Structural Study of the Existing Lateral System

Transcription

1 Daniel Chwastyk F.W. Olin Center Louis Geschwindner Advisor Structural Study of the Existing Lateral System Executive Summary Technical assignment 3 deals with discussing the lateral system used in the F.W. Olin Science Center. The building has only 4 above ground stories so there is relatively low shear loads that must be dealt with. The shear resisting system found in the building is shear walls. These walls are built on top of a continous footing. This assignment mainly concentrates on the east west loading of the building since the east and west face of the building is much longer than the north and south faces. Through analysis i discovered that the earthquake load governs the design of the shear walls. This is understandable since the building s region has a low wind speed. Stiffnesses of the shear loads were found by assuming a unit load on the building. These stiffnesses are extremely high, but this is due to the very low height / length ratio of the walls. Story drift has also been checked in this report. The drift was found to be acceptable on all stories, which was expected. The northern most shear wall has been checked to make sure it could hold the shear loadings. This wall was checked since it has the highest load on it. It can be determined that if this wall is able to withstand the shear loads, then all walls can withstand the loads. -1-

2 Existing System The existing lateral system used in the F.W. Olin Center is shear walls. These shear walls are constructed out of 8"x8"x16" concrete masonry units (CMU). The cores of these CMU units are grouted solid. There is no vertical reinforcement in the CMU above ground. The CMU in the shear walls have a compressive strength of 1900 psf, and are normal weight (145 pcf). I calculated the modulus of elasticity to be 2,512 ksi based on this weight and strength. There are a total of 6 shear walls spanning in the east west direction, and 7 shear walls spanning in the north south direction. Most of this report deals with analyzing the east west shear walls since there is a great east west shear force than north south due mainly to the layout of the building. The east west shear walls are labeled with their above ground heights on the following picture. -2-

3 1a " +/ /8" 30 10" /8" 9 4" SHEAR WALL A a BUILDING CENTROID B /4" b C D E /4" c /4" F BUILDING FOOTPRINT W/ SHEAR WALLS d Wind Loads Wind analysis was performed using ASCE 7 98 for reference. The wind speed listed on the plans is 70 MPH. this wind speed is the fastest mile wind speed. To calculate wind loads in the IBC you must use the three second gust wind speed. This three second gust speed is found from a table in the IBC, and depends entirely on the fastest mile wind speed. The three second gust speed for the f.w. olin center is 85 MPH. The drawing below -3-

4 shows the adjusted loading diagram, with the actual loading numbers listed. Wind pressure windward (psf) Wind pressure leeward psf Seismic Loads Seismic loads were determined by using the international building code 2000 edition. The response modification coefficient was found from a table in the IBC and equals 2.5. The seismic factor is found in a table in the ibc -4-

5 and is In my first technical report i overestimated the overall weight of the building. These caused my Seismic loads to be extremely high. For this report the building weight used is 10,000 kips and produces a seismic base shear of 833 kips. based on ibc 2000 the overall base shear is distributed to each floor based on the heighth of the floor as well as the overall weight of each floor. These floor shears have been calculated and listed below, Assuming that each floor is relatively the same weight save the smaller mechanical penthouse kips kips MECH. PENTHOUSE kips SECOND FLOOR kips FIRST FLOOR GROUND FLOOR BASEMENT SEISMIC SHEAR According to LRFD eq A4 6 the maximum lateral load factors are either 1.3 wind or 1.0 Earthquake. After applying these factors seismic loads are still the controlling shear forces. -5-

6 Distribution of Forces The f.w. Olin center has a rigid diapram. What this basically means is that under shear forces the building has an overall displacement, instead of having a greater displacement in between shear walls and virtually no displacement at the shear walls. In a rigid diaphram the lateral loads are distributed from the exterior walls to through the floor slab into the shear walls. The size of lateral load in each shear wall is directly related to the stiffness (k) of each wall walls. The stiffness of the main shear walls are listed in the following table. These stiffnesses were determined by taking the inverse of the wall deflection under a unit load of 1 KIP. This deflection as well as the true deflections under loading were calculated from the equation delta = (ph 3 / 3Ei) + (2.78 Ph / A w E) Since CMU walls are not completely homogenous boxes a factor of 0.75 was applied to the calculated moment of inertias. Many of these shear walls have exceptionally high stiffness values due to the fact that these walls have very low height over length ratios. These stiffnesses are listed below. Wall Height Area (in 2 ) Inertia(t*l^3/12) (in 4 ) flexure shear total k (kip/in) A E E E B E E E C E D E E E E E F E E E Due to the relative low stiffness of shear walls C and D, they -6-

7 support an miniscule amount of the shear force on the first 3 levels. This percentage of force they need to support is so low that i have just assumed they carry zero force on the first three stories. These shear walls are neccessary since they are the only shear walls that rise all the way to the roof. This means that though shear walls c and d do not play a factor in the lower story shear loads, they must support the entire roof shear load. The shear walls and the amount of shear load they support for each floor are shown in the following table. First Floor Second Floor Mech. Penthouse Roof Wall % of (kips) % of (kips) % of (kips) % of (k) A B C D E F Torsional Forces must also be considered due to the fact that the center of the building is not the same as the centroid of the buidling (which is a center point based on the stiffnesses of the shear walls). From assuming a homogenous building weight the center of mass has been calculted to be 12 3" away from the previously determinded centroid. The Delta torsion determined below is the ultimate displacement of each shear wall due to the base shear load of 833 kips. This load is well above the actual loads on the walls, so the real displacements due to torsional bending are much smaller and negligible. -7-

8 Wall Dist. From Centroid (d i ) Dist. From Centroid to Center of Mass (e) k i *d i 2 k i *d i torsion k i *d i /Σ(k i *d 2 i ) F itorsion A (+) 62' 2" 12' 3" E E-05 B (+) 7' 9" 12' 3" E E E-06 C (-) 9' 2" 12 '3" E E-06 D (-) 17' 11" 12 '3" E E-06 E (-) 32' 2" 12 '3" E E E-05 F (-) 92' 7" 12 '3" -8.9E E E-05 Story Drift Story Drift was checked much in the same way used to previously find the stiffnesses. The stiffnesses for each floor was summed. Then the story shear, which is the controlling seismic forces, was divided by this sum of stiffnesses. This number yeilded the story displacement. These displacements are very low and acceptable in all cases due to the large stiffnesses. This makes sense, since the F.W. Olin Center is a shorter building with very long shear walls. The table below shows the calculated story drifts. Story Drift Story Sum of Stiffnesses Story (kips) Story Drift Allowable Drift (h/400) Acceptable First YES Second YES Mech. Penthouse YES Roof YES -8-

9 Strength Check the strength of the shear walls was checked based on the the stress in the walls caused by the shear loadings. I chose to check the north shear wall a since it holds the greatest shear loads. A interesting feature of CMU walls is that in shear the grout fails before the concrete. This occurs so that cracks can be easily repaired. Though the compressive strength of the grout was listed in the building specifications, I was unable to determine any information on the flexure strength or dealing with bond strength or ASTM E 518. From other projects numbers for flexure strength and keeping in mind that type s mortar has a higher flexure strength than other types i estimated the shear strength of the mortar to be 2500 psi. The maximum force in this wall is kips. The area of grout in the wall is found by multiplying the width of the CMU (8") by the depth of the grout (3/8") by the number of CMUs for the story (approx. 20). This gives you an area of grout of 60 in 2. This means that the maximum shear pressure on shear wall A is 2262 psi (135.7 kips / 60 in 2 ). This pressure is less than the allowable shear strength of mortar (2500 psi) so Wall A, and therefore all the shear walls, is able to withstand the shear forces on the building. Concerns A major concern with this building is the central tower and telescope dome. These two entities rise slightly above the rest of the building and therefore would be subjected to extra lateral loads. Since the tower s -9-

10 wall runs the course of the building and is anchored by a continuous footing it could be considered a shear wall. This being the case, i do not know how to analyze a circular shear wall. However, I did simplify the circular tower into a square tower with 40 walls and then check it in the same manner i checked shear wall A. This simplified version of the tower was easily able to withstand the lateral loading, but the ability of shear resistance of the true tower shape must be examined. Conclusion From the above analysis one can conclude that the F.W. Olin Center has a sound lateral system. The building design turned out to be acceptable for the story drift and member check. The major concern is that the mortar shear strength was approximated. If this strength is actually just a few hundred psi lower some walls will not be able to withst and maximum seismic loads. \Note: Many calculations are omitted from this report. These calculations are in the possession of Daniel Chwastyk and may be obtained upon request -10-