Beam Reinforcement 9

"BMREINF9" --- STEEL MEMBER REINFORCEMENT ANALYSIS

Program Description:

"BMREINF9" is a spreadsheet program written in MS-Excel for the purpose of analysis and code checking of

steel members with various types of reinforcement configurations. Specifically, members are analyzed / code

checked per the AISC 9th Edition Allowable Stress Design (ASD) Manual. Both actual and allowable stresses

are computed, with the final result being a computed "stress ratio" of actual stress/allowable stress. Both the

intermediate and end weld requirements for attaching the reinforcing to the existing member are determined.

This program is a workbook consisting of nine (9) worksheets, described as follows:

Worksheet Name DescriptionDoc This documentation sheet

Member+Plate Analysis of Existing Member Reinforced with Plate at Bottom

Member+WT Analysis of Existing Member Reinforced with WT Section at Bottom

Member+4 Angles Analysis of Existing Member Reinforced with 4 Symmetric Angles

Member+4 Bars Analysis of Existing Member Reinforced with 4 Symmetric Round Bars

Member+2 Plates Analysis of Existing Member Reinforced with Top & Bottom Plates

Member+2 WT's(in) Analysis of Existing Member Reinforced with 2 WT's (webs toed in)

Member+2 WT's(out) Analysis of Existing Member Reinforced with 2 WT's (webs toed out)

Program Assumptions and Limitations:

1. This program follows the procedures and guidelines of the AISC 9th Edition Allowable Stress (ASD) Manual

(1989).

2. This program uses the database of member dimensions and section properties from the "AISC Shapes

Database", Version 3.0 (2001) as well as the AISC 9th Edition (ASD) Manual (1989).

3. This program assumes that existing member is overstressed primarily due to X-axis (strong axis) bending.

However, the member may be also subjected to axial load (compression or tension) as well as Y-axis (minor

axis) bending. Thus, the maximum resultant stress in the top flange of the composite section is assumed to

be compression, while it is assumed to be tension at the bottom of the composite section.

4. This program addresses the concern of existing state of stress for original member by providing input of the

"residual" X-axis (major axis) bending moment in the unreinforced member, and in some of the worksheets

the "residual" axial load in the unreinforced member. These "residual" loads can be due to dead load alone

or a combination of dead load and an assumed portion of the live load.

The "residual", the added (new), and the maximum stresses for the beam are calculated as follows:

fa(residual, mem) = P(residual, mem)/A

fa(added) = (P(total)-P(residual, mem))/Ac

fa(reinf) = fa(added)

fa(mem) = fa(residual, mem) + fa(added)

fbx(residual, mem) = Mx(residual, mem)*12/Sx

fbx(added, reinf) = (Mx(total)-Mx(residual, mem))*12/Sxc(reinf)

fbx(added, mem) = (Mx(total)-Mx(residual, mem))*12/Sxc(mem)

fbx(reinf) = fbx(added, reinf)

fbx(mem) = fbx(residual, mem) + fbx(added, mem)

where: A = cross sectional area of existing member alone

Ac = cross sectional area of entire composite section

Sx = section modulus for the existing member alone

Member+I-Beam Analysis of Existing Member Reinforced with I-Beam at Bottom

Sxc(reinf) = section modulus referenced to extremities of new reinforcing element(s)

Sxc(mem) = section modulus referenced to extremities of existing member

5. This program prompts for the X-axis (strong axis) "cut-off" moment for the beam, which may be input as a

value equal to the total design moment, but is more typically equal to the moment capacity of the existing

unreinforced member. The "cut-off" moment is the moment used to determine the end weld requirements

for the member reinforcing.

6. In this program for members subjected to known loadings consisting of axial load (compression or tension)

and/or uniaxial or biaxial bending, both the actual and allowable stress are computed, with the final result

being a computed "stress ratio" of actual stress/allowable stress.

7. For cases when the axial load is compression, then it is considered at the top flange of the section but it is

not considered at the bottom of the section. When the axial load is tension, then it is not considered at the

top flange of the section but it is considered for the bottom of the section.

8. In this program if beam reinforcing runs for full length, then the composite section properties will be used in

computing both the actual and the allowable axial compressive stresses. Otherwise, only the properties of

the original existing beam are used for reinforcing that does not run for full length of beam. (See Note #9.)

9. When a stiffened element (web) of a member subjected to axial compression is classified as a "slender"

element (exceeding non-compact limits) based on local buckling criteria, then the program complies with

AISC Appendix B. If the effective length of web for axial compression as determined by AISC Eqn. A-B5-8 is

found to be less than the actual web length, then the area of the web to be deducted is assumed as one-half

each side of the centroid of the existing beam for determining the effective section properties.

10. This program utilizes an "Allowable Stress Increase Factor" (ASIF) which is a multiplier of any of the

calculated allowable stresses Fa, Fbx, and Fby and also the Euler column buckling stresses F'ex and F'ey.

It is used and appears only in the stress ratio calculation. Typically a value of 1.0 may be used. However, a

value of 1.333 may be used for load combinations which include wind or seismic loads.

11. If a member loaded with axial compression has a maximum slenderness ratio K*L*12/r >200, then a

message will appear. However, this program does not consider or deem a particular member as

"inadequate" based on the slenderness ratio of 200 being exceeded.

12. For the case of combined axial compression with bending, if the calculated value of fa >=F'e (which is not

allowed) then a warning (error!) message will appear.

13. When the values of either 'Lx', 'Ly', or 'Lb' are input = 0' (or actually <= 1.0'), this program will use a value = 1.0'.

14. When the area of the top (assumed compression) flange is <= the area of the bottom (assumed tension)

flange, then this program will not consider the use of AISC Eqn. F1-8 per Section F1.3 in determining

the allowable X-axis bending stress, 'Fbx'.

15. The values of 'Cb', 'Cmx', 'Cmy', 'Kx, and 'Ky' may be calculated (if applicable) by accessing the additional

input data to the right of the main page in each of the calculation worksheets. Then, these calculated values

can be input under the member design parameters on the main page. (Note: there are equations which

very closely approximate the solutions for 'Kx' and 'Ky' obtained using the AISC Code Alignment Charts.)

16. This program does not calculate or check shear or deflection in member.

17. This program does not consider torsion on member.

18. This program does not consider deduction for holes in members subjected to tension.

19. In weld design, this program determines the AISC Code minimum fillet weld as well as the maximum

effective fillet weld size based on the shear strength of the base material. The weld size used will be the

AISC minimum weld size, not to exceed the maximum assumed value for a single pass weld of 5/16". If the

maximum effective fillet weld size is less than the actual size used, then the maximum effective fillet weld

size is used to determine the actual required weld lengths.

20. This program contains numerous “comment boxes” which contain a wide variety of information including

explanations of input or output items, equations used, data tables, etc. (Note: presence of a “comment box”

is denoted by a “red triangle” in the upper right-hand corner of a cell. Merely move the mouse pointer to the

desired cell to view the contents of that particular "comment box".)

"BMREINF9.xls" ProgramVersion 2.0

5 of 28 02/09/2016 10:44:56

STEEL MEMBER REINFORCEMENT ANALYSISExisting Member Reinforced with Cover Plate at Bottom

Per AISC 9th Edition Manual (ASD)Job Name: Subject: ###

Job Number: Originator: Checker: ###Yes

Input Data: NoSingle

Member and Plate Reinf. Sizes: Exist. W24x76 Properties: Y ReverseExist. Member = W24x76 A = 22.40 in.^2 Plate Thk., tp = 1.5000 in. d = 23.900 in. Unbraced

Plate Width, Bp = 13.000 in. tw = 0.440 in. Y4 =bf = 8.990 in. d/2=11.95

Member Loadings: tf = 0.680 in. Y1=17.861 rtc =P(axial) = 10.00 kips rt = 2.290 in. dc/Af =

Mx(total) = 338.00 ft-kips d/Af = 3.91 N.A. Y3=5.911

Mx(cutoff) = 275.24 ft-kips Ix = 2100.00 in.^4 XMx(residual) = 50.00 ft-kips Sx = 176.00 in.^3 Y2=7.539 Y4=6.789

My = 2.00 ft-kips rx = 9.690 in. Sxc(bot/mem) =Rv(shear) = 30.00 kips Iy = 82.50 in.^4 Bp=13 Sxc(bot/pl) =

Sy = 18.40 in.^3 Sxc(bot/pl) req'd. =Design Parameters: ry = 1.920 in. rxc =Fy (mem./plate) = 36.00 ksi Iyc =

Kx = 1.00 1.5''x13'' Plate Properties: Section Ratios & Parameters:Syc(top/mem) =Ky = 1.00 Ap = 19.50 in.^2 bf/(2*tf) = 6.61 c(bot/mem) =Lx = 12.000 ft. Ixp = 3.66 in.^4 d/tw = 57.73 Syc(bot/pl) =Ly = 12.000 ft. Iyp = 274.63 in.^4 h = 22.540 in.

Lb = 12.000 ft. wt./ft.= 66.35 lb./ft. h/tw = 51.23 Section Ratios and Parameters:Cb = 1.00 be = #N/A in.

Cmx = 0.85 Qs = 1.000 d/tw =Cmy = 0.85 Qa = #N/A h =ASIF = 1.000 h/tw =

Full Lgth. Reinf.? Yes #N/A Qs =#N/A #N/A #N/A #N/A #N/A

Results: #N/A #N/A #N/A #N/A #N/A

#N/A #N/A #N/A #N/A #N/A

Combined Section Properties: Sxce(top/mem) =Y1 = 17.861 in. Y1 = d+tp-Y2 Sxce(bot/mem) =Y2 = 7.539 in. Y2 = (Ap*(tp/2)+A*(tp+d/2))/(A+Ap) Sxce(bot/pl) =Y3 = 5.911 in. Y3 = Y1-d/2 Syce(top/mem) =Y4 = 6.789 in. Y4 = Y2-tp/2 Syce(bot/mem) =Ac = 41.90 in.^2 Ac = A+Ap Syce(bot/pl) =rtc = 2.185 in. rtc = SQRT((tf*bf^3/12+(Y1-tf)/3*tw^3/12)/(bf*tf+(Y1-tf)/3*tw)) Check Stresses:

dc/Af = 4.155 dc/Af = (d+tp)/(bf*tf) Axial Compression:Ixc = 3785.08 in.^4 Ixc = Ix+Ixp+A*Y3^2+Ap*Y4^2 KL/rx =

Sxc(top/mem) = 211.92 in.^3 Sxc(top/mem) = Ixc/Y1 (ref. to top/upper flange of exist. member) KL/ry =Sxc(bot/mem) = 626.72 in^3 Sxc(bot/mem) = Ixc/ABS(Y2-tp) (ref. to bot/lower flg. of exist. member) KL/r =

Sxc(bot/pl) = 502.03 in.^3 Sxc(bot/pl) = Ixc/Y2 (referenced to bottom of plate) Cc =rxc = 9.505 in. rxc = SQRT(Ixc/Ac) fa =

(continued)


6 of 28 02/09/2016 10:44:56

Fa =Combined Section Properties (continued): Use: Fa =

Iyc = 357.13 in.^4 Iyc = Iy+Iyp fa/Fa =Syc(top/mem) = 79.45 in.^3 Syc(top/mem) = Iyc/(bf/2) (ref. outside edge of top flange of member) Pa =Syc(bot/mem) = 79.45 in.^3 Syc(bot/mem) = Iyc/(bf/2) (ref. outside edge of bot. flange of member) X-axis Bending:

Syc(bot/pl) = 54.94 in.^3 Syc(bot/pl) = Iyc/(Bp/2) (ref. outside edge of reinforcing plate)Is Top Flg.<Bot. Flg?ryc = 2.919 in. ryc = SQRT(Iyc/Ac) Lc =

Lu =Check Stresses: Lb/rt =For Axial Compression: fa/Fy =

KL/rx = 15.15 KL/rx = Kx*Lx*12/rxc (use rxc for combined section) fbx(residual) =KL/ry = 49.32 KL/ry = Ky*Ly*12/ryc (use ryc for combined section)fbx(added, top/mem) =KL/r = 49.32 KL/r = Max. of: Kx*Lx*12/rxc or Ky*Ly*12/ryc fbx(added, bot/mem) =

Cc = #N/A #N/A fbx(top/mem) =fa = 0.24 ksi fa = P/Ac (use Ac for combined section) fbx(bot/mem) =

Fa = #N/A ksi #N/A fbx(bot/pl) =Pa = #N/A kips Pa = Fa*Ac (use Ac for combined section) Is Lb<=Lc?

#N/A Is dc/tw<=allow?For X-axis Bending: (for combined section) Is b/t<=65/SQRT(Fy)?

Lc = 9.49 ft. Lc = Min. of: 76*bf/SQRT(Fy)/12 or: 20000/(dc/Af*Fy)/12Is b/t>95/SQRT(Fy)?Lu = 11.14 ft. Lu = Max. of: ((20000/(dc/Af*Fy))/12 or: SQRT(102000/Fy)*rtc/12Fbx(top) =

Lb/rt = 65.91 Lb/rt = Lb*12/rtc = unbraced length for bending ratio Fbx(top) =fbx(residual) = 3.41 ksi fbx(residual) = Mx(residual)*12/Sx (at top and bottom of member)Fbx(top) =fbx(add. top) = #N/A ksi #N/A Fbx(top) =fbx(add. bot) = #N/A ksi #N/A Fbx(top) =

fbx(top/mem) = #N/A ksi fbx(top/mem) = fbx(residual)+fbx(added, top/mem) Fbx(top) =fbx(bot/mem) = #N/A ksi fbx(bot/mem) = fbx(residual)+fbx(added, bot/mem) Fbx(top) =

fbx(bot/pl) = #N/A ksi #N/A Use: Fbx(top) =Fbx(top) = 20.32 ksi Fbx(top)=(2/3-Fy*(Lb/rf)^2/(1530000*Cb))*Fy <= Qs*0.60*Fy (Eqn. F1-6)Use: Fbx(bot) =Fbx(bot) = 21.60 ksi Fbx(bot) = allowable tension bending stress at bottom Mrx =

Mrx = #N/A ft-kips #N/A Y-axis Bending:fby(top/mem) =

X-axis Euler Stress: (use rxc for combined section) fby(bot/mem) =F'ex = 650.56 ksi F'ex = 12*(π)^2*29000/(23*(Kx*Lx*12/rxc)^2) fby(bot/pl) =

Fby =For Y-axis Bending: (for combined section) Mry =

fby(top/mem) = #N/A ksi #N/A Euler Stresses:fby(bot/mem) = #N/A ksi #N/A F'ex =

fby(bot/pl) = #N/A ksi #N/A F'ey =Fby = 27.00 ksi Fby = 0.75*Fy (Eqn. F2-1) Stress Ratios:Mry = #N/A ft-kips #N/A At Top Flange of Existing Member:

S.R. =Y-axis Euler Stress: (use ryc for combined section) S.R. =

F'ey = 61.38 ksi F'ey = 12*(π)^2*29000/(23*(Ky*Ly*12/ryc)^2) S.R. =S.R. =

Stress Ratios: At Bottom Flange of Existing Member:S.R.(top/mem) = #N/A #N/A #N/A S.R. =S.R.(bot/mem) = #N/A #N/A #N/A S.R. =

S.R.(bot/pl) = #N/A #N/A #N/A At Bottom of Reinforcing Plate:(continued)


7 of 28 02/09/2016 10:44:56

S.R. =Weld Design: Weld Design:

Interior Welds of Plate Reinf. to Member:Interior Welds of Plate Reinf. to Member: (assume intermittent fillet weld) Fw =

Fw = 0.525 kips/in. Fw = Rv*Ap*Y4/(2*Ixc) (weld force per inch) 5/16 in.

0.6594 in.

5/16 in. Lw(min) =Lw(min) = 1.500 in./ft. Lw(use) =Lw(use) = 1.500 in./ft.

(use) End Welds of Plate Reinf. to Member:Use: 5/16'' Fillet Weld, 1.5'' @ 12'' Each Side of Plate For Case #1:

Fw(total) =End Welds of Plate Reinf. to Member: (assume continuous fillet weld) Lw(end) =

For Case #1: Weld Across End of Plate a'(min) =Fw(total) = 120.18 kips Fw(total) = M(cutoff)*12*Ap*Y4/Ixc+(Ap/Ac)*ABS(P(axial)) a'(req'd) =Lw(end) = 8.990 in. Lw(end) = Min. of: bf or Bp a'(use) =

a'(min) = 19.500 in. a'(min) = 1.5*Bp, for weld size < 0.75*tp Lw(total) =a'(req'd) = 8.455 in.

a'(use) = 19.500 in. a'(use) = Max. of: a'(min) or a'(req'd) For Case #2:Lw(total) = 47.990 in. Lw(total) = 2*a'(use)+Lw(end) (@ each end of plate reinf.)Fw(total) =

(use) Lw(end) =Use: 5/16'' Fillet Weld, 19.5'' Continuous Each Side of Plate and Across End a'(min) =

a'(req'd) = For Case #2: No Weld Across End of Plate a'(use) =

Fw(total) = 120.18 kips Fw(total) = M(cutoff)*12*Ap*Y4/Ixc+(Ap/Ac)*ABS(P(axial)) Lw(total) =Lw(end) = 0.000 in. Lw(end) = 0 (assumed)

a'(min) = 26.000 in. a'(min) = 2*Bp, for no weld across end of plate ###a'(req'd) = 12.950 in. (Determining End Welds Now Including Eccentricity from C.G. of Reinforcing to Welds)

a'(use) = 26.000 in. a'(use) = Max. of: a'(min) or a'(req'd) End Welds of Plate Reinf. to Member:Lw(total) = 52.000 in. Lw(total) = 2*a'(use) (@ each end of plate) For Case #1:

(use) Fw1 =Use: 5/16'' Fillet Weld, 26'' Continuous Each Side of Plate Lw(end) =

a'(use) =Pw =

Comments: Mw =Sw(max) =Sw(min) =

Fw(result) =

### For Case #2:

Fw1 =Lw(end) =

a'(use) =Pw =Mw =Sw =

Fw(result) =

(min) =(min) = (min) = Min. fillet weld size from AISC Table J2.4, page 5-67(max,eff) =

(max,eff) = (max,eff) = Min. of: (tp or tf)*(0.40*Fy/((SQRT(2)/2)*0.30*70))(use) =(use) = (use) = (min) <= 5/16"

Lw(min) = Max. of: 1.5" or 4*(min)Lw(use) = Max. of: (Fw*12)/(*SQRT(2)/2*0.3*70) or Lw(min)Note: =

Note: =

a'(req'd) = (Fw(total)/(*SQRT(2)/2*0.3*70)-Lw(end))/2 Note: =

Note: =

Note: =

a'(req'd) = (Fw(total)/(*SQRT(2)/2*0.3*70))/2

Note: =

(req'd) =

(req'd) =


8 of 28 02/09/2016 10:44:56

STEEL MEMBER REINFORCEMENT ANALYSISExisting Member Reinforced with WT Section at Bottom




Member and WT Reinf. Sizes: Exist. W24x76 Properties: Y ReverseExist. Member = W24x76 A = 22.40 in.^2

WT Reinf. = WT6x15 d = 23.900 in. Unbracedtw = 0.440 in. Sxc(top/mem) =

Member Loadings: bf = 8.990 in. d/2=11.95P(axial) = 10.00 kips tf = 0.680 in. Y1=14.716 Sxc(bot/mem) =

Mx(total) = 338.00 ft-kips rt = 2.290 in. Sxc(bot/WT) =Mx(cutoff) = 275.24 ft-kips d/Af = 3.91 N.A. Y3=2.766

Mx(residual) = 50.00 ft-kips Ix = 2100.00 in.^4 XMy = 0.00 ft-kips Sx = 176.00 in.^3 Iyc =

Rv(shear) = 30.00 kips rx = 9.690 in. Y4=14.084Iy = 82.50 in.^4 Y2=15.354 Syc(bot/mem) =

Design Parameters: Sy = 18.40 in.^3 y=1.27

Fy (mem./WT) = 36.00 ksi ry = 1.920 in. bf=6.52 ryc =Kx = 1.00 Section Ratios and Parameters:Ky = 1.00 WT6x15 Properties: Section Ratios & Parameters:bf/(2*tf) =Lx = 12.000 ft. A = 4.40 in.^2 bf/(2*tf) = 6.61 d/tw =Ly = 12.000 ft. d = 6.170 in. d/tw = 54.32 h =Lb = 12.000 ft. tw = 0.260 in. h = 22.540 in.

Cb = 1.00 bf = 6.520 in. h/tw = 51.23 Qs =Cmx = 0.85 tf = 0.440 in. be = #N/A in.

Cmy = 0.85 Ix = 13.50 in.^4 Qs = 1.000 be =ASIF = 1.000 y = 1.270 in. Qa = #N/A Qa =

Full Lgth. Reinf.? No Iy = 10.20 in.^4 Sxce(top/mem) =Sxce(bot/mem) =

#N/A Sxce(bot/WT) =Results: #N/A #N/A #N/A #N/A #N/A


Combined Section Properties: #N/A #N/A #N/A #N/A #N/A

Y1 = 14.716 in. Y1 = d+dwt-Y2 Check Stresses:Y2 = 15.354 in. Y2 = (A*(dwt+d/2)+(Awt*y))/(A+Awt) Axial Compression:Y3 = 2.766 in. Y3 = Y1-d/2 KL/rx =Y4 = 14.084 in. Y4 = Y2-y KL/ry =Ac = 26.80 in.^2 Ac = A+Awt KL/r =rtc = 2.246 in. rtc = SQRT((tf*bf^3/12+(Y1-tf)/3*tw^3/12)/(bf*tf+(Y1-tf)/3*tw)) Cc =

dc/Af = 4.919 dc/Af = (d+dwt)/(bf*tf) fa =Ixc = 3157.66 in.^4 Ixc = Ix+Ixwt+A*Y3^2+Awt*Y4^2 Fa =

Sxc(top/mem) = 214.57 in.^3 Sxc(top/mem) = Ixc/Y1 (ref. to top/upper flange of exist. member)Fa =Sxc(bot/mem) = 343.84 in.^3 Sxc(bot/mem) = Ixc/ABS(Y2-dwt) (ref. to bot/lower flg. of exist. member)Use: Fa =

Sxc(bot/WT) = 205.66 in.^3 Sxc(bot/WT) = Ixc/Y2 (referenced to bottom of WT) fa/Fa =rxc = 10.855 in. rxc = SQRT(Ixc/Ac) Pa =

(continued)


9 of 28 02/09/2016 10:44:56

Is Top Flg.<Bot. Flg?Combined Section Properties (continued): Lc =

Iyc = 92.70 in.^4 Iyc = Iy+Iywt Lu =Syc(top/mem) = 20.62 in.^3 Syc(top/mem) = Iyc/(bf/2) (referenced to tip of top flange of member) Lb/rt =Syc(bot/mem) = 20.62 in.^3 Syc(bot/mem) = Iyc/(bf/2) (referenced to tip of bot. flange of member) fa/Fy =

Syc(bot/WT) = 28.44 in.^3 Syc(bot/WT) = Iyc/(bfwt/2) (referenced to tip of flange of reinf. WT)fbx(residual) =ryc = 1.860 in. ryc = SQRT(Iyc/Ac) fbx(added, top/mem) =

fbx(added, bot/mem) =Check Stresses: fbx(top/mem) =For Axial Compression: fbx(bot/mem) =

KL/rx = 14.86 KL/rx = Kx*Lx*12/rx (use rx for existing member only) fbx(bot/WT) =KL/ry = 75.00 KL/ry = Ky*Ly*12/ry (use ry for existing member only) Is Lb<=Lc?KL/r = 75.00 KL/r = Max. of: Kx*Lx*12/rx or Ky*Ly*12/ry Is dc/tw<=allow?

Cc = #N/A #N/A Is b/t<=65/SQRT(Fy)?fa = 0.45 ksi fa = P/A (use A for existing member only) Is b/t>95/SQRT(Fy)?

Fa = #N/A ksi #N/A Fbx(top) =Pa = #N/A kips Pa = Fa*A (use A for existing member only) Fbx(top) =

#N/A Fbx(top) =For X-axis Bending: (for combined section) Fbx(top) =

Lc = 9.41 ft. Lc = Min. of: 76*bf/SQRT(Fy)/12 or: 20000/(dc/Af*Fy)/12 Fbx(top) =Lu = 9.96 ft. Lu = Max. of: ((20000/(dc/Af*Fy))/12 or: SQRT(102000/Fy)*rtc/12Fbx(top) =

Lb/rt = 64.13 Lb/rt = Lb*12/rtc = unbraced length for bending ratio Fbx(top) =fbx(residual) = 3.41 ksi fbx(residual) = Mx(residual)*12/Sx (at top and bottom of member)Use: Fbx(top) =fbx(add. top) = #N/A ksi #N/A Use: Fbx(bot) =fbx(add. bot) = #N/A ksi #N/A Mrx =

fbx(top/mem) = #N/A ksi fbx(top/mem) = fbx(residual)+fbx(added, top/mem) Y-axis Bending:fbx(bot/mem) = #N/A ksi fbx(bot/mem) = fbx(residual)+fbx(added, bot/mem) fby(top/mem) =

fbx(bot/WT) = #N/A ksi #N/A fby(bot/mem) =Fbx(top) = 20.52 ksi Fbx(top)=(2/3-Fy*(Lb/rf)^2/(1530000*Cb))*Fy <= Qs*0.60*Fy (Eqn. F1-6)fby(bot/WT) =Fbx(bot) = 21.60 ksi Fbx(bot) = allowable tension bending stress at bottom Fby =

Mrx = #N/A ft-kips #N/A Mry =Euler Stresses:

X-axis Euler Stress: (use rx for existing member only) F'ex =F'ex = 676.20 ksi F'ex = 12*(π)^2*29000/(23*(Kx*Lx*12/rx)^2) F'ey =

Stress Ratios:For Y-axis Bending: (for combined section) At Top Flange of Existing Member:

fby(top/mem) = #N/A ksi #N/A S.R. =fby(bot/mem) = #N/A ksi #N/A S.R. =

fby(bot/WT) = #N/A ksi #N/A S.R. =Fby = 27.00 ksi Fby = 0.75*Fy (Eqn. F2-1) S.R. =Mry = #N/A ft-kips #N/A At Bottom Flange of Existing Member:

S.R. =Y-axis Euler Stress: (use ry for existing member only) S.R. =

F'ey = 26.55 ksi F'ey = 12*(π)^2*29000/(23*(Ky*Ly*12/ry)^2) At Bottom of WT Reinforcing:S.R. =

Stress Ratios: S.R. =S.R.(top/mem) = #N/A #N/A #N/A Weld Design:S.R.(bot/mem) = #N/A #N/A #N/A Interior Welds of WT Reinf. to Member:

S.R.(bot/WT) = #N/A #N/A #N/A Fw =(continued)


10 of 28 02/09/2016 10:44:56

Weld Design:Lw(min) =

Interior Welds of WT Reinf. to Member: (assume intermittent fillet weld) Lw(use) =Fw = 0.294 kips/in. Fw = Rv*Awt*Y4/(2*Ixc) (weld force per inch)

1/ 4 in. End Welds of WT Reinf. to Member:0.1261 in. Fw(total) = 1/ 4 in. a'(min) =

Lw(min) = 1.500 in./ft. a'(req'd) =Lw(use) = 1.887 in./ft. a'(use) =

(max,eff) Lw(total) =Use: 1/4'' Fillet Weld, 2'' @ 12'' Each Side of WT Web

###End Welds of WT Reinf. to Member: (assume continuous fillet weld) (Determining End Welds Now Including Eccentricity from C.G. of Reinforcing to Welds)

Fw(total) = 66.46 kips Fw(total) = M(cutoff)*12*Awt*Y4/Ixc + (Awt/Ac)*ABS(P(axial)) End Welds of WT Reinf. to Member:a'(min) = 3.000 in. a'(min) = 3" (assumed) Fw1 =

a'(req'd) = 17.751 in. a'(use) =a'(use) = 17.751 in. a'(use) = Max. of: a'(min) or a'(req'd) Pw =

Lw(total) = 35.502 in. Lw(total) = 2*a'(use) (@ each end of WT reinf.) Mw =(max,eff) Sw =

Use: 1/4'' Fillet Weld, 18'' Continuous Each Side of WT Web @ Ends Fw(result) =

###Comments: ###

##############################################################################

(max,eff) =(use) =

Note: = (min) = (min) = Min. fillet weld size from AISC Table J2.4, page 5-67

(max,eff) = (max,eff) = tw(wt)*(0.40*Fy/((SQRT(2)/2)*0.30*70*2))(use) = (use) = (min) <= 5/16"

Lw(min) = Max. of: 1.5" or 4*(min)Lw(use) = Max. of: (Fw*12)/(*SQRT(2)/2*0.3*70) or Lw(min)

Note: = Note: =

a' = (Fw(total)/((SQRT(2)/2)*0.3*70*))/2

Note: =

(req'd) =


11 of 28 02/09/2016 10:44:56

STEEL MEMBER REINFORCEMENT ANALYSIS




Exist. W24x76 Properties: Y Exist. Member = W24x76 A = 22.40 in.^2 Braced

W8x15 d = 23.900 in. Unbracedtw = 0.440 in. d/2=11.95

Member Loadings: bf = 8.990 in. Y1=14.598 Sxc(top/mem) req'd. =P(axial) = 200.00 kips tf = 0.680 in. Sxc(bot/mem) =

Mx(total) = 338.00 ft-kips rt = 2.290 in. N.A. Y3=2.648Mx(cutoff) = 275.24 ft-kips d/Af = 3.91 X

Mx(residual) = 50.00 ft-kips Ix = 2100.00 in.^4 rxc =My = 0.00 ft-kips Sx = 176.00 in.^3 Y4=13.357

Rv(shear) = 30.00 kips rx = 9.690 in. Y2=17.412 Syc(top/mem) =Iy = 82.50 in.^4 d2/2=4.055

Design Parameters: Sy = 18.40 in.^3 Syc(bot/bm) =36.00 ksi ry = 1.920 in. bf=4.01 ryc =

Kx = 1.00 Section Ratios and Parameters:Ky = 1.00 New W8x15 Properties: Section Ratios & Parameters:bf/(2*tf) =Lx = 12.000 ft. A2 = 4.44 in.^2 bf/(2*tf) = 6.61 d/tw =Ly = 12.000 ft. d2 = 8.110 in. d/tw = 72.75 h =Lb = 12.000 ft. tw2 = 0.245 in. h = 31.015 in.

Cb = 1.00 bf2 = 4.010 in. h/tw = 70.49 Qs =Cmx = 0.85 tf2 = 0.315 in. be = #N/A in.

Cmy = 0.85 Ix2 = 48.00 in.^4 Qs = 1.000 be =ASIF = 1.000 Iy2 = 3.41 in.^4 Qa = #N/A Qa =

Full Lgth. Reinf.? Yes Sxce(top/mem) =#N/A Sxce(bot/mem) =




Combined Section Properties: #N/A

Y1 = 14.598 in. Y1 = d1+d2-Y2 Check Stresses:Y2 = 17.412 in. Y2 = (A1*(d2+d1/2)+(A2*d2/2))/(A1+A2) Axial Compression:Y3 = 2.648 in. Y3 = Y1-d1/2 KL/rx =Y4 = 13.357 in. Y4 = Y2-d2/2 KL/ry =Ac = 26.84 in.^2 Ac = A1+A2 KL/r =rtc = 2.248 in. rtc = SQRT((tf*bf^3/12+(Y1-tf)/3*tw^3/12)/(bf*tf+(Y1-tf)/3*tw)) Cc =

dc/Af = 5.236 dc/Af = (d1+d2)/(bf*tf) fa =Ixc = 3097.20 in.^4 Ixc = Ix1+Ix2+A1*Y3^2+A2*Y4^2 Fa =

Sxc(top/mem) = 212.17 in.^3 Sxc(top/mem) = Ixc/Y1 (ref. to top/upper flange of exist. member)Fa =Sxc(bot/mem) = 332.95 in^3 Sxc(bot/mem) = Ixc/ABS(Y2-d2) (ref. to bot/lower flg. of exist. member)Use: Fa =

Sxc(bot/pl) = 177.87 in.^3 Sxc(bot/bm) = Ixc/Y2 (referenced to bottom of reinforcing beam)fa/Fa =rxc = 10.742 in. rxc = SQRT(Ixc/Ac) Pa =

(continued)

Existing Member Reinforced with New I-shaped Member at Bottom

Member and I-Beam Reinf. Sizes:

I-Beam Reinf. =

Fy (mem./I-bm.) =


12 of 28 02/09/2016 10:44:56

Is Top Flg.<Bot. Flg?Combined Section Properties (continued): Lc =

Iyc = 85.91 in.^4 Iyc = Iy+Iy2 Lu =Syc(top/mem) = 19.11 in.^3 Syc(top/mem) = Iyc/(bf/2) (referenced to tip of top flange of member) Lb/rt =Syc(bot/mem) = 19.11 in.^3 Syc(bot/mem) = Iyc/(bf/2) (referenced to tip of bot. flange of member) fa/Fy =

Syc(bot/bm) = 42.85 in.^3 Syc(bot/bm) = Iyc/(bf2/2) (referenced to tip of flange of reinf. beam)fbx(residual) =ryc = 1.789 in. ryc = SQRT(Iyc/Ac) fbx(added, top/mem) =

fbx(added, bot/mem) =Check Stresses: fbx(top/mem) =For Axial Compression: fbx(bot/mem) =

KL/rx = 13.41 KL/rx = Kx*Lx*12/rxc (use rxc for combined section) fbx(bot/bm) =KL/ry = 80.49 KL/ry = Ky*Ly*12/ryc (use ryc for combined section) Is Lb<=Lc?KL/r = 80.49 KL/r = Max. of: Kx*Lx*12/rxc or Ky*Ly*12/ryc Is dc/tw<=allow?

Cc = #N/A #N/A Is b/t<=65/SQRT(Fy)?fa = 7.45 ksi fa = P/Ac (use Ac for combined section) Is b/t>95/SQRT(Fy)?

Fa = #N/A ksi #N/A Fbx(top) =Pa = #N/A kips Pa = Fa*Ac (use Ac for combined section) Fbx(top) =


Lc = 8.84 ft. Lc = Min. of: 76*bf/SQRT(Fy)/12 or: 20000/(dc/Af*Fy)/12 Fbx(top) =Lu = 9.97 ft. Lu = Max. of: ((20000/(dc/Af*Fy))/12 or: SQRT(102000/Fy)*rtc/12Fbx(top) =

Lb/rt = 64.06 Lb/rt = Lb*12/rtc = unbraced length for bending ratio Fbx(top) =fbx(residual) = 3.41 ksi fbx(residual) = Mx(residual)*12/Sx (at top and bottom of member)Use: Fbx(top) =fbx(add. top) = #N/A ksi #N/A Use: Fbx(bot) =fbx(add. bot) = #N/A ksi #N/A Mrx =

fbx(top/mem) = #N/A ksi fbx(top/mem) = fbx(residual)+fbx(added, top/mem) Y-axis Bending:fbx(bot/mem) = #N/A ksi fbx(bot/mem) = fbx(residual)+fbx(added, bot/mem) fby(top/mem) =

fbx(bot/bm) = #N/A ksi #N/A fby(bot/mem) =Fbx(top) = 20.52 ksi Fbx(top)=(2/3-Fy*(Lb/rf)^2/(1530000*Cb))*Fy <= Qs*0.60*Fy (Eqn. F1-6)fby(bot/bm) =Fbx(bot) = 21.60 ksi Fbx(bot) = allowable tension bending stress at bottom Fby =


X-axis Euler Stress: (use rxc for combined section) F'ex =F'ex = 831.02 ksi F'ex = 12*(π)^2*29000/(23*(Kx*Lx*12/rxc)^2) F'ey =

Stress Ratios:For Y-axis Bending: (for combined section) At Top Flange of Existing Member:

fby(top/mem) = #N/A ksi #N/A S.R. =fby(bot/mem) = #N/A ksi #N/A S.R. =

fby(bot/bm) = #N/A ksi #N/A S.R. =Fby = 27.00 ksi Fby = 0.75*Fy (Eqn. F2-1) S.R. =Mry = #N/A ft-kips #N/A At Bottom Flange of Existing Member:

S.R. =Y-axis Euler Stress: (use ryc for combined section) S.R. =

F'ey = 23.05 ksi F'ey = 12*(π)^2*29000/(23*(Ky*Ly*12/ryc)^2)S.R. =

Stress Ratios: S.R. =S.R.(top/mem) = #N/A #N/A #N/A Weld Design:S.R.(bot/mem) = #N/A #N/A #N/A

S.R.(bot/bm) = #N/A #N/A #N/A Fw =(continued)


13 of 28 02/09/2016 10:44:56

Weld Design:Lw(min) =

(assume intermittent fillet weld) Lw(use) =Fw = 0.287 kips/in. Fw = Rv*A2*Y4/(2*Ixc) (weld force per inch)

1/ 4 in.

0.3055 in. For Case #1: 1/ 4 in. Fw(total) =

Lw(min) = 1.500 in./ft. Lw(end) =Lw(use) = 1.500 in./ft. a'(min) =

(use) a'(req'd) =Use: 1/4'' Fillet Weld, 1.5'' @ 12'' Each Side of Beam Reinf. a'(use) =

Lw(total) =(assume continuous fillet weld)

For Case #1: Weld Across End of Beam Reinf. For Case #2:Fw(total) = 96.33 kips Fw(total) = M(cutoff)*12*A2*Y4/Ixc + (A2/Ac)*ABS(P(axial)) Fw(total) =Lw(end) = 4.010 in. Lw(end) = Min. of: bf1 or bf2 Lw(end) =

a'(min) = 4.010 in. a'(min) = bf2, for weld size >= 0.75*tf2 a'(min) =a'(req'd) = 10.969 in. a'(req'd) =

a'(use) = 10.969 in. a'(use) = Max. of: a'(min) or a'(req'd) a'(use) =Lw(total) = 25.949 in. Lw(total) = 2*a'(use)+Lw(end) (@ each end of beam reinf.)Lw(total) =

(use)Use: 1/4'' Fillet Weld, 11'' Continuous Each Side of Beam Reinf. and Across End ###

(Determining End Welds Now Including Eccentricity from C.G. of Reinforcing to Welds) For Case #2: No Weld Across End of Beam Reinf.

Fw(total) = 96.33 kips Fw(total) = M(cutoff)*12*A2*Y4/Ixc + (A2/Ac)*ABS(P(axial)) For Case #1:Lw(end) = 0.000 in. Lw(end) = 0 (assumed) Fw1 =

a'(min) = 8.020 in. a'(min) = 2*bf2, for no weld across end of beam reinf. Lw(end) =a'(req'd) = 12.974 in. a'(use) =

a'(use) = 12.974 in. a'(use) = Max. of: a'(min) or a'(req'd) Pw =Lw(total) = 25.949 in. Lw(total) = 2*a'(use) (@ each end of beam reinf.) Mw =

(use) Sw(max) =Use: 1/4'' Fillet Weld, 13'' Continuous Each Side of Beam Reinf. Sw(min) =

Fw(result) =

Comments: ### For Case #2:

Fw1 =Lw(end) =

a'(use) =Pw =Mw =Sw =

Fw(result) =

###############

(max,eff) =(use) =

Interior Welds of I-Beam Reinf. to Exist. Member:Note: =

(min) = (min) = Min. fillet weld size from AISC Table J2.4, page 5-67(max,eff) = (max eff) = Min. of: (tf1 or tf2)*(0.40*Fy/((SQRT(2)/2)*0.30*70))

(use) = (use) = (min) <= 5/16"Lw(min) = Max. of: 1.5" or 4*(min)Lw(use) = Max. of: (Fw*12)/(*SQRT(2)/2*0.3*70) or Lw(min)

Note: =

End Welds of I-Beam Reinf. to Exist. Member: Note: =

a'(req'd) = (Fw(total)/(*SQRT(2)/2*0.3*70)-Lw(end))/2

Note: = Note: =


Note: =

(req'd) =

(req'd) =


14 of 28 02/09/2016 10:44:56

STEEL MEMBER REINFORCEMENT ANALYSISExisting Member Reinforced with Four Symmetric Angles

Per AISC 9th Edition Manual (ASD)Job Name: Subject: SLV

Job Number: Originator: Checker: LLV###

Input Data: ######

Member and Angle Reinf. Sizes: Exist. W24x76 Properties: Y YesExist. Member = W24x76 A = 22.40 in.^2

Angle Size = L3x3x1/4 d = 23.900 in. SingleAngles Oriented = N.A. tw = 0.440 in. Reverse

bf = 8.990 in. d/2=11.95Member Loadings: tf = 0.680 in. Y1=11.95 Unbraced

P(total) = 10.00 kips rt = 2.290 in. N.A. XP(residual) = 2.00 kips d/Af = 3.91 xa=2.38 Section Ratios and Parameters:

Mx(total) = 338.00 ft-kips Ix = 2100.00 in.^4 Y2=11.95 ya=9.106Mx(cutoff) = 275.24 ft-kips Sx = 176.00 in.^3 (Typ.)

Mx(residual) = 50.00 ft-kips rx = 9.690 in. h =My = 0.00 ft-kips Iy = 82.50 in.^4 bf=8.99 h/tw =

Rv(shear) = 30.00 kips Sy = 18.40 in.^3 Qs =ry = 1.920 in. P(be) =

Design Parameters: be =Fy (mem./L's) = 36.00 ksi L3x3x1/4 Properties: Qa =

Kx = 1.00 Aa = 1.440 in.^2/ea. bf/(2*tf) = 6.61 Sxce =Ky = 1.00 da = 3.000 in. d/tw = 54.32 Syce =Lx = 12.000 ft. ba = 3.000 in. h = 22.540 in.

Ly = 12.000 ft. ta = 0.2500 in. h/tw = 51.23 Axial Compression:Lb = 12.000 ft. Ixa = 1.230 in.^4/ea. be = #N/A in.

Cb = 1.00 y = 0.836 in. Qs = 1.000 KL/ry =Cmx = 0.85 Iya = 1.230 in.^4/ea. Qa = #N/A KL/r =Cmy = 0.85 x = 0.836 in. Cc =ASIF = 1.000 wt./ft.= 19.56 lb./ft./4 L's fa(residual) =

Full Lgth. Reinf.? Yes fa(added) =#N/A fa =

Results: #N/A #N/A #N/A

#N/A #N/A #N/A

Combined Section Properties: Use: Fa =Y1 = 11.950 in. Y1 = d/2 fa/Fa =Y2 = 11.950 in. Y2 = d/2 Pa =xa = 2.384 in. xa = tw/2+(ba-x) X-axis Bending:ya = 9.106 in. ya = d/2-tf-(da-y) Lc =Ac = 28.16 in.^2 Ac = A+4*Aa Lu =Ixc = 2582.53 in.^4 Ixc = Ix+4*(Ixa+Aa*ya^2) Lb/rt =

Sxc = 216.11 in.^3 Sxc = Ixc/(d/2) (for both top and bottom of existing member) fa/Fy =Sxc(req'd) = 193.03 in^3 Sxc(req'd) = (Mx(total)-Mx(residual))*12/(Fbx(total)-fbx(resid.))fbx(residual) =

rxc = 9.576 in. rxc = SQRT(Ixc/Ac) fbx(added) =Iyc = 120.16 in.^4 Iyc = Iy+4*(Iya+Aa*xa^2) fbx(top) =

Syc = 26.73 in.^3 Syc = Iyc/(bf/2) fbx(bot) =ryc = 2.066 in. ryc = SQRT(Iyc/Ac) Is Lb<=Lc?

(continued)

Section Ratios & Parameters:


15 of 28 02/09/2016 10:44:56

Is b/t<=65/SQRT(Fy)?Check Stresses: Is b/t>95/SQRT(Fy)?

Fbx(top) =For Axial Compression: (use Ac for combined section) Fbx(top) =

KL/rx = 15.04 KL/rx = Kx*Lx*12/rxc (use rxc for combined section) Fbx(top) =KL/ry = 69.71 KL/ry = Ky*Ly*12/ryc (use ryc for combined section) Fbx(top) =KL/r = 69.71 KL/r = Max. of: Kx*Lx*12/rxc or Ky*Ly*12/ryc Fbx(top) =

Cc = #N/A #N/A Fbx(top) =fa(residual) = 0.09 ksi fa(residual) = P(residual)/A Fbx(top) =

fa(added) = 0.28 ksi fa(added) = (P(total)-P(residual))/Ac (use Ac for combined section)Use: Fbx(top) =fa = 0.37 ksi fa = fa(residual)+fa(added) Use: Fbx(bot) =

Fa = #N/A ksi #N/A Mrx =Pa = #N/A kips Pa = Fa*Ac (use Ac for combined section) Y-axis Bending:

#N/A fby =For X-axis Bending: (for combined section) Fby =

Lc = 9.49 ft. Lc = Min. of: 76*bf/SQRT(Fy)/12 or: 20000/(d/Af*Fy)/12 Mry =Lu = 11.84 ft. Lu = Max. of: ((20000/(d/Af*Fy))/12 or: SQRT(102000/Fy)*rt/12 Euler Stresses:

Lb/rt = 62.88 Lb/rt = Lb*12/rt = unbraced length for bending ratio F'ex =fbx(residual) = 3.41 ksi fbx(residual) = Mx(residual)*12/Sx F'ey =

fbx(added) = #N/A ksi #N/A Stress Ratios:fbx(top) = #N/A ksi fbx(top) = fbx(residual)+fbx(added) At Top Flange of Existing Member:fbx(bot) = #N/A ksi fbx(bot) = fbx(residual)+fbx(added) S.R. =

Fbx(top) = 21.31 ksi Fbx(top) = 12000*Cb/(Lb*dc/Af) <= Qs*0.60*Fy (Eqn. F1-8) S.R. =Fbx(bot) = 21.60 ksi Fbx(bot) = allowable tension bending stress at bottom flange S.R. =

Mrx = #N/A ft-kips #N/A S.R. =At Bottom Flange of Existing Member:

X-axis Euler Stress: (use rxc for combined section) S.R. =F'ex = 660.45 ksi F'ex = 12*(π)^2*29000/(23*(Kx*Lx*12/rxc)^2) S.R. =

Weld Design:For Y-axis Bending: (for combined section) Interior Welds of Each Reinf. Angle to Member:

fby = #N/A ksi #N/A Fw =Fby = 27.00 ksi Fby = 0.75*Fy (Eqn. F2-1)Mry = #N/A ft-kips #N/A

Y-axis Euler Stress: (use ryc for combined section) Lw(min) =F'ey = 30.73 ksi F'ey = 12*(π)^2*29000/(23*(Ky*Ly*12/ryc)^2) Lw(use) =

Stress Ratios: End Welds of Each Reinf. Angle to Member:S.R.(top) = #N/A #N/A #N/A Fw(total) =S.R.(bot) = #N/A #N/A #N/A a'(min) =

a'(req'd) =a'(use) =

Lw(total) =

#(Determining End Welds Now Including Eccentricity from C.G. of Reinforcing to Welds)End Welds of Each Reinf. Angle to Member:

Fw1 =a'(use) =

Pw =(continued)

(min) =(max,eff) =

(use) =

Note: =

Note: =


16 of 28 02/09/2016 10:44:56

Sw =Weld Design: Fw(result) =

Interior Welds of Each Reinf. Angle to Member: (assume intermittent fillet weld) ###Fw = 0.038 kips/in. Fw = (Rv/2)*Aa*ya/(2*Ixc) (weld force per inch) ###

1/ 4 in. ###0.2133 in. ### 1/ 4 in. ###

Lw(min) = 1.500 in./ft. ###Lw(use) = 1.500 in./ft. ###

(max eff) ###Use: 1/4'' Fillet Weld, 1.5'' @ 12'' Each Side of Reinf. Angles ###

###End Welds of Each Reinf. Angle to Member: (assume continuous fillet weld) ###

Fw(total) = 8.90 kips/angle Fw(total) = M(cutoff)/2*12*Aa*ya/Ixc + (Aa/Ac)*ABS(P(total)) ###a'(min) = 3.000 in./angle a'(min) = 3" (assumed) ###

a'(req'd) = 1.404 in./angle ###a'(use) = 3.000 in./angle a'(use) = Max. of: a'(min) or a'(req'd) ###

Lw(total) = 6.000 in./angle Lw = 2*a'(use) (@ each end of angle reinf.) ###(max eff) ###

Use: 1/4'' Fillet Weld, 3'' Continuous Each Side of Reinf. Angles @ Ends #########

Comments: ####################################################################################

(req'd) =

(min) = (min) = Min. fillet weld size from AISC Table J2.4, page 5-67(max,eff) = (max,eff) = tw*(0.40*Fy/((SQRT(2)/2)*0.30*70*2))

(use) = (use) = Min. of: (min) or (max,eff) <= 5/16"Lw(min) = Max. of: 1.5" or 4*(min)Lw(use) = Max. of: (Fw*12)/(*SQRT(2)/2*0.3*70) or Lw(min)

Note: =

a'(req'd) = (Fw(total)/((SQRT(2)/2)*0.3*70*))/2

Note: =


17 of 28 02/09/2016 10:44:56

STEEL MEMBER REINFORCEMENT ANALYSISExisting Member Reinforced with Four Symmetric Round Bars




Member and Bar Reinf. Sizes: Exist. W24x76 Properties: Y ReverseExist. Member = W24x76 A = 22.40 in.^2

Bar Dia., Db = 1.0000 in. d = 23.900 in. Unbracedtw = 0.440 in. yb =

Member Loadings: bf = 8.990 in. d/2=11.95P(total) = 10.00 kips tf = 0.680 in. Y1=11.95 Ixc =

P(residual) = 2.00 kips rt = 2.290 in. N.A. XMx(total) = 338.00 ft-kips d/Af = 3.91 xb=0.72 Sxc(req'd) =

Mx(cutoff) = 275.24 ft-kips Ix = 2100.00 in.^4 Y2=11.95 yb=10.77Mx(residual) = 50.00 ft-kips Sx = 176.00 in.^3 (Typ.)

My = 0.00 ft-kips rx = 9.690 in. Syc =Rv(shear) = 30.00 kips Iy = 82.50 in.^4 bf=8.99 ryc =

Sy = 18.40 in.^3 Section Ratios and Parameters:Design Parameters: ry = 1.920 in. bf/(2*tf) =Fy (mem./bars) = 36.00 ksi d/tw =

Kx = 1.00 1'' Dia. Bar Properties: h =Ky = 1.00 Ab = 0.785 in.^2/ea. bf/(2*tf) = 6.61 h/tw =Lx = 12.000 ft. Ixb = 0.049 in.^4/ea. d/tw = 54.32 Qs =Ly = 12.000 ft. Iyb = 0.049 in.^4/ea. h = 22.540 in.

Lb = 12.000 ft. wt./ft.= 10.69 lb./ft./4 bars h/tw = 51.23 be =Cb = 1.00 be = #N/A in.

Cmx = 0.85 Qs = 1.000 Sxce =Cmy = 0.85 Qa = #N/A Syce =ASIF = 1.000 Check Stresses:

Full Lgth. Reinf.? Yes #N/A Axial Compression:#N/A #N/A #N/A

Results: #N/A #N/A #N/A

KL/r =Combined Section Properties: Cc =

Y1 = 11.950 in. Y1 = d/2 fa(residual) =Y2 = 11.950 in. Y2 = d/2 fa(added) =xb = 0.720 in. xb = tw/2+Db/2 fa =yb = 10.770 in. yb = d/2-tf-Db/2 Fa =Ac = 25.54 in.^2 Ac = A+4*Ab Fa =Ixc = 2464.60 in.^4 Ixc = Ix+4*(Ixb+Ab*yb^2) Use: Fa =

Sxc = 206.24 in.^3 Sxc = Ixc/(d/2) (for both top and bottom of existing member) fa/Fa =Sxc(req'd) = 193.03 in^3 Sxc(req'd) = (Mx(total)-Mx(residual))*12/(Fbx(total)-fbx(resid.)) Pa =

rxc = 9.823 in. rxc = SQRT(Ixc/Ac) X-axis Bending:Iyc = 84.32 in.^4 Iyc = Iy+4*(Iyb+Ab*xb^2) Lc =

Syc = 18.76 in.^3 Syc = Iyc/(bf/2) Lu =ryc = 1.817 in. ryc = SQRT(Iyc/Ac) Lb/rt =

(continued)



18 of 28 02/09/2016 10:44:56

fbx(residual) =Check Stresses: fbx(added) =

fbx(top) =For Axial Compression: (use Ac for combined section) fbx(bot) =

KL/rx = 14.66 KL/rx = Kx*Lx*12/rxc (use rxc for combined section) Is Lb<=Lc?KL/ry = 79.25 KL/ry = Ky*Ly*12/ryc (use ryc for combined section) Is dc/tw<=allow?KL/r = 79.25 KL/r = Max. of: Kx*Lx*12/rxc or Ky*Ly*12/ryc Is b/t<=65/SQRT(Fy)?

Cc = #N/A #N/A Is b/t>95/SQRT(Fy)?fa(residual) = 0.09 fa(residual) = P(residual)/A Fbx(top) =

fa(added) = 0.31 fa(added) = (P(total)-P(residual))/Ac (use Ac for combined section)Fbx(top) =fa = 0.40 ksi fa = fa(residual)+fa(added) Fbx(top) =



Lc = 9.49 ft. Lc = Min. of: 76*bf/SQRT(Fy)/12 or: 20000/(d/Af*Fy)/12Use: Fbx(top) =Lu = 11.84 ft. Lu = Max. of: ((20000/(d/Af*Fy))/12 or: SQRT(102000/Fy)*rt/12Use: Fbx(bot) =

Lb/rt = 62.88 Lb/rt = Lb*12/rt = unbraced length for bending ratio Mrx =fbx(residual) = 3.41 ksi fbx(residual) = Mx(residual)*12/Sx Y-axis Bending:

fbx(added) = #N/A ksi #N/A fby =fbx(top) = #N/A ksi fbx(top) = fbx(residual)+fbx(added) Fby =fbx(bot) = #N/A ksi fbx(bot) = fbx(residual)+fbx(added) Mry =

Fbx(top) = 21.31 ksi Fbx(top) = 12000*Cb/(Lb*dc/Af) <= Qs*0.60*Fy (Eqn. F1-8) Euler Stresses:Fbx(bot) = 21.60 ksi Fbx(bot) = allowable tension bending stress at bottom flange F'ex =

Mrx = #N/A ft-kips #N/A F'ey =Stress Ratios:

X-axis Euler Stress: (use rxc for combined section) At Top Flange of Existing Member:F'ex = 694.90 ksi F'ex = 12*(π)^2*29000/(23*(Kx*Lx*12/rxc)^2) S.R. =

S.R. =For Y-axis Bending: (for combined section) S.R. =

fby = #N/A ksi #N/A S.R. =Fby = 27.00 ksi Fby = 0.75*Fy (Eqn. F2-1) At Bottom Flange of Existing Member:Mry = #N/A ft-kips #N/A S.R. =

S.R. =Y-axis Euler Stress: (use ryc for combined section) Weld Design:

F'ey = 23.78 ksi F'ey = 12*(π)^2*29000/(23*(Ky*Ly*12/ryc)^2) Interior Welds of Each Reinf. Bar to Member:Fw =

Stress Ratios:S.R.(top) = #N/A #N/A #N/AS.R.(bot) = #N/A #N/A #N/A

Lw(min) =Lw(use) =

End Welds of Each Reinf. Bar to Member:Fw(total) =

a'(min) =a'(req'd) =

a'(use) =Lw(total) =

(continued)

(min) =(max,eff) =

(use) =

Note: =


19 of 28 02/09/2016 10:44:56

#Weld Design: (Determining End Welds Now Including Eccentricity from C.G. of Reinforcing to Welds)

End Welds of Each Reinf. Bar to Member:Interior Welds of Each Reinf. Bar to Member: (assume intermittent fillet weld) Fw1 =

Fw = 0.026 kips/in. Fw = (Rv/2)*Ab*yc/(2*Ixc) (weld force per inch) a'(use) = 5/16 in. Pw =

0.2133 in. Mw = 5/16 in. Sw =

Lw(min) = 1.500 in./ft. Fw(result) =Lw(use) = 1.500 in./ft.

(max,eff) ###Use: 5/16'' Fillet Weld, 1.5'' @ 12'' Each Side of Reinf. Bars ###

###End Welds of Each Reinf. Bar to Member: (assume continuous fillet weld) ###

Fw(total) = 5.98 kips/bar Fw(total) = M(cutoff)/2*12*Ab*yc/Ixc + (Ab/Ac)*ABS(P(total)) ###a'(min) = 3.00 in./bar a'(min) = 3" (assumed) ###

a'(req'd) = 0.94 in./bar ###a'(use) = 3.00 in./bar a'(use) = Max. of: a'(min) or a'(req'd) ###

Lw(total) = 6.00 in./bar Lw(total) = 2*a'(use) (@ each end of bar reinf.) ###(max,eff) ###

Use: 5/16'' Fillet Weld, 3'' Continuous Each Side of Reinf. Bars @ Ends #########

Comments: #################################################################################

(min) = (min) = Min. fillet weld size from AISC Table J2.4, page 5-67(max,eff) = (max,eff) = tw*(0.40*Fy/((SQRT(2)/2)*0.30*70*2))

(use) = (use) = (min) <= 5/16"Lw(min) = Max. of: 1.5" or 4*(min)Lw(use) = Max. of: (Fw*12)/(*SQRT(2)/2*0.3*70) or Lw(min)(req'd) =

Note: =

a' = (Fw(total)/((SQRT(2)/2)*0.3*70*))/2

Note: =


20 of 28 02/09/2016 10:44:56

STEEL MEMBER REINFORCEMENT ANALYSISExisting Member Reinforced with Equal Cover Plates on Both Flanges




Member and Plate Reinf. Sizes: Exist. W24x68 Properties: Y ReverseExist. Member = W24x68 A = 20.10 in.^2 Bp=10 Plate Thk., tp = 1.0000 in. d = 23.700 in. tp=1

Plate Width, Bp = 10.000 in. tw = 0.415 in. Ixc =bf = 8.970 in. Y1=12.85 d/2=11.85

Member Loadings: tf = 0.585 in. (Typ.)

P(total) = 10.00 kips rt = 2.260 in. XP(residual) = 20.00 kips d/Af = 4.52 rxc =

Mx(total) = 338.00 ft-kips Ix = 1830.00 in.^4 Y2=12.85 yp=12.35Mx(cutoff) = 275.24 ft-kips Sx = 154.00 in.^3 (Typ.)

Mx(residual) = 50.00 ft-kips rx = 9.550 in. Syc(pls) =My = 10.00 ft-kips Iy = 70.40 in.^4 Bp=10 ryc =

Rv(shear) = 30.00 kips Sy = 15.70 in.^3 rtc =ry = 1.870 in. dc/Afc =

Design Parameters: Section Ratios and Parameters:Fy (mem./plts.) = 36.00 ksi 1''x10'' Plate Properties: b/(2*tfc) =

Kx = 1.00 Ap = 10.000 in.^2/ea. b/(2*tfc) = 3.15 dc/tw =Ky = 1.00 Ixp = 0.83 in.^4/ea. dc/tw = 61.93 h =Lx = 12.000 ft. Iyp = 83.33 in.^4/ea. h = 22.530 in.

Ly = 12.000 ft. wt./ft.= 68.06 lb./ft./2 plts. h/tw = 54.29 Qs =Lb = 12.000 ft. be = #N/A in.

Cb = 1.00 Qs = 1.000 be =Cmx = 0.85 Qa = #N/A Qa =Cmy = 0.85 Sxce(mem) =ASIF = 1.000 #N/A Sxce(pls) =

Full Lgth. Reinf.? Yes #N/A #N/A #N/A #N/A #N/A


Results: Check Stresses:Axial Compression:

Combined Section Properties: KL/rx =xp = 0.000 in. xp = 0 (plates assumed centered on existing member) KL/ry =yp = 12.350 in. yp = d/2+tp/2 KL/r =Ac = 40.10 in.^2 Ac = A+2*Ap Cc =Ixc = 4882.12 in.^4 Ixc = Ix+2*(Ixp+Ap*yp^2) fa(residual) =

Sxc(mem) = 411.99 in.^3 Sxc(mem) = Ixc/(d/2) (ref. to outside of flanges of exist. member)fa(added) =Sxc(pls) = 379.93 in.^3 Sxc(pls) = Ixc/(d/2+tp) (referenced to outside of top or bottom plate)fa =

rxc = 11.034 in. rxc = SQRT(Ixc/Ac) Fa =Iyc = 237.07 in.^4 Iyc = Iy+2*Iyp Fa =

Syc(mem) = 52.86 in.^3 Syc(mem) = Iyc/(bf/2) (ref. outside edge of flange of exist. member)Use: Fa =Syc(pls) = 47.41 Syc(pls) = Iyc/(Bp/2) (referenced outside edge of reinforcing plate)fa/Fa =

ryc = 2.431 in. ryc = SQRT(Iyc/Ac) Pa =rtc = 2.656 in. rtc = SQRT((tp*Bp^3/12+tf*bf^3/12+(d/2-tf)/3*tw^3/12)/(Bp*tp+bf*tf+(d/2-tf)/3*tw)) X-axis Bending:

dc/Afc = 1.686 dc/Af = (d+2*tp)/(Bp*tp+bf*tf) Lc =(continued)


N.A.


21 of 28 02/09/2016 10:44:56

Lb/rtc =Check Stresses: fa/Fy =

fbx(residual) =For Axial Compression: (use Ac for combined section) fbx(added, mem) =

KL/rx = 13.05 KL/rx = Kx*Lx*12/rxc (use rxc for combined section) fbx(pls) =KL/ry = 59.22 KL/ry = Ky*Ly*12/ryc (use ryc for combined section) fbx(mem) =KL/r = 59.22 KL/r = Max. of: Kx*Lx*12/rxc or Ky*Ly*12/ryc Is Lb<=Lc?

Cc = #N/A #N/A Is dc/tw<=allow?fa(residual) = 1.00 ksi fa(residual) = P(residual)/A Is b/t<=65/SQRT(Fy)?

fa(added) = 0.25 ksi fa(added) = (P(total)-P(residual))/Ac (use Ac for combined section)Is b/t>95/SQRT(Fy)?fa = 1.24 ksi fa = fa(residual)+fa(added) Fbx(top) =


Fbx(top) =For X-axis Bending: (for combined section) Fbx(top) =

Lc = 10.56 ft. Lc = Min. of: 76*MAX(Bp,bf)/SQRT(Fy)/12 or: 20000/(dc/Afc*Fy)/12Fbx(top) =Lu = 27.47 ft. Lu = Max. of: ((20000/(dc/Afc*Fy))/12 or: SQRT(102000/Fy)*rtc/12Fbx(top) =

Lb/rt = 54.22 Lb/rtc = Lb*12/rtc = unbraced length for bending ratio Use: Fbx(top) =fbx(residual) = 3.90 ksi fbx(residual) = Mx(residual)*12/Sx Use: Fbx(bot) =

fbx(added) = #N/A ksi #N/A Mrx =fbx(pls) = #N/A ksi #N/A Y-axis Bending:

fbx(mem) = #N/A ksi fbx(top&bot/mem) = fbx(residual)+fbx(added, mem) fby(pls) =Fbx(top) = 21.60 ksi Fbx(top) = 12000*Cb/(Lb*dc/Afc) <= Qs*0.60*Fy (Eqn. F1-8) fby(mem) =Fbx(bot) = 21.60 ksi Fbx(bot) = allowable tension bending stress at bottom cover plateFby =


X-axis Euler Stress: (use rxc for combined section) F'ex =F'ex = 876.78 ksi F'ex = 12*(π)^2*29000/(23*(Kx*Lx*12/rxc)^2) F'ey =

Stress Ratios:For Y-axis Bending: (for combined section) At Top of Upper Cover Plate:

fby(pls) = #N/A ksi #N/A S.R. =fby(mem) = #N/A ksi #N/A S.R. =

Fby = 27.00 ksi Fby = 0.75*Fy (Eqn. F2-1) S.R. =Mry = #N/A ft-kips #N/A S.R. =

At Top of Existing Member:Y-axis Euler Stress: (use ryc for combined section) S.R. =

F'ey = 42.57 ksi F'ey = 12*(π)^2*29000/(23*(Ky*Ly*12/ryc)^2) S.R. =S.R. =

Stress Ratios: S.R. =S.R.(top/pl) = #N/A #N/A #N/A At Bottom of Existing Member:

S.R.(top/mem) = #N/A #N/A #N/A S.R. =S.R.(bot/mem) = #N/A #N/A #N/A S.R. =

S.R.(bot/pl) = #N/A #N/A #N/A At Bottom of Lower Cover Plate:S.R. =S.R. =

Weld Design:Interior Welds of Plate Reinf. to Member:

Fw =

(continued)

(min) =(max,eff) =

(use) =


22 of 28 02/09/2016 10:44:56

Lw(use) =Weld Design:

End Welds of Plate Reinf. to Member:Interior Welds of Plate Reinf. to Member: (assume intermittent fillet weld) For Case #1:

Fw = 0.379 kips/in. Fw = Rv*Ap*yp/(2*Ixc) (weld force per inch) Fw(total) = 5/16 in. Lw(end) =

0.5673 in. a'(min) = 5/16 in. a'(req'd) =

Lw(min) = 1.500 in./ft. a'(use) =Lw(use) = 1.500 in./ft. Lw(total) =

(use)Use: 5/16'' Fillet Weld, 1.5'' @ 12'' Each Side of Plates For Case #2:

Fw(total) =End Welds of Plate Reinf. to Member: (assume continuous fillet weld) Lw(end) =

For Case #1: Weld Across End of Plates a'(min) =Fw(total) = 86.04 kips Fw(total) = M(cutoff)*12*Ap*yp/Ixc + (Ap/Ac)*ABS(P(total)) a'(req'd) =Lw(end) = 8.970 in. Lw(end) = Min. of: bf or Bp a'(use) =

a'(min) = 15.000 in. a'(min) = 1.5*Bp, for weld size < 0.75*tp Lw(total) =a'(req'd) = 4.786 in.

a'(use) = 15.000 in. a'(use) = Max. of: a'(min) or a'(req'd) ###Lw(total) = 38.970 in. Lw(total) = 2*a'(use)+Lw(end) (@ each end of plate reinf.) (Determining End Welds Now Including Eccentricity from C.G. of Reinforcing to Welds)

(use) End Welds of Plate Reinf. to Member:Use: 5/16'' Fillet Weld, 15'' Continuous Each Side of Plates and Across Ends For Case #1:

Fw1 = For Case #2: No Weld Across End of Plates Lw(end) =

Fw(total) = 86.04 kips Fw(total) = M(cutoff)*12*Ap*yp/Ixc + (Ap/Ac)*ABS(P(total)) a'(use) =Lw(end) = 0.000 in. Lw(end) = 0 (assumed) Pw =

a'(min) = 20.000 in. a'(min) = 2*Bp, for no weld across end of plate Mw =a'(req'd) = 9.271 in. Sw(max) =

a'(use) = 20.000 in. a'(use) = Max. of: a'(min) or a'(req'd) Sw(min) =Lw(total) = 40.000 in. Lw(total) = 2*a'(use) (@ each end of plate) Fw(result) =

(use)Use: 5/16'' Fillet Weld, 20'' Continuous Each Side of Plates ###

For Case #2:Fw1 =

Comments: Lw(end) =a'(use) =

Pw =Mw =Sw =

Fw(result) =

##############################

Note: =

(min) = (min) = Min. fillet weld size from AISC Table J2.4, page 5-67(max,eff) = (max,eff) = Min. of: (tp or tf)*(0.40*Fy/((SQRT(2)/2)*0.30*70))


Note: = Note: =

a'(req'd) = (Fw(total)/(*SQRT(2)/2*0.3*70)-Lw(end))/2 Note: =

Note: =


Note: = (req'd) =

(req'd) =


23 of 28 02/09/2016 10:44:56

STEEL MEMBER REINFORCEMENT ANALYSISExisting Member Reinforced with WT Sections (toed in) on Each Side of Member Web




Member and Plate Reinf. Sizes: Exist. W24x76 Properties: Y ReverseExist. Member = W24x76 A = 22.40 in.^2

WT Reinf. = WT7x45 d = 23.900 in. Unbracedtw = 0.440 in. Iyc =

Member Loadings: bf = 8.990 in. d/2=11.95P(total) = 10.00 kips tf = 0.680 in. Syc(WT's) =

P(residual) = 2.00 kips rt = 2.290 in. XMx(total) = 338.00 ft-kips d/Af = 3.91 Section Ratios and Parameters:

Mx(cutoff) = 275.24 ft-kips Ix = 2100.00 in.^4 xwt=6.14 d/2=11.95Mx(residual) = 50.00 ft-kips Sx = 176.00 in.^3 (Typ.) bfwt/(2*tfwt) =

Rv(shear) = 30.00 kips rx = 9.690 in. d/tw =Iy = 82.50 in.^4 bf=8.99 h =

Sy = 18.40 in.^3 h/tw =Design Parameters: ry = 1.920 in. Qs(mem) =Fy (mem./WT's) = 36.00 ksi Qs(WT) =

Kx = 1.00 WT7x45 Properties: Qs =Ky = 1.00 A = 13.20 in.^2 b/t(mem) = 6.61 P(be) =Lx = 12.000 ft. d = 7.010 in. b/t(WT) = 10.21 be =Ly = 12.000 ft. tw = 0.440 in. d/tw = 54.32 Qa =Lb = 12.000 ft. bf = 14.500 in. h = 22.540 in.

Cb = 1.00 tf = 0.710 in. h/tw = 51.23 Sxce(WT's) =Cmx = 0.85 Ix = 36.50 in.^4 be = #N/A in.

ASIF = 1.000 y = 1.090 in. Qs(mem) = 1.000 Syce(WT's) =Full Lgth. Reinf.? Yes Iy = 181.00 in.^4 Qs(WT) = 1.000 Check Stresses:

Qa = #N/A Axial Compression:#N/A KL/rx =



Combined Section Properties: Cc =xwt = 6.140 in. xwt =dtw-y+tw/2 fa(residual) =ywt = 0.000 in. ywt = 0 (WT's assumed centered on existing member) fa(added) =Ac = 48.80 in.^2 Ac = A+2*Awt fa =Ixc = 2462.00 in.^4 Ixc = Ix+2*Iywt Fa =

Sxc(mem) = 206.03 in.^3 Sxc(mem) = Ixc/(d/2) (ref. to top and bottom flange of member) Fa =Sxc(WT's) = 339.59 in^3 Sxc(WT's) = Ixc/(bfwt/2) (referenced to tips of flanges of WT's)Use: Fa =

rxc = 7.103 in. rxc = SQRT(Ixc/Ac) fa/Fa =Iyc = 1150.77 in.^4 Iyc = Iy+2*(Ixwt+Awt*xwt^2) Pa =

Syc(mem) = 256.01 in.^3 Syc(mem) = Iyc/(bf/2) (referenced to tips of flanges of member) X-axis Bending:Syc(WT's) = 159.17 in.^3 Syc(WT's) = Iyc/(dwt+tw/2) (ref. to outside edges of flanges of WT)Lc =

ryc = 4.856 in. ryc = SQRT(Iyc/Ac) Lu =Lb/rt =

(continued)


N.A.


24 of 28 02/09/2016 10:44:57

fbx(residual) =Check Stresses: fbx(added, mem) =

fbx(mem) =For Axial Compression: (use Ac for combined section) fbx(WT's) =


Cc = #N/A #N/A Is b/t>95/SQRT(Fy)?fa(residual) = 0.09 ksi fa(residual) = P(residual)/A Fbx(top) =

fa(added) = 0.16 ksi fa(added) = (P(total)-P(residual))/Ac (use Ac for combined section)Fbx(top) =fa = 0.25 ksi fa = fa(residual)+fa(added) Fbx(top) =




Lb/rt = 62.88 Lb/rt = Lb*12/rt = unbraced length for bending ratio Mrx =fbx(residual) = 3.41 ksi fbx(residual) = Mx(residual)*12/Sx Euler Stresses:

fbx(added) = #N/A ksi #N/A F'ex =fbx(mem) = #N/A ksi fbx(top&bot/mem) = fbx(residual)+fbx(added, mem) Stress Ratios:fbx(WT's) = #N/A ksi #N/A At Top Flange of Existing Member:

Fbx(top) = 21.31 ksi Fbx(top) = 12000*Cb/(Lb*dc/Af) <= Qs*0.60*Fy (Eqn. F1-8) S.R. =Fbx(bot) = 21.60 ksi Fbx(bot) = allowable tension bending stress at bottom cover plateS.R. =

Mrx = #N/A ft-kips #N/A S.R. =S.R. =

X-axis Euler Stress: (use rxc for combined section) At Upper Outside Edge of Flange of WT Reinf.:F'ex = 363.32 ksi F'ex = 12*(π)^2*29000/(23*(Kx*Lx*12/rxc)^2) S.R. =

S.R. =Stress Ratios: S.R. =

S.R.(top/mem) = #N/A #N/A #N/A S.R. =S.R.(top/WT) = #N/A #N/A #N/A At Lower Outside Edge of Flange of WT Reinf.:S.R.(bot/WT) = #N/A #N/A #N/A S.R. =

S.R.(bot/mem) = #N/A #N/A #N/A S.R. =At Bottom Flange of Existing Member:

S.R. =S.R. =

Weld Design:Interior Welds of WT Reinf. to Member:

Fw =

Lw(min) =Lw(use) =

End Welds of WT Reinf. to Member:Fw(total) =

a'(min) =(continued)

(min) =(max,eff) =

(use) =

Note: =


25 of 28 02/09/2016 10:44:57

a'(use) =Weld Design: Lw(total) =

Interior Welds of Plate Reinf. to Member: (assume intermittent fillet weld) ###Fw = 0.000 kips/in. Fw = Rv*Awt*ywt/(2*Ixc) (weld force per inch) (Determining End Welds Now Including Eccentricity from C.G. of Reinforcing to Welds)

3/16 in. End Welds of WT Reinf. to Member:0.2133 in. Fw1 = 3/16 in. a'(use) =

Lw(min) = 1.500 in./ft. Pw =Lw(use) = 1.500 in./ft. Mw =

(use) Sw =Use: 3/16'' Fillet Weld, 1.5'' @ 12'' Each Side of Plates Fw(result) =

End Welds of WT Reinf. to Member: (assume continuous fillet weld) ###Fw(total) = 2.70 kips Fw(total) = Mx(cutoff)*12*Awt*ywt/Ixc + (Awt/Ac)*ABS(P(total)) ###

a'(min) = 3.000 in. a'(min) = 2*Lw(use) (assumed) ###a'(req'd) = 0.486 in. a'(req'd) = (Fw(total)/((SQRT(2)/2)*0.3*70*w))/2 ###

a'(use) = 3.000 in. a'(use) = Max. of: a'(min) or a'(req'd) ###Lw(total) = 6.000 in. Lw(total) = 2*a(use)' (@ each end of WT reinf.) ###

(use) ###Use: 3/16'' Fillet Weld, 3'' Continuous Each Side of WT Web @ Ends ###

######

Comments: #################################################################################

Note: =

(min) = (min) = Min. fillet weld size from AISC Table J2.4, page 5-67(max,eff) = (max,eff) = Min. of: (tp or tf)*(0.40*Fy/((SQRT(2)/2)*0.30*70))


Note: =

(req'd) =

Note: =


26 of 28 02/09/2016 10:44:57

STEEL MEMBER REINFORCEMENT ANALYSISExisting Member Reinforced with WT Sections (toed out) on Each Side of Member Web




Member and Plate Reinf. Sizes: Exist. W24x76 Properties: Y ReverseExist. Member = W24x76 A = 22.40 in.^2

WT Reinf. = WT7x11 d = 23.900 in. Unbracedtw = 0.440 in. Iyc =

Member Loadings: bf = 8.990 in. d/2=11.95P(total) = 10.00 kips tf = 0.680 in. Syc(WT's) =

P(residual) = 2.00 kips rt = 2.290 in. XMx(total) = 338.00 ft-kips d/Af = 3.91 Section Ratios and Parameters:

Mx(cutoff) = 275.24 ft-kips Ix = 2100.00 in.^4 xwt=1.98 d/2=11.95Mx(residual) = 50.00 ft-kips Sx = 176.00 in.^3 (Typ.) d/tw =

Rv(shear) = 30.00 kips rx = 9.690 in. dwt/twwt =Iy = 82.50 in.^4 bf=8.99 h =

Sy = 18.40 in.^3 h/tw =Design Parameters: ry = 1.920 in. Qs(mem) =Fy (mem./WT's) = 36.00 ksi Qs(WT) =

Kx = 1.00 WT7x11 Properties: Qs =Ky = 1.00 A = 3.25 in.^2 bf/(2*tf) = 6.61 P(be) =Lx = 12.000 ft. d = 6.870 in. d/tw = 54.32 be =Ly = 12.000 ft. tw = 0.230 in. dwt/twwt = 14.93 Qa =Lb = 12.000 ft. bf = 5.000 in. h = 22.540 in.

Cb = 1.00 tf = 0.335 in. h/tw = 51.23 Sxce(WT's) =Cmx = 0.85 Ix = 14.80 in.^4 be = #N/A in.

ASIF = 1.000 y = 1.760 in. Qs(mem) = 1.000 Syce(WT's) =Full Lgth. Reinf.? Yes Iy = 3.50 in.^4 Qs(WT) = 1.000 Check Stresses:

Qa = #N/A Axial Compression:#N/A KL/rx =



Combined Section Properties: Cc =xwt = 1.980 in. xwt=y+tw/2 fa(residual) =ywt = 0.000 in. ywt = 0 (WT's assumed centered on existing member) fa(added) =Ac = 28.90 in.^2 Ac = A+2*Awt fa =Ixc = 2107.00 in.^4 Ixc = Ix+2*Iywt Fa =

Sxc(mem) = 176.32 in.^3 Sxc(mem) = Ixc/(d/2) (for both top and bottom of existing member)Fa =Sxc(WT's) = 18321.7 in^3 Sxc(WT's) = Ixc/(twwt/2) (referenced to tips of webs of WT's)Use: Fa =

rxc = 8.539 in. rxc = SQRT(Ixc/Ac) fa/Fa =Iyc = 137.58 in.^4 Iyc = Iy+2*(Ixwt+Awt*xwt^2) Pa =

Syc(mem) = 30.61 in.^3 Syc(mem) = Iyc/(bf/2) (ref. to tips of flanges of exist. member) X-axis Bending:Syc(WT's) = 19.41 in.^3 Syc(WT's) = Iyc/(dtw+tw/2) (referenced to tips of web of WT) Lc =

ryc = 2.182 in. ryc = SQRT(Iyc/Ac) Lu =Lb/rt =

(continued)


N.A.N.A.N.A.


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fbx(residual) =Check Stresses: fbx(added, mem) =

fbx(mem) =For Axial Compression: (use Ac for combined section) fbx(WT's) =


Cc = #N/A #N/A Is b/t>95/SQRT(Fy)?fa(residual) = 0.09 ksi fa(residual) = P(residual)/A Fbx(top) =

fa(added) = 0.28 ksi fa(added) = (P(total)-P(residual))/Ac (use Ac for combined section)Fbx(top) =fa = 0.37 ksi fa = fa(residual)+fa(added) Fbx(top) =




Lb/rt = 62.88 Lb/rt = Lb*12/rt = unbraced length for bending ratio Mrx =fbx(residual) = 3.41 ksi fbx(residual) = Mx(residual)*12/Sx Euler Stresses:

fbx(added) = #N/A ksi #N/A F'ex =fbx(mem) = #N/A ksi fbx(top&bot/mem) = fbx(residual)+fbx(added, mem) Stress Ratios:fbx(WT's) = #N/A ksi #N/A At Top Flange of Existing Member:

Fbx(top) = 21.31 ksi Fbx(top) = 12000*Cb/(Lb*dc/Af) <= Qs*0.60*Fy (Eqn. F1-8) S.R. =Fbx(bot) = 21.60 ksi Fbx(bot) = allowable tension bending stress at bottom cover plateS.R. =

Mrx = #N/A ft-kips #N/A S.R. =S.R. =

X-axis Euler Stress: (use rxc for combined section) At Upper Outside Edge of Flange of WT Reinf.:F'ex = 525.04 ksi F'ex = 12*(π)^2*29000/(23*(Kx*Lx*12/rxc)^2) S.R. =

S.R. =Stress Ratios: S.R. =

S.R.(top/mem) = #N/A #N/A #N/A S.R. =S.R.(top/WT) = #N/A #N/A #N/A At Lower Outside Edge of Flange of WT Reinf.:S.R.(bot/WT) = #N/A #N/A #N/A S.R. =

S.R.(bot/mem) = #N/A #N/A #N/A S.R. =At Bottom Flange of Existing Member:

S.R. =S.R. =

Weld Design:Interior Welds of WT Reinf. to Member:

Fw =

Lw(min) =Lw(use) =

End Welds of WT Reinf. to Member: For Case #1:

Fw(total) =(continued)

(min) =(max,eff) =

(use) =

Note: =


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a'(min) =Weld Design: a'(req'd) =

a'(use) =Interior Welds of WT Reinf. to Member: (assume intermittent fillet weld) Lw(total) =

Fw = 0.000 kips/in. Fw = Rv*Awt*ywt/(2*Ixc) (weld force per inch) 3/16 in. For Case #2:

0.3249 in. Fw(total) = 3/16 in. Lw(end) =

Lw(min) = 1.500 in./ft. a'(min) =Lw(use) = 1.500 in./ft. a'(req'd) =

(use) a'(use) =Use: 3/16'' Fillet Weld, 1.5'' @ 12'' Each Side of Plates Lw(total) =

End Welds of WT Reinf. to Member: (assume continuous fillet weld) ### For Case #1: Weld Across End of WT Reinf. (Determining End Welds Now Including Eccentricity from C.G. of Reinforcing to Welds)

Fw(total) = 1.12 kips Fw(total) = Mx(cutoff)*12*Awt*ywt/Ixc + (Awt/Ac)*ABS(P(total)) End Welds of WT Reinf. to Member:Lw(end) = 5.000 in. Lw(end) = bfwt For Case #1:

a'(min) = 5.000 in. a'(min) = bfwt, for weld size >= 0.75*tfwt Fw1 =a'(req'd) = 0.000 in. Lw(end) =

a'(use) = 5.000 in. a'(use) = Max. of: a'(min) or a'(req'd) a'(use) =Lw(total) = 15.000 in. Lw(total) = 2*a'(use)+Lw(end) (@ each end of WT reinf.) Pw =

(use) Mw =Use: 3/16'' Fillet Weld, 5'' Continuous Each Side of Plates and Across Ends Sw(max) =

Sw(min) = For Case #2: No Weld Across End of WT Reinf. Fw(result) =

Fw(total) = 1.12 kips Fw(total) = Mx(cutoff)*12*Awt*ywt/Ixc + (Awt/Ac)*ABS(P(total))Lw(end) = 0.000 in. Lw(end) = 0 (assumed) ###

a'(min) = 10.000 in. a'(min) = 2*bfwt, for no weld across end of WT reinf. For Case #2:a'(req'd) = 0.202 in. Fw1 =

a'(use) = 10.000 in. a'(use) = Max. of: a'(min) or a'(req'd) Lw(end) =Lw(total) = 20.000 in. Lw(total) = 2*a'(use) (@ each end of WT reinf.) a'(use) =

(use) Pw =Use: 3/16'' Fillet Weld, 10'' Continuous Each Side of Plates Mw =

Sw =Fw(result) =

Comments:##########################################

Note: = (min) = (min) = Min. fillet weld size from AISC Table J2.4, page 5-67

(max,eff) = (max,eff) = Min. of: (tp or tf)*(0.40*Fy/((SQRT(2)/2)*0.30*70))(use) = (use) = (min) <= 5/16"

Lw(min) = Max. of: 1.5" or 4*(min)Lw(use) = Max. of: (Fw*12)/(*SQRT(2)/2*0.3*70) or Lw(min)

Note: =

Note: =

a'(req'd) = (Fw(total)/(*SQRT(2)/2*0.3*70)-Lw(end))/2

Note: =

(req'd) =


Note: =

(req'd) =

Beam Reinforcement 9

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