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]
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]
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]
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]
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 1.50. [pause] The effective length factor for a column, --- 15 [ft] E 20 [ft] 18 [ft]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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]
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, 0.95. [pause] For Joint E, --- CD 15 [ft] I=1,830 [in4] E BC 20 [ft] 18 [ft] I=1,350 [in4]
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]
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]
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]
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]
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]
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]
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
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 answerD Gtop K Gbottom