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Developed the moment - curvature relationship for the section bel

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.

f'c 4000 psi * :

fc psi :

Gr 60 :

Fy 60 ksi * :

E.s 29000 ksi * :

r 112 / + (in * :

d.t 38 / in * :

d.b 88 / in * : in 1

A.b π d.b 2^4 / * : in 2^0.79

h 20 in * :

b 12 in * :

n.b 3 :

2) Compare the ultimate strength found above to the nominal strength found using ACI 318 stress block assumptions

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a) Before Cracking

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Before cracking occurs, use the uncracked, transformed section properties:

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E.c 57000 f'c fc / sqrt * fc * : ksi 3605

Concrete Modulus

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n E.s E.c / : 8.04

Modular ratio

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d.e h r - d.t - d.b 2 / - : in 17.62

Depth of steel

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A.s n.b A.b * : in 2^2.36

Area of steel

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y.t ΣA.i y.ti * ΣA.i / :

y.t b h * h 2 / * n 1 - (A.s * d.e * + b h * n 1 - (A.s * + / : in 10.49

Transformed centroid

(from top of section)

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Uncracked, transformed

moment of inertia:

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I.un I.i A.i d.i 2^* + Σ :

Neglect the moment of inertia of the steel itself Is=0

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I.un b h 3^* 12 / b h * y.t h 2 / - (2^* + n 1 - (A.s * y.t d.e - (2^* + : in 4^8.9 103^*

Calculate the cracking moment using these section properties:

f.t 7.5 f'c fc / sqrt * fc * : ksi 0.47

Concrete tensile strenght

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M.cr f.t I.un * h y.t - / : kip m * 11.28

Concrete tensile strength

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Calculate the curvature before cracking

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ϕ.bcr f.t E.c h y.t - (* / : 1 in / 1.38 105 -^*

b)After Cracking

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After cracking occurs, properties are based on the cracked, transformed section properties:

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To find the depth of th neutral axis:

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ΣA.i d.i * ΣA.i d.i * ≡

Compression Tension ≡

α b kd * kd 2 / * :

β n A.s * d.e kd - (* :

α β ≡ kd solve maple in 6.05 9.21 - 21 mat

kd 6.05 in * :

Cracked transformed

moment of inertia

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I.cr I.i A.i d.i 2^* + Σ :

I.cr b kd 3^* 12 / b kd * kd 2 / (2^* + n A.s * d.e kd - (2^* + : in 4^3.43 103^*

The moment will be the same at this point (as stress transfers from concrete to steel)

Calculate the curvature after cracking

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Φacr M.cr E.c I.cr * / : 1 in / 3.6 105 -^*

After cracking curvature

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c)When f'c=0.45f'c

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We will assume that concrete and steel are both still elastic at this point. This means that the neutral axis depth (kd)

does not change

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ε.c0.45 0.45 f'c E.c / * : 4.99 104 -^*

Top concrete strain

at 0.45 f'c

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ϕ.0.45 ε.c0.45 kd / : 1 in / 8.25 105 -^*

Curvature when f'c=0.45fc

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Moment when

f'c=0.45 f'c

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M.0.45 M.cr ϕ.0.45 Φacr / * : kip in * 1.02 103^*

Check to make sure the steel has no yielded

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ε.s ε.c0.45 d.e kd - kd / (* : 0.001

Steel strain when

fct=0.45f'c

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< εy

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d)Yield of Steel

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ε.y 0.0021 :

ε.s ε.y :

We can't use our transform our steel beacuse we are no longer assuming linear behavior. We will assume that the maximum

concrete strain is less than the strain corresponding to the maximum stress. (i.e. the maximum concrete is on the ascending

branch of the strees-straing curve)

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Tension Compression ≡

A.s Fy * b f'c ε.c0.45 * y 0 kd int * ≡

A.s Fy * b f'c 2 ε.c0.45 ε.0 / * ε.c0.45 ε.0 / (2^- (* y 0 kd int * ≡

Using Hognestad

relationship

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Concrete strain at any

location y

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ε.c ε.ct kd / y * :

A.s Fy * b f'c 2 ε.ct ε.0 / * y kd / * ε.ct ε.0 / y kd / * (2^- (* y 0 kd int * ≡

A.s Fy * b f'c * 2 y 2^kd / * ε.ct 2 ε.0 * / * y 3^3 kd 2^* / ε.ct ε.0 / (2^* - (* ≡

Using the integration range

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kd Clear 1

A.s Fy * b f'c * kd * ε.ct ε.0 / 13 / ε.ct ε.0 / (2^* - (* ≡

Top concrete strain can be related to the steel strain

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ε.ct ε.y kd d.e kd - / * ≡

Substituting this into the above expression

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ε.0 2 f'c E.c / * : 0.0022

A.s Fy * b f'c * kd * ε.y ε.0 / kd 2^d.e kd - / * 13 / ε.y ε.0 / (2^* kd 3^d.e kd - (2^/ * - (* ≡

A.s Fy * b f'c * ε.y ε.0 / * kd 2^d.e kd - / * 113 / ε.y ε.0 / kd d.e kd - / * * - kd * ≡ kd solve maple Pila vacía.