Monorail Calculation
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Monorail Design Reference: CSA S1609, CMAA 742004 Given: Beam designation: S310x74 d = 305 mm b = 139 mm t = 16.7 mm w = 17.4 mm Ag = 9470 mm2 Sx = 833000 mm3 Sy = 93300 mm3
Zy = 169000 mm3
Ix = 1.27E+08 mm4 Iy = 6480000 mm4
Cw = 1.35e+11 mm6
J = 1.16e+06 mm4 Dead load of beam: Ws = 0.729 kN/m
= [G . J / (Es . Cw)]0.5 = [77,000 x 1.16e+06 / (2.00e+05 x 1.35e+11)]0.5 =0.0018 (1/mm) Basic Information Yield strength: fy = 350 MPa Span of simple span: Lx = 6000 mm distance from edge of flange to point of wheel load application: a = 25 mm Two wheels at one side of beam: nw = 2 Flange thickness at point of wheel load application: ta = t b / 24 + a / 6 = 16.7 139.0 / 24 + 25.0 / 6 =15.1 mm Impact Factor Vertical impact factor: iv = 0.15 Transverse impact factor: it = 0.2 Longitudinal impact factor: iL = 0.1 Lateral Supports of Simple Span distance from the first support to the left support: s1 = 2000 mm Unsupport length about weak axis: Ly = Lx s1 = 4000 mm Cantilever Length of cantilever: Lc = 1000 mm Lateral support at the tip Unsupport length about weak axis: Lyc = Lx + Lc = 7000 mm Loading Monorail rating capacity: Pa = 1000 kN
Weight of trolley and hoist: Wth = 1.5 kN Vertical loading on monorail: Pv = (1 + iv) .(Pa + Wth) = (1 + 0.15) x (1,000.0 + 1.5) =1,151.7 kN Transerve loading on monorail: Pt = it .(Pa + Wth) = 0.2 x (1,000.0 + 1.5) =200.3 kN Longitudinal loading on monorail: PL = iL .(Pa + Wth) = 0.1 x (1,000.0 + 1.5) =100.2 kN Dead load factor: D = 1.25 Live load factor: L = 1.5 Results: Section Classification Section class = 1 Code Check for Simple Span Critical Forces Axial compression: Cf = L . PL = 1.5 x 100.2 =150.2 kN Moment about strong axis: Mfx = (D . Ws) . Lx2/8 + (L . Pv) . Lx/4 = (1.25 x
0.729) x (6,000.0 x 0.001)2 / 8 + (1.5 x 1,151.7) x 6,000.0 / (4 x 1,000) =2,595.5 kN.m Moment about weak axis: Mfy1 = (L . Pt) . Ly / 4 = (1.5 x 200.3) x 4,000.0 / (4 x 1,000) =300.5 kN.m Torque (Pt. d/2) on half section of monorail, Reduction factor: = tanh(.Lx/2)/(.Lx/2) = tanh(0.0018 x 6,000.0/2) / (0.0018 x 6,000.0 / 2) =0.18 Analogy beam, equivalent moment about weak axis for full section: Mfy2 = . (L . Pt) . Lx / 4 = 0.18 x (1.5 x 200.3) x 6,000.0 / (4 x 1,000) =82.6 kN.m Total moment about weak axis: Mfy = Mfy1 + Mfy2 = 300.5 + 82.6 =383.0 kN.m Axial Compressive Capacity Axial Capacity (Reference: CSA S1609 Cl. 13.3) Designation: S310x74 Assume: kz =1 fy = 350 MPa kx = 1 Lx = 6000 mm ky = 1 Ly = 4000 mm t = 16.7 mm
b = 139 mm A0 = 9470 mm2 rx = 116 mm
Cw = 135 x 109 mm6 J = 1160 x 103 mm4 Ix = 127 x 106 mm4 ry = 26.2 mm
Iy = 6.48 x 106 mm4 d = 305 mm w = 17.4 mm x0 =0 mm y0 =0 mm = ky . Ly / ry = 1.0 x 4,000 / 26.2 = 152.7 r0 = [x02 + y02 + rx2 + ry2]0.5 = [02 + 02 + 116.02 + 26.22]0.5 =118.9 mm
Fex = 2.E / (kx.Lx/rx)2 = 3.142 x 2.00e+05 / (1.0 x 6,000 / 116.0)2 =737.8 MPa Fez = [ 2.E.Cw / (kz.Lz)2 + G.J] / (A0.r02) = (3.142 x 2.00e+05 x 1.35e+11 /
(1.0 x 6,000)2 + 77,000 x 1.16e+06) / (9,470 x 118.92) =722.2 MPa Fey = 2.E / (ky.Ly/ry)2 = 3.142 x 2.00e+05 / (1.0 x 4,000 / 26.2)2 =84.7 MPa n = 1.34 = .[fy / (2.E) ]0.5 = 152.7 x [350 / (3.142 x 2.00e+05)]0.5 =2.03 fs1 = fy.[ 1 + 2n ]1/n = 350 x [1 + 2.03(2x1.34) ](1/1.34) =76.3 MPa = (fy / Fez)0.5 = (350 / 722.2)0.5 =0.7
fs2 = fy.[ 1 + 2n ]1/n = 350 x [1 + 0.7(2x1.34) ](1/1.34) =275.4 MPa fs = Min. [ fs1, fs2 ] = Min. [76.3 , 275.4] =76.3 MPa Ae = A = 9,470 mm2 Cr = . Ae. fs = 0.9 x 9,470 x 76.3 /1,000 =651 kN Flexural Strength about X Axis: Mrx Flexural Capacity: (Reference: CSA S1609) Designation: S310x74 fy = 350 MPa Mf = 2595.5 kN.m L0 = 4000 mm 2 = 1 Section Properties: Ix = 127 x 106 mm4
J = 1160 x 103 mm4 Cw = 135 x 109 mm6
Zx = 1000 x 103 mm3
Sx = 833 x 103 mm3 b = 139 mm t = 16.7 mm A0 = 9470 mm2 d = 305 mm Iy = 6.48 x 106 mm4 w = 17.4 mm h = d 2t = 271.6 mm Laterally Supported Capacity, Cl. 13.5 Mp = fy.Zx = 350 x 1.00e+06 x 106 =350.0 kN.m Laterally Unsupported Capacity, Cl. 13.6 Mu = 2./L0 [ Es.Iy.G.J + (.Es/L0)2.Iy.Cw]0.5 = 1.0 x 3.14 / 4,000.0 x [
2.00e+05 x 6.48e+06 x 77,000 x 1.16e+06 + (3.14 x 2.00e+05 / 4,000.0)2 x 6.48e+06 x 1.35e+11 ]0.5 =291.1 kN.m Mu > 0.67 Mp , Mr0 = 1.15 . Mp.(1 0.28 Mp/Mu) = 1.15 x 0.9 x 3.50e+08 x (1 0.28 x 3.50e+08 / 2.91e+08) =240.3 kN.m Moment capacity: Mr = Min [ 0.9 Mp , Mr0 ] = 240.3 kN.m Unsupported length to mobilize the full capacity Mu = 2.15 Mp = 2.15 x 3.50e+08 =7.53e+08 N.mm a = (Mu/)2 = (7.53e+08 / 3.14)2 =5.74e+16 b = Es.Iy.G.J = 2.00e+05 x 6.48e+06 x 77,000 x 1.16e+06 =1.16e+23
c = (Es.)2.Iy.Cw = (2.00e+05 x 3.14)2 x 6.48e+06 x 1.35e+11 =3.45e+29 Lu = { [ b + (b2 4a.c)0.5 ] / (2a) }0.5 = 1,914 mm Flexual capacity: Mrx = Mr = 240.3 kN.m Flexural Strength about Y Axis: Mry
Flexural Capacity about weak axis: Mry = . Zy. fy = 0.9 x 1.69e+05 x 350 x 10 6 =53.2 kN.m
Beam Column Code Check Cl. 13.8 Cex = 2.E.Ix / Lx2 = 3.142 x 2.00e+05 x 1.27e+08 x 0.001 / 6,000.02 =6,964 kN Cey = 2.E.Iy / Ly2 = 3.142 x 2.00e+05 x 6.48e+06 x 0.001 / 4,000.02 =799 kN U1x = 1x / (1 Cf / Cex) = 0.85 / (1 150.2 / 6,963.6) =0.87 U1y = 1y / (1 Cf / Cey) = 0.85 / (1 150.2 / 799.4) =1.05
Cf / Cr + 0.85 U1x. Mfx / Mrx + 0.6 U1y .Mfy / Mry = 150.2 / 650.6 + 0.85 x 0.87 x 2,595.5 / 240.3 + 0.6 x 1.05 x 383.0 / 53.2 =12.73 Mfx / Mrx + Mfy / Mry = 2,595.5 / 240.3 + 383.0 / 53.2 =18.0 Vertical Deflection Check Vertical deflection: Dv = 5 Ws. Lx4 / (384 Es . Ix) + Pv .Lx3 / (48 Es .Ix) = 5 x
0.729 x 6,000.04 / (384 x 2.00e+05 x 1.27e+08) + 1,151.7 x 1,000 x 6,000.03 / (48 x 2.00e+05 x 1.27e+08) =204.5 mm Lx / Dv = 6,000.0 / 204.5 =29 Lateral Deflection Check Lateral deflection: Dt = Pt .Ly3 / (48 Es .Iy) = 200.3 x 1,000 x 4,000.03 / (48 x 2.00e+05 x 6.48e+06) =206.1 mm Ly / Dt = 4,000.0 / 206.1 =19 Local Stresses for Simple Span Reference: CMAA 742004 Stresses from Axial, Bending, and Torsion Axial compression stress: 1 = PL / Ag = 100.2 x 1,000 / 9,470.0 =10.6 MPa Moment about strong axis: Mx = Ws . Lx2/8 + Pv . Lx/4 = 0.729 x (6,000.0 x
0.001)2 / 8 + 1,151.7 x 6,000.0 / (4 x 1,000) =1,730.9 kN.m Normal stress by Mx: 2 = Mx / Sx = 1,730.9 x 106 / 8.33e+05 =2,077.9 MPa Moment about weak axis: My1 = Pt . Ly / 4 = 200.3 x 4,000.0 / (4 x 1,000) =200.3 kN.m Torque (Pt. d/2) on half section of monorail, Analogy beam, equivalent moment about weak axis for full section: My2 = . Pt . Lx / 4 = 0.18 x 200.3 x 6,000.0 / (4 x 1,000) =55.1 kN.m Total moment about weak axis: My = My1 + My2 = 200.3 + 55.1 =255.4 kN.m Normal stress by My: 3 = My / Sy = 255.4 x 106 / 93,300.0 =2,737.0 MPa Stresses from Local Bending of Flanges Due to Wheel Loads = 2 a / (b w) = 2 x 25.0 / (139.0 17.4) =0.41 Cx0 = 1.096 + 1.095 + 0.192 e(6.0 ) = 1.096 + 1.095 x 0.41 + 0.192 x e(6.0 x 0.41) =0.629 Cx1 = 3.965 4.835 3.965 e(2.675 ) = 3.965 4.835 x 0.41 3.965 x
e(2.675 x 0.41) =0.657 Cy0 = 0.981 1.479 + 1.12 e(1.322 ) = 0.981 1.479 x 0.41 + 1.12 x e(1.322 x 0.41) =0.34
Cy1 = 1.81 1.15 + 1.06 e(7.7 ) = 1.81 1.15 x 0.41 + 1.06 x e(7.7 x 0.41) =1.382 x0 = Cx0 . Pv / (2 nw . ta2) = 0.629 x 1,151.7 x 1,000 / (2 x 2 x 15.12) =797.5 MPa y0 = Cy0 . Pv / (2 nw . ta2) = 0.34 x 1,151.7 x 1,000 / (2 x 2 x 15.12) =430.4
MPa x1 = Cx1 . Pv / (2 nw . ta2) = 0.657 x 1,151.7 x 1,000 / (2 x 2 x 15.12) =832.4 MPa y1 = Cy1 . Pv / (2 nw . ta2) = 1.382 x 1,151.7 x 1,000 / (2 x 2 x 15.12) =1,750.8 MPa x2 = Cx0 . Pv / (2 nw . ta2) = (0.629) x 1,151.7 x 1,000 / (2 x 2 x 15.12) =797.5 MPa y2 = Cy0 . Pv / (2 nw . ta2) = (0.34) x 1,151.7 x 1,000 / (2 x 2 x 15.12) =430.4 MPa Combined Stresses at Point 0 x = 0.75 x0 = 0.75 x (797.5) =598.1 MPa y = 0.75 y0 + 1 + 2 = 0.75 x (430.4) + (10.6) + (2,077.9) =2,390.1 MPa Local max. stress at point 0: p0 = [x2 + y2 x . y]0.5 = [(598.1)2 +
(2,390.1)2 (598.1) x (2,390.1)]0.5 =2,738.6 MPa p0 / (0.66 fy) = 2,738.6 / (0.66 x 350) =11.86 Combined Stresses at Point 2 x = 0.75 x2 = 0.75 x (797.5) =598.1 MPa y = 0.75 y2 + 1 + 2 = 0.75 x (430.4) + (10.6) + (2,077.9) =1,744.5 MPa Local max. stress at point 2: p2 = [x2 + y2 x . y]0.5 = [(598.1)2 +
(1,744.5)2 (598.1) x (1,744.5)]0.5 =1,535.5 MPa p2 / (0.66 fy) = 1,535.5 / (0.66 x 350) =6.65 Combined Stresses at Point 1 x = 0.75 x1 = 0.75 x (832.4) =624.3 MPa y = 0.75 y1 + 1 + 2 + 3 = 0.75 x (1,750.8) + (10.6) + (2,077.9) + (2,737.0) =6,117.4 MPa Local max. stress at point 1: p1 = [x2 + y2 x . y]0.5 = [(624.3)2 + (6,117.4)2 (624.3) x (6,117.4)]0.5 =5,830.3 MPa p1 / (0.66 fy) = 5,830.3 / (0.66 x 350) =25.24 Code Check for Canilever Critical Forces Axial compression: Cf = L . PL = 1.5 x 100.2 =150.2 kN
Moment about strong axis: Mfx = (D . Ws) . Lc2/2 + (L . Pv) .Lc = (1.25 x
0.729) x (1,000.0 x 0.001)2 / 2 + (1.5 x 1,151.7) x 1,000.0 / 1,000 =1,728.0 kN.m Moment about weak axis: Mfy1 = (L . Pt) . Lc / 4 = (1.5 x 200.3) x 1,000.0 / (4 x 1,000) =75.1 kN.m Torque (Pt. d/2) on half section of monorail, Reduction factor: = tanh(.Lc)/(.Lc) = tanh(0.0018 x 1,000.0) / (0.0018 x 1,000.0) =0.52
Analogy beam, equivalent moment about weak axis for full section: Mfy2 = . (L . Pt) . Lc = 0.52 x (1.5 x 200.3) x 1,000.0 / 1,000 =156.7 kN.m Total moment about weak axis: Mfy = Mfy1 + Mfy2 = 75.1 + 156.7 =231.8 kN.m Axial Compressive Capacity Axial Capacity (Reference: CSA S1609 Cl. 13.3) Designation: S310x74 Assume: kz =1 fy = 350 MPa kx = 2 Lx = 1000 mm ky = 1 Ly = 1000 mm t = 16.7 mm b = 139 mm A0 = 9470 mm2 rx = 116 mm Cw = 135 x 109 mm6 J = 1160 x 103 mm4 Ix = 127 x 106 mm4 ry = 26.2 mm
Iy = 6.48 x 106 mm4 d = 305 mm w = 17.4 mm x0 =0 mm y0 =0 mm = ky . Ly / ry = 1.0 x 1,000 / 26.2 = 38.2 r0 = [x02 + y02 + rx2 + ry2]0.5 = [02 + 02 + 116.02 + 26.22]0.5 =118.9 mm
Fex = 2.E / (kx.Lx/rx)2 = 3.142 x 2.00e+05 / (2.0 x 1,000 / 116.0)2 =6,640.3 MPa Fez = [ 2.E.Cw / (kz.Lz)2 + G.J] / (A0.r02) = (3.142 x 2.00e+05 x 1.35e+11 / (1.0 x 1,000)2 + 77,000 x 1.16e+06) / (9,470 x 118.92) =2,656.6 MPa Fey = 2.E / (ky.Ly/ry)2 = 3.142 x 2.00e+05 / (1.0 x 1,000 / 26.2)2 =1,355.0 MPa n = 1.34 = .[fy / (2.E) ]0.5 = 38.2 x [350 / (3.142 x 2.00e+05)]0.5 =0.51 fs1 = fy.[ 1 + 2n ]1/n = 350 x [1 + 0.51(2x1.34) ](1/1.34) =312.7 MPa = (fy / Fez)0.5 = (350 / 2,656.6)0.5 =0.36
fs2 = fy.[ 1 + 2n ]1/n = 350 x [1 + 0.36(2x1.34) ](1/1.34) =333.7 MPa fs = Min. [ fs1, fs2 ] = Min. [312.7 , 333.7] =312.7 MPa Ae = A = 9,470 mm2 Cr = . Ae. fs = 0.9 x 9,470 x 312.7 /1,000 =2,665 kN Flexural Strength about X Axis: Mrx Flexural Capacity: (Reference: CSA S1609) Designation: S310x74 fy = 350 MPa Mf = 1728 kN.m L0 = 7000 mm 2 = 1 Section Properties: Ix = 127 x 106 mm4 J = 1160 x 103 mm4 Cw = 135 x 109 mm6
Zx = 1000 x 103 mm3
Sx = 833 x 103 mm3 b = 139 mm t = 16.7 mm A0 = 9470 mm2 d = 305 mm Iy = 6.48 x 106 mm4 w = 17.4 mm h = d 2t = 271.6 mm Laterally Supported Capacity, Cl. 13.5 Mp = fy.Zx = 350 x 1.00e+06 x 106 =350.0 kN.m Laterally Unsupported Capacity, Cl. 13.6 Mu = 2./L0 [ Es.Iy.G.J + (.Es/L0)2.Iy.Cw]0.5 = 1.0 x 3.14 / 7,000.0 x [
2.00e+05 x 6.48e+06 x 77,000 x 1.16e+06 + (3.14 x 2.00e+05 / 7,000.0)2 x 6.48e+06 x 1.35e+11 ]0.5 =157.3 kN.m Mu ≤ 0.67 Mp , Mr0 = 0.9 Mu = 0.9 x 1.57e+08 =141.5 kN.m Moment capacity: Mr = Min [ 0.9 Mp , Mr0 ] = 141.5 kN.m Unsupported length to mobilize the full capacity Mu = 2.15 Mp = 2.15 x 3.50e+08 =7.53e+08 N.mm a = (Mu/)2 = (7.53e+08 / 3.14)2 =5.74e+16 b = Es.Iy.G.J = 2.00e+05 x 6.48e+06 x 77,000 x 1.16e+06 =1.16e+23
c = (Es.)2.Iy.Cw = (2.00e+05 x 3.14)2 x 6.48e+06 x 1.35e+11 =3.45e+29 Lu = { [ b + (b2 4a.c)0.5 ] / (2a) }0.5 = 1,914 mm
Flexual capacity: Mrx = Mr = 141.5 kN.m Beam Column Code Check Cl. 13.8 Cex = 2.E.Ix / (2 Lc)2 = 3.142 x 2.00e+05 x 1.27e+08 x 0.001 / (2 x 1,000.0)2 =62,672 kN Cey = 2.E.Iy / Lc2 = 3.142 x 2.00e+05 x 6.48e+06 x 0.001 / 1,000.02 =12,791 kN U1x = 1x / (1 Cf / Cex) = 0.85 / (1 150.2 / 62,672.0) =0.85 U1y = 1y / (1 Cf / Cey) = 0.85 / (1 150.2 / 12,791.0) =0.86 Cf / Cr + 0.85 U1x. Mfx / Mrx + 0.6 U1y .Mfy / Mry = 150.2 / 2,665.1 + 0.85 x 0.85 x 1,728.0 / 141.5 + 0.6 x 0.86 x 231.8 / 53.2 =11.15 Mfx / Mrx + Mfy / Mry = 1,728.0 / 141.5 + 231.8 / 53.2 =16.56 Vertical Deflection Check Vertical deflection: Dv = Ws. Lc4 / (8 Es . Ix) + Pv .Lc2. (Lx + Lc) / (3 Es .Ix) = 0.729 x 1,000.04 / (8 x 2.00e+05 x 1.27e+08) + 1,151.7 x 1,000 x 1,000.02 x (6,000.0 + 1,000.0) / (3 x 2.00e+05 x 1.27e+08) =105.8 mm 2 Lc / Dv = 2 x 1,000.0 / 105.8 =19 Lateral Deflection Check Lateral deflection: Dt = Pt .Lc3 / (48 Es .Iy) = 200.3 x 1,000 x 1,000.03 / (48 x 2.00e+05 x 6.48e+06) =3.2 mm Lc / Dt = 1,000.0 / 3.2 =311 Local Stresses for Cantilever Reference: CMAA 742004 Stresses from Axial, Bending, and Torsion Axial compression stress: 1 = PL / Ag = 100.2 x 1,000 / 9,470.0 =10.6 MPa Moment about strong axis: Mx = Ws . Lc2/2 + Pv .Lc = 0.729 x (1,000.0 x
0.001)2 / 2 + 1,151.7 x 1,000.0 / 1,000 =1,152.1 kN.m Normal stress by Mx: 2 = Mx / Sx = 1,152.1 x 106 / 8.33e+05 =1,383.1 MPa Moment about weak axis: My1 = Pt . Lc / 4 = 200.3 x 1,000.0 / (4 x 1,000) =50.1 kN.m Torque (Pt. d/2) on half section of monorail, Analogy beam, equivalent moment about weak axis for full section: My2 = . Pt . Lc = 0.52 x 200.3 x 1,000.0 / 1,000 =104.5 kN.m Total moment about weak axis: My = My1 + My2 = 50.1 + 104.5 =154.6 kN.m Normal stress by My: 3 = My / Sy = 154.6 x 106 / 93,300.0 =1,656.5 MPa Combined Stresses at Point 0 x = 0.75 x0 = 0.75 x (797.5) =598.1 MPa
y = 0.75 y0 + 1 + 2 = 0.75 x (430.4) + (10.6) + (1,383.1) =1,695.3 MPa Local max. stress at point 0: p0 = [x2 + y2 x . y]0.5 = [(598.1)2 +
(1,695.3)2 (598.1) x (1,695.3)]0.5 =2,060.5 MPa p0 / (0.66 fy) = 2,060.5 / (0.66 x 350) =8.92 Combined Stresses at Point 2 x = 0.75 x2 = 0.75 x (797.5) =598.1 MPa y = 0.75 y2 + 1 + 2 = 0.75 x (430.4) + (10.6) + (1,383.1) =1,049.7 MPa Local max. stress at point 2: p2 = [x2 + y2 x . y]0.5 = [(598.1)2 +
(1,049.7)2 (598.1) x (1,049.7)]0.5 =912.0 MPa p2 / (0.66 fy) = 912.0 / (0.66 x 350) =3.95 Combined Stresses at Point 1 x = 0.75 x1 = 0.75 x (832.4) =624.3 MPa y = 0.75 y1 + 1 + 2 + 3 = 0.75 x (1,750.8) + (10.6) + (1,383.1) + (1,656.5) =4,342.1 MPa Local max. stress at point 1: p1 = [x2 + y2 x . y]0.5 = [(624.3)2 +
(4,342.1)2 (624.3) x (4,342.1)]0.5 =4,066.0 MPa p1 / (0.66 fy) = 4,066.0 / (0.66 x 350) =17.6 Web Bending under Lateral Loading (FYI) Mf = (L . Pt) .(d 2 t) = (1.5 x 200.3) x (305.0 2 x 16.7) / 1,000 =81.6 kN.m Web flexural capacity of effective section: Mr = . (24 w .w2/4) .fy = 0.9 x (24 x 17.4 x 17.42/4) x 350 x 106 =10.0 kN.m Mf / Mr = 81.6 / 10.0 =8.2 Flange Bending under Vertical Loading (FYI) Mf = (L . Pv) .(b w)/(4 nw) = (1.5 x 1,151.7) x (139.0 17.4) / (1,000 x 4 x 2) =26.3 kN.m tm = t + (b w) / 24 = 16.7 + (139.0 17.4) / 24 =21.8 mm Flange flexural capacity of effective section: Mr = . (24 tm .tm2/6) .fy = 0.9 x (24 x 21.8 x 21.82/6) x 350 x 106 =13.0 kN.m Mf / Mr = 26.3 / 13.0 =2.02
Zy = 169000 mm3
Ix = 1.27E+08 mm4 Iy = 6480000 mm4
Cw = 1.35e+11 mm6
J = 1.16e+06 mm4 Dead load of beam: Ws = 0.729 kN/m
= [G . J / (Es . Cw)]0.5 = [77,000 x 1.16e+06 / (2.00e+05 x 1.35e+11)]0.5 =0.0018 (1/mm) Basic Information Yield strength: fy = 350 MPa Span of simple span: Lx = 6000 mm distance from edge of flange to point of wheel load application: a = 25 mm Two wheels at one side of beam: nw = 2 Flange thickness at point of wheel load application: ta = t b / 24 + a / 6 = 16.7 139.0 / 24 + 25.0 / 6 =15.1 mm Impact Factor Vertical impact factor: iv = 0.15 Transverse impact factor: it = 0.2 Longitudinal impact factor: iL = 0.1 Lateral Supports of Simple Span distance from the first support to the left support: s1 = 2000 mm Unsupport length about weak axis: Ly = Lx s1 = 4000 mm Cantilever Length of cantilever: Lc = 1000 mm Lateral support at the tip Unsupport length about weak axis: Lyc = Lx + Lc = 7000 mm Loading Monorail rating capacity: Pa = 1000 kN
Weight of trolley and hoist: Wth = 1.5 kN Vertical loading on monorail: Pv = (1 + iv) .(Pa + Wth) = (1 + 0.15) x (1,000.0 + 1.5) =1,151.7 kN Transerve loading on monorail: Pt = it .(Pa + Wth) = 0.2 x (1,000.0 + 1.5) =200.3 kN Longitudinal loading on monorail: PL = iL .(Pa + Wth) = 0.1 x (1,000.0 + 1.5) =100.2 kN Dead load factor: D = 1.25 Live load factor: L = 1.5 Results: Section Classification Section class = 1 Code Check for Simple Span Critical Forces Axial compression: Cf = L . PL = 1.5 x 100.2 =150.2 kN Moment about strong axis: Mfx = (D . Ws) . Lx2/8 + (L . Pv) . Lx/4 = (1.25 x
0.729) x (6,000.0 x 0.001)2 / 8 + (1.5 x 1,151.7) x 6,000.0 / (4 x 1,000) =2,595.5 kN.m Moment about weak axis: Mfy1 = (L . Pt) . Ly / 4 = (1.5 x 200.3) x 4,000.0 / (4 x 1,000) =300.5 kN.m Torque (Pt. d/2) on half section of monorail, Reduction factor: = tanh(.Lx/2)/(.Lx/2) = tanh(0.0018 x 6,000.0/2) / (0.0018 x 6,000.0 / 2) =0.18 Analogy beam, equivalent moment about weak axis for full section: Mfy2 = . (L . Pt) . Lx / 4 = 0.18 x (1.5 x 200.3) x 6,000.0 / (4 x 1,000) =82.6 kN.m Total moment about weak axis: Mfy = Mfy1 + Mfy2 = 300.5 + 82.6 =383.0 kN.m Axial Compressive Capacity Axial Capacity (Reference: CSA S1609 Cl. 13.3) Designation: S310x74 Assume: kz =1 fy = 350 MPa kx = 1 Lx = 6000 mm ky = 1 Ly = 4000 mm t = 16.7 mm
b = 139 mm A0 = 9470 mm2 rx = 116 mm
Cw = 135 x 109 mm6 J = 1160 x 103 mm4 Ix = 127 x 106 mm4 ry = 26.2 mm
Iy = 6.48 x 106 mm4 d = 305 mm w = 17.4 mm x0 =0 mm y0 =0 mm = ky . Ly / ry = 1.0 x 4,000 / 26.2 = 152.7 r0 = [x02 + y02 + rx2 + ry2]0.5 = [02 + 02 + 116.02 + 26.22]0.5 =118.9 mm
Fex = 2.E / (kx.Lx/rx)2 = 3.142 x 2.00e+05 / (1.0 x 6,000 / 116.0)2 =737.8 MPa Fez = [ 2.E.Cw / (kz.Lz)2 + G.J] / (A0.r02) = (3.142 x 2.00e+05 x 1.35e+11 /
(1.0 x 6,000)2 + 77,000 x 1.16e+06) / (9,470 x 118.92) =722.2 MPa Fey = 2.E / (ky.Ly/ry)2 = 3.142 x 2.00e+05 / (1.0 x 4,000 / 26.2)2 =84.7 MPa n = 1.34 = .[fy / (2.E) ]0.5 = 152.7 x [350 / (3.142 x 2.00e+05)]0.5 =2.03 fs1 = fy.[ 1 + 2n ]1/n = 350 x [1 + 2.03(2x1.34) ](1/1.34) =76.3 MPa = (fy / Fez)0.5 = (350 / 722.2)0.5 =0.7
fs2 = fy.[ 1 + 2n ]1/n = 350 x [1 + 0.7(2x1.34) ](1/1.34) =275.4 MPa fs = Min. [ fs1, fs2 ] = Min. [76.3 , 275.4] =76.3 MPa Ae = A = 9,470 mm2 Cr = . Ae. fs = 0.9 x 9,470 x 76.3 /1,000 =651 kN Flexural Strength about X Axis: Mrx Flexural Capacity: (Reference: CSA S1609) Designation: S310x74 fy = 350 MPa Mf = 2595.5 kN.m L0 = 4000 mm 2 = 1 Section Properties: Ix = 127 x 106 mm4
J = 1160 x 103 mm4 Cw = 135 x 109 mm6
Zx = 1000 x 103 mm3
Sx = 833 x 103 mm3 b = 139 mm t = 16.7 mm A0 = 9470 mm2 d = 305 mm Iy = 6.48 x 106 mm4 w = 17.4 mm h = d 2t = 271.6 mm Laterally Supported Capacity, Cl. 13.5 Mp = fy.Zx = 350 x 1.00e+06 x 106 =350.0 kN.m Laterally Unsupported Capacity, Cl. 13.6 Mu = 2./L0 [ Es.Iy.G.J + (.Es/L0)2.Iy.Cw]0.5 = 1.0 x 3.14 / 4,000.0 x [
2.00e+05 x 6.48e+06 x 77,000 x 1.16e+06 + (3.14 x 2.00e+05 / 4,000.0)2 x 6.48e+06 x 1.35e+11 ]0.5 =291.1 kN.m Mu > 0.67 Mp , Mr0 = 1.15 . Mp.(1 0.28 Mp/Mu) = 1.15 x 0.9 x 3.50e+08 x (1 0.28 x 3.50e+08 / 2.91e+08) =240.3 kN.m Moment capacity: Mr = Min [ 0.9 Mp , Mr0 ] = 240.3 kN.m Unsupported length to mobilize the full capacity Mu = 2.15 Mp = 2.15 x 3.50e+08 =7.53e+08 N.mm a = (Mu/)2 = (7.53e+08 / 3.14)2 =5.74e+16 b = Es.Iy.G.J = 2.00e+05 x 6.48e+06 x 77,000 x 1.16e+06 =1.16e+23
c = (Es.)2.Iy.Cw = (2.00e+05 x 3.14)2 x 6.48e+06 x 1.35e+11 =3.45e+29 Lu = { [ b + (b2 4a.c)0.5 ] / (2a) }0.5 = 1,914 mm Flexual capacity: Mrx = Mr = 240.3 kN.m Flexural Strength about Y Axis: Mry
Flexural Capacity about weak axis: Mry = . Zy. fy = 0.9 x 1.69e+05 x 350 x 10 6 =53.2 kN.m
Beam Column Code Check Cl. 13.8 Cex = 2.E.Ix / Lx2 = 3.142 x 2.00e+05 x 1.27e+08 x 0.001 / 6,000.02 =6,964 kN Cey = 2.E.Iy / Ly2 = 3.142 x 2.00e+05 x 6.48e+06 x 0.001 / 4,000.02 =799 kN U1x = 1x / (1 Cf / Cex) = 0.85 / (1 150.2 / 6,963.6) =0.87 U1y = 1y / (1 Cf / Cey) = 0.85 / (1 150.2 / 799.4) =1.05
Cf / Cr + 0.85 U1x. Mfx / Mrx + 0.6 U1y .Mfy / Mry = 150.2 / 650.6 + 0.85 x 0.87 x 2,595.5 / 240.3 + 0.6 x 1.05 x 383.0 / 53.2 =12.73 Mfx / Mrx + Mfy / Mry = 2,595.5 / 240.3 + 383.0 / 53.2 =18.0 Vertical Deflection Check Vertical deflection: Dv = 5 Ws. Lx4 / (384 Es . Ix) + Pv .Lx3 / (48 Es .Ix) = 5 x
0.729 x 6,000.04 / (384 x 2.00e+05 x 1.27e+08) + 1,151.7 x 1,000 x 6,000.03 / (48 x 2.00e+05 x 1.27e+08) =204.5 mm Lx / Dv = 6,000.0 / 204.5 =29 Lateral Deflection Check Lateral deflection: Dt = Pt .Ly3 / (48 Es .Iy) = 200.3 x 1,000 x 4,000.03 / (48 x 2.00e+05 x 6.48e+06) =206.1 mm Ly / Dt = 4,000.0 / 206.1 =19 Local Stresses for Simple Span Reference: CMAA 742004 Stresses from Axial, Bending, and Torsion Axial compression stress: 1 = PL / Ag = 100.2 x 1,000 / 9,470.0 =10.6 MPa Moment about strong axis: Mx = Ws . Lx2/8 + Pv . Lx/4 = 0.729 x (6,000.0 x
0.001)2 / 8 + 1,151.7 x 6,000.0 / (4 x 1,000) =1,730.9 kN.m Normal stress by Mx: 2 = Mx / Sx = 1,730.9 x 106 / 8.33e+05 =2,077.9 MPa Moment about weak axis: My1 = Pt . Ly / 4 = 200.3 x 4,000.0 / (4 x 1,000) =200.3 kN.m Torque (Pt. d/2) on half section of monorail, Analogy beam, equivalent moment about weak axis for full section: My2 = . Pt . Lx / 4 = 0.18 x 200.3 x 6,000.0 / (4 x 1,000) =55.1 kN.m Total moment about weak axis: My = My1 + My2 = 200.3 + 55.1 =255.4 kN.m Normal stress by My: 3 = My / Sy = 255.4 x 106 / 93,300.0 =2,737.0 MPa Stresses from Local Bending of Flanges Due to Wheel Loads = 2 a / (b w) = 2 x 25.0 / (139.0 17.4) =0.41 Cx0 = 1.096 + 1.095 + 0.192 e(6.0 ) = 1.096 + 1.095 x 0.41 + 0.192 x e(6.0 x 0.41) =0.629 Cx1 = 3.965 4.835 3.965 e(2.675 ) = 3.965 4.835 x 0.41 3.965 x
e(2.675 x 0.41) =0.657 Cy0 = 0.981 1.479 + 1.12 e(1.322 ) = 0.981 1.479 x 0.41 + 1.12 x e(1.322 x 0.41) =0.34
Cy1 = 1.81 1.15 + 1.06 e(7.7 ) = 1.81 1.15 x 0.41 + 1.06 x e(7.7 x 0.41) =1.382 x0 = Cx0 . Pv / (2 nw . ta2) = 0.629 x 1,151.7 x 1,000 / (2 x 2 x 15.12) =797.5 MPa y0 = Cy0 . Pv / (2 nw . ta2) = 0.34 x 1,151.7 x 1,000 / (2 x 2 x 15.12) =430.4
MPa x1 = Cx1 . Pv / (2 nw . ta2) = 0.657 x 1,151.7 x 1,000 / (2 x 2 x 15.12) =832.4 MPa y1 = Cy1 . Pv / (2 nw . ta2) = 1.382 x 1,151.7 x 1,000 / (2 x 2 x 15.12) =1,750.8 MPa x2 = Cx0 . Pv / (2 nw . ta2) = (0.629) x 1,151.7 x 1,000 / (2 x 2 x 15.12) =797.5 MPa y2 = Cy0 . Pv / (2 nw . ta2) = (0.34) x 1,151.7 x 1,000 / (2 x 2 x 15.12) =430.4 MPa Combined Stresses at Point 0 x = 0.75 x0 = 0.75 x (797.5) =598.1 MPa y = 0.75 y0 + 1 + 2 = 0.75 x (430.4) + (10.6) + (2,077.9) =2,390.1 MPa Local max. stress at point 0: p0 = [x2 + y2 x . y]0.5 = [(598.1)2 +
(2,390.1)2 (598.1) x (2,390.1)]0.5 =2,738.6 MPa p0 / (0.66 fy) = 2,738.6 / (0.66 x 350) =11.86 Combined Stresses at Point 2 x = 0.75 x2 = 0.75 x (797.5) =598.1 MPa y = 0.75 y2 + 1 + 2 = 0.75 x (430.4) + (10.6) + (2,077.9) =1,744.5 MPa Local max. stress at point 2: p2 = [x2 + y2 x . y]0.5 = [(598.1)2 +
(1,744.5)2 (598.1) x (1,744.5)]0.5 =1,535.5 MPa p2 / (0.66 fy) = 1,535.5 / (0.66 x 350) =6.65 Combined Stresses at Point 1 x = 0.75 x1 = 0.75 x (832.4) =624.3 MPa y = 0.75 y1 + 1 + 2 + 3 = 0.75 x (1,750.8) + (10.6) + (2,077.9) + (2,737.0) =6,117.4 MPa Local max. stress at point 1: p1 = [x2 + y2 x . y]0.5 = [(624.3)2 + (6,117.4)2 (624.3) x (6,117.4)]0.5 =5,830.3 MPa p1 / (0.66 fy) = 5,830.3 / (0.66 x 350) =25.24 Code Check for Canilever Critical Forces Axial compression: Cf = L . PL = 1.5 x 100.2 =150.2 kN
Moment about strong axis: Mfx = (D . Ws) . Lc2/2 + (L . Pv) .Lc = (1.25 x
0.729) x (1,000.0 x 0.001)2 / 2 + (1.5 x 1,151.7) x 1,000.0 / 1,000 =1,728.0 kN.m Moment about weak axis: Mfy1 = (L . Pt) . Lc / 4 = (1.5 x 200.3) x 1,000.0 / (4 x 1,000) =75.1 kN.m Torque (Pt. d/2) on half section of monorail, Reduction factor: = tanh(.Lc)/(.Lc) = tanh(0.0018 x 1,000.0) / (0.0018 x 1,000.0) =0.52
Analogy beam, equivalent moment about weak axis for full section: Mfy2 = . (L . Pt) . Lc = 0.52 x (1.5 x 200.3) x 1,000.0 / 1,000 =156.7 kN.m Total moment about weak axis: Mfy = Mfy1 + Mfy2 = 75.1 + 156.7 =231.8 kN.m Axial Compressive Capacity Axial Capacity (Reference: CSA S1609 Cl. 13.3) Designation: S310x74 Assume: kz =1 fy = 350 MPa kx = 2 Lx = 1000 mm ky = 1 Ly = 1000 mm t = 16.7 mm b = 139 mm A0 = 9470 mm2 rx = 116 mm Cw = 135 x 109 mm6 J = 1160 x 103 mm4 Ix = 127 x 106 mm4 ry = 26.2 mm
Iy = 6.48 x 106 mm4 d = 305 mm w = 17.4 mm x0 =0 mm y0 =0 mm = ky . Ly / ry = 1.0 x 1,000 / 26.2 = 38.2 r0 = [x02 + y02 + rx2 + ry2]0.5 = [02 + 02 + 116.02 + 26.22]0.5 =118.9 mm
Fex = 2.E / (kx.Lx/rx)2 = 3.142 x 2.00e+05 / (2.0 x 1,000 / 116.0)2 =6,640.3 MPa Fez = [ 2.E.Cw / (kz.Lz)2 + G.J] / (A0.r02) = (3.142 x 2.00e+05 x 1.35e+11 / (1.0 x 1,000)2 + 77,000 x 1.16e+06) / (9,470 x 118.92) =2,656.6 MPa Fey = 2.E / (ky.Ly/ry)2 = 3.142 x 2.00e+05 / (1.0 x 1,000 / 26.2)2 =1,355.0 MPa n = 1.34 = .[fy / (2.E) ]0.5 = 38.2 x [350 / (3.142 x 2.00e+05)]0.5 =0.51 fs1 = fy.[ 1 + 2n ]1/n = 350 x [1 + 0.51(2x1.34) ](1/1.34) =312.7 MPa = (fy / Fez)0.5 = (350 / 2,656.6)0.5 =0.36
fs2 = fy.[ 1 + 2n ]1/n = 350 x [1 + 0.36(2x1.34) ](1/1.34) =333.7 MPa fs = Min. [ fs1, fs2 ] = Min. [312.7 , 333.7] =312.7 MPa Ae = A = 9,470 mm2 Cr = . Ae. fs = 0.9 x 9,470 x 312.7 /1,000 =2,665 kN Flexural Strength about X Axis: Mrx Flexural Capacity: (Reference: CSA S1609) Designation: S310x74 fy = 350 MPa Mf = 1728 kN.m L0 = 7000 mm 2 = 1 Section Properties: Ix = 127 x 106 mm4 J = 1160 x 103 mm4 Cw = 135 x 109 mm6
Zx = 1000 x 103 mm3
Sx = 833 x 103 mm3 b = 139 mm t = 16.7 mm A0 = 9470 mm2 d = 305 mm Iy = 6.48 x 106 mm4 w = 17.4 mm h = d 2t = 271.6 mm Laterally Supported Capacity, Cl. 13.5 Mp = fy.Zx = 350 x 1.00e+06 x 106 =350.0 kN.m Laterally Unsupported Capacity, Cl. 13.6 Mu = 2./L0 [ Es.Iy.G.J + (.Es/L0)2.Iy.Cw]0.5 = 1.0 x 3.14 / 7,000.0 x [
2.00e+05 x 6.48e+06 x 77,000 x 1.16e+06 + (3.14 x 2.00e+05 / 7,000.0)2 x 6.48e+06 x 1.35e+11 ]0.5 =157.3 kN.m Mu ≤ 0.67 Mp , Mr0 = 0.9 Mu = 0.9 x 1.57e+08 =141.5 kN.m Moment capacity: Mr = Min [ 0.9 Mp , Mr0 ] = 141.5 kN.m Unsupported length to mobilize the full capacity Mu = 2.15 Mp = 2.15 x 3.50e+08 =7.53e+08 N.mm a = (Mu/)2 = (7.53e+08 / 3.14)2 =5.74e+16 b = Es.Iy.G.J = 2.00e+05 x 6.48e+06 x 77,000 x 1.16e+06 =1.16e+23
c = (Es.)2.Iy.Cw = (2.00e+05 x 3.14)2 x 6.48e+06 x 1.35e+11 =3.45e+29 Lu = { [ b + (b2 4a.c)0.5 ] / (2a) }0.5 = 1,914 mm
Flexual capacity: Mrx = Mr = 141.5 kN.m Beam Column Code Check Cl. 13.8 Cex = 2.E.Ix / (2 Lc)2 = 3.142 x 2.00e+05 x 1.27e+08 x 0.001 / (2 x 1,000.0)2 =62,672 kN Cey = 2.E.Iy / Lc2 = 3.142 x 2.00e+05 x 6.48e+06 x 0.001 / 1,000.02 =12,791 kN U1x = 1x / (1 Cf / Cex) = 0.85 / (1 150.2 / 62,672.0) =0.85 U1y = 1y / (1 Cf / Cey) = 0.85 / (1 150.2 / 12,791.0) =0.86 Cf / Cr + 0.85 U1x. Mfx / Mrx + 0.6 U1y .Mfy / Mry = 150.2 / 2,665.1 + 0.85 x 0.85 x 1,728.0 / 141.5 + 0.6 x 0.86 x 231.8 / 53.2 =11.15 Mfx / Mrx + Mfy / Mry = 1,728.0 / 141.5 + 231.8 / 53.2 =16.56 Vertical Deflection Check Vertical deflection: Dv = Ws. Lc4 / (8 Es . Ix) + Pv .Lc2. (Lx + Lc) / (3 Es .Ix) = 0.729 x 1,000.04 / (8 x 2.00e+05 x 1.27e+08) + 1,151.7 x 1,000 x 1,000.02 x (6,000.0 + 1,000.0) / (3 x 2.00e+05 x 1.27e+08) =105.8 mm 2 Lc / Dv = 2 x 1,000.0 / 105.8 =19 Lateral Deflection Check Lateral deflection: Dt = Pt .Lc3 / (48 Es .Iy) = 200.3 x 1,000 x 1,000.03 / (48 x 2.00e+05 x 6.48e+06) =3.2 mm Lc / Dt = 1,000.0 / 3.2 =311 Local Stresses for Cantilever Reference: CMAA 742004 Stresses from Axial, Bending, and Torsion Axial compression stress: 1 = PL / Ag = 100.2 x 1,000 / 9,470.0 =10.6 MPa Moment about strong axis: Mx = Ws . Lc2/2 + Pv .Lc = 0.729 x (1,000.0 x
0.001)2 / 2 + 1,151.7 x 1,000.0 / 1,000 =1,152.1 kN.m Normal stress by Mx: 2 = Mx / Sx = 1,152.1 x 106 / 8.33e+05 =1,383.1 MPa Moment about weak axis: My1 = Pt . Lc / 4 = 200.3 x 1,000.0 / (4 x 1,000) =50.1 kN.m Torque (Pt. d/2) on half section of monorail, Analogy beam, equivalent moment about weak axis for full section: My2 = . Pt . Lc = 0.52 x 200.3 x 1,000.0 / 1,000 =104.5 kN.m Total moment about weak axis: My = My1 + My2 = 50.1 + 104.5 =154.6 kN.m Normal stress by My: 3 = My / Sy = 154.6 x 106 / 93,300.0 =1,656.5 MPa Combined Stresses at Point 0 x = 0.75 x0 = 0.75 x (797.5) =598.1 MPa
y = 0.75 y0 + 1 + 2 = 0.75 x (430.4) + (10.6) + (1,383.1) =1,695.3 MPa Local max. stress at point 0: p0 = [x2 + y2 x . y]0.5 = [(598.1)2 +
(1,695.3)2 (598.1) x (1,695.3)]0.5 =2,060.5 MPa p0 / (0.66 fy) = 2,060.5 / (0.66 x 350) =8.92 Combined Stresses at Point 2 x = 0.75 x2 = 0.75 x (797.5) =598.1 MPa y = 0.75 y2 + 1 + 2 = 0.75 x (430.4) + (10.6) + (1,383.1) =1,049.7 MPa Local max. stress at point 2: p2 = [x2 + y2 x . y]0.5 = [(598.1)2 +
(1,049.7)2 (598.1) x (1,049.7)]0.5 =912.0 MPa p2 / (0.66 fy) = 912.0 / (0.66 x 350) =3.95 Combined Stresses at Point 1 x = 0.75 x1 = 0.75 x (832.4) =624.3 MPa y = 0.75 y1 + 1 + 2 + 3 = 0.75 x (1,750.8) + (10.6) + (1,383.1) + (1,656.5) =4,342.1 MPa Local max. stress at point 1: p1 = [x2 + y2 x . y]0.5 = [(624.3)2 +
(4,342.1)2 (624.3) x (4,342.1)]0.5 =4,066.0 MPa p1 / (0.66 fy) = 4,066.0 / (0.66 x 350) =17.6 Web Bending under Lateral Loading (FYI) Mf = (L . Pt) .(d 2 t) = (1.5 x 200.3) x (305.0 2 x 16.7) / 1,000 =81.6 kN.m Web flexural capacity of effective section: Mr = . (24 w .w2/4) .fy = 0.9 x (24 x 17.4 x 17.42/4) x 350 x 106 =10.0 kN.m Mf / Mr = 81.6 / 10.0 =8.2 Flange Bending under Vertical Loading (FYI) Mf = (L . Pv) .(b w)/(4 nw) = (1.5 x 1,151.7) x (139.0 17.4) / (1,000 x 4 x 2) =26.3 kN.m tm = t + (b w) / 24 = 16.7 + (139.0 17.4) / 24 =21.8 mm Flange flexural capacity of effective section: Mr = . (24 tm .tm2/6) .fy = 0.9 x (24 x 21.8 x 21.82/6) x 350 x 106 =13.0 kN.m Mf / Mr = 26.3 / 13.0 =2.02
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