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Geotech Exam 1 Formulas, Exams of Geotechnical Engineering

Honolulu Community CollegeGeotechnical Engineering

Intro to Geotech Exam 1 Formula Sheet

Typology: Exams

2021/2022

Uploaded on 12/06/2022

moeriarty_wou
moeriarty_wou 🇺🇸

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Formula Sheet for Exam #1
CEE 3424 – Reinforced Concrete Structures I
Basic Load Combination – U = 1.2 D + 1.6 L
Uncracked Transformed Section
Modulus of Elasticity of Concrete -
cc fE '000,57=
, psi
Modulus of Elasticity of Steel – Es = 29,000,000 psi
Modular Ratio -
c
s
E
E
n=
Uncracked Transformed Area = AT = bh + As(n-1)
Parallel axis theorem -
∑
=
+=
i
niiitotal
dAII
1
2
)(
Stress at any depth in a beam in flexure -
I
My
=
σ
Cracked Transformed Section
Modulus of Rupture of Concrete -
cr
ff '5.7=
, psi
Cracking Moment -
y
If
M
r
cr
=
To solve for depth to neutral axis (c) -
)(
2
2
cdnA
bc
s
−=
stress in concrete for given moment -
−
=
3
2
c
dcb
M
f
c
stress in steel for given moment -
At Nominal Flexural Strength (ultimate)
using Whitney stress block method
for internal equilibrium T = C
Assuming steel yields T = Asfy
For a rectangular beam C = 0.85f’cba
Depth of compression block -
bf
fA
a
c
ys
'85.0
=
pf2

Partial preview of the text

Download Geotech Exam 1 Formulas and more Exams Geotechnical Engineering in PDF only on Docsity!

Formula Sheet for Exam #

CEE 3424 – Reinforced Concrete Structures I

Basic Load Combination – U = 1.2 D + 1.6 L

Uncracked Transformed Section

Modulus of Elasticity of Concrete - E (^) c = 57 , 000 f ' c , psi

Modulus of Elasticity of Steel – Es = 29,000,000 psi

Modular Ratio - c

s

E

E

n =

Uncracked Transformed Area = AT = bh + As(n-1)

Parallel axis theorem - ∑

=

i

n

I (^) total Ii Aidi 1

2 ( )

Stress at any depth in a beam in flexure - I

My

Cracked Transformed Section

Modulus of Rupture of Concrete - f (^) r = 7. 5 f ' c , psi

Cracking Moment - y

f I M

r cr =

To solve for depth to neutral axis (c) - ( ) 2

2

And c

bc = s −

stress in concrete for given moment -

c cb d

M

f (^) c

stress in steel for given moment -

c A d

M

f

s

s

At Nominal Flexural Strength (ultimate)

using Whitney stress block method

for internal equilibrium T = C

Assuming steel yields T = Asf (^) y

For a rectangular beam C = 0.85f’cba

Depth of compression block - f b

A f a c

s y

  1. 85 '

Nominal Moment Capacity (Flexural Strength) - (^)  

a M (^) n TorC d

depth to neutral axis -

a c =

for f’c ≤ 4000 psi - β 1 = 0.

for 4000 < f’c < 8000 psi - (^)  

1 1.^050.^05

f c β , with f’c in psi

for f’c ≥ 8000 psi - β 1 = 0.

Steel Properties

in elastic range – f (^) s = Esεs

if εs ≥ εy , then f (^) s = f (^) y

yield strain - εy = f (^) y/Es

Tension controlled, Compression controlled, Transition

Tension controlled if εt ≥ 0.005 or if ≤ 0. 375

d t

c

φ = 0.

Compression controlled if εt ≤ εy = 0.002 for 60 ksi steel

or if for f ksi d

c y t

φ = 0.65 (for beams with stirrups)

Transition if 0.002 < εt < 0.005 (for 60 ksi steel) or 0. 375 < < 0. 60 d t

c

then φ = 0.65 + 83.3(εt – 0.002) or

d t

c

Mid-span moments for simple loading conditions of simply supported beams

uniformly distributed load – M = w L 2 /

Point load at mid-span - M = P L /

Symmetric point loads at “a” from each support - M = P a