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• CVEN 444 Assignment 4 due 6/16/03 The assignment will be review problems and will not be covered in class. You will need to show free-body diagrams, use an engineering format and be neat!

Problem 1

For the beam cross-section shown, determine

whether the failure of the beam will be initiated by

crushing of concrete or yielding of steel.

a) fc = 3500 psi As = 9 in2

b) fc =7500 psi As = 4.5 in2

Also determine whether the section satisfies ACI

Code requirement

a)

Assume that fy = 60 ksi, fc =3.5 ksi, As =9 in2 1= 0.85

Solve using equilibrium

( )( )( )( )

2

s y

c

9 in 60 ksi

0.85 0.85 3.5 ksi 14 in.

12.97 in.

C T

A fa

f b

=

= =

=

Compute c

1

12.97 in.15.25 in.

0.85

ac

= = =

Check the c/d ratio

15.25 in.0.598 0.6

25.5 in.

c

d= = >/

therefore, the beam is transition zone, the steel will yield

before the concrete crushes..

Check the min of the beam

( )( )

2

s 9.0 in 0.025214 in. 25.5 in.

A

bd = = =

• Compute the minimum

y

min

c

y

200 2000.00333

60000 0.00333

3 3 35000.00296

60000

f

f

f

= =

= = =

0.0252 > 0.00333 satisfies the ACI code.

b)

Assume fy = 60 ksi, fc =7.5 ksi, As =4.5 in2

Compute the 1 value.

c1

40000.85 0.05

1000

7500 40000.85 0.05 0.675

1000

f

=

= =

Solve using equilibrium

( )( )( )( )

2

s y

c

4.5 in 60 ksi

0.85 0.85 7.5 ksi 14 in.

3.03 in.

C T

A fa

f b

=

= =

=

Compute c

1

3.03 in.4.48 in.

0.675

ac

= = =

Check the c/d ratio

4.48 in.0.176 0.375

25.5 in.

c

d= = 0.00433 satisfies the ACI code.

• Problem 2

Calculate the nominal moment strength of the beam

a) fc = 4500 psi b) fc =7000 psi

Also determine whether the section satisfies ACI Code requirement.

a) Singly reinforced beam fy = 60 ksi, fc =4.5 ksi, As =4(1.0 in2 ) =4.0 in

2

c1

40000.85 0.05

1000

4500 40000.85 0.05 0.825

1000

f

=

= =

Solve using equilibrium

( )( )( )( )

2

s y

c

4 in 60 ksi

0.85 0.85 4.5 ksi 12 in.

5.23 in.

C T

A fa

f b

=

= =

=

Compute c

1

5.23 in.6.34 in.

0.825

ac

= = =

• Check the c/d ratio

6.34 in.0.309 0.375

20.5 in.

c

d= = 0.00335 satisfies the ACI code.

Compute the nominal moment

( )( )

( )

n s y

2

u n

2

5.23 in.4 in 60 ksi 20.5 in. 4292.4 k-in

2

0.9 4292.4 k-in 3863.16 k-in

aM A f d

M M

=

= =

=

= =

b) Singly reinforced beam fy = 60 ksi, fc =7 ksi, As =4(1.0 in2 ) =4.0 in

2

Compute the 1 value.

c1

40000.85 0.05

1000

7000 40000.85 0.05 0.70

1000

f

=

= =

• Solve using equilibrium

( )( )( )( )

2

s y

c

4 in 60 ksi

0.85 0.85 7 ksi 12 in.

3.36 in.

C T

A fa

f b

=

= =

=

Compute c

1

3.36 in.4.80 in.

0.7

ac

= = =

Check the c/d ratio

4.80 in.0.234 0.375

20.5 in.

c

d= = 0.00418 satisfies the ACI code.

Compute the nominal moment

( )( )

( )

n s y

2

u n

2

3.36 in.4 in 60 ksi 20.5 in. 4516.8 k-in

2

0.9 4516.8 k-in 4065.12 k-in

aM A f d

M M

=

= =

=

= =

• b)For the l-beam use fy = 60 ksi, fc =4.5 ksi, As =6(0.79 in2 )= 4.74 in

2 and 1 = 0.825.

Assume that the 12-in section is going to work, however if it does not need to use area

concrete = 4-in.*12-in. +16-in.*x Solve using equilibrium

( )( )( )( )

2

s y

c

4.74 in 60 ksi

0.85 0.85 4.5 ksi 12 in.

6.20 in.

C T

A fa

f b

=

= =

=

Therefore, the uniform distributed load extends beyond the 4 in segment, so rework the

problem.

( )( ) ( )( )

( )( )( )

( )( ) ( ) ( )( )( )( )

s y c c

s y c

c

2

c

0.85 4 in. 12 in. 0.85 16 in.

0.85 4 in. 12 in.

0.85 16 in.

4.74 in 60 ksi 0.85 4.5 ksi 4 in. 12 in.

0.85 4.5 ksi 16 in.

1.65 in.

T C

A f f f x

A f fx

f

=

= +

=

=

=

Compute c

1

4.0 in. 1.65 in. 5.65 in.

5.65 in.6.85 in.

0.825

a

ac

= + =

= = =

Check the c/d ratio

6.85 in.0.304 0.375

22.5 in.

c

d= = 0.00335 so it satisfies the ACI standards.

• Compute the nominal moment can be done by either solving for location of the center of

the compression zone to compute the moment or break the moment into two sections one

for the (4 in.)(12 in.) area and the other for (1.65 in.) (16 in.) area.

( )( )( )

( )( )( )

1 2n c c1 c c2

u n

0.85 0.852 2

4.0 in.0.85 4.5 ksi 4 in. 12 in. 22.5 in.

2

1.65 in.0.85 4.5 ksi 1.65 in. 16 in. 22.5 in. 4.0 in.

2

3763.8 k-in. 1784.8 k-in.

5548.6 k-in.

a aM f A d f A d

M M

= +

=

+ +

= +

=

=

= ( )0.9 5548.6 k-in. 4993.8 k-in=

b)For an L-beam use fy = 60 ksi, fc =7 ksi, As =6(0.79 in2 )= 4.74 in

2 and 1 =0.7.

Assume that the 12-in section is going to work, however if it does not need to use area

concrete = 4-in.*12-in. +16-in.*x Solve using equilibrium

( )( )( )( )

2

s y

c

4.74 in 60 ksi

0.85 0.85 7 ksi 12 in.

3.98 in.

C T

A fa

f b

=

= =

=

Compute c

1

3.98 in.5.69 in.

0.7

ac

= = = a 0.00418 so it satisfies the ACI standards.

• Compute the nominal moment

( )( )

( )

n s y

2

u n

2

3.98 in.4.74 in 60 ksi 22.5 in. 5833.0 k-in.

2

0.9 5833.0 k-in. 5249.7 k-in

aM A f d

M M

=

= =

=

= =

• Problem 3

Calculate the safe distributed load intensity that the beam can carry. Assume that the

only dead load is the weight of the beam (use 150 lb/ ft3). Solve for a distributed load for

u DL LL1.2 1.6w w w= +

Use

fc = 4000 psi

fy =60000 psi

fy = 60 ksi, fc =4 ksi, As =4(1.0 in2 ) =4.0 in

2 1 =0.85

Solve using equilibrium

( )( )( )( )

2

s y

c

4 in 60 ksi

0.85 0.85 4 ksi 12 in.

5.88 in.

C T

A fa

f b

=

= =

=

Compute c

1

5.88 in.6.92 in.

0.85

ac

= = =

Check the c/d ratio

6.92 in.0.308 0.375

22.5 in.

c

d= = 0.00333 satisfies the ACI code.

Compute the nominal moment

( )( )

( )

n s y

2

u n

2

5.88 in.4 in 60 ksi 22.5 in. 4694.4 k-in

2

0.9 4694.4 k-in 4224.96 k-in

aM A f d

M M

=

= =

=

= =

Compute the maximum moment.

( )

2

uu 2

2

8

8

8 4224.96 k-in0.4075 k/in. 4.89 k/ft

12 in24 ft

1 ft

MwlM w

l

w w

= =

= = =

( )( )2

3 1 ft0.15 k/ft 12 in. 25 in.12 in.

0.3125 k/ft.

DL

=

=

( )

u DLu DL LL LL

LL

1.21.2 1.6

1.6

4.89 k/ft. 1.2 0.3125 k/ft2.822 k/ft

1.6

w ww w w w

w

= + =

= =

• CVEN 444 Assignment 5 due 6/18/03 The assignment will be review problems and will not be covered in class. You will need to show free-body diagrams, use an engineering format and be neat!

Compute the stresses in the compression steel, fs, for the cross-sections. Compute the

nominal moment strength and determine the reduction factor for determining the ultimate

moment.

fc = 5000 psi

fy =60000 psi

a)

fy = 60 ksi, fc =5 ksi, As =3(1.0 in2 ) =3.0 in

2 As =2(0.31 in

2 ) =0.62 in

2

c1

40000.85 0.05

1000

5000 40000.85 0.05 0.80

1000

f

=

= =

Assume the compression steel yields

( ) ( )( )( )( )

2 2s s y

c

3 in 0.62 in 60 ksi

0.85 0.85 5 ksi 10 in.

3.36 in.

A A fa

f b

= =

=

• Compute c

1

3.36 in.4.2 in.

0.8

ac

= = =

Check the strain

s cu

4.2 in. 2.5 in.0.003 0.00121

4.2 in.

c d

c

= = =

Check the yield strain of steel is

y

y

s

60 ksi0.00207

29000 ksi

f

E = = =

Therefore 0.00121 is not greater than 0.00207 so the compression steel does not yield.

Use the equation for finding the c either by iterations or use

( )

( )

s y s s cu c 1

s s cu s y2 s s cu

c 1 c 1

E 0.85

E E0

0.85 0.85

c dA f A f b c

c

A A f A dc c

f b f b

= +

+ =

Plug in to the equati

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