Please show solutions on how to get the 27.57 value. The circle green in the figure. Civil engineering- steel design Note: I thought the value is 86.617 . Feel free to correct me . What should put in circle in the figure.
Please show solutions on how to get the 27.57 value. The circle green in the figure. Civil engineering- steel design Note: I thought the value is 86.617 . Feel free to correct me . What should put in circle in the figure.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Please show solutions on how to get the 27.57 value. The circle green in the figure.
Civil engineering- steel design
Note: I thought the value is 86.617 . Feel free to correct me . What should put in circle in the figure.
![Step 2: Calculation of least
moment of inertia
7- (100×25) × 12-5+ (150 x 25)- (25+75)
(100x25)+ (150x25)
65 mm
Las = [100x216² (160 (25) +52,5"] + [25-150) + (rm)))
= 1-864x10² mm
NEW,
= (100x25)x 50+ (150x25) x)2-5
2500+ 3750
= 295 mm
lyy = [25> 100³ (25-10-22:5] [15 (1501)-15]
= 4388020-833 mm
From above two cases. Imin 1388020-833 mm"
Step 3: Load capacity
calculation
Now, Euler stress (Fe) = ².E
Also, 4.71×
*** JE
2510
Total cross-sectional area = A₁ + A₂ = 2500+ 3750 = 6250 mm 2
Least radius of gyration (romain) = √√Imin
(K4/²
= 26-197
Effective length (KL) = 1x4 = 4m = 4000 mm
:: Slendernes ration (KL/min) =
S 4-71x
Theoritical critical stren
100
R
4000
26.497
for = (0.658fa/fe), fy
(6.658 21/7-5) 218
= 5.734 MPa
nominal strength > P₁ = fer. Ag
(ü) ASD:- Desige strength
= 5.734 x 6250
100
(1) LRFD- Design strength = 4.Pa
43 88020-833
6250
It'v 200x10
150.96
= 35839-73 N
= 35.84 KN
200×10
√218
=133-755
is given by AlSC Equ²¹ F3-2 as
= 0.9 x35-84
= 32-256 KN
175
175
= 150.96 < 200 (ex)
Pr/52
=
= 35-81/1-67
= 21.461 KN
= 27.57 N/mm²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa492bd26-ee44-4470-930f-28e915fd1410%2F58d9ff37-cae4-49df-aba6-3f3999c3fdf1%2Fcwhuzx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Step 2: Calculation of least
moment of inertia
7- (100×25) × 12-5+ (150 x 25)- (25+75)
(100x25)+ (150x25)
65 mm
Las = [100x216² (160 (25) +52,5"] + [25-150) + (rm)))
= 1-864x10² mm
NEW,
= (100x25)x 50+ (150x25) x)2-5
2500+ 3750
= 295 mm
lyy = [25> 100³ (25-10-22:5] [15 (1501)-15]
= 4388020-833 mm
From above two cases. Imin 1388020-833 mm"
Step 3: Load capacity
calculation
Now, Euler stress (Fe) = ².E
Also, 4.71×
*** JE
2510
Total cross-sectional area = A₁ + A₂ = 2500+ 3750 = 6250 mm 2
Least radius of gyration (romain) = √√Imin
(K4/²
= 26-197
Effective length (KL) = 1x4 = 4m = 4000 mm
:: Slendernes ration (KL/min) =
S 4-71x
Theoritical critical stren
100
R
4000
26.497
for = (0.658fa/fe), fy
(6.658 21/7-5) 218
= 5.734 MPa
nominal strength > P₁ = fer. Ag
(ü) ASD:- Desige strength
= 5.734 x 6250
100
(1) LRFD- Design strength = 4.Pa
43 88020-833
6250
It'v 200x10
150.96
= 35839-73 N
= 35.84 KN
200×10
√218
=133-755
is given by AlSC Equ²¹ F3-2 as
= 0.9 x35-84
= 32-256 KN
175
175
= 150.96 < 200 (ex)
Pr/52
=
= 35-81/1-67
= 21.461 KN
= 27.57 N/mm²
Transcribed Image Text:Determine the maximum safe capacity of an axially
loaded hinged ends column having an unsupported
length of 4m. Fy = 248 MPa and E = 200 GPa. Use
AISC Specifications.
25
CIVIL ENGINEERING:
STEEL DESIGN
100
25
175
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