Editing Structural analysis (section)

==Strength of materials methods (classical methods)==
The simplest of the three methods here discussed, the mechanics of materials method is available for simple structural members subject to specific loadings such as axially loaded bars, prismatic [[Beam (structure)|beams]] in a state of [[pure bending]], and circular shafts subject to torsion. The solutions can under certain conditions be superimposed using the [[superposition principle]] to analyze a member undergoing combined loading. Solutions for special cases exist for common structures such as thin-walled pressure vessels.

For the analysis of entire systems, this approach can be used in conjunction with statics, giving rise to the ''method of sections'' and ''method of joints'' for [[truss]] analysis, [[moment distribution method]] for small rigid frames, and ''portal frame'' and ''cantilever method'' for large rigid frames. Except for moment distribution, which came into use in the 1930s, these methods were developed in their current forms in the second half of the nineteenth century. They are still used for small structures and for preliminary design of large structures.

The solutions are based on linear isotropic infinitesimal elasticity and Euler–Bernoulli beam theory. In other words, they contain the assumptions (among others) that the materials in question are elastic, that stress is related linearly to strain, that the material (but not the structure) behaves identically regardless of direction of the applied load, that all [[deformation (engineering)|deformation]]s are small, and that beams are long relative to their depth. As with any simplifying assumption in engineering, the more the model strays from reality, the less useful (and more dangerous) the result.

===Example===
There are 2 commonly used methods to find the truss element forces, namely the method of joints and the method of sections. Below is an example that is solved using both of these methods. The first diagram below is the presented problem for which the truss element forces have to be found. The second diagram is the loading diagram and contains the reaction forces from the joints.
:[[Image:Truss Structure Analysis, Full Figure2.jpg|center|border|750x450px|A simple triangular truss with loads imposed .]]
Since there is a pin joint at A, it will have 2 reaction forces. One in the x direction and the other in the y direction. At point B, there is a roller joint and hence only 1 reaction force in the y direction. Assuming these forces to be in their respective positive directions (if they are not in the positive directions, the value will be negative).
:[[Image:Truss Structure Analysis, FBD2.jpg|center|border|750x450px]]
Since the system is in static equilibrium, the sum of forces in any direction is zero and the sum of moments about any point is zero.
Therefore, the magnitude and direction of the reaction forces can be calculated.

:<math>\sum M_A=0=-10*1+2*R_B \Rightarrow R_B=5</math>
:<math>\sum F_y=0=R_{Ay}+R_B-10 \Rightarrow R_{Ay}=5</math>
:<math>\sum F_x=0=R_{Ax}</math>

====Method of joints====
This type of method uses the force balance in the x and y directions at each of the joints in the truss structure. 
:[[Image:Truss Structure Analysis, Method of Joints2.png|border|350x450px]]

At A,
:<math>\sum F_y=0=R_{Ay}+F_{AD}\sin(60)=5+F_{AD}\frac{\sqrt{3} }{2} \Rightarrow F_{AD}=-\frac{10}{\sqrt{3}}</math>
:<math>\sum F_x=0=R_{Ax}+F_{AD}\cos(60)+F_{AB}=0-\frac{10}{\sqrt{3} }\frac{1}{2}+F_{AB} \Rightarrow F_{AB}=\frac{5}{\sqrt{3}}</math>
At D,
:<math>\sum F_y=0=-10-F_{AD}\sin(60)-F_{BD}\sin(60)=-10-\left(-\frac{10}{\sqrt{3}}\right)\frac{\sqrt{3} }{2}-F_{BD}\frac{\sqrt{3}}{2} \Rightarrow F_{BD}=-\frac{10}{\sqrt{3}}</math>
:<math>\sum F_x=0=-F_{AD}\cos(60)+F_{BD}\cos(60)+F_{CD}=-\frac{10}{\sqrt{3}}\frac{1}{2}+\frac{10}{\sqrt{3} }\frac{1}{2}+F_{CD} \Rightarrow F_{CD}=0</math>
At C,
:<math>\sum F_y=0=-F_{BC} \Rightarrow F_{BC}=0</math>

Although the forces in each of the truss elements are found, it is a good practice to verify the results by completing the remaining force balances.
:<math>\sum F_x=-F_{CD}=-0=0 \Rightarrow verified</math>
At B,
:<math>\sum F_y=R_B+F_{BD}\sin(60)+F_{BC}=5+\left(-\frac{10}{\sqrt{3}}\right)\frac{\sqrt{3} }{2}+0=0 \Rightarrow verified</math>
:<math>\sum F_x=-F_{AB}-F_{BD}\cos(60)=\frac{5}{\sqrt{3}}-\frac{10}{\sqrt{3}}\frac{1}{2}=0 \Rightarrow verified</math>

====Method of sections====
This method can be used when the truss element forces of only a few members are to be found. This method is used by introducing a single straight line cutting through the member whose force has to be calculated. However this method has a limit in that the cutting line can pass through a maximum of only 3 members of the truss structure. This restriction is because this method uses the force balances in the x and y direction and the moment balance, which gives a maximum of 3 equations to find a maximum of 3 unknown truss element forces through which this cut is made. Find the forces FAB, FBD and FCD in the above example

=====Method 1: Ignore the right side=====
:[[Image:Truss Structure Analysis, Method of Sections Left2.jpg|border|550x450px]]
:<math>\sum M_D=0=-5*1+\sqrt{3}*F_{AB} \Rightarrow F_{AB}=\frac{5}{\sqrt{3} }</math>
:<math>\sum F_y=0=R_{Ay}-F_{BD}\sin(60)-10=5-F_{BD}\frac{\sqrt{3}}{2}-10 \Rightarrow F_{BD}=-\frac{10}{\sqrt{3}}</math>
:<math>\sum F_x=0=F_{AB}+F_{BD}\cos(60)+F_{CD}=\frac{5}{\sqrt{3}}-\frac{10}{\sqrt{3}}\frac{1}{2}+F_{CD} \Rightarrow F_{CD}=0</math>

=====Method 2: Ignore the left side=====
:[[Image:Truss Structure Analysis, Method of Sections Right2.jpg|border|600x450px]]
:<math>\sum M_B=0=\sqrt{3}*F_{CD} \Rightarrow F_{CD}=0</math>
:<math>\sum F_y=0=F_{BD}\sin(60)+R_B=F_{BD}\frac{\sqrt{3}}{2}+5 \Rightarrow F_{BD}=-\frac{10}{\sqrt{3}}</math>
:<math>\sum F_x=0=-F_{AB}-F_{BD}\cos(60)-F_{CD}=-F_{AB}-\left(-\frac{10}{\sqrt{3}}\right)\frac{1}{2}-0 \Rightarrow F_{AB}=\frac{5}{\sqrt{3}}</math>

The truss elements forces in the remaining members can be found by using the above method with a section passing through the remaining members.