Ubc

UBC '97

The Uniform Building Code, 1997 revision

Force = A x P

A = the projected area of the item.

P, Wind pressure (Psf), = Ce x Cq x Qs

Ce, combined height, exposure and gust response factor is taken from table 16-G

3 terrain exposures termed "B" "C" & "D", are cited in the table. For each one, a variety of heights are listed and a corresponding value for Ce.

Cq, pressure coefficient (same as drag, Cd), is taken from table 16-H

Cq = 1.3 for flat plates, and Cq = .8 for cylinders over 2" in diameter, 1.0 for cylinders 2" or less in diameter.

No differences due to aspect ratio are cited. The cylinder is either 1/1.3 = .769, or .8/1.3 = .615 of the flat plate value. Not quite 2/3 either.

Qs, wind stagnation pressure, is taken from table 16-F

The values in the table are the same as one would get from using the .00256V^2 formula.  

UBC 97 thinks the wind speed is the "fastest mile basic wind speed" at 33 feet above the ground, not the actual peak sustained wind speed, and is obtained from a map that is part of the spec. It also cites the current EIA spec as a suitable method.

The UBC exposure definitions are as follows:

EXPOSURE B has terrain with buildings, forest or surface irregularities, covering at least 20 percent of the ground level area extending 1 mile (1.61 km) or more from the site.

EXPOSURE C has terrain that is flat and generally open, extending 1/2 mile (.81km) or more from the site in any full quadrant.

EXPOSURE D represents the most severe exposure in areas with basic wind speeds of 80 miles per hour (mph) (129 km/h) or greater and has terrain that is flat and unobstructed facing large bodies of water over 1 mile (1.61km) or more in width relative to any quadrant of the building site. Exposure D extends inland from the shoreline 1/4 mile (.40km) or 10 times the building height, which ever is greater.  

 CENE 437: The Class: Loads: Lateral Loads: Seismic Loads: UBC '97: Introduction

1997 UBC Earthquake DesignIntroduction

Seismic forces are a particularly important consideration for engineers working in the Western U.S. where the frequency of earthquake occurrences is common.

Seismic building forces are the result of the sudden movement and rupturing of crustal plates along fault lines.

There are more than 160 known active faults in California alone.

New faults continued to be discovered, usually when an unexpected earthquake occurs.

When a fault slip occurs suddenly, it generates seismic shock waves that travel through the ground in a manner unlike that of tossing a pebble onto the surface of calm water.

These seismic waves cause the ground to shake.

The effect of this dynamic ground motion can be simply modeled using a cereal box standing upon a piece of sand paper.

Upon yanking the paper, the box topples in the direction opposite of the yank, as if a pushing force had been applied to the box.

The heavier the box, the greater the apparent applied force which is called an inertia force.

As the ground moves suddenly, the building attempts to remain stationary, generating the inertia induced seismic forces that are approximated by the static lateral force procedure covered here.

This procedure is introduced in UBC '97 1629.8.3 and discussed in detail in UBC '97 1630.

The static force procedure is limited to use with regular structures less than 240 feet in height.

And, also to irregular structures 65 feet or 5 stories in height.

o See UBC '97 1629.8.3 for exact definition of limitations.

o Regular structures are symmetric, without discontinuities in plan or elevation.

The building plan is generally rectangular.

The mass is reasonably uniform throughout the building's height.

The shear walls line up from story to story.

o Irregular structures include both vertical irregularities (UBC Table 16-L) or plan irregularities (UBC Table 16-M). These irregular features include:

Reentrant corners.

Large openings in diaphragms.

Non-uniform distribution of mass or stiffness over building height (e.g. soft story).

Basic premise of seismic code provisions:Earthquake Damage to StructureMinor NoneModerate Some damage to non-structural elementsMajor Maybe severe damage, but not collapse.

Seismic zones in U.S. (UBC '97 Figure No. 16-2):

Zones Damage to Structure MMI* Scale0 No Damage -----1 Minor V, VI2 Moderate VII3 Major VII4 Major -----

*MMI = Modified Mercalli Intensity scale of 1933.

Modeling Forces

1997 UBC static lateral method considers both horizontal movement and vertical ground movement.

The vertical component may be taken as zero, however, when using the allowable stress design procedure.

We statically model the inertial effects using Newton's 2nd law of motion:

Rewrite equation (1) as:

Compare (2) to UBC base shear design equations, as given below, where each equation is a function of the building weight and some form of an acceleration factor.

o Each acceleration factor is somewhat equivalent to a/g, except they account for factors like underlying soil, the structural system, and building occupancy.

Where:

o V= base shear force. The horizontal seismic force acting at the base of the structure as modeled by the "yank" of the paper in the previous cereal

box example. It is important to note that this force was developed for the strength design methodology and not the allowable stress approach.

o W = the dead weight of the building plus a percentage of the live load that is thought to be present during a seismic event. See UBC '97 1630.1.1 for details about this live load addition.

o (Cv I / R T) = acceleration factor (also known as a seismic base shear coefficient). This coefficient will govern V for buildings with medium to long fundamental period of vibrations. The forces in these buildings are induced by the velocity component of the bedrock motion. Hence the "v" subscript.

o (2.5 Ca I/R) = this coefficient is independent of the period of vibration. It will govern V for buildings with short fundamental periods of vibrations, like the buildings being studied in this class. The forces in these stiff buildings are generated by the acceleration component of the bedrock motion. Hence the "a" subscript.

o (0.11 Ca I) = this coefficient is also independent of the period of vibration. It is a lower bound value, keeping V at some minimum value.

o (0.82 N v I / R) = this lower bound coefficient is only applicable to structures located in seismic zone 4 and within 9.3 miles (15 km) of a known seismic fault.

o The difference in building response can be simply demonstrated by "shaking" the base of two different "structures".

It is common practice to express the base shear design force as a percentage of W; calculating only the coefficient term.

o The following are some typical base shear coefficient values for a regular, single-story masonry building not located near a fault. In addition, we conservatively assumed that a geotechnical site investigation was not completed. Because this type of building is so stiff, the (2.5 Ca I / R) coefficient governs V.

Zone Coefficient

1 V = .067W

2a V = .122W

2b V = .156W

3 V = .200W

4 V = .244W

Base Shear Terms

In this section, the various terms of the static base shear equation are examined in more detail.

Z = seismic zone factor. o Effective peak ground accelerations with 10% probability of being

exceeded in 50 yrs.

o Given as a percentage of acceleration due to gravity.

For example, consider zone 4, where Z = .4 horizontal ground acceleration is predicted at .4g at bedrock.

o Doesn't account for building dynamic properties or local soil conditions.

o '97 UBC Figure 16.2 seismic zone map.

o Table 16.I Z values as given below:

Zone Z0 01 .0752A .152B .203 .304 .40

I = importance factor.

o Classifying buildings according to use and importance.

Essential facilities, hazardous facilities, special occupancy structures, standard occupancy structures, miscellaneous structures.

Essential facilities mean that the building must remain functioning in a catastrophe.

Essential facilities include: hospitals, communication centers, fire and police stations.

Design for greater safety.

o '97 UBC Table 16-K.

I = 1.25 for essential and hazardous facilities.

I = 1.0 all others.

T = building's fundamental period of vibration.

o Fundamental period of vibration is the length of time, in seconds, it takes a structure to move through one complete cycle of free vibration in the first mode.

o There are two methods to estimate T:

Method A:

Method B: (an iterative approach not generally used in regular structures)

Using Method A, the fundamental period of vibrations for masonry buildings is estimated at:

Height (ft) Period (seconds)20 .1940 .3260 .43120 .73160 .90

Ca and Cv = seismic dynamic response spectrum values.

o Accounts for how the building and soil can amplify the basic ground acceleration or velocity.

o Ca and Cv are determined from respectively '97 UBC tables 16-Q and 16-R as a function of Z, underlying soil conditions, and proximity to a fault.

Using method A,

o Soil profile type:

The soil layers beneath a structure effects the way that structure responds to the earthquake motion.

When the period of vibration of the building is close to the period of vibration of the underlying soil, the bedrock motion is amplified. The building experiences larger motions than that predicted by Z alone. The following are generalizations about building response as a function of building flexibility and underlying soil stiffness.

Building Description Soil Description Induced Seismic ForceFlexible (Large T's) Soft (big S) Higher

Flexible Stiff LowerStiff Soft Higher

Flexible Stiff Lower

The soil profile types are:

Description Type

Hard Rock SA

Rock SB

Very dense soil and soft rock SC

Stiff soil SD

Soft soil SE

See '97 UBC 1629.3.1 SF

Specific details about each type can be found in '97 UBC Table 16-J and '97 UBC 1629.3.1.

o In the absence of a geotechnical site investigation, use SD. This is in accordance with '97 UBC 1629.3

Do not confuse this requirement with the one stated in '97 UBC 1630.2.3.2 which applies ONLY when using the simplified design base shear procedures of '97 UBC 1630.2.3. This web site is NOT using these simplified procedures, but is using 1630.2.1.

R = response modification factor. o A judgement factor that accounts for building ductiltiy, damping, and

over-strength.

Ductility = ability to deform in the inelastic range prior to fracture:

Damping = resistance to motion provided by internal material friction.

Over-strength = the extra or reserve strength in the structural system. It comes from the practice of designing every member in a group according to the forces in the most critical member of that group.

o Structural systems with larger R = better seismic performance.

o In '97 UBC Table 16-N, R range from 2.8 (light steel frame bearing walls with tension bracing) to 8.5 (special SMRFS of steel or concrete and some dual systems).

o For bearing wall systems where the wall elements resist both lateral and vertical loads:

Wood shear panel buildings with 3 or less stories: R = 5.5

Masonry shear walls: R = 4.5.

Nv and Na = near source factors that are applicable in only seismic zone 4. They account for the very large ground accelerations that occur near the seismic source (the fault).

o Nv is generally used with Cv for structures located < 9.3 miles (15km) from the fault.

Nv is found in '97 UBC Table 16-T

o Na is used with Ca for structures located < 6.2 miles (10 km) from the fault.

Na is found in '97 UBC Table 16-S.

o Both Na and Nv are based upon the type of seismic source, A-C. This source type, and location of fault, must be established using approved geotechnical data like a current USGS survey.

Distribution of Seismic Forces to Primary LFRS

Now that we have the base shear force, what type of induced forces act through the height of the building?

o How to model the inertial force that acts opposite to yank of paper on the cereal box?

Recall for wind loads

o First, calculate loads/pressures over the height of building.

o Then developed base values.

o These values are at the allowable stress level.

In contrast, with seismic -

o First, determine base force.

o Then determine and distribute forces over the height of the building, called story forces, Fx.

o There are two different sets of story forces distributed to the primary LFRS:

For vertical elements, use Fx.

For horizontal elements, use Fpx.

Recall that the primary LFRS for a box building = horizontal diaphragms and vertical shear walls.

o Then adjust these strength level forces by a redundancy/reliability factor, , and an allowable stress factor of 1.4 discussed further in item d, below.

a. Story forces for vertical elements.

o Used in design of shear walls and shear wall anchorage at the foundation.

o Determined before Fpx's.

o Applied simultaneously at all levels.

o Results in a triangular distribution of forces over a multi-story building that has approximately equal floor masses.

and

Where:

Ft = roof level force accounting for whiplash effect.

Ft.07TV .25V

or

{ 0 if T .7 sec.

wx, wi = tributary weights at levels x and i.hx, hi = height above base to levels x and i.

o further detail can be found in '97 UBC 1630.5.

b. Story forces for horizontal elements.

o At roof level, Fpx = Fx.

o At other levels, Fpx > Fx.

o Accounting for the possibility that larger instantaneous forces can occur on individual diaphragms.

o Applied individually to each level for the design of that diaphragm.

where wpx = weight of diaphragm and elements tributary to it at level x.

o For masonry buildings (and concrete) supported by flexible diaphragms, the R factor used to determine V must be reduced to 4.0 from 4.5 ('97 UBC 1633.2.9.3).

o For more information see '97 UBC 1630.6.

c. The single story building is a special case.

o In most cases, T .7 and Ft then is taken as zero.

o From equation 30-15:

o From equation 33-1:

o Consequently, F1 = Fp1 = V for the case of wood frame buildings.

o For masonry buildings, Fp, is based upon a slightly larger V due to R changing from 4.5 to 4.0 according to '97 UBC 1633.2.9.3. In this case, then: F1 = V and Fp1 = 1.125 V.

d. Redundancy/reliability factor and the 1.4 ASD adjustment:

o In the load combination equations as discussed in the last sub-module in the load module of this site, all earthquake forces are generically called E.

Where:Eh = load developed from V, (like Fx or Fpx) or Fp, (the design force on a part of a structure).Ev = 0 for ASD = redundancy/reliability factor, discussed below.

o E is at strength level and must be divided by 1.4 for use in allowable stress design.

The application of 1.4 and p are shown in example one of this sub-module.

o The redundancy/reliability factor penalizes structures in seismic zones 3 and 4 that do not have a reasonable number and distribution of lateral force resisting elements, such as shear walls. These structures with a limited number of shearwalls are referred to as non-redundant structures where the failure of one wall loads to the total collapse of the structure.

o

o Where:

AB = the ground floor area of the structure in ft2.rmax = maximum element-story shear ratio, ri, occurring at any story level in bottom 2/3 of the structure. rmax identifies the least redundant story.ri = Rwall/Rstory(10/lw)Where:

Rwall = shear in most heavily loaded wallRstory = total story force, Fxlw = length of most heavily loaded shear wall.

= 1 when in seismic zones 0, 1, or 2. = 1 when calculating drift.

Upon careful inspection of the r and ri equation with application to a single story, regular building, we see:

o To maintain a = 1.0, the minimum length of the most heavily loaded shear wall is fixed as:

o If a flexible diaphragm, a common controlling case will be when Rwall/Rstory = .5.

In this case then to keep = 1.0.

o Although the Breyer, et al book uses the subscript "u" to distinguish strength-level vs. allowable stress-level loads, I have opted for a different convention that I believe is simpler.

Upon modifying the various Eh values by and 1.4, Eh becomes E'h. For our single story building, the shear wall forces and diaphragm forces at ASD level would look like:

F'1 = F1 (1/1.4)F'1 = Fp1 (1/1.4)

Example 1

Develop the applicable seismic forces for a one-story, box-type industrial building located in Southern California. Assume partially grouted CMU walls weighing 61 lb/ft2,

a roof dead load of 9 psf, and the building is not located near (further than 9.3 miles) a seismic source. No geotechnical investigation was completed.

1. Base shear coefficient, V. o The base shear equation(s) are quite cumbersome to use, unless on

knows beforehand which equation governs.

Recall that middle equation is for buildings medium to long fundamental T's. The left-hand equations are lower bound values. The right-hand equation is for short (stiff) T buildings.

You can determine if its the right-hand equation quickly by comparing the building's T to Ts:

o TS is a limiting period of vibration that is used to differentiate between stiff and flexible buildings.

o The seismically-induced forces in stiff buildings are related to the bedrock acceleration. The forces in flexible buildings are related more to bedrock velocity.

o Calculate T and Ts:

Zone = 4 (figure 16-2)Z = .4 (Table 16-I)Soil profile type = SD

NV = 1.0 (Table 16-T)CV = .64 (Table 16-R)Na = 1.0 (Table 16-S)Ca = .44 (Table 16-Q)

TS = .64/(2.5(.44)) = .582 sec

Therefore, use short T base shear equation.

o Calculate V:

Zone = 4 (Figure 16-2)Z = .4 (Table 16-I)Na = 1.0 (Table 16-S)Ca = .44 (Table 16-Q)I = 1.0(Table 16-K)R = 4.5 (Table 16-N)

2. Story force for vertical elements, F1

o Recall that for a one story structurewhere T .7, the vertical story force and the base shear are equal.

F1 = V = .244W

3. Story force for horizontal elements, Fp1

o In a one story masonry building with flexible diaphragms, recall that:

Fp1 = 1.125VFp1 = 1.125(.244)W = .275W

4. Calculate the diaphragm design forces in the transverse loading direction due to seismic.

o To do this, need to first determine W, the weight of the structure that is supported by the diaphragm.

Consider this weight (and resulting force) on a per foot basis.

A 1' strip of dead load = the mass that causes the inertial forces on a per foot basis in the diaphragm.

Similarly in the longitudinal direction:

It is customary to ignore wall openings in these diaphragm force calculations, as the added accuracy is generally not warranted.

o This uniform diaphragm force is at the strength level, and has not yet been adjusted by and 1.4.

In other words, Fp in the transverse direction could be generically labeled Eh from the equation:

E = Eh + 0

Recall that r = 1.0 for this type of building when

o On the right-hand side of our building, a window occurs; lessening the length of wall to resist shear to 2 - 17.5 ft segments.

Adjusting to ASD:transverse:Fp = 476/1.4 = 340 lb/ftlongitudinal:Fp = 600/1.4 = 429 lb/ft

o Comparing wind vs. seismic forces, it is apparent that seismic will govern the lateral design of the diaphragm in both directions.

Transverse: 340 plf > 93 plf.

Longitudinal: 429 plf > 93 plf.

2. Calculate the unit shear forces in the flexible diaphragm. o Flexible diaphragms are like deep, thin, uniformly-loaded beams that are

simply supported by the shear walls.

Consider the transverse direction:

Longitudinal force direction:

The transverse direction is often the more critical direction for rectangular buildings due to the longer diaphragm span and the shorter shear walls.

o In summary: the diaphragm spans between the supporting shear walls, transferring the inertial affect of the perpendicular walls and itself to those walls located parallel to force direction.

Wall Forces

1. Shear walls (in-plane lateral forces):

The shear walls support the diaphragm by receiving the diaphragm reaction, R, through proper connections at the diaphragm boundary.

o See lateral load path for a review of the basic building's behavior.

o If you have a masonry or concrete building supported by a flexible diaphragm, the shear wall load (e.g. the diaphragm reaction) could be recalculated to reflect the lower design requirement.

This comes from the R = 4.5 for vertical LFRS elements vs. R = 4.0 for the horizontal LFRS.

In addition, the wall itself develops inertial forces that also act parallel to the wall and must be accounted for.

o This additional seismic force is assumed to be generated from the weight of the top half of the wall

The shear wall is then usually evaluated for shear stress capacity at mid-height.

o According to '97 UBC 2107.1.7 if in seismic zones 3 or 4, the shear wall must be designed to resist 1.5 times v.

2. Out-of-plane bending of walls (lateral forces perpendicular to wall): Refer to '97 UBC 1632 - Lateral forces on elements of structures.

o Applied forces on elements, (vs. the LFRS), may be larger in magnitude because these elements respond dynamically to the motion of the structure instead of the ground.

The strength-level design equation for elements and components has changed considerably from earlier (pre-1997) versions of the code. It now is:

Where:

o the subscript "p" refers to elements or components (e.g. parts) of the structure.

o ap = in-structure component amplification factor found in '97 UBC Table 16-O

ap 2.5

o Ca = seismic response spectrum value found in '97 UBC Table 16-Q.

o Rp = component response modification factor form '97 UBC Table 16-O

o hx = the location (elevation) of the attachment point of the part taken with respect to grade

o hr = the structure's roof elevation with respect to grade

o wp = the weight of the element or component under consideration

o Fp is at strength-level and must be adjusted by 1.4 to reduce it to ASD level.

o = 1.0 for elements and components.

A common calculation that makes use of this element's provision is to determine the seismic force normal to a wall as shown in the following figure. A sample calcualtion is given in Example 2.

o Please note that there appears to be a height mistake and a code interpretation problem in Breyer's Example 2.17

3. Diaphragm anchorage: Lateral forces acting perpendicular to the wall will tend to separate the

wall from the horizontal diaphragm.

o Must provide a positive anchorage system connecting masonry walls to diaphragms, shown above as the "specially designed anchor".

This anchorage must resist:

o Wind forces on wall element.

o Seismic force normal to the wall using UBC '97 Eqn 32-2. According to '97 UBC 1633.2.81:

In seismic zones 3 and 4 with a flexible diaphragm, Rp = 3.0 and ap = 1.5. This ap factor essentially increases the design forces at the wall to diaphragm by 50%.

In seismic zone 4, the Fp for wall anchorage 420 lb/ft.

o Regardless of governing lateral force (wind vs. seismic) the code specifies a minimum, strength-level, anchorage force of 280 lb/ft for concrete and masonry walls ('97 UBC 1605.2.3 and 1611.4)

Requirements about anchorage detailing for concrete or masonry walls:

o In seismic zones 3 and 4 ('97 UBC 1605.2.3)

Use embedded straps that attach or hook around reinforcing steel or ensure effective transfer of forces to steel.

Limit anchor spacing to 4' unless wass are designed to resist bending between anchors.

o In seismic zones 2, 3, and 4('97 UBC 1633.2.9.5)

Anchorage shall not use nails in withdrawal or toe nails.

Ledgers or framing shall not be used in cross-grain bending/tension, which is shown in the following figure.

To avoid these problems, use specially designed seismic ties that are known as purlin anchors by Simpson Strong-Tie, a manufacturer of prefabricated, light-gauge, steel connectors for wood construction.

Example 2

Continue with the previous example, Example 1, and examine the following: 6. Shear wall design force.

7. Lateral forces normal to the wall.

8. Shear wall overturning.

9. Shear wall drift.

6. Shear wall forces

a. Consider seismic loading in the transverse direction.

From example 1, step 5, the diaphragm force (at ASD level) supported by one of the 50' endwall was determined to be 17000 lbs.

Recall, however, that this was developed from Fpx based upon a R = 4.0, and not Fx with a R = 4.5. Since the shear wall is a vertical element of the LFRS, it is permissible to reduce this reaction force by 89%; accounting for this difference between 4.0 and 4.5.

Calculate the top half of the wall's inertia force assuming no openings in the wall:

Total unit shear, applied at midheight:v` = (15111 + 5582) / 50 = 414 lb/ft.

Since this building is located in seismic zone 4, increase v by 50 %:v` = 414 (1.5) = 621 lb/ft

b. Similarly, for seismic movement in the longitudinal direction:

c. The above unit shear shear wall forces were developed for the 50' end wall that had no openings. What happens when the wall has openings, like th 15' on on the other end wall?

The lateral load must be carried by the effective wall segments known as shear panels if wood walls or piers if concrete or masonry walls.

Different procedures are used to distribute the horizontal diaphragm reaction to the effective wall segments, depending if wood or masonry walls.

o In shear panels, the unit shear is the same in every panel due to the assumption that the panel force is inversely proportional to the panel length.

o In piers, the pier force in inversely proportional to pier rigidity. The unit shear in wider pier will be greater than the unit shear in a narrow pier.

In this example, however, the endwall piers are the same with the same rigidities. The unit shear in each 17.5' pier will be:

7. Lateral forces normal to walls:

o It is assumed that the walls perpendicular to the ground motion span vertically between the roof diaphragm and foundation.

a. Force on main wall using '97 UBC Eqn 32-2:

Also note: Fp = .587Wp is greater than .7(.44) (1.0)Wp = .308Wp and less than 4.0(.44)(1.0)Wp = 1.76 Wp according to '97 UBC 1632.2

b. Force on cantilvered parapet:

c. Anchorage design force at diaphragm:

Recall that if in seismic zones 3 or 4, ap for the main wall is increased to 1.5. This increases the main wall force by 50% for use in anchorage force calculation.

This exceeds the code minimum requirement of 420/1.4 = 300 lb/ft (at ASD level).

Therefore, provide an anchorage system capable of resisting 520.8 lb/ft.

7. Overturning check on shearwalls:

o A lower factor of safety is permitted in seismic design vs. that used in wind because of the transient and reversing nature of the seismic forces.

The factor of safety is accounted for by reducing the resisting moment to dead load by .9.

o Consider overturning of the short walls due to seismic forces acting in the transverse building direction.

7. Story drift, :

o According to '97 UBC 1630.10.2, the maximum in elastic displacement M, should not exceed .025h for structures with T < .7 seconds.

A serviceability consideration.

M is a strength level inelastic displacement due to a design earthquake.

o M = .7RS where R is from Tabe 16-N and S is static story drift due to strength-level forces.

o Consider the following calculation for the 50' shear wall: