UBC Shear Wall Design
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1997 UBC Earthquake Design 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,
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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. 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. 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: It is assumed that the walls perpendicular to the ground motion span vertically between the roof diaphragm and foundation.
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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.
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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. 8. Overturning check on shearwalls: 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. Consider overturning of the short walls due to seismic forces acting in the transverse building direction.
9. Story drift, D: According to '97 UBC 1630.10.2, the maximum in elastic displacement DM, should not exceed .025h for structures with T < .7 seconds. A serviceability consideration. DM is a strength level inelastic displacement due to a design earthquake. DM = .7RDS where R is from Tabe 16-N and DS is static story drift due to strength-level forces. Consider the following calculation for the 50' shear wall:
9/16/2010 12:13 PM
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9/16/2010 12:13 PM
1997 UBC Earthquake Design 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,
9/16/2010 12:13 PM
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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. 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. 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: It is assumed that the walls perpendicular to the ground motion span vertically between the roof diaphragm and foundation.
9/16/2010 12:13 PM
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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.
9/16/2010 12:13 PM
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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. 8. Overturning check on shearwalls: 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. Consider overturning of the short walls due to seismic forces acting in the transverse building direction.
9. Story drift, D: According to '97 UBC 1630.10.2, the maximum in elastic displacement DM, should not exceed .025h for structures with T < .7 seconds. A serviceability consideration. DM is a strength level inelastic displacement due to a design earthquake. DM = .7RDS where R is from Tabe 16-N and DS is static story drift due to strength-level forces. Consider the following calculation for the 50' shear wall:
9/16/2010 12:13 PM
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