Introduction to Structural Design

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Nov 26, 2013 (3 years and 8 months ago)

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CE
496

Introduction to
Structural Design

Winter 2011

Howard
Lum

February 17, 2011

Agenda


Loads


Tributary Area


Dead Load


Live Load & LL Reduction


Wind Load


Seismic Load


Steel Tension, Compression and Flexural Design


Concrete Basics


Q&A


Loads


CBC Chapter 16 provides the requirements


Dead Loads:


Wt. of Steel = 490
pcf


Wt. of Concrete = 150
pcf


Wt. of Masonry = 115
pcf


Density of Water = 62.4
pcf


Density of Wood = 40
pcf

Live Loads


Live Loads depend on the group of occupancy


Live Loads: See CBC Table 1607.1 (2 pages)


For the County project, special conditions
include:


Handrail design:

Lateral Load = 50
plf


Vehicle Barrier:

Lateral Load = 6,000 lb

Loads are given in pressure (psf)


To convert psf to uniform load W (pounds/ft):


W (#/ft) = Load (psf) x Tributary Width

Example:


LL=60
psf
, beams spaced at 12 ft on center,
span = 40 ft



w
LL

= (60)*12 = 720 lb/ft


A
T
= 12 * 40 = 480 SF


Trib. Width
for Beam A

Beam A

PLAN

Beam & Column Loads

Beam Tributary Area

Column Tributary Area

Live Load Reduction


Floor Live Load Reduction


CBC 1607.9


Method 1 (1607.9.1)


LL Reduction if
K
LL

A
T
>= 400 SF


L = Lo (0.25 + 15/√K
LL

A
T
)


Lo: unreduced LL (
psf
)


K
LL:
1
-
4 (Table 1607.9.1)


A
T

Tributary Area (sq ft)


LL > 100
psf

shall not be reduced EXCEPT:


Member supports 2 or more floors


LL


no reduction in public assembly areas

LL Distribution


DL always applies to the entire structure


LL applies to areas of maximum stress:


Simple Span Beam




Overhang Beam



LL

LL

LL

DL

DL

DL

Load Combinations


CBC Section 1605


Strength Design: U <=
Ф
*(Strength)


U = 1.2 (D+F) + 1.6 (L+H)


U = 1.2 D + 1.6 W + 0.5 L


U = 0.9 D + 1.6 W + 1.6H


Working Stress Design:


D + L < (Allowable stress)


Not used


Strength Design is based on probabilistic
approach of loads and strength variance


Working Stress Design is based on conventional
elastic stress less than allowable values

Deflection Limits


Allowable deflections in CBC Table 1604.3


Max Deflection = L/360 or L/240 where:


L
=
beam span


Floor LL: L/360

Floor DL+LL: L/240


If L=10 ft, max allow
defl
. = L/240 = 0.5 inch


Use working loads for all deflection calculations



L

Wind Load


CBC Section 1609 and ASCE 7
-
05 Ch 6


Simplified method: Ps =
λ

*
Kzt

*I * p
s30


where:


λ

is height/exposure factor


Kzt

is
topgraphic

factor (
Eq

6
-
3)


I is importance factor


P
s30
is pressure at 30 ft, I=1


Other
structures ASCE 6.5.15


F = q
z
GC
f
A
f


where q
z

= velocity pressure, G = 0.85 (rigid
structure), C
f
from Fig. 6
-
21, A
f

=
proj
. area

Wind Load


Importance Factor I is based on Occupancy
Table 1
-
1 and Table 6
-
1


California: basic wind speed = 85 mph


Exposure B, C, D as defined in CBC 1609.4.3


City of Long Beach has special wind provisions
based on geographical locations

General approach

Earthquake Load Design


Maximum Considered Earthquake (MCE)


2% probability of exceedance in 50 years ( or 2,500
years return period)


RP = 1/Pe = 1/(0.02/50) = 2500


All MCE’s are given in the CBC with the latest
update in USGS website


http://earthquake.usgs.gov/hazards/


Spectral Accelerations characterized by:


Ss (short T) and S
1
(long T)


T
L

(transition T)


CBC Fig. 22
-
16




S
S

(short T)


MCE
Acceleration
for T=0.2 sec

S
1

(Long T)


MCE
Acceleration
for T=1 sec

Site Class per CBC

Earthquake Load


Site Class Modifications to MCE


Site Class (A
-
F) determined by Table 1613.5.2


Hard rock to soft soil


Fa = Short Period Mod. Factor
-

1613.5.3(1)


Fv = Long Period Mod. Factor
-

1613.5.3(2)


Design Earthquake:


S
DS

= 2/3 * Fa * Ss


S
D1

= 2/3 * Fv * S
1

Earthquake behavior of Structures

Effective Seismic Weight


W (in seismic analysis):


Weight (DL) of the Diaphragm


Weight (DL) of the Exterior Walls


+ 25% floor LL for Storage Areas


+10 psf floor LL for Partitions


+ weight of permanent equipment


+ 20% of flat roof snow load (> 30 psf)


Reference: ASCE 12.7.2 and 12.14.8.1

Tributary Weights

Base Shear (ASCE 12.8)


V= Cs * W:


Cs = S
DS
/ (R/I)


Max Cs = S
D1
/T(R/I) or S
D1
T
L
/T^2(R/I)


Min Cs = 0.01


Min Cs = 0.5S
1
/(R/I) for S
1

> 0.6


T calculation:


Ta = Ct * h
n
x


Min. T = Cu * Ta where Cu from Table 12.8
-
1


Reference: ASCE 12.8.1 to 12.8.3

TABLE 11.6
-
1 SEISMIC DESIGN
CATEGORY BASED ON SHORT PERIOD
RESPONSE ACCELERATION PARAMETER

Value of SDS

Occupancy Category


I or II

III

IV


SDS < 0.167

A

A

A


0.167 ≤
SDS < 0.33

B

B


C


0.33 ≤
SDS < 0.50

C

C


D


0.50 ≤ SDS

D

D


D



Vertical distribution of Forces


Reference: ASCE 12.8.3


F
x

= C
vx

* V


where C
vx

=
w
x
h
x
k
/∑
w
i
h
i
k






V

W

Seismic Load Combination


Load combination w/seismic


Strength Design (12.4.2.3):


(1.2 + 0.2 S
DS
)D +

*Q
E
+ L + 0.2S


(0.9


0.2 S
DS
)D +

*Q
E
+ 1.6H


Check both downward seismic and uplift seismic
forces in combination with dead load and live
load


Note: L can be 0.5L (if Lo <= 100psf)

Steel Properties

Es=29,000
ksi

Fy

Fu

1

Steel: Availability of ASTM Grades

58

50

58

Tension Members


Strength Design:
Pu

<=

t *
Pn



Tensile Capacity:

t *
Pn



Failure Modes:


Deformation at yield (gross area)


t = 0.90,
Pn

=
Fy
*Ag


Fracture at tensile strength (net area)



t = 0.75,
Pn

= Fu*
Ae

Yield

Fracture

Compression


Stability controls


Elastic Buckling Stress (Euler)
given in AISC
Eq

E3
-
4


Fe: critical compressive

stress
above which column buckles


Fe

is independent of Fy or Fu


Fe =
π
2
*E/(KL/r)
2


Fcr

= 0.877*Fe




K=1.0

Compression


Strength Controls


KL/r <=4.71 √E/
Fy
, column
strength governs (not stability)


Fcr

= [0.658
(Fy/Fe)
]*Fy


where Fe =
π
2
*E/(KL/r)
2






AISC Compression E3

0.00
10.00
20.00
30.00
40.00
50.00
60.00
70.00
0
20
40
60
80
100
120
140
160
180
200
Fe (Euler Critical Stress)

KL/r

Fcr

Fy
=36
ksi

Fy
=50
ksi

Double Sym. Beam Design


AISC Chapter F



b
*
Mn

= 0.9*
Mn


Mn

is dependent on
lateral
unbraced

length
L
b

of the compression
flange


Lateral
-
torsional

buckling governs
design if L
b
>
Lp


Compression
Flange


L
b

-

Unbraced

Length


L
b

is independent of the span length


L
b

can be 0 if the compression flange is
continuously braced


Example: Span = 50 ft, L
b

= 25 ft

Steel Beam Design Curve


Concrete Design


ACI 318




Mn

>= Mu where


is 0.9




Vn

=


(
Vc

+Vs) >= Vu where


is 0.75




Pn

>=
Pu

where


is 0.65 to 0.90


f’c
: 28
-
day compressive strength (3000


8000 psi)


Fy
: Yield strength of reinforcement (60
ksi
)





Questions & Answers