1030 - Embedded Plate Design - 4_s7.00 2l4r8
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- Pages: 9
1030 - UD ISE Building
EMBEDDED PLATE DESIGN - 4/S7.00 (REF: ANCHORING TO CONCRETE - ACI Appendix D) LAY-OUT:
1 1/2"
Connection Plate/Angle to be designed separately
7@6 "
dm = 24"
Ax2 d = 11.14"
V
Ax1 Ay
1 1/2"
A
Stress distribution per Case II (AISC 13th Ed. Manual p7-12) This Connection plate/Angle justifies the assumed stress distribution as it stiffens the embed plate PL1" - A36
1 1/2"
6"
1 1/2"
16 - 3/4in. dia. ASTM A108 Grade 1015 or 1020 (6" long min)
Page 1 of 9
1030 - UD ISE Building
REQUIREMENT: A, Axial Load = V, Shear Load = Ax2, Axial Load =
80.00 35.00 60.00
kips <-- From brace kips <-- From beam kips <-- From beam
CALCULATION: Brace Slope: Ay
=
1.0
80 (1) √ [ (1) ² + (0.82) ² ]
= Ax1
=
/ 0.82
61.9
kips
80 (0.82) √ [ (1) ² + (0.82) ² ] 50.73
=
kips
Discussion: Since the diagonal and the horizontal share the same connection, (forces are concurrent) there will be a force interaction. Thus, Ax Ax
= =
max (Ax1, Ax2) kips 60.00
Total Shear, VT = Total Shear, VT =
Ay + V 96.86
kips <- See attached sheet for the calculation.
Per AISC, bolts/stud above the neutral axis are subjected to the tensile force and the effect of the prying action Tensile Load, T =
Ax * d / dm where: Ax = d = dm =
Tensile Load, T =
27.85
60.0 11.14 24
kips in in
kips <- See attached sheet for the calculation.
The thickness required to eliminate prying action is tmin
=
√
4.44 T b' p Fu
where: T= T=
required strength 27.85 kips
b' = g - d - tp where: g = stud's gage = 6.0 in
Page 2 of 9
1030 - UD ISE Building
d = stud diameter = in 0.75 tp = connection plate thickness = 0.50 in b' = p= p=
2.375
<-- assumed
in
tributary length per pair of studs for a tee 6.0 in
Fu = Tensile strength of embed plate Fu = 65.0 in
tmin
=
0.868
in
<
1.00
in.
(OK)
Page 3 of 9
1030 - UD ISE Building
SHEAR CAPACITY OF STUDS (REF: ANCHORING TO CONCRETE - ACI Appendix D)
Stud Diameter, do :
3/4
in. dia. ASTM A108 Grade 1015 or 1020
Fy
= =
Yield Strength of Stud 51 ksi
Fu
= =
Tensile Strength of Stud 65 ksi
L
= =
Nominal Length of Stud in 6.00
tp
= =
Embed Plate Thickness in 1.00
Ase
= =
Shank Area 0.442 in²
ths
= =
Head Thickness, in 0.375 in
hef
= = =
effective embedment depth L + tp - ths - 1/8 6.500 in
fc'
= =
concrete compresive strength 5 ksi
Steel Strength of Studs in Shear Vsa
=
n * 0.6 * Ase * futa
( ø = 0.65 )
where:
øVsa
=
Ase
= =
effective cross-sectional area of studs in² 0.442
futa
= =
ultimate tensile strength of studs 65 ksi
n
= =
no. of studs 16.00 in
179.19
kips
Concrete Breakout Strength of Studs in Shear Studs are not located near free edge perpendicular to the application of the load. Therefore, breakout for shear is not applicable.
Concrete Pryout Strength of Anchor in Shear
Page 4 of 9
1030 - UD ISE Building
Vg
=
k * Ncbg
ø = 0.70
where: k k
= = =
1.0 for hef < 2.5 in 2.0 for hef > 2.5 in 2.00
Ng
=
(ANC/ANCO) * ψec,N * ψed,N * ψc,N * ψ,N * Nb
ψec,N
= =
eccentricity factor 1 <1 1 + (2eN'/3hef) eN'
=
in
0.00
ψec,N
=
1.000
ψed,N
=
1.0
=
0.7 + 0.3
if (Ca,min > 1.5hef)
ca,min
Ca,min 1.5hef
if Ca,min < 1.5hef
15.00
=
in
ψed,N
=
1.000
ψc,N
=
1.000
(cracked concrete)
ψ,N
=
1.000
(for cast-in anchors in cracked concrete)
Nb
=
kc
√fc' (hef)1.5 kc
Nb
=
ANCO
= = =
=
28123.3
24
(for cast-in anchors)
lbs
(2*1.5hef) * (2*1.5hef) 9*hef² 380.25 in² x
sx
x
y
sy
y
Page 5 of 9
1030 - UD ISE Building
ANC
=
(2y + sy) (2x + sx) where: y = =
ANC Ng
=
=
min (1.5hef , c'a1) ; c'a1 = 180 in 9.75 in
x
= =
sy
=
42.00
in
sx
=
6.00
in
1568.25
115988
min (1.5hef , c'a2) 9.75 in
c'a2 =
15
in
in²
lbs
therefore, øVg øVg
= =
162383 162.383
lbs kips
SUMMARY: Steel Strength of Studs in Shear Concrete Pryout Strength of Anchor in Shear
Controlling limit state, øV =
162.38
kips
>
179.19 162.38
kips kips
96.86
kips
(OK)
Page 6 of 9
1030 - UD ISE Building
TENSILE CAPACITY OF STUDS (REF: ANCHORING TO CONCRETE - ACI Appendix D)
Stud Diameter, do :
3/4
in. dia. ASTM A108 Grade 1015 or 1020
Fy
= =
Yield Strength of Stud 51 ksi
Fu
= =
Tensile Strength of Stud 65 ksi
L
= =
Nominal Length of Stud in 6.00
tp
= =
Embed Plate Thickness in 1.00
Ase
= =
Shank Area 0.442 in²
Abrg
= =
Bearing Area 0.790 in²
ths
= =
Head Thickness, in 0.375 in
hef
= = =
effective embedment depth L + tp - ths - 1/8 6.500 in
fc'
= =
concrete compresive strength 5 ksi
Steel Strength of Studs in Tension =
Nsa
n * Ase * futa
( ø = 0.75 )
where:
øNsa
=
Ase
= =
effective cross-sectional area of studs 0.442 in²
futa
= =
ultimate tensile strength of studs 65 ksi
n
= =
no. of studs 8.00 in
172.30
kips
Concrete Breakout Strength of Studs in Tension Ng
=
(ANC/ANCO) * ψec,N * ψed,N * ψc,N * ψ,N * Nb
ø = 0.70
where:
Page 7 of 9
1030 - UD ISE Building
=
ψec,N
=
eccentricity factor 1 <1 1 + (2eN'/3hef) eN'
=
0.00
ψec,N
=
ψed,N
=
1.0
=
0.7 + 0.3
in
1.000 if (Ca,min > 1.5hef)
ca,min
Ca,min 1.5hef
=
if Ca,min < 1.5hef
15.00
in
ψed,N
=
1.000
ψc,N
=
1.000
(cracked concrete)
ψ,N
=
1.000
(for cast-in anchors in cracked concrete)
Nb
=
kc
√fc' (hef)1.5 kc
Nb
=
ANCO
= = =
=
28123.3
24
(for cast-in anchors)
lbs
(2*1.5hef) * (2*1.5hef) 9*hef² in² 380.25 sx
x
x
y
sy
y
ANC
=
(2y + sy) (2x + sx)
where: y = = x
=
min (1.5hef , c'a1) ; c'a1 = 180 in in 9.75 min (1.5hef , c'a2)
c'a2 =
15
in
Page 8 of 9
1030 - UD ISE Building
ANC øNg øNg
9.75
in
sy
=
18.00
in
sx
=
6.00
in
956.25
=
49507 49.51
= =
=
in²
lbs kips
Concrete Pullout Strength of Studs in Tension øNpn
=
n * ψc,P * NP
ø = 0.70
where:
øNpn
=
n
= =
ψc,P
=
Abrg
= =
bearing area of studs 0.79 in²
Np
= =
8*Abrg *f'c 31.6
252.8
no. of studs in 8.00 (cracked concrete)
1.0
(for 0.75'' dia studs)
kips
kips
Concrete side-face blowout strength of studs in tension The side-face blowout failure mode must be investigated when the edge distance is less than 0.4 hef edge dist. =
in
15
>
0.4 hef =
2.60
in
(OK)
SUMMARY: Steel Strength of Studs in Tension Concrete Breakout Strength of Studs in Tension Concrete Pullout Strength of Studs in Tension
Controlling limit state, øV =
49.51
kips
>
1.20
(OK)
172.30 49.51 252.80
kips kips kips
27.85
kips
(OK)
Check Interaction Equation Na Nn/Ω
+
Va vn/Ω
<
1.2
27.9 49.5
+
96.9 162
=
1.16
<
Page 9 of 9
EMBEDDED PLATE DESIGN - 4/S7.00 (REF: ANCHORING TO CONCRETE - ACI Appendix D) LAY-OUT:
1 1/2"
Connection Plate/Angle to be designed separately
7@6 "
dm = 24"
Ax2 d = 11.14"
V
Ax1 Ay
1 1/2"
A
Stress distribution per Case II (AISC 13th Ed. Manual p7-12) This Connection plate/Angle justifies the assumed stress distribution as it stiffens the embed plate PL1" - A36
1 1/2"
6"
1 1/2"
16 - 3/4in. dia. ASTM A108 Grade 1015 or 1020 (6" long min)
Page 1 of 9
1030 - UD ISE Building
REQUIREMENT: A, Axial Load = V, Shear Load = Ax2, Axial Load =
80.00 35.00 60.00
kips <-- From brace kips <-- From beam kips <-- From beam
CALCULATION: Brace Slope: Ay
=
1.0
80 (1) √ [ (1) ² + (0.82) ² ]
= Ax1
=
/ 0.82
61.9
kips
80 (0.82) √ [ (1) ² + (0.82) ² ] 50.73
=
kips
Discussion: Since the diagonal and the horizontal share the same connection, (forces are concurrent) there will be a force interaction. Thus, Ax Ax
= =
max (Ax1, Ax2) kips 60.00
Total Shear, VT = Total Shear, VT =
Ay + V 96.86
kips <- See attached sheet for the calculation.
Per AISC, bolts/stud above the neutral axis are subjected to the tensile force and the effect of the prying action Tensile Load, T =
Ax * d / dm where: Ax = d = dm =
Tensile Load, T =
27.85
60.0 11.14 24
kips in in
kips <- See attached sheet for the calculation.
The thickness required to eliminate prying action is tmin
=
√
4.44 T b' p Fu
where: T= T=
required strength 27.85 kips
b' = g - d - tp where: g = stud's gage = 6.0 in
Page 2 of 9
1030 - UD ISE Building
d = stud diameter = in 0.75 tp = connection plate thickness = 0.50 in b' = p= p=
2.375
<-- assumed
in
tributary length per pair of studs for a tee 6.0 in
Fu = Tensile strength of embed plate Fu = 65.0 in
tmin
=
0.868
in
<
1.00
in.
(OK)
Page 3 of 9
1030 - UD ISE Building
SHEAR CAPACITY OF STUDS (REF: ANCHORING TO CONCRETE - ACI Appendix D)
Stud Diameter, do :
3/4
in. dia. ASTM A108 Grade 1015 or 1020
Fy
= =
Yield Strength of Stud 51 ksi
Fu
= =
Tensile Strength of Stud 65 ksi
L
= =
Nominal Length of Stud in 6.00
tp
= =
Embed Plate Thickness in 1.00
Ase
= =
Shank Area 0.442 in²
ths
= =
Head Thickness, in 0.375 in
hef
= = =
effective embedment depth L + tp - ths - 1/8 6.500 in
fc'
= =
concrete compresive strength 5 ksi
Steel Strength of Studs in Shear Vsa
=
n * 0.6 * Ase * futa
( ø = 0.65 )
where:
øVsa
=
Ase
= =
effective cross-sectional area of studs in² 0.442
futa
= =
ultimate tensile strength of studs 65 ksi
n
= =
no. of studs 16.00 in
179.19
kips
Concrete Breakout Strength of Studs in Shear Studs are not located near free edge perpendicular to the application of the load. Therefore, breakout for shear is not applicable.
Concrete Pryout Strength of Anchor in Shear
Page 4 of 9
1030 - UD ISE Building
Vg
=
k * Ncbg
ø = 0.70
where: k k
= = =
1.0 for hef < 2.5 in 2.0 for hef > 2.5 in 2.00
Ng
=
(ANC/ANCO) * ψec,N * ψed,N * ψc,N * ψ,N * Nb
ψec,N
= =
eccentricity factor 1 <1 1 + (2eN'/3hef) eN'
=
in
0.00
ψec,N
=
1.000
ψed,N
=
1.0
=
0.7 + 0.3
if (Ca,min > 1.5hef)
ca,min
Ca,min 1.5hef
if Ca,min < 1.5hef
15.00
=
in
ψed,N
=
1.000
ψc,N
=
1.000
(cracked concrete)
ψ,N
=
1.000
(for cast-in anchors in cracked concrete)
Nb
=
kc
√fc' (hef)1.5 kc
Nb
=
ANCO
= = =
=
28123.3
24
(for cast-in anchors)
lbs
(2*1.5hef) * (2*1.5hef) 9*hef² 380.25 in² x
sx
x
y
sy
y
Page 5 of 9
1030 - UD ISE Building
ANC
=
(2y + sy) (2x + sx) where: y = =
ANC Ng
=
=
min (1.5hef , c'a1) ; c'a1 = 180 in 9.75 in
x
= =
sy
=
42.00
in
sx
=
6.00
in
1568.25
115988
min (1.5hef , c'a2) 9.75 in
c'a2 =
15
in
in²
lbs
therefore, øVg øVg
= =
162383 162.383
lbs kips
SUMMARY: Steel Strength of Studs in Shear Concrete Pryout Strength of Anchor in Shear
Controlling limit state, øV =
162.38
kips
>
179.19 162.38
kips kips
96.86
kips
(OK)
Page 6 of 9
1030 - UD ISE Building
TENSILE CAPACITY OF STUDS (REF: ANCHORING TO CONCRETE - ACI Appendix D)
Stud Diameter, do :
3/4
in. dia. ASTM A108 Grade 1015 or 1020
Fy
= =
Yield Strength of Stud 51 ksi
Fu
= =
Tensile Strength of Stud 65 ksi
L
= =
Nominal Length of Stud in 6.00
tp
= =
Embed Plate Thickness in 1.00
Ase
= =
Shank Area 0.442 in²
Abrg
= =
Bearing Area 0.790 in²
ths
= =
Head Thickness, in 0.375 in
hef
= = =
effective embedment depth L + tp - ths - 1/8 6.500 in
fc'
= =
concrete compresive strength 5 ksi
Steel Strength of Studs in Tension =
Nsa
n * Ase * futa
( ø = 0.75 )
where:
øNsa
=
Ase
= =
effective cross-sectional area of studs 0.442 in²
futa
= =
ultimate tensile strength of studs 65 ksi
n
= =
no. of studs 8.00 in
172.30
kips
Concrete Breakout Strength of Studs in Tension Ng
=
(ANC/ANCO) * ψec,N * ψed,N * ψc,N * ψ,N * Nb
ø = 0.70
where:
Page 7 of 9
1030 - UD ISE Building
=
ψec,N
=
eccentricity factor 1 <1 1 + (2eN'/3hef) eN'
=
0.00
ψec,N
=
ψed,N
=
1.0
=
0.7 + 0.3
in
1.000 if (Ca,min > 1.5hef)
ca,min
Ca,min 1.5hef
=
if Ca,min < 1.5hef
15.00
in
ψed,N
=
1.000
ψc,N
=
1.000
(cracked concrete)
ψ,N
=
1.000
(for cast-in anchors in cracked concrete)
Nb
=
kc
√fc' (hef)1.5 kc
Nb
=
ANCO
= = =
=
28123.3
24
(for cast-in anchors)
lbs
(2*1.5hef) * (2*1.5hef) 9*hef² in² 380.25 sx
x
x
y
sy
y
ANC
=
(2y + sy) (2x + sx)
where: y = = x
=
min (1.5hef , c'a1) ; c'a1 = 180 in in 9.75 min (1.5hef , c'a2)
c'a2 =
15
in
Page 8 of 9
1030 - UD ISE Building
ANC øNg øNg
9.75
in
sy
=
18.00
in
sx
=
6.00
in
956.25
=
49507 49.51
= =
=
in²
lbs kips
Concrete Pullout Strength of Studs in Tension øNpn
=
n * ψc,P * NP
ø = 0.70
where:
øNpn
=
n
= =
ψc,P
=
Abrg
= =
bearing area of studs 0.79 in²
Np
= =
8*Abrg *f'c 31.6
252.8
no. of studs in 8.00 (cracked concrete)
1.0
(for 0.75'' dia studs)
kips
kips
Concrete side-face blowout strength of studs in tension The side-face blowout failure mode must be investigated when the edge distance is less than 0.4 hef edge dist. =
in
15
>
0.4 hef =
2.60
in
(OK)
SUMMARY: Steel Strength of Studs in Tension Concrete Breakout Strength of Studs in Tension Concrete Pullout Strength of Studs in Tension
Controlling limit state, øV =
49.51
kips
>
1.20
(OK)
172.30 49.51 252.80
kips kips kips
27.85
kips
(OK)
Check Interaction Equation Na Nn/Ω
+
Va vn/Ω
<
1.2
27.9 49.5
+
96.9 162
=
1.16
<
Page 9 of 9
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