6. A proposed 5-story health care facility with 110 people maximum occupancy is

Question

6. A proposed 5-story health care facility with 110 people maximum occupancy is to be constructed in Southbend, Indiana. The ground motion periods are expected to be short at this location. The soil report indicates that the soil at the construction site is very dense silty-clay with soft rock profile. Use the same building dimensions and floor weights as in problem 5. Use the IBC 2003 Code to calculate: A. The design base shear. B. The respective vertical distribution of base shear at each floor.

Explanation / Answer

Solution:-

Base shear(VB ) = Ah *W

W = weight of building

Building weight is not given.

Let floor weight =250 kips

And roof weight = 100 kips

Ah = Z/2 *I/R * Sa/g

   Z= 0.36 (for Indiana)

I =1.5 (for medical building)

R = 3

Time period (T) = 0.075*(h)0.75

= 0.075*(20.1168)0.75 = 0.71 second

h = 20.1168 meter (building height)

For Medium soil sites

Sa/g = 1.36/T = 1.36/0.71 = 1.91

Ah = 0.36/2 * 1.5/3 * 1.91 =0.1719

Total weight of building = 4*250 +100 = 1100 kips

VB = Ah*W

Design base shear(VB)= 0.1719 *1100 = 189.09 kips   Answer

Distribution of base shear on each floor

For first floor(Q1) = VB *W1 h12/(W1h12 +W2h22 + W3h32 +W4h42 +W5h52)

Q1 = 189.09 * 250 * 142/(250*142 + 250 *282 +250 * 412 +250 *542 + 100 *662 )

= 5.06 kips   Answer

For second floor (Q2) =189.09 *250 *282/(1.82985 *106)

                                      = 20.25 kips

For third floor (Q3 ) = 189.09 * 250 *412/(1.82985 * 106)

                                   = 43.427 kips

For fourth floor (Q4) = 189.09 *250 *542/(1.82985*106)

                                    = 75.33 kips

For roof

Q5 = 189.09 *100 *662/(1.82985 *106)

      = 45.01 kips    Answer

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