...line for the cantilever beam with ascending triangular load is thus completely defined. It is \[ \tag{11} w = \frac{\frac{q_0x^5}{20}-\frac{l^2q_0x^3}{2}+l^3q_0x^2}{6EIl} \] Don't miss the other exercises regarding Engineering Mechanics II!...
...x}{72 E I}... \] \[ \tag{53} ...-\frac{{{l}^{4}} q+32 F\, {{l}^{3}}}{24 E I} \] The determination of the bending lines is now complete. It can be clearly seen that using the...
...{F_{\mathit{Az}}}=\frac{11 l\, {q_o}-5 l\, {q_0}}{16} \] \[ \tag{33} {F_{\mathit{Bz}}}=-\frac{11 l\, {q_o}-21 l\, {q_0}}{8} \] \[ \tag{34} {F_{\mathit{Cz}}}=\frac{11 l\, {q_o}-5 l\, {q_0}}{16} \] And for the sake of completeness \[ \tag{35}...
...and forces in a comfortable way, we introduce the angle β. This results in: \[ \tag{1} \sum F_x = = F_{Ax} - S \cdot \cos \beta \] \[ \tag{2} \sum...
This exercise shows how the safety against flow is calculated according to the TRESCA shear stress hypothesis. The equivalent stress is calculated from the principal stresses and then compared with...
...bearing reactions are drawn. The horizontal component of the fixed bearing is not discussed here, as it is obviously zero. The line load q, even if it is given here...
Contact: Christoph Mahlke Sieker Landstraße 95, 22927 Großhansdorf, Germany E-Mail: service@pickedshares.com Phone: 0172 2309784 Content responsibility: Christoph Mahlke...
...side is sufficient. It is a system of concurrent forces in a plane. Rope forces for exercise 4 (only one side) The vertical component of the rope force equals to...
