Shear lug verification example
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This document provides a summary of the design and verification of anchor bolts and a shear lug for a column base connection. It includes the geometry, loads, materials, and design calculations for the base plate, anchor bolts, and shear lug plate. The calculations show the base plate and anchor bolts satisfy strength requirements for bearing, tension, and shear. The shear lug plate is designed to resist the portion of shear load not resisted by friction, and calculations verify it satisfies strength requirements for bearing and shear.
![PIP STE05121 TECHNICAL CORRECTION
Anchor Bolt Design Guide October 2006
Page 20 of 24 Process Industry Practices
The shear lug should be designed for the applied shear portion not resisted by friction
between the base plate and concrete foundation. Grout must completely surround the lug
plate or pipe section and must entirely fill the slot created in the concrete. When using a
pipe section, a hole approximately 2 inches in diameter should be drilled through the
base plate into the pipe section to allow grout placement and inspection to assure that
grout is filling the entire pipe section.
9.1 Calculating Shear Load Applied to Shear Lug
The applied shear load, Vapp, used to design the shear lug should be computed as
follows:
Vapp = Vua - Vf
9.2 Design Procedure for Shear Lug Plate
Design of a shear lug plate follows (for an example calculation, see Appendix
Example 3, this Practice):
a. Calculate the required bearing area for the shear lug:
Areq = Vapp / (0.85 * φ * fc’) φ = 0.65
b. Determine the shear lug dimensions, assuming that bearing occurs only on
the portion of the lug below the grout level. Assume a value of W, the lug
width, on the basis of the known base plate size to find H, the total height
of the lug, including the grout thickness, G:
H = (Areq /W) + G
c. Calculate the factored cantilever end moment acting on a unit length of the
shear lug:
Mu = (Vapp/W) * (G + (H-G)/2)
d. With the value for the moment, the lug thickness can be found. The shear
lug should not be thicker than the base plate:
t = [(4 * Mu)/(0.9*fya)]0.5
e. Design weld between plate section and base plate.
f. Calculate the breakout strength of the shear lug in shear. The method
shown as follows is from ACI 349-01, Appendix B, section B.11:
Vcb = AVc*4*φ*[fc’]0.5
where
AVc = the projected area of the failure half-truncated pyramid defined
by projecting a 45-degree plane from the bearing edges of the
shear lug to the free edge. The bearing area of the shear lug
shall be excluded from the projected area.
φ = concrete strength reduction factor = 0.85](https://image.slidesharecdn.com/shearlugverificationexample-180815092728/85/Shear-lug-verification-example-2-320.jpg)
![TECHNICAL CORRECTION
October 2006
PIP STE05121
Anchor Bolt Design Guide
EXAMPLE 3 - Shear Lug Plate Section Design
Design a shear lug plate for a 14-in. square base plate, subject to a factored axial dead load
of 22.5 kips, factored live load of 65 kips, and a factored shear load of 40 kips. The base
plate and shear lug have fya = 36 ksi and fc' = 3 ksi. The contact plane between the grout and
base plate is assumed to be 1 in. above the concrete. A 2-ft 0-in. square pedestal is
assumed. Ductility is not required.
Vapp = Vua – Vf = 40 – (0.55)(22.5) = 27.6 kips
Bearing area = Areq = Vapp / (0.85 φ fc') = 27.6 kips / (0.85*0.65*3 ksi) = 16.67 in.
2
On the basis of base plate size, assume the plate width, W, will be 12 in.
Height of plate = H = Areq / W + G = 16.67 in.
2
/12 in. + 1 in. = 2.39 in. Use 3 in.
Ultimate moment = Mu = (Vapp / W) * (G + (H – G)/2)
= (27.6 kips / 12 in.) * (1 in. + (3 in.-1 in.)/2) = 4.61 k-in. / in.
Thickness = t = [(4 * Mu)/(φ* fya)]
½
= ((4*4.61 kip-in.)/(0.9*36 ksi))
½
= 0.754 in. Use 0.75 in.
This 12-in. x 3-in. x 0.75-in. plate will be sufficient to carry the applied shear load and resulting
moment. Design of the weld between the plate section and the base plate is left to the
engineer.
Check concrete breakout strength of the shear lug in shear.
Distance from shear lug to edge of concrete = (24 - 0.75) / 2 = 11.63 in.
AV = 24 * (2+11.63) – (12 * 2) = 303 in.
2
Vcb = AVc*4*φ*[fc']
0.5
= 303 * 4 * 0.85 * [3000]
0.5
= 56400 lb = 56.4 kips > 27.3 kips OK
Process Industry Practices Page A-26](https://image.slidesharecdn.com/shearlugverificationexample-180815092728/85/Shear-lug-verification-example-4-320.jpg)
![Project:
Engineer:
Descrip:
Verification Example
Javier Encinas, PE
Shear Lug Verification
Page # ___
7/20/2014
ASDIP Steel 3.2.5 STEEL BASE PLATE DESIGN www.asdipsoft.com
Column Section .................
Width Length
Column ..........
Plate ..............
Concrete
Support
Rod Offset .....
Thickness of Grout ............
Wp1
Wp2
Lp1
Lp2
in
in
in
in
in
in
OK
OK
OK
OK
Vertical Load P ................
Bending Moment M .........
Horizontal Load V ............
Design Eccentricity e .......
Design Eccentricity Is < L/6
0.0
kip
k-ft
kip
in
Plate Steel Strength Fy ....
Pier Concrete Strength f'c
ksi
ksi
Bearing stress 22.5 / (14.0 * 14.0) = 0.1 ksi
Bearing strength = 0.85 * 3.0 * = 4.4 ksi ACI 10.14.1
Under-strength factor ϕ = 0.65 ACI 9.3.2.4
Bearing strength ratio = =
0.1
0.65 * 2.8
= 0.04 < 1.0 OK
Critical section m =
Critical section n =
0.5 * (14.0 - 0.95 *8.0) = 3.2 in
0.5 * (14.0 - 0.80 *8.0) = 3.8 in
AISC-DG#1 3.1.2
[
4 * 8.0 * 8.0
(8.0 + 8.0)²
] * 0.04 = 0.04 AISC-DG#1 3.1.2
=
+ -
= 0.20
= = 2.0 in
Controlling section Max (3.2, 3.8, 0.20 * 2.0) = 3.8 in
Plate moment 0.1 * 3.8² / 2 = 0.8 k-in/in
Plate thickness = 3.8 * = 0.32 in AISC-DG#1 3.1.2
1](https://image.slidesharecdn.com/shearlugverificationexample-180815092728/85/Shear-lug-verification-example-8-320.jpg)
![Project:
Engineer:
Descrip:
Verification Example
Javier Encinas, PE
Shear Lug Verification
Page # ___
7/20/2014
ASDIP Steel 3.2.5 STEEL BASE PLATE DESIGN www.asdipsoft.com
- Concrete side-face blowout strength of anchors in tension ACI D.5.4
Side-face blowout Nsbg = N.A. (Embed < 2.5 Ca₁ , 12.0 < 2.5 * 6.5 = 16.3) ACI D.5.4.1
Tension Design Ratio = = 0.00 < 1.0 OK ACI D.4.1.1
ACI D.5
Shear resisted by Shear Lug + Friction
Total shear force Vu = 40.0 kip , Compression force C = 22.5 kip , Friction coeff. = 0.20
Friction strength 0.75 * 22.5 * 0.20 = 3.4 kip
Friction strength ratio = =
40.0
0.70 * 3.4
= 1.00 <= 1.0 OK
Shear lug width Wl = 12.0 in , Shear lug height Hl = 3.0 in , Shear lug thickness tl = 1.0 in
- Steel strength of lug in flexure
Lug moment 36.6 * (1.0 + (3.0 - 1.0) / 2) = 73.3 k-in
Lug flexural strength 12.0 * 36 * 1.0 ² / 4 = 108.0 k-in AISC F.11
Under-strength factor ϕ = 0.90 AISC F.1
Flexural strength ratio = =
73.3
0.90 * 108.0
= 0.75 < 1.0 OK AISC B3.4
- Steel strength of lug in shear
Shear force in lug 40.0 - 3.4 = 36.6 kip
Lug shear strength 0.6 * 36 * 12.0 * 1.0 = 259.2 kip AISC Eq. (G2.1)
Under-strength factor ϕ = 0.90 AISC G.1
Shear strength ratio = =
36.6
0.90 * 259.2
= 0.16 < 1.0 OK AISC B3.4
- Weld strength in shear lug
Shear lug fillet weld size a = 0.250 in) (Min size = 0.313 in) NG AISC J2.2a
Shear per unit width 36.6 / (2 * 12.0) = 1.5 kip
Tension per unit width 73.3 / [(1.0 + 2 * 0.250 / 3) * 12.0] = 5.2 kip
Resultant per lug width R = = = 130.8 kip
Weld stress 0.6 * 70 * (1 + 0.5) = 63.0 ksi AISC Eq. (J2.5)
Weld strength 63.0 * 0.707 * 0.250 * 2 * 12.0 = 267.3 kip
Under-strength factor ϕ = 0.75 AISC J3b.4
Weld strength ratio = =
130.8
0.75 * 267.3
= 0.65 AISC B3.3< 1.0 OK
- Concrete bearing strength of lug in shear
Lug bearing area (3.0 - 1.0) * 12.00 = 24.0 in²
Lug bearing strength 1.3 * 3.0 * 24.0 = 93.6 kip ACI 349 D.4.6
Under-strength factor ϕ = 0.65 ACI 9.3.2.4
Bearing strength ratio = =
36.6
0.65 * 93.6
= 0.60 < 1.0 OK ACI 4.1.1
4](https://image.slidesharecdn.com/shearlugverificationexample-180815092728/85/Shear-lug-verification-example-11-320.jpg)
Shear lug verification example
- 1. TECHNICAL CORRECTION PIP STE05121 October 2006 Anchor Bolt Design Guide Process Industry Practices Page 19 of 24 will develop friction even when the column or vertical vessel is in uplift. This downward load can be considered in calculating frictional resistance. Care shall be taken to assure that the downward load that produces frictional resistance occurs simultaneously with the shear load. In resisting horizontal loads, the friction resistance attributable to downward force from overturning moment may be used. The frictional resistance can also be used in combination with shear lugs to resist the factored shear load. The frictional resistance should not be used in combination with the shear resistance of anchors unless a mechanism exists to keep the base plate from slipping before the anchors can resist the load (such as welding the washer to the base plate). Note: If the design requires welding the washer to the base plate, plain washers or steel plate (rather than hardened washers) must be specified to ensure that a good weld can be produced. 8.2 Calculating Resisting Friction Force The resisting friction force, Vf, may be computed as follows: Vf = μP P = normal compression force μ = coefficient of friction The materials used and the embedment depth of the base plate determine the value of the coefficient of friction. (Refer to Figure F for a pictorial representation.) a. μ = 0.90 for concrete placed against as-rolled steel with the contact plane a full plate thickness below the concrete surface. b. μ = 0.70 for concrete or grout placed against as-rolled steel with the contact plate coincidental with the concrete surface. c. μ = 0.55 for grouted conditions with the contact plane between grout and as-rolled steel above the concrete surface. 9. Shear Lug Design Normally, friction and the shear capacity of the anchors used in a foundation adequately resist column base shear forces. In some cases, however, the engineer may find the shear force too great and may be required to transfer the excess shear force to the foundation by another means. If the total factored shear loads are transmitted through shear lugs or friction, the anchor bolts need not be designed for shear. A shear lug (a plate or pipe stub section, welded perpendicularly to the bottom of the base plate) allows for complete transfer of the force through the shear lug, thus taking the shear load off of the anchors. The bearing on the shear lug is applied only on the portion of the lug adjacent to the concrete. Therefore, the engineer should disregard the portion of the lug immersed in the top layer of grout and uniformly distribute the bearing load through the remaining height. Shear Lug Verification Example PIP STE05121 Anchor Bolt Design Guide PIP - Oct 2006
- 2. PIP STE05121 TECHNICAL CORRECTION Anchor Bolt Design Guide October 2006 Page 20 of 24 Process Industry Practices The shear lug should be designed for the applied shear portion not resisted by friction between the base plate and concrete foundation. Grout must completely surround the lug plate or pipe section and must entirely fill the slot created in the concrete. When using a pipe section, a hole approximately 2 inches in diameter should be drilled through the base plate into the pipe section to allow grout placement and inspection to assure that grout is filling the entire pipe section. 9.1 Calculating Shear Load Applied to Shear Lug The applied shear load, Vapp, used to design the shear lug should be computed as follows: Vapp = Vua - Vf 9.2 Design Procedure for Shear Lug Plate Design of a shear lug plate follows (for an example calculation, see Appendix Example 3, this Practice): a. Calculate the required bearing area for the shear lug: Areq = Vapp / (0.85 * φ * fc’) φ = 0.65 b. Determine the shear lug dimensions, assuming that bearing occurs only on the portion of the lug below the grout level. Assume a value of W, the lug width, on the basis of the known base plate size to find H, the total height of the lug, including the grout thickness, G: H = (Areq /W) + G c. Calculate the factored cantilever end moment acting on a unit length of the shear lug: Mu = (Vapp/W) * (G + (H-G)/2) d. With the value for the moment, the lug thickness can be found. The shear lug should not be thicker than the base plate: t = [(4 * Mu)/(0.9*fya)]0.5 e. Design weld between plate section and base plate. f. Calculate the breakout strength of the shear lug in shear. The method shown as follows is from ACI 349-01, Appendix B, section B.11: Vcb = AVc*4*φ*[fc’]0.5 where AVc = the projected area of the failure half-truncated pyramid defined by projecting a 45-degree plane from the bearing edges of the shear lug to the free edge. The bearing area of the shear lug shall be excluded from the projected area. φ = concrete strength reduction factor = 0.85
- 3. TECHNICAL CORRECTION October 2006 PIP STE05121 Anchor Bolt Design Guide Example 3 - Shear Lug Plate Section Design V = 40 (ULTIMATE) PLAN SECTION u K Process Industry Practices Page A-25
- 4. TECHNICAL CORRECTION October 2006 PIP STE05121 Anchor Bolt Design Guide EXAMPLE 3 - Shear Lug Plate Section Design Design a shear lug plate for a 14-in. square base plate, subject to a factored axial dead load of 22.5 kips, factored live load of 65 kips, and a factored shear load of 40 kips. The base plate and shear lug have fya = 36 ksi and fc' = 3 ksi. The contact plane between the grout and base plate is assumed to be 1 in. above the concrete. A 2-ft 0-in. square pedestal is assumed. Ductility is not required. Vapp = Vua – Vf = 40 – (0.55)(22.5) = 27.6 kips Bearing area = Areq = Vapp / (0.85 φ fc') = 27.6 kips / (0.85*0.65*3 ksi) = 16.67 in. 2 On the basis of base plate size, assume the plate width, W, will be 12 in. Height of plate = H = Areq / W + G = 16.67 in. 2 /12 in. + 1 in. = 2.39 in. Use 3 in. Ultimate moment = Mu = (Vapp / W) * (G + (H – G)/2) = (27.6 kips / 12 in.) * (1 in. + (3 in.-1 in.)/2) = 4.61 k-in. / in. Thickness = t = [(4 * Mu)/(φ* fya)] ½ = ((4*4.61 kip-in.)/(0.9*36 ksi)) ½ = 0.754 in. Use 0.75 in. This 12-in. x 3-in. x 0.75-in. plate will be sufficient to carry the applied shear load and resulting moment. Design of the weld between the plate section and the base plate is left to the engineer. Check concrete breakout strength of the shear lug in shear. Distance from shear lug to edge of concrete = (24 - 0.75) / 2 = 11.63 in. AV = 24 * (2+11.63) – (12 * 2) = 303 in. 2 Vcb = AVc*4*φ*[fc'] 0.5 = 303 * 4 * 0.85 * [3000] 0.5 = 56400 lb = 56.4 kips > 27.3 kips OK Process Industry Practices Page A-26
- 5. Project: Engineer: Descrip: Verification Example Javier Encinas, PE Shear Lug Verification Page # ___ 7/20/2014 ASDIP Steel 3.2.5 STEEL BASE PLATE DESIGN www.asdipsoft.com GEOMETRY Column Section ................. Width Length Column .......... Plate .............. Concrete Support Rod Offset ..... Thickness of Grout ............ Wp1 Wp2 Lp1 Lp2 W8X31 8.0 8.0 14.0 14.0 12.0 12.0 12.0 12.0 4.0 5.5 1.0 in in in in in in OK OK OK OK FACTORED LOADS (LRFD) Vertical Load P ................ Bending Moment M ......... Horizontal Load V ............ Design Eccentricity e ....... Design Eccentricity Is < L/6 22.5 0.0 40.0 0.0 kip k-ft kip in MATERIALS Plate Steel Strength Fy .... Pier Concrete Strength f'c 36.0 3.0 ksi ksi AXIALLY LOADED PLATES Cantilever Model Bearing Stress fp ............. Critical Section @ Long m Critical Section @ Short n Plate Thickness tp .......... 0.11 3.20 3.80 0.32 ksi in in in OK Thornton Model Bearing Strength ϕFp ....... Critical Section @ Int λn' . Design Moment @ Plate ... Plate Thickness tp ............ 2.84 0.41 0.01 0.03 ksi in k-in/in in BASE PLATES WITH MOMENT Blodgett Method Max. Bearing Stress fp ...... Bearing @ Critical Section Moment @ Critical Section Moment due to Rod Tension Design Moment @ Plate .... Plate Thickness tp ............. 0.11 0.11 0.59 0.00 0.59 0.27 ksi ksi k-in/in k-in/in k-in/in in OK DeWolf Method Max. Bearing Stress fp ...... Bearing @ Critical Section Moment @ Critical Section Moment due to Rod Tension Design Moment @ Plate .... Plate Thickness tp ............. 0.11 0.11 0.59 0.00 0.59 0.27 ksi ksi k-in/in k-in/in k-in/in in OK 1
- 6. Project: Engineer: Descrip: Verification Example Javier Encinas, PE Shear Lug Verification Page # ___ 7/20/2014 ASDIP Steel 3.2.5 STEEL BASE PLATE DESIGN www.asdipsoft.com ANCHORAGE DESIGN Rod Material Specification ......... Anchor Rod Size .. F1554-36 (4) Rods , fya = 36.0 ksi, futa = 58.0 ksi 1" diam. x 12.0 in emb. Concrete Is Uncracked at Service Load Level Tension Analysis (kip) Total Tension Force Nug ....... Tension Force per Rod Nui ... No Reinforcing Bars Provided 0.0 0.0 kip kip Failure Mode ϕ Nn Nu / ϕNn Steel Strength Nsa Rebars Strength Nrg Conc. Breakout Ncbg Pullout Strength Npn Side Blowout Nsbg Nu / ϕNn Tension Design Ratio .... OK 0.75 0.75 0.70 0.70 0.70 35.1 N.A. 25.0 50.4 N.A. 0.00 N.A. 0.00 0.00 N.A. 0.00 Shear Analysis (kip) Shear Taken by Shear Lug + Friction Total Shear Force V ........... 40.0 kip Compression Force C ........ Friction Coefficient ............... Friction Strength ϕFr ........... 22.5 0.20 3.4 kip kip V / ϕFr Shear Friction Ratio ....... 1.00 OK Shear Force in Lug ............... Shear Lug Width Wl ............ Shear Lug Height Hl ............ Shear Lug Thickness tl ........ 36.6 12.0 3.0 1.0 kip in in in Failure Mode ϕ Vn Vu / ϕVn Conc. Bearing Vcbr Conc. Breakout Vcb 0.65 0.75 93.6 57.8 0.60 0.84 0.84V / ϕVn Shear Design Ratio ...... OK SUMMARY OF RESULTS Design Moment @ Plate ... Plate Thickness tp ............ Max. Bearing Stress fp ..... Bearing Strength ϕFp ....... fp / ϕFp Design Ratio ............... 0.8 0.32 0.11 2.84 0.04 k-in/in in ksi ksi OK DESIGN IS DUCTILE DESIGN CODES Steel design ............. Base plate design .... Anchorage design ... AISC 360-10 (14th Ed.) AISC Design Series # 1 ACI 318-11 Appendix D 2
- 7. Project: Engineer: Descrip: Verification Example Javier Encinas, PE Shear Lug Verification Page # ___ 7/20/2014 ASDIP Steel 3.2.5 STEEL BASE PLATE DESIGN www.asdipsoft.com Tension Breakout Shear Breakout 3
- 8. Project: Engineer: Descrip: Verification Example Javier Encinas, PE Shear Lug Verification Page # ___ 7/20/2014 ASDIP Steel 3.2.5 STEEL BASE PLATE DESIGN www.asdipsoft.com Column Section ................. Width Length Column .......... Plate .............. Concrete Support Rod Offset ..... Thickness of Grout ............ Wp1 Wp2 Lp1 Lp2 in in in in in in OK OK OK OK Vertical Load P ................ Bending Moment M ......... Horizontal Load V ............ Design Eccentricity e ....... Design Eccentricity Is < L/6 0.0 kip k-ft kip in Plate Steel Strength Fy .... Pier Concrete Strength f'c ksi ksi Bearing stress 22.5 / (14.0 * 14.0) = 0.1 ksi Bearing strength = 0.85 * 3.0 * = 4.4 ksi ACI 10.14.1 Under-strength factor ϕ = 0.65 ACI 9.3.2.4 Bearing strength ratio = = 0.1 0.65 * 2.8 = 0.04 < 1.0 OK Critical section m = Critical section n = 0.5 * (14.0 - 0.95 *8.0) = 3.2 in 0.5 * (14.0 - 0.80 *8.0) = 3.8 in AISC-DG#1 3.1.2 [ 4 * 8.0 * 8.0 (8.0 + 8.0)² ] * 0.04 = 0.04 AISC-DG#1 3.1.2 = + - = 0.20 = = 2.0 in Controlling section Max (3.2, 3.8, 0.20 * 2.0) = 3.8 in Plate moment 0.1 * 3.8² / 2 = 0.8 k-in/in Plate thickness = 3.8 * = 0.32 in AISC-DG#1 3.1.2 1
- 9. Project: Engineer: Descrip: Verification Example Javier Encinas, PE Shear Lug Verification Page # ___ 7/20/2014 ASDIP Steel 3.2.5 STEEL BASE PLATE DESIGN www.asdipsoft.com Eccentricity 0.0 * 12 / 22.5 = 0.0 in < L / 6 = 14.0 / 6 = 2.3in Max bearing stress = 22.5 14.0 * 14.0 6 * 0.0 * 12 14.0 * 14.0² = 0.1 ksi Min bearing stress = = 22.5 14.0 * 14.0 6 * 0.0 * 12 14.0 * 14.0² = 0.1 ksi Bearing at critical section 0.1 - 3.2 / 14.0 * (0.1 - 0.1) = 0.1 ksi Moment due to bearing Mb = 0.1 * 3.2² / 2 * (0.1 - 0.1) * 3.2² / 3 = 0.6 k-in/in Plate thickness = = 0.27 in AISC-DG#1 3.1.2 2
- 10. Project: Engineer: Descrip: Verification Example Javier Encinas, PE Shear Lug Verification Page # ___ 7/20/2014 ASDIP Steel 3.2.5 STEEL BASE PLATE DESIGN www.asdipsoft.com Rod Material Specification ...... F1554-36 , Use (4) Rods , fya = 36.0 ksi, futa = 58.0 ksi Anchor Rod Size .... 1" diam. x 12.0 in emb. , Ase = 0.61 in² , Abrg = 1.50 in² ACI D.5 Total tension force Nu = 0.0 kip , # of tension rods = 0 , Tension force per rod Nui = 0.0 kip - Steel strength of anchors in tension ACI D.5.1 Steel strength 0.606 * 58.0 = 35.1 kip ACI Eq. (D-2) Under-strength factor ϕ = 0.75 ACI D.4.3 Steel strength ratio = = 0.0 0.75 * 35.1 = 0.00 ACI D.4.1.1< 1.0 OK - Concrete breakout strength of anchors in tension ACI D.5.2 No Reinforcing Bars Provided Effective embedment 17.50 / 1.5 = 11.67 in ACI D.5.2.3 Anchor group area Anc = (17.5 + 6.5) * (8.0 + 8.0 + 6.5) = 576.0 in² ACI D.5.2.1 Single anchor area 9 * (11.7)² = 1225.0 in² Eq. (D-5) Single anchor strength = 24 = 52.4 kip Eq. (D-6) Eccentricity factor 1.00 (No eccentric load) ACI D.5.2.4 Edge effects factor = 0.7 + 0.3 6.5 1.5 * 11.7 = 0.81 ACI D.5.2.5 Cracking factor 1.25 (Uncracked concrete at service load level) ACI D.5.2.6 Breakout strength 576.0 1225.0 1.00 * 0.81 * 1.25 * 52.4 = 25.0 kip Eq. (D-4) Under-strength factor ϕ = 0.70 ACI D.4.3 Breakout strength ratio = = 0.0 0.70 * 25.0 = 0.00 ACI D.4.1.1< 1.0 OK Breakout strength ratio controls (0.00 < 0.00) ACI D.5.2.9 - Concrete pullout strength of anchors in tension ACI D.5.3 Single anchor strength 8 * 1.50 * 3.0 = 36.0 kip ACI Eq. (D-14) Cracking factor 1.25 (Uncracked concrete at service load level) ACI D.5.3.6 Pullout strength 1.25 * 36.0 = 50.4 kip ACI Eq. (D-13) Under-strength factor ϕ = 0.70 ACI D.4.3 Pullout strength ratio = = 0.0 0.70 * 50.4 = 0.00 ACI D.4.1.1< 1.0 OK 3
- 11. Project: Engineer: Descrip: Verification Example Javier Encinas, PE Shear Lug Verification Page # ___ 7/20/2014 ASDIP Steel 3.2.5 STEEL BASE PLATE DESIGN www.asdipsoft.com - Concrete side-face blowout strength of anchors in tension ACI D.5.4 Side-face blowout Nsbg = N.A. (Embed < 2.5 Ca₁ , 12.0 < 2.5 * 6.5 = 16.3) ACI D.5.4.1 Tension Design Ratio = = 0.00 < 1.0 OK ACI D.4.1.1 ACI D.5 Shear resisted by Shear Lug + Friction Total shear force Vu = 40.0 kip , Compression force C = 22.5 kip , Friction coeff. = 0.20 Friction strength 0.75 * 22.5 * 0.20 = 3.4 kip Friction strength ratio = = 40.0 0.70 * 3.4 = 1.00 <= 1.0 OK Shear lug width Wl = 12.0 in , Shear lug height Hl = 3.0 in , Shear lug thickness tl = 1.0 in - Steel strength of lug in flexure Lug moment 36.6 * (1.0 + (3.0 - 1.0) / 2) = 73.3 k-in Lug flexural strength 12.0 * 36 * 1.0 ² / 4 = 108.0 k-in AISC F.11 Under-strength factor ϕ = 0.90 AISC F.1 Flexural strength ratio = = 73.3 0.90 * 108.0 = 0.75 < 1.0 OK AISC B3.4 - Steel strength of lug in shear Shear force in lug 40.0 - 3.4 = 36.6 kip Lug shear strength 0.6 * 36 * 12.0 * 1.0 = 259.2 kip AISC Eq. (G2.1) Under-strength factor ϕ = 0.90 AISC G.1 Shear strength ratio = = 36.6 0.90 * 259.2 = 0.16 < 1.0 OK AISC B3.4 - Weld strength in shear lug Shear lug fillet weld size a = 0.250 in) (Min size = 0.313 in) NG AISC J2.2a Shear per unit width 36.6 / (2 * 12.0) = 1.5 kip Tension per unit width 73.3 / [(1.0 + 2 * 0.250 / 3) * 12.0] = 5.2 kip Resultant per lug width R = = = 130.8 kip Weld stress 0.6 * 70 * (1 + 0.5) = 63.0 ksi AISC Eq. (J2.5) Weld strength 63.0 * 0.707 * 0.250 * 2 * 12.0 = 267.3 kip Under-strength factor ϕ = 0.75 AISC J3b.4 Weld strength ratio = = 130.8 0.75 * 267.3 = 0.65 AISC B3.3< 1.0 OK - Concrete bearing strength of lug in shear Lug bearing area (3.0 - 1.0) * 12.00 = 24.0 in² Lug bearing strength 1.3 * 3.0 * 24.0 = 93.6 kip ACI 349 D.4.6 Under-strength factor ϕ = 0.65 ACI 9.3.2.4 Bearing strength ratio = = 36.6 0.65 * 93.6 = 0.60 < 1.0 OK ACI 4.1.1 4
- 12. Project: Engineer: Descrip: Verification Example Javier Encinas, PE Shear Lug Verification Page # ___ 7/20/2014 ASDIP Steel 3.2.5 STEEL BASE PLATE DESIGN www.asdipsoft.com - Concrete breakout strength of lug in shear ACI 349 D.11.2 Lug breakout area Avc = (10.0 + 3.0 - 1.0) * (6.0 + 12.0 + 6.0) - 24.0 = 264.0 in² Breakout strength = 264.0 * 4 = 57.8 kip Under-strength factor ϕ = 0.75 ACI D.4.3 Breakout strength ratio = = 36.6 0.75 * 57.8 = 0.84 < 1.0 OK ACI D.4.1.1 Shear Design Ratio = = 0.84 < 1.0 OK ACI D.4.1.1 Anchorage design is ductile Steel design ................... Base plate design .......... Anchorage design ......... AISC 360-10 (14th Ed.) AISC Design Series # 1 ACI 318-11 Appendix D 5