Theory and design of Structures

The power of singularity functions isn't observed with a beam supporting a concentrated load at some distance from its support just like the above one. It is usually useful when multiple loads are assigned over the span of the beam and it would help avoid the numerous sections that we take in order to write the shear and moment equations.

Substituting the boundary conditions into the above equations: y'(0)=0, C4=0 and y'(L)=0, C3=-P(1-a/L)L^2 /3EI Therefore the equation of slope and deflection with singularity functions become: y'=(P<x-a>^(2)/2+P(1-a/L)*x^2/2)/EI -P(1-a/L)L^2 /3EI y=(P<x-a>^(3)/6+P(1-a/L)*x^(3)/6)/EI-(P(1-a/L)L^2 /3EI )*x

We can proceed further with it and determine the slope and deflection equations. Since y"( I was tempted to write d^2y/dx2 gene see how messy it is ) =M/EI : y"=(P<x-a>^(1)+P(1-a/L)*x)/EI Integrating the above equation gives y'=(P<x-a>^(2)/2+P(1-a/L)*x^2/2)/EI +C3 And integration of the above equation would be: y=(P<x-a>^(3)/6+P(1-a/L)*x^(3)/6)/EI+C3*x+C4 In order to calculate the constants of integration we need to identify the boundary conditions. Since the system is made of pin and roller support we know the deflection at the two supports is zero. Meaning: y(0)=0 and y(L)=0

Now substituting into (1) V(0)=0, we can get C1=Ay=P(1-a/L) And substituting into (2) M(0)=0, we can get C2=0 Now the equations would simplify to: V(x)= P^(0)+p(1-a/L) M(x)=P^(1)+p(1-a/L)*x Here we use <> ( macauleys brackets after the person who developed them) because it has meaning. If x<=a then the value in <> is zero automatically. And if x>a then the value in <> becomes the value that is evaluated from the term in it.

Here is a beam with a concentrated load of 'P' at 'a' distance from the left support. Let's call the left support A and the right support B. From equilibrium of the simple beam the reactions at the left and the right supports are P(1-a/L) and Pa/L. Now the singularity function for the load would be, P^(-1). The shear force on the beam is the integral of the load singularity function. And by definition of integral of singularity function( I have a reason for saying this, u can ask want to understand) the integral of ^(-1) is just ^(0). So the shear force equation would be V(x)=P^(0)+c1 And the moment equation can be obtained by simply integration of the shear function obtained above. M(X)=P^(1)+c1*x+c2 Now that we have the equations all we need to worry about is the constants of integration. For them we have boundary conditions which goes: The shear at x=0 is the reaction at the support and the moment at x=0 is 0 since the support is pin. Two boundary conditions would give two equation

loads with singularity functions and respective shear and moment.

Some of the singularity functions with their respective integrals are listed in this picture.

Singularity functions. When dealing with bending moment, shear force and beam deflection a more generalized and powerful way is using singularity functions. This functions besides their application in computer programming, they give a huge advantage for students and practitioners dealing with the above mentioned problems. The process starts with defining the function representing the loading. Once that is done successfully and the boundary conditions are set successive integarals are done to arrive at the conservative shear, moments , slope and deflection equations. and each integral will yeild constants. Using boundary conditions each constant is solved for and the equations are set. Then the resulting functions can be used cautiously since rules are associated with the functions. One such rule is shown In the above picture.

Continuing the above discussion for RCC beams, we can understand why we place reinforcement bar close to the bottom and top fibers. The first way to understand this is following again bending stress distribution for ELASTIC stress distribution. Since the maximum tensile stress for a given beam is found near the top or bottom fibers, It can assist(before cracking concrete can carry a very small amount of tensile stress) in carrying the developed tensile stress the more stressed concrete fibers. Once the section is cracked, all the tension is carried by the reinforcement. At this stage the advantage of placing the rebar near the bottom or top fibers is in gaining more lever arm. Since the moment resistance of the section is dependent on the lever arm( the distance between the centroid of compression carrying concrete and the tension carrying steel) more lever arm means, more moment resistance of the section. One more advantage is in gaining more second moment of area about the centroid of the section.

The reason why I section is made in such way, besides economy, is to place more material away from the neutral axis which leads to more second moment of area following the parallel axis theorem. The gain in second moment of area would lead to better resistance in bending stress. Other way to understand why I section is in such way Is following bending stress distribution. Since elastic bending stress for a given beam linearly increases from zero at neutral axis to the maximum at the top and bottom fibers, we can argue that more material should be placed near the top and bottom fibers than near the neutral axis.

...Meaning in order to maintain joint and the whole structure equilibrium, the web members follow patterned tension and compression, I.e, we can state that the diagonal membranes would be in compression( the term diagonal compression strut that we use for concrete...Vrd,max😊) and the vertical members would be tension members. Even this discussion is subjected to wind caused stress reversal.

😁 Trusses....we can say a lot about this miraculous Structures. One thing we must know, trusses are used when the span of a structure is very large and beam sections become uneconomical to justify. So in some way trusses are beams( for real dude, are u comparing flexural and axial members 😳...not at all, gene one can be used to understand the behavior of the other). Simply supported Beams under gravity load, would experience tension at the bottom fibers and compression at the top fibers. Following this One can say with confidence, top chord truss members are compression members while bottom chord members are tension members. So even in preliminary design we can understand what kind of section to use for this members. But, there is this thing called wind load which usually leads to stress reversal. Meaning members which act as compression elements would become tension elements under wind load( since wind has both pressure and uplift effect). The diagonal or web members act like shear resisting elements

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Criteria to consider second order effect according to EBCS-2,1995 1. If θ<0.1 the structure is classified as non-sway. 2. If 0.1<θ<=0.2 the structure is classified as sway and Approximate analysis can be carried out by multiplying relevant seismic action by a factor of 1/(1-θ). 3. The value of θ shall not be more than 0.25. The only difference in this section between EBCS-8 and ES EN 1998:2015 is the maximum allowable value of θ which is 0.25 and 0.3 for the old and new codd respectively.

Back to the old code, EBCS-8,1995, we can see a different formula yet similar principle of displacement analysis. EBCS-8,1995 doesn’t state, in this section, how the displacement from static and dynamic non-linear analysis is to be considered. @theoryanddesignofstructures

Related to displacement analysis, the design story drift in the calculation of the interstory drift sensitivity coefficient(θ) should follow the principles stated in ES EN 1998:2015, Art. 4.3.4( which is discussed above). @theoryanddesignofstructures

⬆️ Displacement analysis(ES EN 1998:2015): 2. If static and dynamic non-linear analysis is performed, the displacement obtained directly from the static analysis without further modification can be used. @theoryanddesignofstructures

⬆️Displacement analysis(ES EN 1998:2015): 1. If elastic analysis is performed the displacement induced by the design seismic action shall be calculated on the basis of the elastic deformation of the structural system by: dₛ=qd*dₑ (definition of terms can be seen from the above picture) What one should consider is the value of qd. Two things should be keept in mind about the value of qd: I. It can be assumed equal to q II. qd is larger than q if the fundamental period if the structure is less than Tc. @theoryanddesignofstructures