Civil Engineering Association

If the beam is made of reinforced concrete then it is obvious that the stress-block (this is the compressed portion of the concrete section) of inverted "T" has smaller static-moment than the regular "T". In simpler words: for the inverted T the compressed portion of the concrete is in the "stem" and reaches far closer to the "zero line", compared to the compressed portion for the upwards T, where it is in the web, not in the stem.

What is the problem with your eye-vision? Or it is matter of taste: bold, italic, red, no capital letters.

deadlord, the deflection member is dependent on mom of I, E and load span keeping all constant and finding mom of I for t and inverted T will give U the exact answer

Ssamis, you're right, but only for homogeneous material of the beam. So if you have steel T-beam, then it doesn't matter if it is inverted or not, because the Inertial moment is the same if you rotate the section at 180 degrees, because you will have the same axis. That means that for steel beam the deflection will be the same. We will have different stability, but this is another topic.

Let's see for reinforced-concrete where we have changeable stressed cross section, because the concrete cracks when it is in tension (and the steel bars are take this tension).

In the reversed T the concrete is not used optimally, because the centroid of the compressed concrete goes much more (actually with the ration of the top to bottom width of the T) closely to the centroid of the steel bars. So for RC beam you will have different stiffness, depending on the width-distribution of the concrete.

deadlord wrote:-
Hi,how u all doing??
our professor recently stated that with experience they've discovered that inverted beams(T-beams) deflect more than the usual T-beam(the web is downward)

and he made it as an open challenge to whom brings him the answer,and that he'll give him a full mark in the 2nd exam lol
anyway,it's not for marks that i'm asking it's just to know,cuz if searched the internet and thought of it a lot actually ,and i can't find a certain explanation to what he have told us .

so if any of our forum's beloved members have any idea about how inverted beams behave please state it in here.
i'm just so thrilled to know what is happening :)

have a nice day
Comment:- I think that reverse of what you wrote is the fact i.e. that the usual T-beam(the web is downward) deflect more than inverted beam (T-beams-the web is upward).
REASON:-The conventional or the classical equations in the codes are results of cumulative contributions to deflection that result from different sources such as bending stresses, shearing stresses, creep, fatigue etc. The equations also made some compromises by eliminating the parts of the equation that are “troublesome” but which has little effect on the resultant net deflection (like the square of the differential dy/dz i.e. (dy/dz)^2). Though these may be assumed negligible (as to facilitate computation), in actual fact, the theory of deflection is based on the assumption of small deflections as such it becomes doubtful if eliminating the parts of the equation that (I would prefer to refer to as “troublesome”) does not compromise the whole undertaking (by reducing what is already small further thus rendering it much smaller).
The codes took care of this by building into the equations some factors of safety (either in the loads or in the materials). So the classical equations (like the ones in the codes) that relate deflection to load or moment, modulus of elasticity and the second moment of inerter do not tell the whole stories (though they are adequate for all practical purposes).
The equation of deflection has the general form y = CwL^n/EI. n could assume a value of 3 or 4, depending on whether w is a point load or distributed load. For 2 beams of the same material, same end conditions, same loading, what can determine their deferent behaviors in relation to deflection is the moment of inerter (I). In the case of a T-beam, whether standing up-right or inverted, the moment of inerter (I) remains the same (since the moment of inerter (I) will be taken about the axis that passes through the centroid of the section, which remains same in both situations) as such the classical equation will not solve the case. So let’s get back to basics. The basic equation for deflection is derived from the theories of elastic bending and curvature. Curvature is the inverse of the radius of the circle (the elastic curve) traced by the beam’s axis when subject to load- thus when deflected. This is given as 1/R = M/EI = d^2y/dz^2. Double integral of d^2y/dz^2 = y = deflection. This implies that deflection is directly proportional to curvature (1/R).
Stress (s) = Mx/I which implies that M = sI/x, where M = moment imposed on the structural member, x = depth of the neutral axis and I is as defined previously. Substituting for M, 1/R = M/EI = s/Ex = d^2y/dz^2. Thus deflection = y = double intgration of the above equation) = ff(s/Ex )dzdz = sz^2/2Ex + Az + C. Since the 2 beams have same end conditions, we can ignore the constants, s, A, and C. We can also ignore the distance of the point at which we are calculating deflection (z) from the supports (since it is same for the 2 conditions. This implies that the depth of the neutral axis is the only variable that can influences deflection. Based on this equation, it could be inferred that deflection =y is inversely proportional to the depth of neutral axis i.e. deflection increases with decrease in the depth of the neutral axis and verse versa. For the case under consideration (up-right T-section and the inverted T-section), if we represent deflection for the up-right case as D1 and that for the inverted as D2, this implies that the ratio of the deflections = D1/D2 = (sz^2/2Ex1)/(sz^2/2Ex2) = x2/x1. This demonstrates that the deflection in both cases are not same but will be governed by the location of the neutral axis.
Take for instance, a steel T-section of dimension 191mm x229mm (of flange thicknesses = 19.6mm and web = 11.4mm). The depth of the neutral axis while standing up-right (x1) = 55.5mm, whilst its depth when inverted (x2) = 233.7 – 55.5 = 178.2. So D1/D2 = 178.2/55.5 = 3.2; indicating that the up-right T-section will deflect more than three times the inverted T-section!!!. The reasons for this improved performance (of the inverted T-section) over the common bean configuration (of the up-right T-section), apart from the reason demonstrated above, are:- at the supports, the direction of the moments are reversed, as such the whole flange is subjected to compression. This presents an enlarged area in compression with net stiffening effect. Again, the centroid of the section is closer to the base as such the mass of the cross section is concentrated within a much limited area around the base line. This gives it a more robust structure.
The above tells only a part of the whole story. Since deflection is a complex combination of deflections from various sources, this aspect may not govern the overall deflection, though in most normal cases, the major contributor to deflection is the bending stress (as calculated above)
For the inverted concrete T-section, following the fact that the centroid is lower (closer to the base of the section), as such the vertical (depth of the beam) portion subjected to compression is greater, this sets up the creep problem. In all cases (irrespective of the beam material), the web is not restrained in compression (unlike the up-right T-section for which the web is restrained by the flange while subjected to compression) This could create buckling problem; as such it will be necessary to check this arrangement for buckling. Another problem with this configuration (the inverted T-section) is that it is not very practicable, in that the web of the beam becomes an obstacle to movement (except for remote places like the roof of buildings where serve areas are of no importance or if they are so arrenged that they are located in the positions of walls i.eare under the walls).
Regards
Teddy