1. Find: Fcr [lb/in2] Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2]
fixed-fixed connection
W 10 x 22
L=15 [ft]
A) 24,000
B) 28,000
C) 32,000
D) 36,000
Find the critical stress, in pounds per square inch. [pause] In this problem, ----
Fcr

2. Find: Fcr [lb/in2] Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2]
fixed-fixed connection
W 10 x 22
L=15 [ft]
A) 24,000
B) 28,000
C) 32,000
D) 36,000
a 15 foot long I beam is connected in a fixed-fixed orientation. The Young’s Modulus and ---
Fcr

3. Find: Fcr [lb/in2] Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2]
fixed-fixed connection
W 10 x 22
L=15 [ft]
A) 24,000
B) 28,000
C) 32,000
D) 36,000
yield stress of the steel are provided. [pause] The critical stress of a steel member in compression ---
Fcr

4. Find: Fcr [lb/in2] ≤ 4.71 > 4.71 Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
(1) Fcr=0.658 fy/fe * fy
K * L E
≤ 4.71 if *
L=15 [ft] r fy
(2) Fcr=0.877 * fe
is computed using 1 of 2 equations. The particular equation to use, depends on, the terms ---
K * L E
> 4.71 if *
W10x22 r fy
Fcr

5. Find: Fcr [lb/in2] ≤ 4.71 > 4.71 Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
(1) Fcr=0.658 fy/fe * fy
K * L E
≤ 4.71 if *
L=15 [ft] r fy
(2) Fcr=0.877 * fe
K L over r and 4.71 times root E over f y. Therefore, we’ll first evaluate these two terms. The former term, ---
K * L E
> 4.71 if *
W10x22 r fy
Fcr

6. Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy
L=15 [ft]
K * L r
K L over r, is the effective length factor, K, ---
W10x22
Fcr

7. Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy
L=15 [ft]
length
K * L r
times the length of the member, L, divided by the radius of gyration, ---
effective
W10x22
length
Fcr
factor

8. Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy
L=15 [ft]
length
K * L r
radius of
r. [pause] This term, as a whole, ---
gyration
effective
W10x22
length
Fcr
factor

9. Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy
L=15 [ft]
length
K * L r
radius of
is referred to as the slenderness ratio. [pause] From the problem statement, ---
gyration
effective
W10x22
slenderness
length
Fcr
ratio
factor

10. Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy
L=15 [ft]
length
K * L r
radius of
we know the length is 15 feet, which is the same as, ---
gyration
effective
W10x22
slenderness
length
Fcr
ratio
factor

11. Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy
L=15 [ft]
length
12[in/ft]
K * L
L = 180 [in] * r
radius of
180 inches. [pause] The effective length factor, for fixed-fixed ---
gyration
effective
W10x22
slenderness
length
Fcr
ratio
factor

12. Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 *
fixed r fy
K=0.65
L=15 [ft]
L = 180 [in]
K * L
fixed r
radius of
end connections, equals And the radius of gyration, r, ---
gyration
effective
W10x22
slenderness
length
Fcr
ratio
factor

13. Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 *
fixed r fy
K=0.65
L=15 [ft]
L = 180 [in]
K * L
fixed r
radius of
equals the smaller of the radii of gyration, in the x and y directions.
gyration
r = min (rx, ry)
W10x22
Fcr

14. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
K=0.65
L=15 [ft]
L = 180 [in]
K * L
fixed r
radius of
If we take a cross section of the beam, A A prime, ---
gyration
r = min (rx, ry)
W10x22
Fcr

15. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
K=0.65 y
L=180 [in]
K * L r
we can look up the radii of gyration values as being, ---
W10x22 x
r = min (rx, ry)
section A-A’

16. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
K=0.65 y
L=180 [in]
K * L r
4.27 inches and 1.33 inches, for a W 10 by 22 I beam. Therefore, ---
W10x22 x
rx= 4.27 [in]
r = min (rx, ry)
ry= 1.33 [in]
section A-A’

17. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
K=0.65 y
L=180 [in]
K * L r
we’ll use 1.33 inches for the radius of gyration, r. And our slenderness ratio, --- x
rx= 4.27 [in]
r = min (rx, ry)
ry= 1.33 [in]
r = 1.33 [in]
section A-A’

18. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
88.0
K=0.65 y
L=180 [in]
K * L r
computes to [pause] To solve the right hand side of the equation, --- x
rx= 4.27 [in]
r = min (rx, ry)
ry= 1.33 [in]
r = 1.33 [in]
section A-A’

19. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
88.0
K=0.65 y
L=180 [in]
K * L r
we can plug in the given Young’s Modulus and yield stress of the steel, and the right hand side, --- x
rx= 4.27 [in]
r = min (rx, ry)
ry= 1.33 [in]
r = 1.33 [in]
section A-A’

20. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
88.0
K=0.65
133.7
L=180 [in]
K * L r
equals, [pause] Since 88.0 is less than 133.7, ---- x
rx= 4.27 [in]
r = min (rx, ry)
ry= 1.33 [in]
r = 1.33 [in]
section A-A’

21. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
88.0
133.7
(1) Fcr=0.658 fy/fe * fy
we’ll use the first equation to compute the critical stress, F cr. [pause] In this equation, --- x
section A-A’

22. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
88.0
133.7
(1) Fcr=0.658 fy/fe * fy
we already know the yield stress, f y. [pause] And the variable f e, --- x
section A-A’

23. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
88.0
133.7
(1) Fcr=0.658 fy/fe * fy
represents the buckling stress, which equals, PI squared, ---
buckling x
stress
section A-A’

24. Find: Fcr [lb/in2] π2 * E ≤ or > 4.71 A A’ Fcr
E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
88.0
133.7
(1) Fcr=0.658 fy/fe * fy
times the Young’s Modulus, divided by the slenderness ratio, squared. [pause]
buckling
π2 * E x
Fe=
stress
K * L 2 r
section A-A’

25. Find: Fcr [lb/in2] π2 * E ≤ or > 4.71 A A’ Fcr
E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
88.0
133.7
(1) Fcr=0.658 fy/fe * fy
Since we’ve be given the Young’s modulus, and already solved for ---
buckling
π2 * E x
Fe=
stress
K * L 2 r
section A-A’

26. Find: Fcr [lb/in2] π2 * E ≤ or > 4.71 A A’ Fcr
E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
88.0
133.7
(1) Fcr=0.658 fy/fe * fy
the slenderness ratio, we compute the buckling stress as ---
buckling
π2 * E x
Fe=
stress
K * L 2 r
section A-A’

27. Find: Fcr [lb/in2] π2 * E ≤ or > 4.71 A A’ Fcr
E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
88.0
133.7
(1) Fcr=0.658 fy/fe * fy
36,960 pounds per square inch. [pause] After plugging in the yield stress, and, ---
buckling
π2 * E x
Fe=
stress
K * L 2 r
Fe= 36,960 [lb/in2]
section A-A’

28. Find: Fcr [lb/in2] π2 * E ≤ or > 4.71 A A’ Fcr
E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
88.0
133.7
(1) Fcr=0.658 fy/fe * fy
buckling stress, the critical stress on the steel member, equals, ---
π2 * E x
Fe=
K * L 2 r
Fe= 36,960 [lb/in2]
section A-A’

29. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
133.7
(1) Fcr=0.658 fy/fe * fy
23,947 pounds per square inch. [pause]
Fe= 36,960 [lb/in2] x
Fcr = 23,947 [lb/in2]
section A-A’

30. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
133.7
(1) Fcr=0.658 fy/fe * fy
A) 24,000
B) 28,000
C) 32,000
D) 36,000
When reviewing the possible solutions, ---
Fe= 36,960 [lb/in2]
Fcr = 23,947 [lb/in2]

31. Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2]
fy = 36,000 [lb/in2]
K * L E
≤ or > 4.71 * r fy A A’
133.7
(1) Fcr=0.658 fy/fe * fy
A) 24,000
B) 28,000
C) 32,000
D) 36,000
the answer is A.
Fe= 36,960 [lb/in2]
Fcr = 23,947 [lb/in2]
answerA