4-1 Figure P4-1 shows a simply supported beam and the ...



Chapter 4

4-1 Figure P4-1 shows a simply supported beam and the cross-section at midspan. The beam supports a uniform service (unfactored) dead load consisting of its own weight plus 1.4 kips/ft and a uniform service (unfactored) live load of 1.5 kip/ft. The concrete strength is 3500 psi, and the yield strength of the reinforcement is 60,000 psi. The concrete is normal-weight concrete. Use load and strength reduction factors from ACI Code Sections 9.2 and 9.3. For the midspan section shown in part (b) of Fig. P4-1, compute [pic] and show that it exceeds[pic].

1. Calculate the dead load of the beam.

Weight/ft = [pic]kips/ft

2. Compute the factored moment,[pic].

Factored load/ft: [pic] = 1.2(0.30 + 1.40) + 1.6(1.50) = 4.44 k/ft

[pic] kips-ft

3. Compute the nominal moment capacity of the beam, [pic]and the strength reduction factor,[pic].

Tension steel area: As = 3 No. 9 bars = [pic]in.2 = 3.00 in.2

Compute the depth of the equivalent rectangular stress block,[pic], assuming that tension steel is yielding. From equilibrium (using Eq. (4-14)):

[pic] in.

For [pic]psi, [pic]. Therefore, [pic] in.

Check whether tension steel is yielding:

using Eq.(4-18) [pic]

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Compute the nominal moment strength, using Eq. (4-21):

[pic]kips-ft

Since, [pic] the section is clearly tension-controlled and [pic]=0.9. Then,

[pic]kip-ft [pic]kip-ft. Clearly, [pic]

4-2 A cantilever beam shown in Fig. P4-2. The beam supports a uniform service (unfactored) dead load of 1 kip/ft plus its own dead load and it supports a concentrated service (unfactored) live load of 12 kips as shown. The concrete is normal-weight concrete with [pic]psi and the steel is Grade 60. Use load and strength-reduction factors form ACI Code Section 9.2 and 9.3. For the end section shown in part (b) of Fig. P4-2, compute [pic] and show it exceeds[pic].

1. Calculate the dead load of the beam.

Weight/ft = [pic]kips/ft

2. Compute the factored moment,[pic].

Factored distributed load/ft: [pic] = 1.2(0.563 + 1.0) = 1.88 k/ft

Factored live load is a concentrated load: [pic]kips

[pic]k-ft

3. Compute the nominal moment capacity of the beam, [pic]and the strength reduction factor,[pic].

Tension steel area: As = 6 No. 8 bars = [pic]in.2 =4.74 in.2

Compute the depth of the equivalent rectangular stress block,[pic], assuming that tension steel is yielding. From equilibrium (using Eq. (4-14)):

[pic] in.

For [pic]psi, [pic]. Therefore, [pic] in.

Check whether tension steel is yielding:

using. Eq.(4-18) [pic]> 0.0021

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Compute the nominal moment strength, using Eq. (4-21):

[pic]kip-ft

Since, [pic] the section is clearly tension-controlled and [pic]=0.9. Then,

[pic] kip-ft [pic]kip-ft. Clearly, [pic]

4-3 (a) Compare [pic] for singly reinforced rectangular beams having the following properties. Use loads and strength reduction factors from ACI Code Sections 9.2 and 9.3.

|Beam |b |d | |[pic] |[pic] |

| | | |Bars | | |

|No. |(in.) |(in.) | |(psi) |(psi) |

|1 |12 |22 |3 No. 7 |3,000 |60,000 |

|2 |12 |22 |2 No. 9 plus 1 No. 8 |3,000 |60,000 |

|3 |12 |22 |3 No. 7 |3,000 |40,000 |

|4 |12 |22 |3 No. 7 |4,500 |60,000 |

|5 |12 |33 |3 No. 7 |3,000 |60,000 |

Beam No.1

Compute the depth of the equivalent rectangular stress block,[pic], assuming that tension steel is yielding.

[pic] in.

For [pic]psi, [pic]. Therefore, [pic] in.

[pic]

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Since, [pic] the section is clearly tension-controlled and [pic]=0.9.

[pic]kip-ft

For Beam 1, [pic] kip-ft

Beam No.2

Compute the depth of the equivalent rectangular stress block,[pic], assuming that tension steel is yielding.

[pic] in.

For [pic]psi, [pic]. Therefore, [pic] in.

[pic]

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Since, [pic] the section is clearly tension-controlled and [pic]=0.9.

[pic]kip-ft

For Beam 2, [pic] kip-ft

Beam No.3

Compute the depth of the equivalent rectangular stress block,[pic], assuming that tension steel is yielding.

[pic] in.

For [pic]psi, [pic]. Therefore, [pic] in.

[pic]

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Since, [pic] the section is clearly tension-controlled and [pic]=0.9.

[pic]kip-ft

For Beam 3, [pic] kip-ft

Beam No.4

Compute the depth of the equivalent rectangular stress block,[pic], assuming that tension steel is yielding.

[pic] in.

For [pic]psi, [pic]. Therefore, [pic] in.

[pic]

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Since, [pic] the section is clearly tension-controlled and [pic]=0.9.

[pic]kip-ft

For Beam 4, [pic] kip-ft

Beam No.5

Compute the depth of the equivalent rectangular stress block,[pic], assuming that tension steel is yielding.

[pic] in.

For [pic]psi, [pic]. Therefore, [pic] in.

[pic]

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Since, [pic] the section is clearly tension-controlled and [pic]=0.9.

[pic]kip-ft

For Beam 5, [pic] kip-ft

(b) Taking beam 1 as the reference point, discuss the effects of changing [pic],[pic], [pic]and [pic]on [pic]. (Note that each beam has the same properties as beam 1 except for the italicized quantity.)

|Beam |[pic] |

| | |

|No. |(kips-ft) |

|1 |164 |

|2 |242 |

|3 |113 |

|4 |169 |

|5 |253 |

Effect of [pic](Beams 1 and 2)

An increase of 55% in [pic](from 1.80 to 2.79 in.2) caused on increase of 48% in [pic]. It is clear that increasing the tension steel area causes a proportional increase in the strength of the section, with a loss of ductility. Note that in this case, the strength reduction factor was 0.9 for both sections.

Effect of [pic] (Beams 1 and 3)

A decrease of 33% in [pic] caused a decreased of 31% in [pic]. A decrease in the steel yield strength has essentially the same effect as decreasing the tension steel area.

Effect of [pic] (Beams 1 and 4)

An increase of 50% in [pic] caused an increase of 3% in [pic]. It is clear that changes in the concrete strength have a much smaller effect on moment strength compared with changes in the tension steel area and steel yield strength.

Effect of [pic] (Beams 1 and 5)

An increase of 50% in [pic]caused an increase of 54% in [pic].It is clear that increasing the effective flexural depth of the section increases the section moment strength (without decreasing the section ductility).

(c) What is the most effective way of increasing [pic]? What is the least effective way?

Disregarding any other effects of increasing [pic] or [pic]such as changes in cost, etc., the most effective way to increase[pic] is the increase the effective flexural depth of the section,[pic], followed by increasing [pic]and [pic]. Note that increasing[pic]and [pic]too much may make the beam over-reinforced and thus will result in a decrease in ductility.

The least effective way of increasing [pic]is to increase [pic].Note that increasing [pic]will cause a significant increase in curvature at failure.

4-4 A 12-ft-long cantilever supports its own dead load plus an additional uniform service (unfactored) dead load of 0.5 kip/ft. The beam is made from normal-weight 4000-psi concrete and has [pic]in., [pic]in., and [pic]in. It is reinforced with four No. 7 Grade-60 bars. Compute the maximum service (unfactored) concentrated live load that can be applied at 1ft from the free end of the cantilever. Use load and strength –reduction factors from ACI Code Sections 9.2 and 9.3. Also check[pic].

1. Compute the nominal moment capacity of the beam, [pic]and the strength reduction factor,[pic].

Tension steel area: [pic] = 4 No. 7 bars = [pic]in.2 =2.40 in.2

Compute the depth of the equivalent rectangular stress block,[pic], assuming that tension steel is yielding. From equilibrium (using Eq. (4-14)):

[pic] in.

For [pic]psi, [pic]. Therefore, [pic] in.

Check whether tension steel is yielding:

using Eq.(4-18) [pic]

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Compute the nominal moment strength, using Eq. (4-21):

[pic]kips-ft

Since, [pic] the section is clearly tension-controlled and,

[pic] kips-ft = [pic] kips-ft

2. Compute Live Load

Set [pic]kips-ft

Weight/ft of beam = [pic]kips/ft

Factored dead load = [pic]kips/ft

Factored dead load moment = [pic]kips-ft

Therefore the maximum factored live load moment is: 153 kips-ft – 69.1 kip-ft = 83.9 kips-ft

Maximum factored load at 1 ft from the tip = 83.9 kips-ft / 11 ft = 7.63 kips

Maximum concentrated service live load = 7.63 kips / 1.6 = 4.77 kips

3. Check of [pic]

The section is subjected to positive bending and tension is at the bottom of this section, so we should use [pic] in Eq. (4-11). Also, [pic]is equal to 189 psi, so use 200 psi in the numerator:

[pic] in.2 < [pic] (o.k.)

4-5 Compute [pic]and check [pic] for the beam shown in Fig. P4-5. Use [pic]psi and [pic] psi.

1. Compute the nominal moment capacity of the beam, [pic]and the strength reduction factor,[pic].

Tension steel area: As = 6 No. 8 bars = [pic]in.2 =4.74 in.2

The tension reinforcement for this section is provided in two layers, where the distance from the tension edge to the centroid of the total tension reinforcement is given as [pic]19 in.

Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the compression flange,[pic] and that the tension steel is yielding,[pic], using Eq. (4-16):

[pic] in. [pic] in. (o.k.)

For [pic]psi, [pic]. Therefore, [pic] in.

Comparing the calculated depth to the neutral axis,[pic], to the values for [pic]and[pic], it is clear that the tension steel strain, [pic], easily exceeds the yield strain (0.00207) and the strain at the level of the extreme layer of tension reinforcement, [pic], exceeds the limit for tension-controlled sections (0.005). Thus, [pic]=0.9 and we can use Eq. (4-21) to calculate [pic]:

[pic]kips-ft

[pic] kips-ft = [pic] kips-ft

2. Check of [pic]

The section is subjected to positive bending and tension is at the bottom of this section, so we should use [pic] in Eq. (4-11). Also, [pic]is equal to 201 psi, so use [pic]in the numerator:

[pic] in.2 < [pic] (o.k.)

4-6 Compute [pic]and check [pic] for the beam shown in Fig. P4-6. Use [pic]psi and [pic] psi.

1. Compute the nominal moment capacity of the beam, [pic]and the strength reduction factor,[pic].

Tension steel area: [pic] = 6 No. 8 bars = [pic]in.2 =4.74 in.2

The tension reinforcement for this section is provided in two layers, where the distance from the tension edge to the centroid of the total tension reinforcement is given as [pic]18.5 in.

Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the compression flange,[pic] and that the tension steel is yielding,[pic], using Eq. (4-16):

[pic] in. [pic] in. (o.k.)

For [pic]psi, [pic]. Therefore, [pic] in.

Check whether tension steel is yielding:

using Eq.(4-18) [pic]

Thus, [pic]> 0.002 and it is clear that the steel is yielding in both layers of reinforcement.

It is also clear that the section is tension-controlled ([pic]=0.9), but just for illustration the value of[pic] can be calculated as:

[pic]

We can use Eq. (4-21) to calculate [pic]:

[pic]kips-ft

[pic] kips-ft = [pic] kips-ft

2. Check of [pic]

The section is subjected to positive bending and tension is at the bottom of this section, so we should use [pic] in Eq. (4-11). Also, [pic]is equal to 190 psi, so use 200 psi in the numerator:

[pic] in.2 < [pic] (o.k.)

4-7 Compute the negative-moment capacity,[pic], and check [pic] for the beam shown in Fig. P4-7. Use [pic]psi and [pic] psi.

1. Calculation of [pic]

This section is subjected to negative bending and tension will develop in the top flange and the compression zone is at the bottom of the section. ACI Code Section 10.6.6 requires that a portion of the tension reinforcement be distributed in the flange, so assuming that the No. 6 bars in the flange are part of the tension reinforcement:

[pic]in.2

The depth of the Whitney stress block can be calculated using Eq. (4-16) , using [pic] in., since the compression zone is at the bottom of the section:

[pic] in.

For [pic]psi, [pic]. Therefore, [pic] in.

We should confirm that the steel is yielding:

using Eq.(4-18) [pic]

Clearly, the steel is yielding [pic]and this is tension-controlled section[pic].

We can use Eq. (4-21) to calculate [pic]:

[pic]kips-ft

[pic] kips-ft = [pic] kips-ft

2. Check of [pic]

The beam is subjected to negative bending and since the flanged portion of the beam section is in tension, the value of [pic]will depend on the use of that beam.

Assuming that the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using [pic] in Eq. (4-11). Also, [pic]is equal to 177 psi, so use 200 psi in the numerator:

[pic] in.2 < [pic] (o.k.)

However, for a statically determinate beam, [pic] should be replaced by the smaller of [pic]or [pic]. Given that [pic] is 48 in. for this beam section,

[pic]< [pic] (o.k.)

4-8 For the beam shown in Fig. P4-8, [pic]psi and [pic] psi.

(a) Compute the effective flange width at midspan.

The limits given in ACI Code Section 8.12 for determining the effective compression flange,[pic], for a flanged section that is part of a continuous floor system are:

[pic]

Assuming that the columns are [pic], the longitudinal span is approximated as:

[pic]

The clear transverse distance for the [pic]span is: [pic]

and for the [pic] span is: [pic]

So, the average clear transverse distance is [pic]

The effective compression flange can now be computed as:

[pic]

The first limit governs for this section, so [pic]

(b) Compute [pic] for the positive- and negative-moment regions and check [pic] for both sections. At the supports, the bottom bars are in one layer; at midspan, the No. 8 bars are in the bottom, the No. 7 bars in a second layer.

Positive moment region

1. Calculation of [pic]

Tension steel area: [pic] = 3 No. 8 bars + 2 No. 7 bars = [pic]in.2

The tension reinforcement for this section is provided in two layers. Assuming the section will include a No. 3 or No. 4 stirrup, it is reasonable to assume that the distance from the extreme tension edge of the section to the centroid of the lowest layer of steel is approximately 2.5 in. Thus the distance from the top of the section to the extreme layer of tension reinforcement,[pic], can be calculated to be:

[pic]21 in. – 2.5 in. =18.5 in.

The minimum spacing required between layers of reinforcement is 1 in. (ACI Code Section 7.6.2). Thus the spacing between the centers of the layers is approximately 2 in. So the distance from the tension edge to the centroid of the total tension reinforcement is:

[pic]in.

Therefore, the effective flexural depth, [pic], is:

[pic]21 in. – 3.17 in. =17.8 in.

Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the compression flange,[pic] and that the tension steel is yielding,[pic], using Eq.(4-16):

[pic] in. [pic] in. (o.k.)

For [pic]psi, [pic]. Therefore, [pic] in.

Comparing the calculated depth to the neutral axis,[pic], to the values for [pic]and[pic], it is clear that the tension steel strain, [pic], easily exceeds the yield strain (0.00207) and the strain at the level of the extreme layer of tension reinforcement, [pic], exceeds the limit for tension-controlled sections (0.005). Thus, [pic]=0.9 and we can use Eq. (4-21) to calculate [pic]:

[pic]kips-ft

[pic] kips-ft = [pic] kips-ft

2. Check of [pic]

The section is subjected to positive bending and tension is at the bottom of this section, so we should use [pic] in Eq. (4-11). Also, [pic]is equal to 177 psi, so use 200 psi in the numerator:

[pic] in.2 < [pic] (o.k.)

Negative moment region

The tension and compression reinforcement for this section is provided in single layers. Assuming the section will include a No. 3 or No. 4 stirrup, it is reasonable to assume that the distance from the extreme tension or compression edge of the section to the centroid of the tension or compression layer of steel is approximately 2.5 in.

[pic] = 7 No. 7 bars = [pic]in.2 , [pic]18.5 in.

[pic] = 2 No. 8 bars = [pic]in.2 , [pic] in.

Because this is a doubly reinforced section, we will initially assume the tension steel is yielding and use the trial and error procedure described in Section 4-7 to find the neutral axis depth, c.

Try [pic]

[pic]

[pic]

[pic]

[pic]

[pic]

Because [pic], we should increase [pic] for the second trial.

Try [pic]

[pic]

[pic]

[pic]

[pic]

[pic]

With section equilibrium established, we must confirm the assumption that the tension steel is yielding.

using Eq.(4-18) [pic]

Thus, the steel is yielding [pic] and it is a tension-controlled section [pic].

So, using[pic], use Eq. (4-21) to calculate [pic].

[pic]

[pic] kips-ft = [pic] kips-ft

2. Check of [pic]

The flanged portion of the beam section is in tension and the value of [pic]will depend on the use of that beam. Since the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using [pic] in Eq. (4-11). Also, [pic]is equal to 177 psi, so use 200 psi in the numerator:

[pic] in.2 < [pic] (o.k.)

4-9 Compute [pic] and check [pic]for the beam shown in Fig. P4-9. Use [pic]psi and [pic] psi, and

(a) the reinforcement is six No. 8 bars.

1. Compute the nominal moment capacity of the beam, [pic]and the strength reduction factor,[pic].

Tension steel area: [pic] = 6 No. 8 bars = [pic]in.2 = 4.74 in.2

Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the top flange,[pic] and that the tension steel is yielding,[pic], using Eq. (4-16) with [pic]:

[pic] in. [pic] in. (o.k.)

For [pic]psi, [pic]. Therefore, [pic] in.

Check whether tension steel is yielding:

using Eq.(4-18) [pic]

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Since, [pic] the section is clearly tension-controlled and [pic]=0.9.

We can use Eq. (4-21) to calculate [pic]:

[pic]kips-ft

[pic] kips-ft = [pic] kips-ft

2. Check of [pic]

The flanged portion of the beam section is in tension and the value of [pic]will depend on the use of that beam.

Assuming that the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using [pic] in Eq. (4-11). Also, [pic]is equal to 177 psi, so use 200 psi in the numerator:

[pic] in.2 < [pic] (o.k.)

However, for a statically determinate beam, [pic] should be replaced by the smaller of [pic]or [pic]. Given that [pic] is 30 in. for this beam section,

[pic] in.2 < [pic] (o.k.)

(b) the reinforcement is nine No. 8 bars.

1. Compute the nominal moment capacity of the beam, [pic]and the strength reduction factor,[pic].

Tension steel area: As = 9 No. 8 bars = [pic]in.2 =7.11 in.2

Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the compression flange,[pic] and that the tension steel is yielding,[pic], using Eq. (4-16) with [pic]:

[pic] in. [pic] in. (o.k.)

For [pic]psi, [pic]. Therefore, [pic] in.

Check whether tension steel is yielding:

using Eq.(4-18) [pic]

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Since, [pic] the section is clearly tension-controlled and [pic]=0.9.

We can use Eq. (4-21) to calculate [pic]:

[pic]kips-ft

[pic] kips-ft = [pic] kips-ft

2. Check of [pic]

[pic] is the same as in part (a).

4-10 Compute [pic] and check [pic]for the beam shown in Fig. P4-10. Use [pic]psi and [pic] psi, and

1. Compute the nominal moment capacity of the beam, [pic]and the strength reduction factor,[pic].

Tension steel area: As = 8 No. 7 bars = [pic]in.2 =4.8 in.2

Tension will develop in the bottom flange and the compression zone is at the top of the section. Thus, assuming that the tension steel is yielding,[pic], in Eq. (4-16) we should use [pic] and we find the depth of the Whitney stress block as:

[pic] in.

For [pic]psi, [pic]. Therefore, [pic] in.

Check whether tension steel is yielding:

using Eq.(4-18) [pic]

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Since, [pic] the section is tension-controlled and [pic]=0.9.

We can use Eq. (4-21) to calculate[pic]:

[pic]kips-ft

[pic] kips-ft = [pic] kips-ft

2. Check of [pic]

The flanged portion of the beam section is in tension and the value of [pic]will depend on the use of that beam.

Assuming that the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using [pic] in Eq. (4-11). Also, note that [pic]is equal to 212 psi:

[pic] in.2 < [pic] (o.k.)

However, for a statically determined beam, [pic] should be replaced by the smaller of [pic]or [pic]. Given that [pic] is 42 in. for this beam section,

[pic] in.2 < [pic] (o.k.)

4-11 (a) Compute [pic] for the three beams shown in Fig. P4-11. In each case, [pic]psi and [pic]ksi, [pic]

Beam No. 1

Tension steel area: As = 6 No. 9 bars = [pic]in.2 =6.00 in.2

The tension reinforcement for this section is provided in two layers. Assuming the section will include a No. 3 or No. 4 stirrup, it is reasonable to assume that the distance from the extreme tension edge of the section to the centroid of the lowest layer of steel is approximately 2.5 in. Thus the distance from the top of the section to the extreme layer of tension reinforcement, [pic], can be calculated to be:

[pic]36 in. – 2.5 in. =33.5 in.

The effective flexural depth, [pic], is given as : [pic]32.5 in.

Assuming that the tension steel is yielding,[pic], using Eq. (4-16):

[pic] in.

For [pic]psi, [pic]. Therefore, [pic] in.

We need to check whether tension steel is yielding:

using Eq.(4-18) [pic]

Thus, [pic]> 0.002 and the steel is yielding ([pic]).

Also, clearly [pic], the section is tension-controlled and [pic]=0.9.

Thus, [pic]=0.9 and we can use Eq. (4-21) to calculate [pic]:

[pic]kips-ft

[pic] kips-ft = [pic] kips-ft

Beam No. 2

Tension steel area: As = 6 No. 9 bars = [pic]in.2 =6.00 in.2

Compression steel area: [pic] = 2 No. 9 bars = [pic]in.2 =2.00 in.2

As was discussed for beam No. 1, [pic]32 in., [pic]33.5 in. and [pic]is given as [pic]in.

Because this is a doubly reinforced section, we will initially assume the tension steel is yielding and use the trial and error procedure described in Section 4-7 to find the neutral axis depth, c.

Try [pic]

[pic]

[pic]

[pic]

[pic]

[pic]

Because [pic], we should decrease [pic] for the second trial.

Try [pic]

[pic]

[pic]

[pic]

[pic]

[pic]

With section equilibrium established, we must confirm the assumption that the tension steel is yielding.

using Eq.(4-18) [pic]

Clearly, the steel is yielding [pic] and it is a tension-controlled section [pic].

So, using[pic], use Eq. (4-21) to calculate [pic].

[pic]

[pic] kips-ft = [pic] kips-ft

Beam No. 3

Tension steel area: As = 6 No. 9 bars = [pic]in.2 =6.00 in.2

Compression steel area: [pic] = 4 No. 9 bars = [pic]in.2 =4.00 in.2

As was discussed for beam No. 1, [pic]32.5 in., and [pic]33.5 in.

The compression reinforcement for this beam section is provided in two layers and [pic]is given as 3.5 in.

Because this is a doubly reinforced section, we will the same procedure as for beam No. 2 (assuming that the tension steel is yielding).

The depth of the neutral axis for this section should be smaller compared with beam section No. 2, since the compression reinforcement is increased for this section.

Try [pic]

[pic]

[pic]

[pic]

[pic]

[pic]

Because [pic], we should decrease [pic] for the second trial.

Try [pic] (Note that both layers of the compression steel will actually be in the compression zone)

[pic]

[pic]

[pic]

[pic]

[pic]

With section equilibrium established, we must confirm the assumption that the tension steel is yielding.

using Eq.(4-18) [pic]

Clearly, the steel is yielding [pic] and it is a tension-controlled section[pic].

So, using [pic], use Eq. (4-21) to calculate [pic].

[pic]

[pic] kips-ft = [pic] kips-ft

(b) From the results of part (a), comment on weather adding compression reinforcement is a cost-effective way of increasing the strength, [pic], of a beam.

Comparing the values of[pic]for the three beams, it is clear that for a given amount of tension reinforcement, the addition of compression steel has little effect on the nominal moment capacity, provided the tension steel yields in the beam without compression reinforcement. As a result, adding compression reinforcement in not a cost effective way of increasing the nominal moment capacity of a beam. However, adding compression reinforcement improves the ductility and might be necessary when large amounts of tension reinforcement are used to change the mode of failure.

4-12 Compute [pic] for the beam shown in Fig. P4-12. Use [pic]psi and [pic]psi. Does the steel yield in this beam at nominal strength?

[pic]= 6 No. 8 bars = [pic]in.2 =4.74 in.2 , [pic]

[pic] = 2 No. 7 bars = [pic]in.2 =1.2 in.2 , [pic]

Because this is a doubly reinforced section, we will initially assume the tension steel is yielding and use the trial and error procedure described in Section 4-7 to find the neutral axis depth, c.

Try [pic]

For [pic]psi, [pic]. Thus, [pic]

Since the depth of the Whitney stress block is less than 5.0 in. ,[pic],the width of the compression zone is constant and equal to 10 in., i.e. [pic]

[pic]

[pic]

[pic]

[pic]

[pic]

Because [pic], we should increase [pic] for the second trial.

Try [pic]

and find[pic]

[pic]

[pic]

[pic][pic]

In this case, the width of the compression zone is not constant. Using a similar reasoning as in the case of flanged sections, where the depth of the Whitney stress block is in the web of the section, the compression force can be calculated from the following equations (refer to Fig. S4-12):

[pic]

[pic]

[pic]

[pic], we should increase [pic] for the third trial.

Try [pic]

and find[pic]

[pic]

[pic]

[pic][pic]

[pic]

[pic]

[pic]

[pic]

With section equilibrium established, we must confirm the assumption that the tension steel is yielding.

using Eq.(4-18) [pic]

Thus, the steel is yielding [pic] and it is a tension-controlled section[pic].

Summing the moments about the level of the tension reinforcement:

[pic]

[pic] kips-ft = [pic] kips-ft

[pic]

Fig. S4-12.1 Beam section and internal forces for the case of [pic].

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download