Application to Truss Problems Assignment Help

Assignment Help: >> Application of Castiglianos theorem - Application to Truss Problems

Application to Truss Problems:

In this section, we will apply Castigliano's theorem to calculate the deflections in a truss.

Let a truss within m members be subjected to a system of loads P1, P2, . . . , etc. The forces in the members of the truss might be obtained by standard methods. Assume Fi be the force in the ith member because of the above system of forces. Let Li & Ai be respectively the length & area of cross-section of the member. After that the strain energy stored in this member shall be

1273_Application to Truss Problems.png

where E is the Young's modulus of the material.

Then, the total strain energy stored in the truss might be written as follows :

573_Application to Truss Problems1.png

Now, as per Castigliano's theorem the deflection δn under the load Pn in its direction shall be

1691_Application to Truss Problems2.png

Let us observe this equation closely.

∂Fi /∂Pn is the rate at which F varies along with P with all other loads remaining constant or in other terms it is the force in the member while a unit load is applied rather then Pn, while all of other loads are zero.

The above concept may still be applied to discover the deflection in any direction at any joint of the truss even while no load is working in that direction or at that point. In this case, we introduce a fictitious force Q at the pint in the direction in which we need the deflection. After that we determine the member forces for the load system that now includes Q also.

Then,

      1158_Application to Truss Problems3.png  (now Fi includes the effect of Q also)

Therefore,

2466_Application to Truss Problems4.png

However, in real problem there is no Q or its value is zero. Thus we put Q = 0 in the above expression and get δQ.

The concept of unit load might be brought in even here. Now while we include Q in the load system Fi may be written as

Fi  = FiP  + FiQ  (principle of superposition)

where FiP & FiQ are the member forces only because of the Ps and Q, respectively.

Then,

U  =∑ (FiP  + FiQ )2  Li/2 Ai Ei

And

662_Application to Truss Problems5.png

 (As FiP does not involve Q and, hence,   ∂F iP /∂Q = 0 ).

While we put Q = 0 in the above expression it decrease to

1905_Application to Truss Problems6.png

∂FiQ /∂Q is nothing but the member force because of a unit load in the direction of Q at the point of application of Q while all other loads are zero.

Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd