Example 1
Determine the stresses of a T-section subjected to combined forces.
| Dimensions of section | Geometric Properties | Forces |
 |
Moment of Inertia
Distans to Max and Min
Shear Properties
Torsion
|
|
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 | \(\frac{r_{max}}{J}=\frac{0.1·1000000·13.5357}{1.46870·10^5} = 9.216\)
(7.43%) |
9.570
(3.87%) |
Example 2
Determine the stresses of a section subjected to combined forces.
| Dimensions of section | Geometric Properties | Forces |
 |
Moment of Inertia
Distans to Max and Min
Shear Properties Shear Y
Shear Z
Torsion
|
|
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%) |
18.73
(0.00%) |
| min | 18.729 | \(\frac{Area}{Axial}=\frac{10·1000}{533.9368} = 18.729\)
(0.00%) |
18.793
(0.00%) |
|
| Bending Y | max | 166.538 | \(\frac{M_y·\cos(\alpha)}{\frac{I_y}{z_{max}}}+\frac{M_y·\sin(\alpha)}{\frac{I_z}{y_{max}}}=\frac{1000000·\cos(-0.1562^\circ)}{\frac{3.84955·10^5}{-42.0526}}+\frac{1000000·\sin(-0.1562^\circ)}{\frac{9.59281·10^4}{14.1016}}=166.694\)
(0.00%) |
166.5
(0.00%) |
| min | -165.951 | \(\frac{M_y·\cos(\alpha)}{\frac{I_y}{z_{min}}}+\frac{M_y·\sin(\alpha)}{\frac{I_z}{y_{min}}}=\frac{1000000·\cos(-0.1562^\circ)}{\frac{3.84955·10^5}{30.7351}}+\frac{1000000·\sin(-0.1562^\circ)}{\frac{9.59281·10^4}{-15.9392}}=166.045\)
(0.00%) |
-166.0
(0.00%) |
|
| Bending Z | max | 97.189 | \(\frac{M_z·\cos(\alpha)}{\frac{I_z}{y_{max}}}+\frac{M_z·\sin(\alpha)}{\frac{I_y}{z_{max}}}=\frac{1000000·\cos(-0.1562^\circ)}{\frac{3.84955·10^5}{37.2424}}+\frac{1000000·\sin(-0.1562^\circ)}{\frac{9.59281·10^4}{-15.7027}}=97.19\)
(0.00%) |
97.19
(0.00%) |
| min | -109.639 | \(\frac{M_z·\cos(\alpha)}{\frac{I_z}{y_{min}}}+\frac{M_z·\sin(\alpha)}{\frac{I_y}{z_{min}}}=\frac{1000000·\cos(-0.1562^\circ)}{\frac{3.84955·10^5}{-42.0526}}+\frac{1000000·\sin(-0.1562^\circ)}{\frac{9.59281·10^4}{14.1016}}=-109.64\)
(0.00%) |
-109.6
(0.00%) |
|
| Resultant Shear Y | max | 4.302 | \(\frac{ShearY·\cos(\alpha)·Qz}{Izp·t}+\frac{ShearZ·\cos(\alpha)·Qy}{Iyp·t}=\frac{1000·\cos(-0.1562^\circ)·6533.7159}{{3.84955·10^5·3.9624}}+\frac{1000·\sin(-0.1562^\circ)·4.2994}{9.59281·10^4·3.9624}=4.283\)
(0.44%) |
4.297
(0.12%) |
| Resultant Shear Z | max | 16.629 | \(\frac{ShearZ·\sin(\alpha)·Qz}{Izp·t}+\frac{ShearZ·\cos(\alpha)·Qy}{Iyp·t}=\frac{1000·\sin(-0.1562^\circ)·929.3201}{{3.84955·10^5·2.8145}}+\frac{1000·\cos(-0.1562^\circ)·3337.6406}{9.59281·10^4·2.8145}=12.36\)
(25.67%) |
17.37
(4.46%) |
| Torsion | max | 30.418 | \(\frac{r_{max}}{J}=\frac{0.1·1000000·4.6293}{1513.65} = 30.584\)
(0.55%) |
31.98
(5.14%) |
