Internal Stresses in a Cracked Reinforced Concrete Beam Updated

Continuing from before ... (here)

At the risk of getting thoroughly worn out with calculations, let's look at the stresses in the steel and concrete in the cracked section as we load the beam above the Cracking Moment, theoretically.

The stress are assumed to increase linearly from the neutral axis outward. The concrete carries no stress in tension. In reality, probably some concrete carries a bit of tension. And, once above about 1/2 of f '_c, concrete departs from linear elastic behavior (E changes) ... and thus the triangular stress distribution, use of I, etc., break down.

The moment in the beam at any stage in loading is, in this cracked (but still linear elastic) condition,

... M_a = Td* = Cd*,

where T = the tension force carried entirely by the steel,

C is the resultant compressive force (concrete in compression), and

... d* is the distance (`moment arm') between C and T.

So, the compression block is 0.88 in. tall (deep) by 11 in. wide. In the supposed under-reinforced beam the stress in the concrete varies linearly outward. The resultant C occurs 2/3 of 0.88 in. from the neutral axis = 0.59 in. up from the neutral axis (or 0.29 in. down from the top). The distance between C and T is therefore d - 0.29 in. = 3.0 in. - 0.29 in. = 2.71 in.

At any stage of loading, M _a, then,

... T = M_a / 2.71 in.,

... and ...

C = M_a / 2.71 in.

The stress in the steel is ... σ _t = T / A_s = T / 0.22 in.².

(Note that now I am calculating stresses in the steel directly, not the stresses in `steel transformed to concrete'. We could calculate the stresses in the transformed material, and then multiply them by `n' to get the real stresses.)

The stresses in the concrete vary from 0 psi at the neutral axis to σ _{c, max} = 2 σ _{c, ave} = 2(C/A_c) at the extreme fiber, where A_c = the area of the concrete in compression = 0.88 in. x 11 in. = 9.68 in.².

Now we can calculate the stress in the steel and the extreme fiber stress in the concrete as long as things stay linear (up until steel yields).

Here goes ...

P = 967 lb ... Ma = 1696 lb-ft = 20,346 lb-in. ... T = C = 7508 lb ... σ _t = 34,100 psi ... σ _c = 1551 psi.

P = 1000 lb ... Ma = 20,940 lb-in. ... σ _t = 35,100 psi ... σ _c = 1596 psi.

P = 1200 lb ... Ma = 24,540 lb-in. ... σ _t = 41,161 psi ... σ _c = 1800 psi ???

P = 1400 lb ... Ma = 28,140 lb-in. ... σ _t = 47,200 psi ... σ _c = ...

P = 1600 lb ... Ma = 31,740 lb-in. ... σ _t = 53,237 psi ... σ _c = ...

P = 1800 lb ... Ma = 35,340 lb-in. ... σ _t = 59,275 psi ... σ _c = ...

P = 1824 lb ... Ma = 35,772 lb-in. ... σ _t = 60,000 psi ... σ _c = ...

Above about 1/2 of f 'c I don't even show the concrete stress (the `...'); we can't calculate it with our `linear' analysis. And, the relationship between Ma and σ _t is also suspect. But note that we are approaching steel yield faster than a theoretical (linear) f '_c; at P=1200 lb we are 2/3 the way to f_y and only half way to f '_c. And, as the concrete becomes nonlinear, we will approach f '_c even slower.

All this to say that we can anticipate that the steel will yield before the concrete crushes ... GOOD!

And again I remind the reader that this is just a DRAFT!

So, let's say at about P = 1800 lb (Ma = 36,000 lb-ft) ... we expect steel yield.

References

Deflections of an Uncracked Reinforced Concrete Beam, Jeff Filler, Associated Content.

Cracked Section Neutral Axis and Moment of Inertia, Jeff Filler, Associated Content.

Effective Moment of Inertia and Deflections of a Cracked Concrete Beam, Jeff Filler, Associated Content.

Design of Concrete Structures, Christian Meyer, Prentice Hall, 1996.