"MONORAIL" --- MONORAIL BEAM ANALYSIS

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"MONORAIL" --- MONORAIL BEAM ANALYSIS Program Description: "MONORAIL" is a spreadsheet program written in MS-Excel for the purpose of analysis of either S-shape or W-shape underhung monorail beams analyzed as simple-spans with or without overhangs (cantilevers). Specifically, the x-axis and y-axis bending moments as well as any torsion effects are calculated. The actual and allowable stresses are determined, and the effect of lower flange bending is also addressed by two different approaches. This program is a workbook consisting of three (3) worksheets, described as follows: Worksheet Name Doc S-shaped Monorail Beam W-shaped Monorail Beam Description This documentation sheet Monorail beam analysis for S-shaped beams Monorail beam analysis for W-shaped beams Program Assumptions and Limitations: 1. The following references were used in the development of this program: a. Fluor Enterprises, Inc. - Guideline 000.215.1257 - "Hoisting Facilities" (August 22, 2005) b. Dupont Engineering Design Standard: DB1X - "Design and Installation of Monorail Beams" (May 2000) c. American National Standards Institute (ANSI): MH27.1 - "Underhung Cranes and Monorail Syatems" d. American Institute of Steel Construction (AISC) 9th Edition Allowable Stress Design (ASD) Manual (1989) e. "Allowable Bending Stresses for Overhanging Monorails" - by N. Stephen Tanner - AISC Engineering Journal (3rd Quarter, 1985) f. Crane Manufacturers Association of America, Inc. (CMAA) - Publication No. 74 - "Specifications for Top Running & Under Running Single Girder Electric Traveling Cranes Utilizing Under Running Trolley Hoist" (2004) g. "Design of Monorail Systems" - by Thomas H. Orihuela Jr., PE (www.pdhengineer.com) h. British Steel Code B.S. 449, pages 42-44 (1959) i. USS Steel Design Manual - Chapter 7 "Torsion" - by R. L. Brockenbrough and B.G. Johnston (1981) j. AISC Steel Design Guide Series No. 9 - "Torsional Analysis of Structural Steel Members" - by Paul A. Seaburg, PhD, PE and Charlie J. Carter, PE (1997) k. "Technical Note: Torsion Analysis of Steel Sections" - by William E. Moore II and Keith M. Mueller - AISC Engineering Journal (4th Quarter, 2002) 2. The unbraced length for the overhang (cantilever) portion, 'Lbo', of an underhung monorail beam is often debated. The following are some recommendations from the references cited above: a. Fluor Guideline 000.215.1257: Lbo = Lo+L/2 b. Dupont Standard DB1X: Lbo = 3*Lo c. ANSI Standard MH27.1: Lbo = 2*Lo d. British Steel Code B.S. 449: Lbo = 2*Lo (for top flange of monorail beam restrained at support) British Steel Code B.S. 449: Lbo = 3*Lo (for top flange of monorail beam unrestrained at support) e. AISC Eng. Journal Article by Tanner: Lbo = Lo+L (used with a computed value of 'Cbo' from article) 3. This program also determines the calculated value of the bending coefficient, 'Cbo', for the overhang (cantilever) portion of the monorail beam from reference "e" in note #1 above. This is located off of the main calculation page. Note: if this computed value of 'Cbo' is used and input, then per this reference the total value of Lo+L should be used for the unbraced length, 'Lbo', for the overhang portion of the monorail beam. 4. This program ignores effects of axial compressive stress produced by any longitudinal (traction) force which is usually considered minimal for underhung, hand-operated monorail systems. 5. This program contains 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".)

MONORAIL BEAM ANALYSIS For S-shaped Underhung Monorails Analyzed as Simple-Spans with / without Overhang Per AISC 9th Edition ASD Manual and CMAA Specification No. 74 (2004) Job Name: Subject: Job Number: Originator: Checker: Input: RL(min)=-0.73 RR(max)=9.13 Monorail Size: L=17 Lo=3 Select: S12x50 x=8.313 Design Parameters: S=0.75 Beam Fy = 36 ksi Beam Simple-Span, L = 17.0000 ft. S12x50 Unbraced Length, Lb = 17.0000 ft. Bending Coef., Cb = 1.00 Pv=7.4 Overhang Length, Lo = 3.0000 ft. Nomenclature Unbraced Length, Lbo = 11.5000 ft. Bending Coef., Cbo = 1.00 S12x50 Member Properties: Lifted Load, P = 6.000 kips A = 14.60 in.^2 d/af = 3.32 Trolley Weight, Wt = 0.400 kips d = 12.000 in. Ix = 303.00 in.^4 Hoist Weight, Wh = 0.100 kips tw = 0.687 in. Sx = 50.60 in.^3 Vert. Impact Factor, Vi = 15 % bf = 5.480 in. Iy = 15.60 in.^4 Horz. Load Factor, HLF = 10 % tf = 0.659 in. Sy = 5.69 in.^3 Total No. Wheels, Nw = 4 k= 1.438 in. J = 2.770 in.^4 Wheel Spacing, S = 0.7500 ft. rt = 1.250 in. Cw = 502.0 in.^6 Distance on Flange, a = 0.3750 in. Support Reactions: (with overhang) Results: RR(max) = 9.13 = Pv*(L+(Lo-S/2))/L+w/1000/(2*L)*(L+Lo)^2 RL(min) = -0.73 = -Pv*(Lo-S/2)/L+w/1000/(2*L)*(L^2-Lo^2) Parameters and Coefficients: Pv = 7.400 kips Pv = P*(1+Vi/100)+Wt+Wh (vertical load) = 1.850 kips/wheel = Pv/Nw (load per trolley wheel) Ph = 0.600 kips Ph = HLF*P (horizontal load) ta = 0.493 in. ta = tf-bf/24+a/6 (for S-shape) l = 0.156 l = 2*a/(bf-tw) Cxo = -0.850 Cxo = -1.096+1.095*l+0.192*e^(-6.0*l) Cx1 = 0.600 Cx1 = 3.965-4.835*l-3.965*e^(-2.675*l) Czo = 0.165 Czo = -0.981-1.479*l+1.120*e^(1.322*l) Cz1 = 1.948 Cz1 = 1.810-1.150*l+1.060*e^(-7.70*l) Bending Moments for Simple-Span: x = 8.313 ft. x = 1/2*(L-S/2) (location of max. moments from left end of simple-span) Mx = 31.88 ft-kips Mx = (Pv/2)/(2*L)*(L-S/2)^2+w/1000*x/2*(L-x) My = 2.44 ft-kips My = (Ph/2)/(2*L)*(L-S/2)^2 Lateral Flange Bending Moment from Torsion for Simple-Span: (per USS Steel Design Manual, 1981) e = 6.000 in. e = d/2 (assume horiz. load taken at bot. flange) at = 21.662 at = SQRT(E*Cw/(J*G)), E=29000 ksi and G=11200 ksi Mt = 0.29 ft-kips Mt = Ph*e*at/(2*(d-tf))*TANH(L*12/(2*at))/12 X-axis Stresses for Simple-Span: fbx = 7.56 ksi fbx = Mx/Sx Lb/rt = 163.20 Lb/rt = Lb*12/rt Fbx = 17.72 ksi Fbx = 12000*Cb/(Lb*12*(d/Af)) <= 0.60*Fy fbx <= Fbx, O.K. 2 of 7 7/22/2013 7:29 AM

Y-axis Stresses for Simple-Span: fby = 5.14 ksi fby = My/Sy fwns = 1.21 ksi fwns = Mt*12/(Sy/2) (warping normal stress) fby(total) = 6.35 ksi fby(total) = fby+fwns Combined Stress Ratio for Simple-Span: S.R. = 0.662 S.R. = fbx/fbx+fby(total)/fby S.R. <= 1.0, O.K. Vertical Deflection for Simple-Span: D(max) = 0.1412 in. D(max) = Pv/2*(L-S)/2/(24*E*I)*(3*L^2-4*((L-S)/2)^2)+5*w/12000*L^4/(384*E*I) D(ratio) = L/1445 D(ratio) = L*12/D(max) D(allow) = 0.4533 in. D(allow) = L*12/450 Defl.(max) <= Defl.(allow), O.K. Bending Moments for Overhang: Mx = 19.65 ft-kips Mx = (Pv/2)*(Lo+(Lo-S))+w/1000*Lo^2/2 My = 1.58 ft-kips My = (Ph/2)*(Lo+(Lo-S)) Lateral Flange Bending Moment from Torsion for Overhang: (per USS Steel Design Manual, 1981) e = 6.000 in. e = d/2 (assume horiz. load taken at bot. flange) at = 21.662 at = SQRT(E*Cw/(J*G)), E=29000 ksi and G=11200 ksi Mt = 0.57 ft-kips Mt = Ph*e*at/(d-tf)*TANH(Lo*12/at)/12 X-axis Stresses for Overhang: fbx = 4.66 ksi fbx = Mx/Sx Lbo/rt = 110.40 Lbo/rt = Lbo*12/rt Fbx = 21.60 ksi Fbx = 12000*Cbo/(Lbo*12*(d/Af)) <= 0.60*Fy fbx <= Fbx, O.K. Y-axis Stresses for Overhang: fby = 3.32 ksi fby = My/Sy fwns = 2.42 ksi fwns = Mt*12/(Sy/2) (warping normal stress) fby(total) = 5.74 ksi fby(total) = fby+fwns Combined Stress Ratio for Overhang: S.R. = 0.428 S.R. = fbx/fbx+fby(total)/fby S.R. <= 1.0, O.K. Vertical Deflection for Overhang: (assuming full design load, Pv without impact, at end of overhang) D(max) = 0.0715 in. D(max) = Pv*Lo^2*(L+Lo)/(3*E*I)+w/12000*Lo*(4*Lo^2*L-L^3+3*Lo^3)/(24*E*I) D(ratio) = L/503 D(ratio) = Lo*12/D(max) D(allow) = 0.0800 in. D(allow) = Lo*12/450 Defl.(max) <= Defl.(allow), O.K. Bottom Flange Bending (simplified): be = 7.908 in. Min. of: be = 12*tf or S*12 (effective flange bending length) tf2 = 0.859 in. tf2 = tf+(bf/2-tw/2)/2*(1/6) (flange thk. at web based on 1:6 slope of flange) am = 1.818 in. am = (bf/2-tw/2)-(k-tf2) (where: k-tf2 = radius of fillet) Mf = 3.363 in.-kips Mf = *am Sf = 0.572 in.^3 Sf = be*tf^2/6 fb = 5.88 ksi fb = Mf/Sf Fb = 27.00 ksi Fb = 0.75*Fy fb <= Fb, O.K. 3 of 7 7/22/2013 7:29 AM

Bottom Flange Bending per CMAA Specification No. 74 (2004): Local Flange Bending Stress @ Point 0: sxo = -6.46 ksi sxo = Cxo*/ta^2 szo = 1.25 ksi szo = Czo*/ta^2 Local Flange Bending Stress @ Point 1: sx1 = 4.56 ksi sx1 = Cx1*/ta^2 sz1 = 14.82 ksi sz1 = Cz1*/ta^2 (Note: torsion is neglected) (Sign convention: + = tension, - = compression) S-shape Trolley Wheel Local Flange Bending Stress @ Point 2: sx2 = 6.46 ksi sx2 = -sxo sz2 = -1.25 ksi sz2 = -szo Resultant Biaxial Stress @ Point 0: sz = 13.65 ksi sz = fbx+fby+0.75*szo sx = -4.85 ksi sx = 0.75*sxo sto = 16.61 ksi sto = SQRT(sx^2+sz^2-sx*sz+3*txz^2) <= Fb = 0.66*Fy = 23.76 ksi, O.K. Resultant Biaxial Stress @ Point 1: sz = 23.82 ksi sz = fbx+fby+0.75*sz1 sx = 3.42 ksi sx = 0.75*sx1 st1 = 22.30 ksi st1 = SQRT(sx^2+sz^2-sx*sz+3*txz^2) <= Fb = 0.66*Fy = 23.76 ksi, O.K. Resultant Biaxial Stress @ Point 2: sz = 11.76 ksi sz = fbx+fby+0.75*sz2 sx = 4.85 ksi sx = 0.75*sx2 st2 = 10.24 ksi st2 = SQRT(sx^2+sz^2-sx*sz+3*txz^2) <= Fb = 0.66*Fy = 23.76 ksi, O.K. Y tw Z X tf Point 2 Point 0 tw/2 Point 1 bf/4 bf ta 4 of 7 7/22/2013 7:29 AM

MONORAIL BEAM ANALYSIS For W-shaped Underhung Monorails Analyzed as Simple-Spans with / without Overhang Per AISC 9th Edition ASD Manual and CMAA Specification No. 74 (2004) Job Name: Subject: Job Number: Originator: Checker: Input: RL(min)=-0.73 RR(max)=9.13 Monorail Size: L=17 Lo=3 Select: W12x50 x=8.313 Design Parameters: S=0.75 Beam Fy = 36 ksi Beam Simple-Span, L = 17.0000 ft. W12x50 Unbraced Length, Lb = 17.0000 ft. Bending Coef., Cb = 1.00 Pv=7.4 Overhang Length, Lo = 3.0000 ft. Nomenclature Unbraced Length, Lbo = 11.5000 ft. Bending Coef., Cbo = 1.00 W12x50 Member Properties: Lifted Load, P = 6.000 kips A = 14.60 in.^2 d/af = 2.36 Trolley Weight, Wt = 0.400 kips d = 12.200 in. Ix = 391.00 in.^4 Hoist Weight, Wh = 0.100 kips tw = 0.370 in. Sx = 64.20 in.^3 Vert. Impact Factor, Vi = 15 % bf = 8.080 in. Iy = 56.30 in.^4 Horz. Load Factor, HLF = 10 % tf = 0.640 in. Sy = 13.90 in.^3 Total No. Wheels, Nw = 4 k= 1.140 in. J = 1.710 in.^4 Wheel Spacing, S = 0.7500 ft. rt = 2.170 in. Cw = 1880.0 in.^6 Distance on Flange, a = 0.3750 in. Support Reactions: (with overhang) Results: RR(max) = 9.13 = Pv*(L+(Lo-S/2))/L+w/1000/(2*L)*(L+Lo)^2 RL(min) = -0.73 = -Pv*(Lo-S/2)/L+w/1000/(2*L)*(L^2-Lo^2) Parameters and Coefficients: Pv = 7.400 kips Pv = P*(1+Vi/100)+Wt+Wh (vertical load) = 1.850 kips/wheel = Pv/Nw (load per trolley wheel) Ph = 0.600 kips Ph = HLF*P (horizontal load) ta = 0.640 in. ta = tf (for W-shape) l = 0.097 l = 2*a/(bf-tw) Cxo = -1.903 Cxo = -2.110+1.977*l+0.0076*e^(6.53*l) Cx1 = 0.535 Cx1 = 10.108-7.408*l-10.108*e^(-1.364*l) Czo = 0.192 Czo = 0.050-0.580*l+0.148*e^(3.015*l) Cz1 = 2.319 Cz1 = 2.230-1.490*l+1.390*e^(-18.33*l) Bending Moments for Simple-Span: x = 8.313 ft. x = 1/2*(L-S/2) (location of max. moments from left end of simple-span) Mx = 31.88 ft-kips Mx = (Pv/2)/(2*L)*(L-S/2)^2+w/1000*x/2*(L-x) My = 2.44 ft-kips My = (Ph/2)/(2*L)*(L-S/2)^2 Lateral Flange Bending Moment from Torsion for Simple-Span: (per USS Steel Design Manual, 1981) e = 6.100 in. e = d/2 (assume horiz. load taken at bot. flange) at = 53.354 at = SQRT(E*Cw/(J*G)), E=29000 ksi and G=11200 ksi Mt = 0.67 ft-kips Mt = Ph*e*at/(2*(d-tf))*TANH(L*12/(2*at))/12 X-axis Stresses for Simple-Span: fbx = 5.96 ksi fbx = Mx/Sx Lb/rt = 94.01 Lb/rt = Lb*12/rt Fbx = 21.60 ksi Fbx = 12000*Cb/(Lb*12*(d/Af)) <= 0.60*Fy fbx <= Fbx, O.K. 5 of 7 7/22/2013 7:29 AM

Y-axis Stresses for Simple-Span: fby = 2.11 ksi fby = My/Sy fwns = 1.16 ksi fwns = Mt*12/(Sy/2) (warping normal stress) fby(total) = 3.27 ksi fby(total) = fby+fwns Combined Stress Ratio for Simple-Span: S.R. = 0.397 S.R. = fbx/fbx+fby(total)/fby S.R. <= 1.0, O.K. Vertical Deflection for Simple-Span: D(max) = 0.1094 in. D(max) = Pv/2*(L-S)/2/(24*E*I)*(3*L^2-4*((L-S)/2)^2)+5*w/12000*L^4/(384*E*I) D(ratio) = L/1865 D(ratio) = L*12/D(max) D(allow) = 0.4533 in. D(allow) = L*12/450 Defl.(max) <= Defl.(allow), O.K. Bending Moments for Overhang: Mx = 19.65 ft-kips Mx = (Pv/2)*(Lo+(Lo-S))+w/1000*Lo^2/2 My = 1.58 ft-kips My = (Ph/2)*(Lo+(Lo-S)) Lateral Flange Bending Moment from Torsion for Overhang: (per USS Steel Design Manual, 1981) e = 6.100 in. e = d/2 (assume horiz. load taken at bot. flange) at = 53.354 at = SQRT(E*Cw/(J*G)), E=29000 ksi and G=11200 ksi Mt = 1.41 ft-kips Mt = Ph*e*at/(d-tf)*TANH(Lo*12/at)/12 X-axis Stresses for Overhang: fbx = 3.67 ksi fbx = Mx/Sx Lbo/rt = 63.59 Lbo/rt = Lbo*12/rt Fbx = 21.60 ksi Fbx = 12000*Cbo/(Lbo*12*(d/Af)) <= 0.60*Fy fbx <= Fbx, O.K. Y-axis Stresses for Overhang: fby = 1.36 ksi fby = My/Sy fwns = 2.43 ksi fwns = Mt*12/(Sy/2) (warping normal stress) fby(total) = 3.79 ksi fby(total) = fby+fwns Combined Stress Ratio for Overhang: S.R. = 0.310 S.R. = fbx/fbx+fby(total)/fby S.R. <= 1.0, O.K. Vertical Deflection for Overhang: (assuming full design load, Pv without impact, at end of overhang) D(max) = 0.0554 in. D(max) = Pv*Lo^2*(L+Lo)/(3*E*I)+w/12000*Lo*(4*Lo^2*L-L^3+3*Lo^3)/(24*E*I) D(ratio) = L/650 D(ratio) = Lo*12/D(max) D(allow) = 0.0800 in. D(allow) = Lo*12/450 Defl.(max) <= Defl.(allow), O.K. Bottom Flange Bending (simplified): be = 7.680 in. Min. of: be = 12*tf or S*12 (effective flange bending length) am = 3.355 in. am = (bf/2-tw/2)-(k-tf) (where: k-tf = radius of fillet) Mf = 6.207 in.-kips Mf = *am Sf = 0.524 in.^3 Sf = be*tf^2/6 fb = 11.84 ksi fb = Mf/Sf Fb = 27.00 ksi Fb = 0.75*Fy fb <= Fb, O.K. 6 of 7 7/22/2013 7:29 AM

Bottom Flange Bending per CMAA Specification No. 74 (2004): Local Flange Bending Stress @ Point 0: sxo = -8.60 ksi sxo = Cxo*/ta^2 szo = 0.87 ksi szo = Czo*/ta^2 (Note: torsion is neglected) (Sign convention: + = tension, - = compression) Local Flange Bending Stress @ Point 1: sx1 = 2.42 ksi sx1 = Cx1*/ta^2 sz1 = 10.47 ksi sz1 = Cz1*/ta^2 Local Flange Bending Stress @ Point 2: sx2 = 8.60 ksi sx2 = -sxo sz2 = -0.87 ksi sz2 = -szo Resultant Biaxial Stress @ Point 0: sz = 8.72 ksi sz = fbx+fby+0.75*szo sx = -6.45 ksi sx = 0.75*sxo sto = 13.18 ksi sto = SQRT(sx^2+sz^2-sx*sz+3*txz^2) <= Fb = 0.66*Fy = 23.76 ksi, O.K. Resultant Biaxial Stress @ Point 1: sz = 15.92 ksi sz = fbx+fby+0.75*sz1 sx = 1.81 ksi sx = 0.75*sx1 st1 = 15.09 ksi st1 = SQRT(sx^2+sz^2-sx*sz+3*txz^2) <= Fb = 0.66*Fy = 23.76 ksi, O.K. Resultant Biaxial Stress @ Point 2: sz = 7.41 ksi sz = fbx+fby+0.75*sz2 sx = 6.45 ksi sx = 0.75*sx2 st2 = 6.98 ksi st2 = SQRT(sx^2+sz^2-sx*sz+3*txz^2) <= Fb = 0.66*Fy = 23.76 ksi, O.K. Y Z X tw Point 2 Point 0 tf Point 1 bf 7 of 7 7/22/2013 7:29 AM