SkyCiv Documentation

Your guide to SkyCiv software - tutorials, how-to guides and technical articles

SkyCiv Section Builder

  1. Home
  2. SkyCiv Section Builder
  3. Analysis
  4. Calculation Verification

Calculation Verification

Example 1

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

Dimensions of section Geometric Properties Forces
section builder analysis verification T-section
  • 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
section builder analysis verification aluminium section
  • 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%)
Was this article helpful to you?
Yes No

How can we help?

Go to Top