Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
5th Edition
ISBN: 9781305084766
Author: Saeed Moaveni
Publisher: Cengage Learning
Question
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Chapter 20, Problem 1P

(a)

To determine

Find the future value for $10000at6.75% compounding annually for 10 years.

(a)

Expert Solution
Check Mark

Answer to Problem 1P

The future value for $10000at6.75% compounding annually for 10 years is $19216.7.

Explanation of Solution

Given data:

The present cost (P) is 10000,

The normal interest rate (i) is 6.75%,

The number of years (n) is 10years,

The number of compound period (m) is 1.

Formula used:

Formula to calculate the future value is,

F=P(1+i)n (1)

Here,

P is the present cost,

i is the normal interest rate,

n is the number of years.

Calculation:

To calculate the future value:

Substitute 10000 for P, 6.75% for i, and 10years for n in equation (1) to find F.

F=(10000)(1+(6.75%))10=(10000)(1+0.0675)10=$19216.7

Therefore, the future value for $10000at6.75% compounding annually for 10 years is $19216.7.

Conclusion:

Thus, the future value for $10000at6.75% compounding annually for 10 years is $19216.7.

(b)

To determine

Find the future value for $10000at6.75% compounding quarterly for 10 years.

(b)

Expert Solution
Check Mark

Answer to Problem 1P

The future value for $10000at6.75% compounding quarterly for 10 years is $19530.

Explanation of Solution

Given data:

The present cost (P) is 10000,

The normal interest rate (i) is 6.75%,

The number of years (n) is 10years,

The number of compound period (m) is 4.

Formula used:

Formula to calculate the future value is,

F=P(1+im)nm (2)

Here,

P is the present cost,

i is the normal interest rate,

n is the number of years,

m is the number of compound periods per year.

Calculation:

To calculate the future value:

Substitute 10000 for P, 6.75% for i, 10years for n, and 4 for m in equation (2) to find F.

F=(10000)(1+0.06754)(10)(4)=(10000)(1+0.06754)(40)=$19530

Therefore, the future value for $10000at6.75% compounding quarterly for 10 years is $19530.

Conclusion:

Thus, the future value for $10000at6.75% compounding quarterly for 10 years is $19530.

(c)

To determine

Find the future value for $10000at6.75% compounding monthly for 10 years.

(c)

Expert Solution
Check Mark

Answer to Problem 1P

The future value for $10000at6.75% compounding monthly for 10 years is $19603.21.

Explanation of Solution

Given data:

The present cost (P) is 10000,

The normal interest rate (i) is  6.75%,

The number of years (n) is 10years,

The number of compound periods per year (m) is 12.

Formula used:

Formula to calculate the future value is,

F=P(1+im)nm (3)

Here,

P is the present cost,

i is the normal interest rate,

n is the number of years,

m is the number of compound periods per year.

Calculation:

To calculate the future value:

Substitute 10000 for P, 6.75% for i, 10years for n, and 12 for m in equation (3) to find F.

F=(10000)(1+0.067512)(10)(12)=(10000)(1+0.067512)(120)=$19603.21

Therefore, the future value for $10000at6.75% compounding monthly for 10 years is $19603.21.

Conclusion:

Thus, the future value for $10000at6.75% compounding monthly for 10 years is $19603.21.

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WU Example 6 For the exterior transverse frame of the flat slab floor shown in figure, and by using the Direct Design Method, find: a. Longitudinal distribution of the total static moment at factored loads. b. Lateral distribution of moment at exterior panel (column and middle strip moments at exterior support) D= 6.5 kN/m² L= 5.0 kN/m² تفكر وکھل flat slap ما لا يوجد bon حامل . 3000 1000 5000 160 + 2000+ +2000+ 5000 2608 300 2000 Drop Panal السعف
Example 4 For the transverse interior frame (Frame C) of the flat plate floor with edge beams shown in Figure, by using the Direct Design Method, find: 1) Longitudinal distribution of total static moment at factored loads. 2) Lateral distribution of moment at interior panel (column and middle strip moments atnegative and positive moments). 3) Lateral distribution of moment at exterior panel (column and middle strip moments atnegative and positive moments). Plat 5000-5000 5000 -Frame C لا بوجود deen 0009 0009 Slab thickness = 180 mm, d = 150 mm q₁ = 16.0 kN/m² All edge beams = 250x 500 mm All columns = 500x 500 mm 6000
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