Geometry Essay

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Calculation of a beam’s moment capacity:


The triangular beam indicated is made of 4ksi concrete and carries three
#9 rebars of fy=60ksi.


Note that geometry offers itself for calculations regarding the Compression, the lever-arm, and the depth of a.

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Calculation of a beam’s moment capacity:


The triangular beam indicated is made of f`c=4ksi concrete and carries three (3) #9 rebars of fy=60ksi.


Note that geometry offers itself for calculations regarding the Compression, the lever-arm, and the depth of α.

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We begin by the fundamental assumption that fs=fy (we verify this at

a later step) and we calculate the area of steel As:



3 #9 rebars give us a total area of 3 sq. in.
T=As(fy) = (3in^2)(60ksi) = 180 kips fs=Calculated stress of reinforcement at service loads, fy=specified yield strength of non prestressed reinforcement.

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We compute the compression block so that we have C=T:



Given C=T = 180 kips
The specific geometry of the beam allowed us to generate the formula for C. That can be reversed to solve for a:
…which gives us a value of 10.29 in.

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Verifying our assumption that fs=fy:

As mentioned earlier, εcu will carry the value of 0.003




c=α/β1…thus c=10.29in/0.85…..= 12.11in
Using the similar triangles method as seen above… thus εs is 0.00394 > εy which is f`c/fy that is 60ksi/29000ksi yielding a value of 0.00207, proving that fs=fy εs=strain in steel, εy= yield strain. fs=Calculated stress of reinforcement at service loads, fy=specified yield strength of non prestressed reinforcement.

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Computing the Mn:


We can derive from the geometry that jd=d-(2a/3)



Thus it would be safe to claim that



..which yields the result of 285.4 k`

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That wonderful property of structural materials to bend, crack and yet not break, is one of the possible characteristics of RC.
When flexural forces surpass the limit My, steel reinforcement continues to elongate. Resistance increases slightly, related to the increase of distance between C and T. That distance increases as the depth of the concrete stress block decreases until the concrete fractures. Although the stress of the steel remains constant, the strain at the point of failure is several times greater than the steel yield strain εy, ..approximately εy=fy/Es ≈.002

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On a section that fractures when the strain of steel is 0.006, if the As was multiplied by a certain factor (let’s say doubled), then the Whitney block would take a similar magnification (double in this case). The strain at the tension could only be 0.003. The stretching of the steel at the range between yield and beam failure would only be 0.001 instead of
0.004 as it was for the section with half as much tension reinforcement.

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There are three methods of flexural failure of
concrete
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