1. Find: KCE λc>1.50 E = 2.9 * 107 [lb/in2] AC, CE  W12x120 BC 
unbraced
CD 
W24x68 A
12 [ft]
A) 0.83
B) 0.95
C) 1.30
D) 1.85 B C D
Find the effective length factor, K, for column C E. [pause] In this problem, ---
15 [ft] E
20 [ft]
18 [ft]

2. Find: KCE λc>1.50 E = 2.9 * 107 [lb/in2] AC, CE  W12x120 BC 
unbraced
CD 
W24x68 A
12 [ft]
A) 0.83
B) 0.95
C) 1.30
D) 1.85 B C D
an unbraced, 2-story structure is connected with pin connections. The lengths of the ---
15 [ft] E
20 [ft]
18 [ft]

3. Find: KCE λc>1.50 E = 2.9 * 107 [lb/in2] AC, CE  W12x120 BC 
unbraced
CD 
W24x68 A
12 [ft]
A) 0.83
B) 0.95
C) 1.30
D) 1.85 B C D
columns and girders are provided, and the specific I beams are specified for ---
15 [ft] E
20 [ft]
18 [ft]

4. Find: KCE λc>1.50 E = 2.9 * 107 [lb/in2] AC, CE  W12x120 BC 
unbraced
CD 
W24x68 A
12 [ft]
A) 0.83
B) 0.95
C) 1.30
D) 1.85 B C D
4 of the members. We also know, the slenderness parameter, ---
15 [ft] E
20 [ft]
18 [ft]

5. Find: KCE λc>1.50 E = 2.9 * 107 [lb/in2] AC, CE  W12x120 BC 
unbraced
CD 
W24x68 A
12 [ft]
A) 0.83
B) 0.95
C) 1.30
D) 1.85 B C D
lambda c, is greater than [pause] The effective length factor for a column, ---
15 [ft] E
20 [ft]
18 [ft]

6. Find: KCE λc>1.50 KCE = f (GC, GE) E = 2.9 * 107 [lb/in2] AC, CE 
W12x120
unbraced
BC 
W24x55 A
CD 
W24x68
12 [ft]
A) 0.83
B) 0.95
C) 1.30
D) 1.85 B C D
as part of a structural frame, is a function of ----
15 [ft] E
20 [ft]
18 [ft]

7. Find: KCE λc>1.50 KCE = f (GC, GE) E = 2.9 * 107 [lb/in2]
end condition
unbraced
coefficient A
12 [ft]
A) 0.83
B) 0.95
C) 1.30
D) 1.85 B C D
the end condition coefficients, at the 2 ends, of the column. So for column C E, ---
15 [ft] E
20 [ft]
18 [ft]

8. Find: KCE λc>1.50 KCE = f (GC, GE) E = 2.9 * 107 [lb/in2]
end condition
unbraced
coefficient A
12 [ft]
A) 0.83
B) 0.95
C) 1.30
D) 1.85 B C D
we’ll focus on joints C and E. [pause] Once we know the 2 ---
15 [ft] E
20 [ft]
18 [ft]

9. Find: KCE λc>1.50 KCE = f (GC, GE) E = 2.9 * 107 [lb/in2]
end condition
unbraced
coefficient A
12 [ft]
A) 0.83
B) 0.95
C) 1.30
D) 1.85 B C D
end condition coefficients, we’ll use an alignment chart ---
15 [ft] E
20 [ft]
18 [ft]

10. Find: KCE λc>1.50 KCE = f (GC, GE) E = 2.9 * 107 [lb/in2]
end condition
unbraced
coefficient A GE B C D GC
to find the effective length factor for the column, ---- E
Gtop K
Gbottom
20 [ft]
18 [ft]

11. Find: KCE λc>1.50 KCE = f (GC, GE) E = 2.9 * 107 [lb/in2]
end condition
unbraced
coefficient A
KCE GE B C D GC
K. [pause] The end condition coefficient, G, equals, --- E
Gtop K
Gbottom
20 [ft]
18 [ft]

12. Find: KCE Σ Ec * Ic / Lc Σ Eg * Ig / Lg E = 2.9 * 107 [lb/in2]
KCE = f (GC, GE)
Σ Ec * Ic / Lc G=
Σ Eg * Ig / Lg A
KCE GE B C D GC
the sum of the E I over L, for the adjoining column members, divided by, --- E
Gtop K
Gbottom
20 [ft]
18 [ft]

13. Find: KCE Σ Ec * Ic / Lc Σ Eg * Ig / Lg E = 2.9 * 107 [lb/in2]
KCE = f (GC, GE)
Σ Ec * Ic / Lc
adjoining columns G=
Σ Eg * Ig / Lg
adjoining girders A
KCE GE B C D GC
the sum of the E I over L, for the adjoining girders. E
Gtop K
Gbottom
20 [ft]
18 [ft]

14. Find: KCE Σ Ec * Ic / Lc Σ Eg * Ig / Lg E = 2.9 * 107 [lb/in2]
KCE = f (GC, GE)
Σ Ec * Ic / Lc
adjoining columns
GC=
Σ Eg * Ig / Lg
adjoining girders A
KCE GE B C D GC
For joint C, the column members will be --- E
Gtop K
Gbottom
20 [ft]
18 [ft]

15. Find: KCE Σ Ec * Ic / Lc Σ Eg * Ig / Lg E = 2.9 * 107 [lb/in2]
KCE = f (GC, GE)
Σ Ec * Ic / Lc
adjoining columns
GC=
Σ Eg * Ig / Lg
adjoining girders A
KCE GE B C D GC
A C and C E, and the girder members will be --- E
Gtop K
Gbottom
20 [ft]
18 [ft]

16. Find: KCE Σ Ec * Ic / Lc Σ Eg * Ig / Lg E = 2.9 * 107 [lb/in2]
KCE = f (GC, GE)
Σ Ec * Ic / Lc
adjoining columns
GC=
Σ Eg * Ig / Lg
adjoining girders A B C D
KCE GE GC
B C, and C D. [pause] From the problem statement, --- E
Gtop K
Gbottom
20 [ft]
18 [ft]

17. Find: KCE Σ Ec * Ic / Lc Σ Eg * Ig / Lg E = 2.9 * 107 [lb/in2]
KCE = f (GC, GE)
Σ Ec * Ic / Lc
adjoining columns
GC=
Σ Eg * Ig / Lg
adjoining girders A
AC, CE 
W12x120 B C D
we know the specific I beams corresponding to these adjoining members. And we can ---
CD 
W24x68 E
BC 
W24x55
20 [ft]
18 [ft]

18. Find: KCE Σ Ec * Ic / Lc Σ Eg * Ig / Lg E = 2.9 * 107 [lb/in2]
KCE = f (GC, GE)
Σ Ec * Ic / Lc
adjoining columns
GC=
Σ Eg * Ig / Lg
adjoining girders A
AC, CE 
W12x120 B C D
I=1,070 [in4]
look up their corresponding area moment of inertia values, in the strong direction. [pause] Next we’ll write out the ---
CD 
W24x68
I=1,830 [in4] E
BC 
W24x55
20 [ft]
18 [ft]
I=1,350 [in4]

19. Find: KCE E = 2.9 * 107 [lb/in2] KCE = f (GC, GE) joint C
Ec*IAC/LAC + Ec*ICE/LCE
GC=
Eg*IBC/LBC + Eg*ICD/LCD A
AC, CE 
W12x120 B C D
I=1,070 [in4]
the equation for G, specific to joint C. Unless otherwise noted, we’ll assume ---
CD 
W24x68
I=1,830 [in4] E
BC 
W24x55
20 [ft]
18 [ft]
I=1,350 [in4]

20. Find: KCE E = 2.9 * 107 [lb/in2] KCE = f (GC, GE) joint C
Ec*IAC/LAC + Ec*ICE/LCE
GC=
Eg*IBC/LBC + Eg*ICD/LCD A
AC, CE 
W12x120 B C D
I=1,070 [in4]
the modulus of elasticity is constant for all members, therefore, ---
CD 
W24x68
I=1,830 [in4] E
BC 
W24x55
20 [ft]
18 [ft]
I=1,350 [in4]

21. Find: KCE E = 2.9 * 107 [lb/in2] KCE = f (GC, GE) joint C
Ec*IAC/LAC + Ec*ICE/LCE
GC=
Eg*IBC/LBC + Eg*ICD/LCD A
AC, CE 
W12x120 B C D
I=1,070 [in4]
it cancels out of the equation, and we’ll revise our equation for ---
CD 
W24x68
I=1,830 [in4] E
BC 
W24x55
20 [ft]
18 [ft]
I=1,350 [in4]

22. Find: KCE = E = 2.9 * 107 [lb/in2] KCE = f (GC, GE)
Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD A
AC, CE 
W12x120 B C D
I=1,070 [in4]
the end condition coefficient, at Joint C. Next, we’ll plug in ---
CD 
W24x68
I=1,830 [in4] E
BC 
W24x55
20 [ft]
18 [ft]
I=1,350 [in4]

23. Find: KCE = E = 2.9 * 107 [lb/in2] KCE = f (GC, GE)
Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD A
AC, CE 
W12x120 B C D
I=1,070 [in4]
the area moment of inertia terms, as well as the length terms, ---
CD 
W24x68
I=1,830 [in4] E
BC 
W24x55
20 [ft]
18 [ft]
I=1,350 [in4]

24. Find: KCE = E = 2.9 * 107 [lb/in2] KCE = f (GC, GE)
Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD A
AC,CE
12 [ft] B C D
I=1,070 [in4]
and our value of G, at joint C, --- CD
15 [ft]
I=1,830 [in4] E BC
20 [ft]
18 [ft]
I=1,350 [in4]

25. Find: KCE = GC=0.95 KCE = f (GC, GE) Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD A
AC,CE
12 [ft] B C D
I=1,070 [in4]
equals, [pause] For Joint E, --- CD
15 [ft]
I=1,830 [in4] E BC
20 [ft]
18 [ft]
I=1,350 [in4]

26. Find: KCE = GC=0.95 KCE = f (GC, GE) Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD A
AC,CE
12 [ft] B C D
I=1,070 [in4]
since pin connections do not resist moments, pinned ends are assigned --- CD
15 [ft]
I=1,830 [in4] E BC
20 [ft]
18 [ft]
I=1,350 [in4]

27. Find: KCE = GC=0.95 KCE = f (GC, GE) Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD A
AC,CE
12 [ft] B C D
I=1,070 [in4]
a value of 10, for their end condition coefficient. [pause] Returning to the --- CD
GE=10
15 [ft]
I=1,830 [in4] E BC
20 [ft]
18 [ft]
I=1,350 [in4]

28. Find: KCE = GC=0.95 KCE = f (GC, GE) Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD A B C D
alignment chart. We’ll mark the values for G C and ----
GE=10 E
Gtop K
Gbottom
20 [ft]
18 [ft]

29. Find: KCE = GC=0.95 KCE = f (GC, GE) Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD A GE B C D GC
G E. [pause] Drawing a straight line between these points helps us ---
GE=10 E
Gtop K
Gbottom
20 [ft]
18 [ft]

30. Find: KCE = GC=0.95 KCE = f (GC, GE) Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD A GE B C D GC
identify the effective length factor for column C E, which equals, ---
GE=10
KCE E
Gtop K
Gbottom
20 [ft]
18 [ft]

31. Find: KCE = GC=0.95 KCE = f (GC, GE) Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD A GE B C D GC
1.85. [pause]
GE=10
KCE=1.85 E
Gtop K
Gbottom
20 [ft]
18 [ft]

32. Find: KCE = GC=0.95 KCE = f (GC, GE) Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD
A) 0.83
B) 0.95
C) 1.30
D) 1.85 GE GC
When looking back at the possible solutions, ---
KCE=1.85
Gtop K
Gbottom

33. Find: KCE = GC=0.95 KCE = f (GC, GE) Ec*IAC/LAC + Ec*ICE/LCE
IAC/LAC+ICE/LCE
GC= =
Eg*IBC/LBC + Eg*ICD/LCD
IBC/LBC+ICD/LCD
A) 0.83
B) 0.95
C) 1.30
D) 1.85 GE GC
the answer is D.
KCE=1.85
answerD
Gtop K
Gbottom