To save steel-handling costs, an alternative design is proposed for the beam in Problem 1 using two No. 9 Grade 75 bars to provide approximately the same steel strength as the originally proposed four No. 7

Grade 60 bars. Check to determine if the redesigned beam is satisfactory with respect to cracking according to the ACI Code. What modification could you suggest that would minimize the number of bars to reduce cost, yet satisfy requirements of crack control?

See the explanation for the answer

Explanation:

Check to determine the redesigned beam is satisfactory for cracking:

crack width is controlled by establishing a minimum spacing.

Steps followed to check for cracking in the beams

i) According to aci code 10.6.7 if depth of beam isgreater than 36 in then skin reinforcement has to provide.the skin reinforcement to be provided should be such that it should not be greater than the actual main tension reinforcement.

ii) In second step steel stress is determined:

steel stressfs=Ms/(As*(d-hf/2)) or fs= 0.60fy

where Ms=service load moment

As*(d-hf/2)=area of reinforcement *moment area

iii) S=540/fs-2.5Cc is less than or equal to 12*(36/fs)

here Cc = clear spacing

if S=center to center spacing is with in the limit as specified above then the cracking is with in the control if not then redesign has to done.

In the given problem the data given is :

grade of steel in desigened beam is 75 and in redesigned beam is 60 so the stress in steel is 75*0.6=45ksi and 0.6*60=36ksi respectively

now the spacing is calculated for the two design and redesigned beams

the center to center spacing is given by S=540/fs-2.5Cc

For designed beam

S=540/45-(2.5*2.25)=6.375in which is less than 12*36/fs=12*36/45=9.6in hence it is safe

For redesigned beam

S=540/36-(2.5*2.25)=9.375in and it is less than the maximum spacing which is given by 12*36/fs=12*36/36=12in

Hence, the beam is within the limits and the beam is safe against the cracking.

Modifications to reduce the number of reinforcing bars

The addition of steel does not prevent cracking due to restrained shrinkage but it limits the width of crack by causing the formation of the number of narrow cracks rather than single wide crack.

Larger size bars leads to fewer cracks but wider cracks while smaller size bars leads to number of narrow cracks hence it is advisable to provide number of smaller diameter bars of equal strength of designed bars rather than larger  bars.

explanation:

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