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# Calculation Verification

### Example 1

Determine the stresses of a T-section subjected to combined forces.

 Dimensions of section Geometric Properties Forces Area = 3579 mm2 Moment of Inertia Iz = 1.05786·107 mm4              Iy = 6.32306·106 mm4 Distans to Max and Min ymax = 39.3877 mm ymin = -150.6123 mm zmax = 90 mm zmin = -90 mm Shear Properties Qz = 7.93943·104 mm3 Qy = 5.25658·104 mm3 Torsion J = 1.46870·105 mm4 rmax = 13.5357 mm Axial = 10.0 kN Shear Y = 1.0 kN Shear Z = 1.0 kN Torsion = 0.1 kN·m Bending Y = 1 .0 kN·m Bending Z =  1 .0 kN·m

### Comparison of Results

 Result Location SkyCiv SB Analysis Manual Third-Party Primary Stresses (MPa) Axial max 2.794 $$\frac{Area}{Axial}=\frac{10·1000}{3579} = 2.794$$ (0.00%) 2.794 (0.00%) min 2.794 $$\frac{Area}{Axial}=\frac{10·1000}{3579} = 2.794$$ (0.00%) 2.794 (0.00%) Bending Y max 14.234 $$\frac{Bending Y}{I_y/y_{max}}=\frac{1·1000000}{6.32306·10^6/90} =14.234$$ (0.00%) 14.234 (0.00%) min -14.234 $$\frac{Bending Y}{I_y/y_{min}}=\frac{1·1000000}{6.32306·10^6/-90} =-14.234$$ (0.00%) -14.234 (0.00%) Bending Z max 3.723 $$\frac{Bending Z}{I_z/z_{max}}=\frac{1·1000000}{1.05786·10^7/39.3877} =3.723$$ (0.00%) 3.723 (0.00%) min -14.237 $$\frac{Bending Z}{I_z/z_{min}}=\frac{1·1000000}{1.05786·10^7/-150.6123} =-14.237$$ (0.00%) -14.237 (0.00%) Resultant Shear Y max 1.123 $$\frac{Shear Y·Q_z}{I_z·t}=\frac{1·1000·7.93943·10^4}{1.05786·10^7·7} = 1.072$$ (4.54%) 1.120 (0.26%) Resultant Shear Z max 0.698 $$\frac{Shear Z·Q_y}{I_y·t}=\frac{1·1000·5.25658·10^4}{6.32306·10^6·13} = 0.639$$ (8.45%) 0.709 (1.57%) Torsion max 9.956 $$\frac{Torsion·r_{max}}{J}=\frac{0.1·1000000·13.5357}{1.46870·10^5} = 9.216$$ (7.43%) 9.570 (3.87%)

### Example 2

Determine the stresses of a L-section subjected to combined forces.

 Dimensions of section Geometric Properties Forces Area = 1701.0 mm2 Moment of Inertia Iz = 4.02578·106 mm4              Iy = 1.47813·106 mm4 α = -24.2186° Izp = 4.67198·106 mm4              Iyp = 8.31922·105 mm4 Distans to Max and Min Bending Y Pointmax = (53.8855, -68.7955) mm Bending Y Pointmin = (-40.1846, -34.1575) mm Bending Z Pointmax = (27.7320, 99.7689) mm Bending Z Pointmin = (-40.1846, -34.1575) mm Shear Properties Qz = ????? mm3 Qy = ????? mm3 Torsion J = 27433.1 mm4 rmax = 8.7506 mm Axial = 10.0 kN Shear Y = 1.0 kN Shear Z = 1.0 kN Torsion = 0.01 kN·m Bending Y = 1 .0 kN·m Bending Z =  1 .0 kN·m

### Comparison of Results

 Result Location SkyCiv SB Analysis Manual Third-Party Primary Stresses (MPa) Axial max 5.879 $$\frac{Area}{Axial}=\frac{10·1000}{1701.00} = 5.879$$ (0.00%) 5.879 (0.00%) min 5.879 $$\frac{Area}{Axial}=\frac{10·1000}{1701.00} = 5.879$$ (0.00%) 5.879 (0.00%) Bending Y max 65.112 $$\frac{Bending Y·\cos(\alpha)}{I_yp/z_{max}}-\frac{Bending Y·\sin(\alpha)}{I_zp/y_{max}}=\frac{1·1000000·\cos(24.2186^\circ)}{8.31922·10^6/-68.7955}-\frac{1·1000000·\sin(24.2186^\circ)}{4.67198·10^6/53.8855}=65.112$$ (0.00%) 65.112 (0.00%) min -41.053 $$\frac{Bending Y·\cos(\alpha)}{I_yp/z_{min}}-\frac{Bending Y·\sin(\alpha)}{I_zp/y_{min}}=\frac{1·1000000·\cos(24.2186^\circ)}{8.31922·10^6/-40.1846}-\frac{1·1000000·\sin(24.2186^\circ)}{4.67198·10^6/-34.1575}=-41.053$$ (0.00%) -41.053 (0.00%) Bending Z max 33.15 $$\frac{Bending Z·\cos(\alpha)}{I_zp/y_{max}}+\frac{Bending Z·\sin(\alpha)}{I_yp/z_{max}}=\frac{1·1000000·\cos(24.2186^\circ)}{4.67198·10^6/99.7689}+\frac{1·1000000·\sin(24.2186^\circ)}{8.31922·10^5/27.7320}=33.15$$ (0.00%) 33.15 (0.00%) min -26.483 $$\frac{Bending Z·\cos(\alpha)}{I_zp/y_{min}}+\frac{Bending Z·\sin(\alpha)}{I_yp/z_{min}}=\frac{1·1000000·\cos(24.2186^\circ)}{4.67198·10^6/-34.1575}+\frac{1·1000000·\sin(24.2186^\circ)}{8.31922·10^5/-40.1846}=-26.483$$ (0.00%) -26.483 (0.00%) Resultant Shear Y max 1.123 $$\frac{Shear Y·Q_z}{I_z·t}=\frac{1·1000·7.93943·10^4}{1.05786·10^7·7} = 1.072$$ (4.54%) 1.120 (0.26%) Resultant Shear Z max 0.698 $$\frac{Shear Z·Q_y}{I_y·t}=\frac{1·1000·5.25658·10^4}{6.32306·10^6·13} = 0.639$$ (8.45%) 0.709 (1.57%) Torsion max 9.956 $$\frac{Torsion·r_{max}}{J}=\frac{0.1·1000000·13.5357}{1.46870·10^5} = 9.216$$ (7.43%) 9.570 (3.87%

### Example 3

Determine the stresses of a section subjected to combined forces.

 Dimensions of section Geometric Properties Forces Area = 533.9368 mm2 Moment of Inertia Iz = 3.84955·105 mm4              Iy = 9.59303·104 mm4 α = -0.1562° Izp = 3.84957·105 mm4              Iyp = 9.59281·104 mm4 Distans to Max and Min Pointmax = (-15.7027, 37.3265) mm Pointmin =(14.1017, -42.0526) mm Shear Properties Qz = 6533.7159 mm3 Qy = 3440.8698 mm3 Torsion J = 1513.8 mm4 rmax = 5.1417 mm Axial = 10.0 kN Shear Y = 1.0 kN Shear Z = 1.0 kN Torsion = 0.01 kN·m Bending Y = 1 .0 kN·m Bending Z =  1 .0 kN·m

### Comparison of Results

 Result Location SkyCiv SB Analysis Manual Third-Party Primary Stresses (MPa) Axial max 18.729 $$\frac{Area}{Axial}=\frac{10·1000}{533.9368} = 18.729$$ (0.00%) 18.793 (0.00%) min 18.729 $$\frac{Area}{Axial}=\frac{10·1000}{533.9368} = 18.729$$ (0.00%) 18.793(0.00%) Bending Y max 14.234 $$\frac{Bending Y·\cos(\alpha)}{I_yp/z_{max}}+\frac{Bending Z·\sin(\alpha)}{I_zp/y_{max}}=\frac{1·1000000·\cos(24.2186^\circ)}{8.31922·10^5/-15.7027}+\frac{1·1000000·\sin(24.2186^\circ)}{4.67198·10^6/37.3265}=$$ (0.00%) 14.234 (0.00%) min -14.234 $$\frac{Bending Y}{I_y/y_{min}}=\frac{1·1000000}{6.32306·10^6/-90} =-14.234$$ (0.00%) -14.234 (0.00%) Bending Z max 3.723 $$\frac{Bending Z}{I_z/z_{max}}=\frac{1·1000000}{1.05786·10^7/39.3877} =3.723$$ (0.00%) 3.723 (0.00%) min -14.237 $$\frac{Bending Z}{I_z/z_{min}}=\frac{1·1000000}{1.05786·10^7/-150.6123} =-14.237$$ (0.00%) -14.237 (0.00%) Resultant Shear Y max 1.123 $$\frac{Shear Y·Q_z}{I_z·t}=\frac{1·1000·7.93943·10^4}{1.05786·10^7·7} = 1.072$$ (4.54%) 1.120 (0.26%) Resultant Shear Z max 0.698 $$\frac{Shear Z·Q_y}{I_y·t}=\frac{1·1000·5.25658·10^4}{6.32306·10^6·13} = 0.639$$ (8.45%) 0.709 (1.57%) Torsion max 9.956 $$\frac{Torsion·r_{max}}{J}=\frac{0.1·1000000·13.5357}{1.46870·10^5} = 9.216$$ (7.43%) 9.570 (3.87%)