CVEN 444 - Homework 5 - Concrete problems

Problem 8.1A cantilever beam with b =12 in containing 3 No. 7 bars which are anchored in the column by standard 90 hooks. fc = 5 ksi and fs= 60 ksi. Assume that the steel is stressed to fy at the face of the column, can these bars

a be anchored by hooks into the column. The clear cover to the side of the hooks is 2.75 in. The clear cover to the bar extension beyond the bend is 1.875 in. The joint is enclosed by ties at 6 in off center.

b be developed in the beam. The bar ends 2 in. from the end of the beam. The beam has No. 3 double leg stirrups at 7.5 in.

From the size of the column the room available for a hook

The hook distance

We need to calculate the cover factor from coverage factors

1. Section 12.5.3.1 does not apply so factor =12. Section 12.5.3.2 does not apply so factor =13. Drawing shows two ties in joint, spacing of ties is 6 in. therefor spacing exceeds 3db

and 12.5.3.3 does not apply so factor =1.

Therefore, ldh = (1.0)*(1.0)*(1.0)*14.8 in. The hook can develop within the 16.125 in.

(b)

The development length available in the beam is l = 48 in – 2 in. = 46 in.

Clear spacing of the bars exceeds db and stirrups exceed the code minimum, therefore this is case 1 and the bars are No. 7 so

The bars are “top bars” because they have 15.5 in of concrete below them (d = 18 in. – 2.5 in = 15.5 in), therefore = 1.3.

Bars are uncoated, therefore = 1.0.

Normal weight concrete, therefore =1.0.

Since ld = 48.3 in. is greater than 46 in. the No. 7 bars can not be developed in the beam. Either hook the ends or try a No. 5 or No. 6 bar. For a No. 6 bar:

No. 6 bar can be developed

Problem 8.4 A simply supported rectangular beam with b = 14 in. and d = 17.5 in. an No. 3 minimum stirrups spans 14 ft and supports a total factored uniform load of 6.5 kip/ft, including its own dead load. It is built of 3000 psi concrete and contains 2 No. 10 grade 50 bars, which extend 5 in. past centers of the supports at each end of No. 3 minimum stirrups Does this beam satisfy ACI Sec. 12.11.3? If not, what is the largest size bar, which can be used?

Clear spacing of the bars exceeds db, and beam has minimum stirrups, therefore this is Case 1. Bars are No. 10, thus

The bars are bottom bars, therefore is not enough cover = 1.0.

Bars are uncoated, therefore = 1.0.

Normal weight concrete, therefore =1.0.

The moment capacity of the beam is defined as:

The maximum shear force acting at the end Vu =l/ 2 = (6.5 k/ft*14 ft.) / 2 = 45.5 kips.

From the given material, la = 5 in.

Look at the development length:

Since ld = 69.5 in exceeds 56.5 in. Section 12.11.3 is not satisfied. The No. 10 bars are too large. We need to find another set of bars, which will fit the cross-sectional area.

Try 1 # 8 and 3 # 7 As = 0.79 in2 +3(0.6 in2) =2.59 in2

The ld for #8 bottom bar:

The moment capacity of the beam is defined as:

The maximum shear force acting at the end Vu =l/ 2 = (6.5 k/ft*14 ft.) / 2 = 45.5 kips.

From the given material, la = 5 in.

Look at the development length:

Since ld = 54.8 in. < 57.3 in.

The problem will work. Use 3 #7 and 1#8.

Design a 14 ft simply supported slab to carry a uniform dead load (exclude self-weight) of 140 lb/ft2 and a uniform live load of 160 lb/ft2. Use fc = 3 ksi and fy =60 ksi.

Need to find the thickness of the slab for a simply supported slab from Table 9.5a ACI code

The weight of the concrete

The design load will be

The value for d = h – cover – 0.5 dia. bar = 8.5 in. – 0.75 in – 0.5 in. = 7.25 in.

Compute the reinforcement for the slab for a 1 ft slab section.

The area of steel = d b = 0.00510*(7.25 in)(12 in)= 0.44 in2 # 5 (0.31 in2) and use spacing from ACI 7.6.4

The minimum area of steel = b h = 0.00180*(8.5 in)(12 in)= 0.184 in2 / ft # 4 (0.20 in2) at 12 in spacing

Check one-way shear capacity,

A 17 ft span simply supported beam has a clear span of 16 ft and carries a uniformly distributed dead and live loads of 5.0 k/ft and 4.0 k/ft, respectively. The dimensions beam section are 25 in. x 14 in., reinforced with 4 #10 bars in single row (d = 22.5 in.) and 2 #4 at top of the beam. Check the section for shear and design the necessary shear reinforcement. Given fc = 3 ksi and fy = 60 ksi.

Calculate the load for self-weight of the beam

The design load will be

There are two cases for the shear forces.

The maximum Vu = 121.635 k and Vu at center = -14.45 k from each case. We do not need to use 1.15 Vu on simply supported beam

Slope for the problem

Calculate the moment capacity of the concrete

The maximum capacity of the steel is

The maximum spacing for #4 bars are:

It would be the same for #3 bar

Starting point is at 0.5 ft + d = 0.5 ft +22.5 in (1 ft/12 in) = 2.375 ft

So the maximum Vn = 107.87 k < (34.5 k + 138 k) = 172.5 k OK Use either a #3(0.11 in2) or # 4 bar (0.20 in2) so to find the maximum spacing is going to be:

The shear capacity of the stirrups

The first bar is going to be 1.5 in. in from the edge first bar is 3 spacing and then start 6 in spacing (8 stirrups) x = 1.5 in + 3 in. + 6(6 in) = 40.5 in (3.375 ft) Plug in and find Vn = 93.1 k

The shear capacity of the stirrups

Use 4 stirrups so that x = 40.5 in + 4(9 in) = 76.5 in (6.375 ft) Vn = 48.525 k

The shear capacity of the stirrups

Use 3 #3 stirrups at 8.5 in spacing so that x = 76.5 in + 3(8.5 in) = 108 in (8.5 ft)

The spacing can vary for your design.