Types of Shear Walls. • Shear Wall Stiffness. • Maximum Reinforcement
Requirements. • Shear Strength. • Example: simple building. Example: simple
building.
Shear Walls • • • • • •
Types of Shear Walls Shear Wall Stiffness Maximum Reinforcement Requirements Shear Strength Example: simple building Shear walls with openings
Shear Walls
1
Shear Walls _____________ (unreinforced) shear wall (1.17.3.2.2): Unreinforced wall _____________ (unreinforced) shear wall (1.17.3.2.3): Unreinforced wall with ith prescriptive i ti reinforcement. i f t Structurally connected floor and roof levels
Within 16 in. of top of wall
40db or 24 in. ≤ 8 in.
≤ 8 iin. Corners and end of walls
≤ 16 iin.
Control joint
≤ 10 ft.
Joint reinforcement at 16 in. o.c. or bond beams at 10 ft. Reinforcement not required at openings smaller than 16 in. in in either vertical or horizontal direction
Reinforcement of at least 0.2 in2 Shear Walls
2
Shear Walls _____________ reinforced shear wall (1.17.3.2.4): Reinforced wall with prescriptive reinforcement of detailed plain shear wall. ______________ reinforced shear wall (1.17.3.2.5): Reinforced wall with prescriptive reinforcement of detailed plain shear wall. Spacing of vertical reinforcement reduced to 48 inches. ___________ reinforced shear wall (1.17.3.2.6): 1. 2. 3. 4. 5.
Maximum spacing of vertical and horizontal reinforcement is min{1/3 length of g of wall,, 48 in. [24 [ in. for masonryy in other than running g bond]}. ]} wall,, 1/3 height Minimum area of vertical reinforcement is 1/3 area of shear reinforcement Shear reinforcement anchored around vertical reinforcing with standard hook Sum of area of vertical and horizontal reinforcement shall be 0.002 times gross cross sectional area of wall cross-sectional Minimum area of reinforcement in either direction shall be 0.0007 times gross cross-sectional area of wall [0.0015 for horizontal reinforcement for masonry in other than running bond].
Shear Walls
3
Minimum Reinforcement of Special Shear W ll Walls Reinforcement Ratio 0.0007
0.0010
0.0013
8 in. CMU wall As (in2/ft)
Possibilities
0.064
#4@32 #5@56
12 in. CMU wall As (in2/ft)
Possibilities
0.098
#4@24 #5@32 #6@48
0.092
@ #4@24 #5@32 #6@40
0.140
#4@16 @ #5@24 #6@32
0.119
#4@16 #5@32 #6@40
0.181
#5@16 #6@24
Use specified dimensions, e.g. 7.625 in. for 8 in. CMU walls. Shear Walls
4
Special Shear Walls: Design Minimum strength, strength design (1.17.3.2.6.1.1): Design shear strength, Vn, greater than 1.25 times nominal flexural strength strength, Mn, except Vn need not be greater than 2.5Vu.
Shear Walls
5
Seismic Design Category Seismic Design Category
Allowed Shear Walls
A or B C D and higher Shear Wall
Response modification factor: Seismic design force divided by response modification factor, which accounts for ductility and energy absorption. Knoxville, Tennessee Category C for Use Group I and II Category D for Use Group III (essential facilities
R
Ordinary plain
1.5
Detailed plain
2
Ordinary reinforced
2
Intermediate reinforced
3.5
Special reinforced
Shear Walls
5
6
Shear Walls: Stiffness
h
d
h/d < 0.25
0.25 < h/d < 4.0
_____ stiffness predominates
Both shear and bending stiffness are important
h/d >4.0
_______ stiffness predominates
Shear Walls
7
Coupled Shear Walls
________ Shear Sh W Wallll
___________ Shear Wall Shear Walls
8
Shear Wall Stiffness 3.1.5.1 Deflection calculations shall be based on cracked section properties. Assumed properties shall not exceed half of gross section properties, unless a cracked-section analysis is performed. Cantilever wall
cant
kcant
Et 2 h h 4 3 L L
Fi d wallll (fixed against rotation at top) Fixed
fixed
k fixed
h = height of wall Av = shear area; 5/6A for a rectangle G = shear modulus; given as 0.4E (1.8.2.2.2) t = thickness hi k off wallll L = length of wall
Et 2 h h 3 L L
Real wall is probably between two cases; diaphragm provides some rotational restraint, but not full fixity. Shear Walls
9
T- or L- Shaped Shear Walls Section 1.9.4 Wall intersections designed either to: a) ____________________: b) ____________________ Connection that transfers shear: (must be in running bond) a) Fifty percent of masonry units interlock b) Steel connectors at max 4ft. c) Intersecting bond beams at max 4 ft. ft Reinforcing of at least 0.1in 0 1in2 per foot of wall
Metal lath or wire screen to support grout
2-#4’s
1/ in. x 11/ in. x 28in. 4 2 with 2in. long 90 deg bends at each end to form U or Z shape
Shear Walls
11
Effective Flange Width (1.9.4.2.3) Effective flange width on either side of web shall be smaller of actual flange width, distance to a movement joint, or: • Flange in compression: 6t • Flange in tension: • Unreinforced masonry: 6t • Reinforced masonry: 0.75 times floor-to-floor wall height
Shear Walls
12
Maximum reinforcing (3.3.3.5) No limits on maximum reinforcing for following case (3.3.3.5.4): Mu 1 and R 1.5 Squat walls, not designed for ductility Vu d v In other cases, can design by either providing boundary elements or limiting reinforcement. Boundary element design (3.3.6.5): More difficult with masonry than concrete Boundary elements not required if:
Pu 0.1 f m A g
geometrically symmetrical sections
Pu 0.05 f m A g
geometrically unsymmetrical sections
AND Mu 1 Vu lw
OR
Vu 3 An f m
Shear Walls
AND
Mu 3 Vu lw 13
Maximum reinforcing (3.3.3.5) Reinforcement limits: Calculated using Maximum stress in steel of fy Axial forces taken from load combination D+0.75L+0.525Q D 0.75L 0.525QE Compression reinforcement, with or without lateral ties, permitted to be included for calculation of maximum flexural tensile reinforcement
Uniformly distributed reinforcement = 1.5 ordinary walls = 3 intermediate walls = 4 special walls
mu P 0.64 f m b d As y v mu dv f y y mu y mu mu P 0.64 f m b bd As mu y d bd f y min mu mu y , y Es d
Compression steel with area equal to tension steel
Shear Walls
14
Maximum reinforcing Consider a wall with uniformly distributed steel: y Cm Cs Ts P y mu
Strain fy
0.8f’m
Steel in tension
Stress
Steel in compression
mu Cm 0.8 f m 0.8 d v b mu y y y y 1 y Ts f y As 2 mu y y y Portion of steel in tension
Yielded steel
Elastic steel
mu mu y 1 y Cs f y As 2 mu mu y mu mu mu 0.5 y f y As mu y mu
As taken as total steel dv is actual depth of masonry Shear Walls
mu 0.5 y f y As mu y
15
Maximum reinforcing Ts Cs Cm P y 0.5 y 0.5 y mu f y As mu f y As y Ts C s f y As mu y mu y mu y y mu 0.8 f m 0.8 mu Ts Cs f y As d v b P Cm P y mu mu y
mu P 0.64bf m d As y v mu dv mu fy y mu y Shear Walls
16
Maximum reinforcing Maximum steel on a per foot length of wall basis given for 8 in. wall
0.03
0.06
3 2.5
CMU f'm =2ksi
Clay f'm =2ksi
0.015
1.5
CMU f'm =1.5ksi
0.01 0.005
4
0.04 2
0.02
2
As/(bdv)
Clay f'm =4ksi
3.5 3
0.03
2.5
CMU f'm =2ksi
1
0.02
0.5
0.01
2 15 1.5 1 0.5
CMU f'm =1.5ksi
0
0
0
0.05
0.1
0.15
0
0.2
0
0
[P/(bd)]/f'm
0.05
0.1
0.15
0.2
0.25
0.3
[P/(bd)]/f'm
Shear Walls
5 4.5
Clay f'm =2ksi
As (in /ft)
As/(bdv)
0.025
Clay f'm =4ksi
s=2 y 0.05
17
2
s=4 y
As (in /ft)
0.035
Shear Strength (3.3.4.1.2) 0.8
Vn ___ ____
M Vm 4.0 1.75 u An Vu d v A Vs 0.5 v f y d v s Maximum Vn is:
Mu/Vudv need d nott be b ttaken k >1 1.0 0 Pu = axial load
f m 0.25Pu
Vertical reinforcement shall not be less than onethird horizontal reinforcement; reinforcement shall be uniformly distributed, max spacing of 8 ft (3.3.6.2)
Vn 6 An
f m
( M u / Vu d v ) 0.25
Vn 4 An
f m
( M u / Vu d v ) 1.0
Interpolate for values of Mu/Vudv between 0 0.25 25 and 1 1.0 0 M 4 Vn 5 2 u An f m Vu d v 3 Shear Walls 18
Example Given: 10ft. high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, f’m=1500psi; 2-#5 each end; #5 at 48in. (one space will actually be 40 in.); superimposed dead load of 1kip/ft. R Required: i d Maximum M i wind i d lload d based b d on fl flexurall capacity; it shear h steel t l required to achieve this capacity. Solution: Check capacity in flexure using 0.9D+1.0W load combination. Weight of wall: 38 psf(10ft) = 0.38 k/ft (Wall weight from lightweight units, grout at 48 in. o.c.)
Pu=0.9(1k/ft+0.38k/ft)=1.24k/ft 0 9(1k/ft 0 38k/ft) 1 24k/ft
T t l 1.24k/ft(16ft) Total: 1 24k/ft(16ft) = 19 19.88kips 88ki
Shear Walls
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Example Solve for stresses, forces and moment in terms of c, depth to neutral axis. T1
T2
T3
T4
C2
c
C1
8 in
52 in 92 in 140 in 188 in
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Example Solve for stresses, forces and moment in terms of c, depth to neutral axis. T1
T2
T3
T4
C2
c
C1
8 in
52 in 92 in 140 in 188 in
• Use trial and error to find c such that Pn = 19.88 kips/0.9 = 22.09 kips • Pn = C1 + C2 – T1 – T2 – T3 – T4 • c = 28.2 inches • Sum moments about middle of wall (8 ft from end) to find Mn • Mn = 1123 kip kip-ft; ft; Mu = 0.9(1123 0 9(1123 kip kip-ft) ft) = 1010 kip kip-ft ft • Wind = Mu/h = (1010 kip-ft)/(10ft) = 101 kips Shear Walls
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Example Calculate net area, An, including grouted cells.
Net area
Net area:
M u / Vu d v Vn 4 An
Maximum Vn
f m
M u / Vu d v 0.25
M / V d 0.25 Vn 6 2 u u v An 1.0 0.25
Vn 6 An
f m
M u / Vu d v 1.0
f m
Vu 0.8Vn 0.8132.6kips 106.1kips
Maximum Vu
Shear Walls
23
Example M Vm 4.0 1.75 u An f m 0.25 Pu Vu d v
Pu = 0.9(1k/ft) = 0.9k/ft Total: 0.9k/ft(16ft) = 14.4 kips
Vs Vn Vm 101k / 0.8 80.7 k 45.6k Use #5 bars in bond beams beams.
A Vs 0.5 v f y d v s
s
Determine spacing spacing.
0.5 Av f y d v Vs
Use 4 bond beams, spaced at 32 in. o.c. vertically. Actual construction has 15 courses: Bond beam in course 4, 8, 12, and 15 from bottom Shear Walls
25
Example: Maximum Reinforcing 3.3.3.5 No limit to reinforcement since Mu/(Vudv) < 1. If we needed to check maximum reinforcing, g, there are at least two ways. y Check on axial load: a) Set steel strain to limit and find c b) Find axial load for this c c) If applied axial load is less than axial load, OK
Steel strain = 1.5y = 0.00310 c = 0.0025/(0.0025+0.00310)*192 0 0025/(0 0025+0 00310)*192 = 85 85.7 7 in in. From spreadsheet, Pu = 181 kips OK
Find steel strain for given axial load: a)) Through Th h ttrial i l and d error ((or solver), l ) find c for given axial load b) Determine steel strain c)) If steel strain g greater than limit,, OK
P = 22 22.08 08 kips (Dead load) c = 28.46 in. s = (192-28.5)/28.5*0.0025 = 0.0143 = 6.93y OK
Note: did not include compression steel which is allowed when checking max reinforcing
Shear Walls
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Example: Issues with Strength Design If this were a special shear wall: • Vn ≥ shear corresponding to 1.25Mn (1.17.3.2.6.1.1) • 1.25Mn = 1.25(1123) = 1404k-ft • Vn = Mn/h = 1404k-ft/10ft = 140.4kips • Vn = 140.4k/0.8 = 175.4kips • But, maximum Vn is 132.6 kips (76% of required). Cannot build wall. Need to take out some flexural steel or fully grout to get greater shear capacity.
Shear Walls
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Partially Grouted Shear Walls Recent research has shown that TMS Code equations overestimate capacity of partially grouted shear walls. T Two possible ibl modifications: difi ti •
Only use face shell area, 2.5in(192in) = 480in2 (30% reduction)
•
Multiply An by An/Ag, an empirical reduction
Shear Walls
(53% reduction)
29
Example Given: Wall system constructed with 8 in. grouted CMU (Type S mortar). Required: Determine the shear distribution to each wall
112in
Solution: S l ti U gross Use properties (will not make any difference for shear distribution). distribution)
40in
40in
64in
40in
40in
6t=48in
56in
Elevation
a
b
c
Plan Shear Walls
30
Example Pier b
kb
Et h h 4 3 L L 2
E 7.62in 0.286inE 2 112 in 112 in 3 4 64 in 64 in
Piers a and c Determine stiffness from basic principles. A Find centroid from outer flange surface
y I
Av Awebb Shear Walls
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Example k a ,c
1 P P 0.157inE 3 3 Ph Ph 112in 112in 3EI Av G 3E 86000in 4 305in 2 0.4 E
Portion of shear load = stiffness of pier over the sum of the stiffnesses.
Va
ka
k
V
0.157inE V 0.26V 0.157inE 0.286inE 0.157inE
Vb = 0.48V
Deflection calculations: k = (0.600in)E = (0.600in)(1350ksi) = 810kip/in Reduce by a factor of 2 to account for cracking: k = 405kip/in
Shear Walls
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Example Required: Design the system for a horizontal earthquake load of 35 kips, a dead load of 1200 lb/ft length of building, a live load of 1000 lb/ft length of building, and a vertical earthquake force of 0.2 times the dead load. Use a special reinforced shear wall (R>1.5). Solution: Use strength design. Check deflection.
P 35k 0.086in k 405k / in
0.086in 0.000768 h 112in 112 in
or
Shear Walls
h 1302
34
Example Wall b: Vub = 0.48Vu = 0.48(35k) = 16.8k D = 1.2k/ft(_____________)(1ft/12in) + 0.075ksf(112in)(64in)(1ft2/144in2) = 14.1kips Tributary width
Use load combination 0.9D+E Mu = 16.8k(112in)(1ft/12in) = 156.8k-ft
Pu = 0.9(14.1k) - 0.2(14.1k) = 9.87k
Minimum prescriptive reinforcement for special shear walls: Sum of area of vertical and horizontal reinforcement shall be 0.002 times gross cross-sectional area of wall Minimum area of reinforcement in either direction shall be 0.0007 times gross cross-sectional area of wall Use prescriptive reinforcement of (0.002-0.0007)Ag = 0.0013Ag in vertical direction As = 0.0013(64in)(7.625in) = 0.63in2
Shear Walls
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Example: Flexural Steel Use load combination 0.9D+E
Mu = 156.8k-ft
Pu = 9.87k
Try y 2-#4 each end;; 1-#4 middle: As=1.00in2 Middle bar is at 28 in. (8 cells, cannot place bar right in center) From spreadsheet: Mn=150.5k-ft NG N t used Note: d solver l tto fifind d value l
Try 1-#6 each end; 1-#6 middle: As=1.32in2 Mn=183.4k-ft
OK
Try 2-#5 each end; 1-#5 middle: As=1.55in2 Mn=214.1k-ft
OK
Use 2-#5 each end; 1-#5 middle: As=1.59in2
Shear Walls
36
Example: Maximum Reinforcing Check maximum reinforcing: Axial force from load combination D+0.75L+0.525QE L 1k/ft(104in)(1ft/12in) = 8.67kip L=1k/ft(104in)(1ft/12in) 8 67kip D+0.75L+0.525QE = 14.1k+0.75(8.67k)+0.525(0) = 20.60kip For maximum reinforcing calculations, compression steel can be included even if nott laterally l t ll supported t d For #5 bars and P = 20.60kip, c = 7.77in. (from solver seek on spreadsheet)
s
d kd 60in 7.77in emu 0.0025 0.0168 8.12 y 4 y kd 7.77in
OK Shear Walls
37
Example: Shear M u Vu 112in 1.75 1 Vu d v Vu 64in
Use Mu/Vudv = 1.0
Find Pu neglecting wall weight Pu = 0.9(10.4k) - 0.2(10.4k) = 7.28k
Vnm 4.0 1.75M u / Vu d v An f m 0.25Pu 2.25(7.625in)(64in) 1500 psi 1k / 1000lb 0.257.28k 44.3kips
Vn 44.3kips p
Ignore any shear steel:
Vn 0.844.3kips 35.5kips
OK
Shear Walls
38
Example: Shear Mn = Mn/ = 214.1k-ft/0.9 = 237.9k-ft V corresponding to Mn: [237.9k-ft/(112in)](12in/ft) = 25.5kips 1.25 times V: 1.25(25.5k) = 31.9kips < Vn=35.5kips
OK
Minimum prescriptive horizontal reinforcement: 0.0007Ag As = 0.0007(64in)(7.625in) = 0.34in2 Maximum spacing is 48 in in. Use 3 bond beams, 40in. o.c. (top spacing is only 32in.)
Wall b: Vertical bars: Horizontal bars:
2-#5 each end; 1-#5 middle #4 in bond beam at 40 in. o.c. (Total 3-#4)
Shear Walls
39
Example Wall a: Vua = 0.26Vu = 0.26(35k) = 9.1k
D = 1.2k/ft(20in+40in)(1ft/12in) + 0.075ksf(112in)(40in+48in)(1ft2/144in2) = 11.1kips
Use load combination 0.9D+E Mu = 9.1k(112in)(1ft/12in) = 84.9k-ft
Pu = 0.9(11.1k) - 0.2(11.1k) = 7.77k
Minimum prescriptive reinforcement: 0.0013Ag As = 0.0013(88in)(7.625in) = 0.87in2
Design issues:
Flange in tension: easy to get steel; problems with max Flange in compression: hard to get steel; max is easy Steel in both flange and web affect max Need to also consider loads in perpendicular direction Need to check shear at interface of flange and web
Shear Walls
40
Example Use load combination 0.9D+E
Mu = 84.9k-ft
Pu = 7.77k
Axial force for max from load combination D+0.75L+0.525QE L=1k/ft(60in)(1ft/12in) = 5.00kip 5 00kip D+0.75L+0.525QE = 11.1k+0.75(5.00k)+0.525(0) = 14.85kip To be consistent with other wall wall, try #5 bars Try 3 in the flange (either distributed as shown, or one in the corner, and two in the jamb Try 2 in the web jamb
Flange in tension: Mn=146.4k-ft
Max reinforcing check: s=0.0101=4.86y
OK
Note: with 2 2-#5 #5 in flange flange, Mn=104.8k-ft, =104 8k ft but might need 3 bars for perpendicular direction
Flange in compression: Mn=124.3k-ft Max reinforcing check: s=0.0417=20.2y Shear Walls
OK 41
Example: Shear M u Vu 112in 2.8 1 Vu d v Vu 40in
Use Mu/Vudv = 1.0
Vm 2.25(7.625in)(40in) 1500 psi 1k / 1000lb 0.254.2k 27.6kips Ignore any shear steel:
Vn 27.6kips
Vn 0.827.6kips 22.1kips
Mn = Mn/ = 146.4k-ft/0.9 = 162.7k-ft V corresponding to Mn: [162.7k-ft/(112in)](12in/ft) = 17.4kips 1 25 times V: 1 1.25 1.25(17.4k) 25(17 4k) = 21 21.8kips 8kips < Vn=22.1kips =22 1kips
OK
OK
Minimum prescriptive horizontal reinforcement: 0.0007Ag Use 3 bond beams beams, 40in 40in. o o.c. c (top spacing is only 32in 32in.))
Walls a,c:
Vertical bars: Hori ontal bars Horizontal bars:
5-#5 as shown in previous slide #4 in bond beam at 40 in in. o.c. o c (Total 3 3-#4) #4) Shear Walls
42
Example Section 1.9.4.2.4 requires shear to be checked at interface. Check intersection of flange and web in walls a and c. Approximate shear force is tension force in flexural steel.
Vu As f y 3 0.31in 2 60ksi 55.8kips
Conservatively take Mu/Vudv=1.0, 1.0, and Pu=0. 0.
Vm 4.0 1.75M / Vd v An f m 0.25 Pu 4.0 1.751.0(7.625in)(112in) 1500 ppsi 1k / 1000lb 74.4kips p
Vn 0.874.4kips 59.5kips
OK
Shear Walls
43
Example: Drag Struts Distributed shear force:
Vu 35k 12in 1.875k / ft L 224in 1 ft
Look at top of wall: 1.88k/ft 9.1k 3 33f 3.33ft
1.88k/ft 88 / 2.8k Drag strut 3 33f 3.33ft
1.88k/ft 88 / 3.4k
16.8k 5.33ft 33f
1.88k/ft 88 / 3.4k
Drag strut 2.8k
3 33f 3.33ft
Shear Walls
1.88k/ft 88 / 9.1k 3 33f 3.33ft
44
Shear Walls: Building Layout 1. All elements either need to be isolated, or will participate in carrying the load 2. Elements that participate in carrying the load need to be properly detailed for seismic requirements 3. Most shear walls will have openings 4. Can design g only yap portion to carry y shear load, but need to detail rest of structure
Shear Walls
45
Shear Walls with Openings Shear walls with openings will be composed of solid wall portions, and portions with piers between openings. Piers: Pi • Requirements only in strength design • Length ≥ 3 x thickness; width ≤ 6 x thickness • Height < 5 x length • Factored axial compression ≤ 0.3Anf’m (3.3.4.3.1) • One bar in each end cell (3.3.4.3.2) • Minimum reinforcement is 0 0.0007bd 0007bd (3.3.4.3.2) (3 3 4 3 2)
Shear Walls
46
f
h 3 2kt kt kb 2kb 3 12h 1 6 EI kt 2kt kb kb 5 EA
kt 1 kb M t Vh k 2 k k k t b b t kb 1 kt M b Vh k t 2k t kb k b As the top spandrel decreases in height, the top approaches a fixed condition against rotation.
kt
kb
h ht h hb
Top spandrel
h
Divide wall into piers. Find flexibility of each pier Stiffness is reciprocal of flexibility Di t ib t load Distribute l d according di tto stiffness tiff
hb
1. 2. 3. 4 4.
ht
Shear Walls with Openings
Bottom spandrel
If hb = 0 h 3 2 kt 12h 1 f 6 EI 2kt 1 5 EA kt M t Vh 2 k 1 t
1 kt M b Vh 2 1 k t
Qamaruddin, M., Al-Oraimi, S., and Hago, A.W. (1996). “Mathematical model for lateral stiffness of shear walls with openings.” Proceedings, Seventh North American Masonry Conference, 605-617. Shear Walls 47
Example - Shear Walls with Openings Required: A. Stiffness of wall B. Forces in each pier under 10 kip horizontal load
4ft 4
4fft
Given:
B
C
8ft
A
2ft
4ft
2ft
6ft
2ft
Shear Walls
48
Example - Shear Walls with Openings Sample calculations: Pier B h = 4ft I
f
ht = 4ft
hb = 8ft kt = h/ht = 4ft/4ft = .0
tL3 12
kb = h/hb = 4ft/8ft = 0.5
A tL 2 ft t
h 3 2kt kt kb 2kb 3 12h 1 6 EI kt 2kt kb kb 5 EA
k
1 1 0.0210 Et f 47.6 / Et
Shear Walls
49
Example - Shear Walls with Openings Pier h (ft) ht (ft) hb (ft) kt
kb
f
I (ft3) A (ft)
s
f
k
A
12
4
0
3
inf 0.667t
2t
308.4 18 326.4/Et 0.0031Et
B
4
4
8
1 0.5 0.667t
2t
41.6
6
47.6/Et 0.0210Et
C
4
4
8
1 0.5 0.667t
2t
41.6
6
47.6/Et 0.0210Et
f is flexural deformation; s is shear deformation Total stiffness is 0.0451Et Average stiffness from finite element analysis 0.0440Et Solid wall stiffness 0 0.1428Et 1428Et (free at top); 0 0.25Et 25Et (fixed at top)
Shear Walls
51
Example - Shear Walls with Openings Sample calculations: Pier B
Vb
kb
k
V
0.0210 Et 10k 4.66kips 0.0451Et
kt 1 kb M t Vh k k k k 2 t b b t kb 1 kt M b Vh kt 2kt k b k b
Pier
V (kips)
Mt (k-ft)
Mb (k-ft)
A
0.68
3.50
4.66
B
4.66
11.19
7.46
C
4.66
11.19
7.46
Shear Walls
52
Example - Shear Walls with Openings To find axial force in piers, look at FBD of top 4ft of wall.
Moment direction on piers
4ft
10k
0.68k 3.50k-ft 2ft
4.66k 11.19k-ft
4ft
2ft
6ft
4.66k 11.19k-ft 2ft
This 65.9k-ft moment is carried by axial forces in the piers. Axial forces are d t determined i d ffrom M Mc/I, /I where h each h pier i iis considered id d a concentrated t t d area. Find centroid from left end.
y
y A A i
i
i
Shear Walls
54
Example - Shear Walls with Openings The axial force in pier A is: 4.45k FBD’s FBD s of piers:
3.50k-ft 0.68k
0.45k
11.19k-ft 4.66k
B A
4.66k 0.45k 7.46k-ft
4.90k
11.19k-ft 4.66k
C 4.66k 4.90k 7.46k-ft
0.68k 4.45k 4.66k-ft Shear Walls
56
Example - Shear Walls with Openings FBD’s of entire wall to obtain forces in bottom right spandrel: 10k
F
x
10k 0.68k Fx 0 Fx 9.32k
F
y
4.45k Fy F 0 Fy 4.45k
A
B
C
y
z
10k (16 ft ) (4.66k ft )
4.45k (1 ft ) 4.45k (11 ft ) Mz 0 Mz 110.8k ft
x
Z
0.68k 4 66k ft 4.45k 4.66k-ft
M
Fy=4.45k
Fx=9.32k M 110 8k ft Mz=110.8k-ft Shear Walls
58