   Chapter 2, Problem 2.1P ### Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
ISBN: 9781337094740

#### Solutions

Chapter
Section ### Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
ISBN: 9781337094740
Textbook Problem

# A column in a building is subjected to the following load effects:9 kips compression from dead load5 kips compression from roof live load6 kips compression from snow7 kips compression from 3 inches of rain accumulated on the roof8 kips compression from winda. If lead and resistance factor design is used, determine the factored load (required strength) to be used in the design of the column. Which AISC load combination controls?b. What is the required design strength of the Column?c. What is the required nominal strength of the column for a resistance factor ϕ of 0.90?d. If allowable strength design is used, determine the required load capacity (required strength) to be used in the design of the column. Which AISC load combination controls?e. What is the required nominal strength of the column for a safety factor Ω of 1.67?

To determine

(a)

The maximum factored load and the controlling AISC load combination by using load and resistance factor design.

The maximum factored load is 26kips.

The AISC combination that controls is 1.2D+1.6(LrorSorR)+(0.5Lor0.5W).

Explanation

Given data:

The compression live load is 0kips.

The compression snow load is 6kips.

The compression roof live load is 5kips.

The compression wind load is 8kips.

The compression rain accumulated on the roof from 3inches is 7kips.

Concept used:

The load and resistance design for load combinations is given below:

1. Combination 1: 1.4D
2. Combination 2: 1.2D+1.6L+0.5(LrorSorR)
3. Combination 3: 1.2D+1.6(LrorSorR)+(0.5Lor0.5W)
4. Combination 4: 1.2D+1.0W+0.5(LrorSorR)
5. Combination 5: 0.9D+1.0W

Here, the dead load is D, the live load is L, the roof live load is Lr, the snow load is S, the rain or ice load is R and the wind load is W.

Determining the maximum factored load, depending upon the 5 load combinations.

Write the expression for load combination 1.

L1=1.4D ...... (I)

Here, load combination 1 is L1.

Substitute 9kips for D in Equation (I).

L1=1.4×9kips=12.6kips

Write the expression for load combination 2.

L2=1.2D+1.6L+0.5(LrorSorR) ...... (II)

Here, load combination 2 is L2.

Substitute 9kips for D, 0kips for L and 7kips for (LrorSorR) in Equation (II).

L2=(1.2×9kips)+(1.6×0kips)+(0.5×7kips)=14.3kips

Write the expression for load combination 3.

L3=1.2D+1.6(LrorSorR)+(0.5Lor0.5W) ...... (III)

Here, load combination 3 is L3.

Substitute 9kips for D, 0kips for L, 8kips for W and 7kips for (LrorSorR) in Equation (III).

L3=(1.2×9kips)+(1.6×7kips)+(0.5×8kips)=26kips

Write the expression for load combination 4.

L4=1.2D+1.0W+0.5(LrorSorR) ...... (IV)

Here, load combination 4 is L4.

Substitute 9kips for D, 8kips for W, 0kips for L and 0kips for (LrorSorR) in Equation (IV).

L4=(1.2×9kips)+(1.0×8kips)+(0.5×7kips)=22.3kips

Write the expression for load combination 5.

L5=0.9D+1.0W ...... (V)

Here, load combination 5 is L5.

Substitute 9kips for D and 8kips for W in Equation (V).

L5=(0.9×9kips)+(1.0×8kips)=16.1kips

From the above 5 load combinations L3 is maximum.

Conclusion:

Thus, the maximum factored load and AISC combination is estimated bythis expression: L1=1.4D

To determine

(b)

The required design strength of the column.

The required design strength of the column is 26kips.

Explanation

The load combination L3 is the required design strength of the column b, because the load combination L3 includes all the criteria like rain, wind, live load and dead load.

Conclusion:

Thus, the design strength of the column is estimated by the load combination and concept.

To determine

(c)

The required nominal strength for a resistance factor of 0.90.

The required nominal strength of the column for a resistance factor is 28.9kips.

Explanation

Concept used:

Write the expression for nominal strength of the column for LRFD.

Rn=Ruϕ ...... (VI)

Here, the maximum factored load is Ru, the nominal strength of the column is Rn and the resistance factor is ϕ.

Substitute 0.90 for ϕ and 26kips for Ru in Equation (VI).

Rn=26kips0.90=28.9kips

Conclusion:

Thus, the required nominal strength of the column for a resistance factor is calculated by the expression for nominal strength of LFRD.

To determine

(d)

The maximum load and the controlling AISC load combination by using allowable strength design (ASD).

The maximum load is 17.85kips.

The AISC combination that controls is D+0.75L+0.75(0.6W)+0.75(LrorSorR).

Explanation

Concept used:

The following are the load combination for ASD:

1. Combination 1: D
2. Combination 2: D+L
3. Combination 3: D+(LrorSorR)
4. Combination 4: D+0.75L+0.75(LrorSorR)
5. Combination 5: D+0.6W
6. Combination 6: D+0.75L+0.75(0.6W)+0.75(LrorSorR)
7. Combination 5: 0.6D+0.6W

Write the expression for load combination 1.

L1=D ...... (VII)

Here, load combination 1 is L1.

Substitute 9kips for D in Equation (VII).

L1=9kips

Write the expression for load combination 2.

L2=D+L ...... (VIII)

Here, load combination 2 is L2.

Substitute 9kips for D and 0kips for L in Equation (VIII).

L2=9kips+0kips=9kips

Write the expression for load combination 3.

L3=D+(LrorSorR) ...... (IX)

Here, load combination 3 is L3.

Substitute 9kips for D and 7kips for (LrorSorR) in Equation (IX).

L3=9kips+7kips=16kips

Write the expression for load combination 4.

L4=D+0.75L+0.75(LrorSorR) ...... (X)

Here, load combination 4 is L4.

Substitute 9kips for D, 0kips for L and 7kips for (LrorSorR) in Equation (X).

L4=9kips+(0.75×0kips)+(0.75×7kips)=14.25kips

Write the expression for load combination 5.

L5=D+0.6W ...... (XI)

Here, load combination 5 is L5.

Substitute 9kips for D and 8kips for W in Equation (XI).

L5=9kips+(0.6×8kips)=13.8kips

Write the expression for load combination 6.

L6=D+0.75L+0.75(0.6W)+0.75(LrorSorR) ...... (XII)

Here, load combination 6 is L6.

Substitute 9kips for D, 0kips for L, 8kips for W and 7kips for (LrorSorR) in Equation (XII).

L6=9kips+(0.75×0kips)+0.75(0.6×8kips)+(0.75×7kips)=17.85kips

Write the expression for load combination 7.

L7=0.6D+0.6W ...... (XIII)

Here, load combination 7 is L7.

Substitute 9kips for D and 8kips for W in Equation (XIII).

L7=(0.6×9kips)+(0.6×8kips)=10.2kips

From the above 7 load combinations L6 is maximum.

Conclusion:

Thus, the maximum load and AISC combination is controlled by the load combination and the expression L1= D

To determine

(e)

The required nominal strength of the column for a safety factor of 1.67.

The required nominal strength of the column for a safety factor is 29.80kips.

Explanation

Concept used:

Write the expression for nominal strength of the column for safety factor.

Ra=RnΩRn=Ra×Ω ...... (XIV)

Here, the required strength is Ra, the nominal strength of the column is Rn and the safety factor is Ω.

Substitute 1.67 for Ω and 17.85kips for Ra in Equation (XIV).

Rn=(17.85kips)×(1.67)=29.80kips

Conclusion:

Thus, the required nominal strength of the column for safety factor is calculated by expression for the nominal strength of the column for safety factor.

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

## Additional Engineering Solutions

#### Determine the base area of an electric steam iron.

Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)

#### Describe four uses of the combination set.

Precision Machining Technology (MindTap Course List)

#### Determine the resultant of the three forces if (a) =30; and (b) =45.

International Edition---engineering Mechanics: Statics, 4th Edition

#### Convert P = 5.00 atm into Pa, bar, and psia.

Fundamentals of Chemical Engineering Thermodynamics (MindTap Course List)

#### How would you balance DFDs?

Systems Analysis and Design (Shelly Cashman Series) (MindTap Course List)

Find all the answers to your study problems with bartleby.
Textbook solutions plus Q&A. Get As ASAP arrow_forward 