nscp2001-ubc97 static earthquake.pdf

Static Earthquake

Tutorial Problem

using

NSCP 2001/UBC 97 CODE

by:

Engr. Arnel R. Aguel

Copyright© 2006

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Static Earthquake Tutorial Problem Copyright 2006

________________________________________________________________________ NSCP 2001/UBC 97 Code by: Engr. Arnel R. Aguel

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Tutorial example of solving problem using NSCP 2001/UBC 97 static earthquake

provisions of the code.

Problem:

A 10-storey RC building which is square in plan has a series of special

moment resisting frames around the perimeter of the building. The floor

loads and the frame dimensions are shown in the figures below.

Fig. 1



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Fig. 2

Design Criteria:

• Seismic Zone 4 Z=0.40 (from NSCP table 208-3 & Figure 208-1 page 2-39)

• Importance Factor I=1.0 (from NSCP table 208-1 page 2-38)

• Soil Profile Type (Dense Soil & Soft Rock) = Sc (from NSCP table 208-2

page 2-39)

• Seismic Source Type A

- Near Source Factor Na=1.0 (from NSCP table 208-4 page 2-39)

- Near Source Factor Nv=1.0 (from NSCP table 208-5 page 2-39)



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Required:

1. Calculate the total base shear (V) of the structure.

2. Calculate the storey shear at each level.

3. Calculate the overturning moment at base level (ground level).

Solution:

I. Solving for the total base shear V:

a) The structure is more than 3 stories in level, hence, the simplified equation for

base shear in eq (208-11) cannot be used.

b) The structure is also less than 75m in height and it has a regular framing system

all throughout the top most level, thus, equivalent Static Force Procedure can be

used in solving the problem.

c) From NSCP table 208-7 page 2-40 it can be found that for seismic zone 4 and soil

type profile Sc the seismic coefficient Ca & Cv are the following:

Ca=0.40Na or simply 0.40 since Na is given as 1.0

Cv=0.56Nv or simply 0.56 since Na is given as 1.0

d) From NSCP table 208-11 page 2-46 it can also be found that for SMRF RC

structure the overstrength factor R=8.5

e) Solving for the period T:

T= Ct (hn)¾ eq (208-8) page 2-44

Where Ct=0.0731 for RC structures (NSCP section 208.5.2.2 page 2-44)

T= (0.0731)(35.0)¾

T=1.05 sec.

f) Solving for the base shear V:

V=CvI W eq (208-4) page 2-44

RT

Where W=total dead weight of the structure in KN

W= (6600)(9)+7000

W=66,400 KN



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V=(0.56)(1.0)(66,400) = 4,166.27 KN

(8.5)(1.05)

Other provisions of the code must be satisfied before using the result of eq. (208-4).

The total design base shear must not exceed the following:

V=2.5CaI W eq (208-5) page 2-44

R

V=(2.5)(0.40)(1.0)(66,400)

(8.5)

V=7,811.76 KN > 4,166.27 KN (ok)

Also the total design base shear must not be less than the following:

V=0.11CaIW eq (208-6) page 2-44

V=(0.11)(0.40)(1.0)(66,400)

V=2,921.60 KN < 4,166.27 (ok)

In addition, for seismic zone 4, the total base shear also shall not be less than the

following:

V=0.8ZNvI W eq (208-7) page 2-44

R

V=(0.80)(0.40)(1.0)(1.0)(66,400)

8.5

V=2,499.76 KN < 4,166.27 KN (ok)

Therefore, the total design base shear of the structure is:

V=4,166.27 KN (Answer for question 1)



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II. Solving for the storey shear at each level:

a) The total force (base shear) shall be distributed over the height of the structure in

accordance with the following equation.

n

V= Ft + ∑ Fi eq (208-13) page 2-47 i=1

Where Ft is the concentrated load located at the top most level and Fi is the

summation of forces distributed at each level of the structure. Thus, in order to

satisfy eq (208-13) we must solve first Ft and Fi from the equations below.

Ft=0.07TV eq (208-14) page 2-47

Ft=(0.07)(1.05)(4,166.27)

Ft=306.25 KN

However, the code says that the value of Ft need not exceed the following.

Ft=0.25V

Ft=(0.25)(4,166.25)

Ft=1,041.56 KN >306.25KN (ok)

Therefore, Ft=306.25 Governs.

Additionally Ft may be considered zero when the period T≤0.70 sec. In our case

since T=1.05 so Ft must be considered acting in full intensity at the top most level.

b) Solving for the distributed force at each level Fx:

Fx=(V-Ft)Wxhx eq (208-15) page 2.47 n

∑Wihi i=1

V-Ft=4,166.27-306.22

V-Ft=3,860.05 KN



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Level V-Ft Wx hx Wxhx Fx

1 (i) 3,860.05 6,600 0 0.00 0.00

2 3,860.05 6,600 3.5 23,100 69.49

3 3,860.05 6,600 7.0 46,200 138.99

4 3,860.05 6,600 10.5 69,300 208.48

5 3,860.05 6,600 14.0 92,400 277.97

6 3,860.05 6,600 17.5 115,500 347.47

7 3,860.05 6,600 21.0 138,600 416.96

8 3,860.05 6,600 24.5 161,700 486.45

9 3,860.05 6,600 28.0 184,800 555.95

10 3,860.05 7,000 31.5 220,500 663.35

Roof (n) 3,860.05 6,600 35.0 231,000 694.94

∑ 1,283,100 3,860.05 √

c) Drawing the Force and Shear diagram at each level.



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III. Solving for the overturning moment at base level (ground level):

n

Mo = Ft hn + ∑ Fi hi i=1

Mo = [(306.25 + 694.94) (35)] + (663.35)(31.5) + (555.95)(28) + (486.45)(24.5) +

(416.96)(21) + (347.47)(17.5) + (277.97)(14) + (208.48)(10.5) + (138.99)(7) +

(69.49)(3.5)

Mo = 105,555.45 KN-M (Answer for question 3)

nscp2001-ubc97 static earthquake.pdf

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