Question

i need like a course work and iam also attaching a model of course work in down please do like that. First page is the question rest pages which i written as model is the model of the question

Technical Mechanics Course work: Problem 1. Input data: Rigid body (a x b) is connected with immovable base by plane bar system (plane truss) and loaded by plane forces. The sizes and load components are given below (table 1 and schemes). Define: Internal force of bars and reaction of supports.

Answer 1

Solution:

1. General cross section (Look at picture)

The right part of system is removed and its acting effect to left part is shown by internal forces in cross section of bars.

2.The form of the equillibrium of this part pf a truss selected.

$N_{2-4}=\frac{(2.4\times 1.7)+4.5+(1.5\times 2.2\left ( \frac{2.2}{2} \right ))}{1 .7}$

Now,

$\sum_{k=1}^{n}mom_{4}F_{k}=0$

$N_{1-3}+M_{1}+qa\left ( \frac{3a}{2} \right )=0$

$N_{1-3}=\frac{-M_{1}-q\left ( \frac{3a^{2}}{2} \right )}{b}$

$N_{1-3}=\frac{-3.5-1.5\left ( \frac{3(2.2)^{2}}{2} \right )}{1.7}$

$N_{1-3}=-8.464kN$

Also,

$\sum_{k=1}^{n}F_{ky}=0$

$N_{2-3}\cos \beta -qa=0$

$\cos \beta =\frac{b}{\sqrt{a^{2}+b^{2}}}$$\sin \beta =\frac{a}{\sqrt{a^{2}+b^{2}}}$

$\cos \beta =\frac{1.7}{\sqrt{2.2^{2}+1.7^{2}}}$   $\sin \beta =\frac{2.2}{\sqrt{2.2^{2}+1.7^{2}}}$

$\cos \beta =0.6114$$\sin \beta =0.791$

Back to Equation,

$N_{2-3}\cos \beta -qa=0$

$N_{2-3}=\frac{qa}{\cos \beta }$

$N_{2-3}=\frac{1.5\times 2.2}{0.6114 }$

$N_{2-3}=5.397kN$

3. Next step: Local cutting of Node 3

Now,

$\sum_{k=1}^{n}F_{kx}=0$

$-N_{2-3}\sin \beta -N_{1-3}+N_{3-5}=0$

$N_{3-5}=N_{2-3}\sin \beta +N_{1-3}$

$N_{3-5}=\left [ (5.397\times 0.791) +(-8.464) \right ]$

$N_{3-5}=-4.195kN$

Also,

$\sum_{k=1}^{n}F_{ky}=0$

$-N_{2-3}\cos \beta +N_{3-4}=0$

$N_{3-4}=N_{2-3}\cos \beta$

$N_{3-4}=5.397\times0.791$

$N_{3-4}=4.269kN$

4. Next step: Local cutting node 4

$\sum_{k=1}^{n}F_{ky}=0$

$-N_{4-5}\cos \beta -N_{3-4}=0$

$N_{4-5}=-\frac{N_{3-4}}{\cos \beta}$

$N_{4-5}=-\frac{4.269}{0.6114}$

$N_{4-5}=-6.98kN$

Now,

$\sum_{k=1}^{n}F_{kx}=0$

$N_{4-5}\sin \beta +N_{4-6}-N_{2-4}=0$

$N_{4-6}=N_{2-4}-N_{4-5}\sin \beta$

$N_{4-6}=7.182-(-6.98\times 0.791)$

$N_{4-6}=12.632kN$

Number	1-3	2-3	2-4	3-4	3-5	4-5	4-6
N,kN	-8.464	5.397	7.182	4.269	-4.195	-6.98	12.632