5)= 35 sq in N.G.
use 2- 6x12 - 49 sq in. or a steel plate 3.5X10.5
LECTURE #4
REVIEW OF SOIL MECHANICS
•VERTICAL STRESSES
• LATERAL STRESSES
BASIC SOIL MECHANICS REVIEW:

= UNIT WEIGHT OF SOIL (PCF, KN/m3)
 sat
b
W
= SATURATED UNIT WEIGHT OF SOIL
= BOUYANT UNIT WEIGHT OF SOIL
= UNIT WEIGHT OF WATER(62.4PCF,9.81 KN/m3)
LECTURE #4(CONT)
BASIC SOIL MECHANICS REVIEW:
 b  sat  W
-
=
VERTICAL STRESSES:
V = VERTICAL STRESS (PSF, TSF,KN/m2)
V  h   (TOTAL STRESS)
 'V  h    W (EFFECTIVE STRESS)
CALCULATE TOTAL AND EFFECTIVE V
3'
 dry  105 pcf
GWT
5'
 sat  120 pcf
Ka=0.5
LECTURE #4(CONT)
BASIC SOIL MECHANICS REVIEW:
VERTICAL STRESSES:
V  (h   )
V  105(3)  120(5)
V  315  600  915 psf
EFFECTIVE VERTICAL STRESSES:
 'V  (h   )
 'V   d h1   b h2
 'V  105(3)  (120  62.4)5
 'V  315  288  603 psf
LECTURE #4(CONT)
BASIC SOIL MECHANICS REVIEW:
LATERAL STRESSES:
h  (h   ) K
 h1   V K  (315)(.5)  157.5 psf
 'h 2   V 2 K  (288)(.5)  144 psf
 w   V K w  (62.4)(5)(1)  312 psf
LATERAL FORCE:
FV  0.5H (h   )
FH  0.5H (h   ) K
LECTURE #4(CONT)
BASIC SOIL MECHANICS REVIEW:
LATERAL FORCES:
3'
 dry  105 pcf
GWT
5'
157.5
 sat  120 pcf
144 312
LECTURE #4(CONT)
BASIC SOIL MECHANICS REVIEW:
LATERAL FORCES:
3'
 dry  105 pcf
GWT
5'
157.5
 sat  120 pcf
144 312
TO FIND RESULTANT SUM FORCES: R
TO FIND RESULTANT LOCATION TAKE MOMENT:y
LECTURE #4(CONT)
BASIC SOIL MECHANICS REVIEW:
LATERAL FORCES:
3'
 dry  105 pcf
GWT
5'
236.25#
315
 sat  120 pcf
1575#
1140#
144312
TO FIND RESULTANT SUM FORCES: R
TO FIND RESULTANT LOCATION TAKE MOMENT:y
LECTURE #4(CONT)
BASIC SOIL MECHANICS REVIEW:
LATERAL FORCES:
R=236.25+1575+1140=2951.25#
LATERAL ARM:
Ry 6(236.25)  2.5(1575)  5 / 3(1140)
y

R
2951.25
y  2.5 ft.
LECTURE #4(CONT)
BASIC STEEL DESIGN REVIEW:
For temporary structures we use primarily involves:
Rolled Sections - W: sections
Angle Sections
Channel Sections
Very Rarely
- Steel Joist
Decking(thin gages)
Trusses
Elements include Girders, Beams, Columns,
& Struts (angles &channels)
LECTURE #4(CONT)
BASIC STEEL DESIGN REVIEW:
Review of AISC Construction Manual:
Allowable Stresses
Most commonly used is A36 Steel
Steel Design Procedures
a) Use ASD - Allowable Stress Design Procedure
(Basic Allowable Stress For A36 Steel)
Tension- Ft=.6 Fy=14,400psi
Shear - Fv= .4 Fy=32,400psi
Bearing - Fp=.9Fy=32,400 psi
Bending - Fb = .66Fy=23,760psi
Based on Compact Section
Compression=Fa (variable depending on
unbraced length)
LECTURE #4(CONT)
BASIC STEEL DESIGN REVIEW:
Review of AISC Construction Manual (cont):
Compact vs. Non- Compact
Compact Beam are rolled beam that can achieve
the plastic moment.
Stress Distribution of I Beam
non- compact
compact
LECTURE #4(CONT)
BASIC STEEL DESIGN REVIEW:
Review of AISC Construction Manual (cont):
Compact Sections• are symmetrical about the y-y axis
• webs/flanges must have certain web thickness ratios
•compression flange must be adequately braced
against lateral buckling
Design Procedure for Bending
1. Determine the Maximum Bending Moment
2. Compute the Required Section Modulus based
on allowable stresses Fb=.66Fy (compact) or
Fb=.60 Fy (non compact)
3.Lightest weight section is the most economical
LECTURE #4(CONT)
BASIC STEEL DESIGN REVIEW:
Review of AISC Construction Manual (cont):
Compression Members:
• lc- distance between spacing of lateral braces
When brace spacing is less than or equal to lc
then Fb=.66Fy
•Lu - maximum unsupported length
When brace spacing is greater than Lc but less than
Lu then Fb=.60Fy
When brace spacing is greater than lu, Fb is not
determined and Total allowable moment has to be
taken from a chart.