Solution Manual For Steel Design 7th Edition by William T. Segui, Sayed Soleimani Copyright 2026

STEEL DESIGN [7TH EDITION]
William T. Segui and Sayed M. Soleimani

CHAPTER 1 – INTRODUCTION

1.5-1

(a) 𝐴 = 𝜋(0.550)2 /4 = 0.2376 in.2

𝑃𝑢 28,500
𝐹𝑢 = = = 120,000 psi = 120 ksi
𝐴 0.2376

Fu = 120 ksi (827.4 MPa)

2.300−2.030
(b) 𝑒 = × 100 = 13.3%
2.030


e = 13.3%

(c) 𝐴𝑓 = 𝜋(0.430)2 /4 = 0.1452 in.2

𝐴𝑓 −𝐴𝑜 0.1452−0.2376
Change = × 100 = × 100 = −38.9%
𝐴𝑜 0.2376


Reduction = 38.9%



1.5-2

𝐴 = 𝜋(0.5)2 /4 = 0.1963 in.2

𝑃 13.5
𝑓= = = 68.77 ksi
𝐴 0.1963

Δ𝐿 4.66 × 10−3
𝜀= = = 2.33 × 10−3
𝐿 2

𝑓 68.77
𝐸= = = 29,500 ksi
𝜀 2.33 × 10−3

E = 29,500 ksi (203.4 GPa)



[1-1]

© 2026 Cengage Learning®. All Rights Reserved. May not be scanned, copied, or duplicated,
or posted to a publicly accessible website, in whole or in part.

,1.5-3

𝐴 = 𝜋(0.510)2 /4 = 0.2043 in.2
𝑃 250
For 𝑃 = 250 lb, 𝑓 = 𝐴 = 0.2043 = 1224 psi

Spreadsheet results:

(a)




(b)




(c)

E = slope = 31,600 ksi (217.9 GPa)



[1-2]

© 2026 Cengage Learning®. All Rights Reserved. May not be scanned, copied, or duplicated,
or posted to a publicly accessible website, in whole or in part.

,1.5-4

𝐴 = 𝜋(0.5)2 /4 = 0.1963 in.2

𝑓 𝑃/𝐴 𝑃 𝐿
𝐸= = = ⋅ = 1,392(4/0.1963) = 28,400 ksi
𝜀 Δ𝐿/𝐿 Δ𝐿 𝐴

E = 28,400 ksi (195.8 GPa)




1.5-5

𝐴 = 𝜋(3/8)2 /4 = 0.1104 in.2

𝑃 550
For 𝑃 = 550 lb, 𝑓 = 𝐴 = 0.1104 = 4,982 psi

Δ𝐿 350 × 10−6
𝜀= = = 175 × 10−6
𝐿 2

Spreadsheet results:

(a)




[1-3]

© 2026 Cengage Learning®. All Rights Reserved. May not be scanned, copied, or duplicated,
or posted to a publicly accessible website, in whole or in part.

, (b)




15,000
(c) 𝐸 = slope = 500×10−6 = 30,000,000 psi = 30,000 ksi


E = 30,000 ksi (206.8 GPa)

(d)

Fy ≈ 44,000 psi ≈ 44 ksi (≈ 303 MPa)




[1-4]

© 2026 Cengage Learning®. All Rights Reserved. May not be scanned, copied, or duplicated,
or posted to a publicly accessible website, in whole or in part.