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Deflection has an inverse relationship with the elastic modulus, which is one parameter of the stiffness (EI). The tolerance of stress on the material is tied to its yield strength (another indicator of stiffness), which is the stress the material ends to deform elastically as seen in the stress-strain diagram below. Note the yield point is at the end of the ...


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When you want to determine the internal reactions to a structure (in a example such as this), then the best course of action is to isolate the structure of interest by creating section around the part. The next step is to draw the internal forces on the points where the section (blue bubble) intersects with the structure. Please notice, that in this ...


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Without access to FEM program, you have to refresh or learn the "Column Analogy Method" developed by Prof. Hardy Cross. Here is a paper provides the basic for using such method to find the fixed end moment, carry over factor and distribution factor, that all are required for solving indeterminate beams with the varying moment of inertia using the ...


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Use graphics to visualize the force flow and determine zero-force members. (Draw member forces from joint D)


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DE and AE are zero are indeed zero. However you need to start from point C to reason about that. If you look at this question, when you have a node without external forces with only two members which are not colinear, the forces on the nodes should be zero. So from this $DC$ and $CB$ are equal to zero. Node D: Now if you take point D and write the equations ...


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In this problem, you have at A $$ F_{Ax}, F_{Ay}, M_{Ax}, M_{Ay},M_{Az} \text{ not equal to zero}$$ only $ F_{Az}$ is equal to zero. and also at point C, one force with magnitude $F_{C}$, and components you can derive from the direction of the wire $$\vec{BC} = \begin{bmatrix} -L_1\\ -L_3-L_2\\ L_4 \end{bmatrix}$$ you can estimate the direction (unit ...


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Resolve the cable by the upward force equals F at B. So do the sin/cos thing to get the cable tension. Then get the horizontal component of the cable which will come a moment about Az. F produces an additional moment about Ax from the force at B. The key is resolving the cable tension via its x and z components, with the z component = F.


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What you need to do is you need to isolate the masses from the environment. I.e. create a section. At the points that section intersects with other objects (i.e. spring) you should place forces there with the correct direction. Additionally you need to add any external forces like weight on the masses.


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The forces are lying on planes defined by the space grid system, u, v & w. You need to work out the given angle between F2 and V-axis first, then from the given condition F1v = F2v, the angle between F1 and V-axis can be found. After that, everything is cleared up then.


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I would say that you can divide them into two groups: At nodes with two member and no external force Node C: CD,CB, Node G: FG, GH Node K: KJ, KL If you write down the equilibrium at node C for x and y you would get: $$x: \quad \sum F_x = 0 \rightarrow CD = 0 $$ $$y: \quad \sum F_y = 0 \rightarrow CB = 0 $$ Same at the other nodes. Note that nodes A, and ...


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Members KJ and KL by inspection are zero force. there may be more but not immediately evident to mho.


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Simple rule of thumb for non-critical situations, where appearance is also important - place each support 20% in from each end


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