Find: KBE PBE=180 [k] AB, BC  W12x14 compression fy= 36 [ksi] BE, EH  W10x33 DE, EF  W14x22 unbraced A B C 12 [ft] A) 1.00 B) 1.37 C) 1.43 D) 1.52 D E F 12 [ft] Find the effective length factor, K, for column B E. [pause] In this problem, --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE PBE=180 [k] AB, BC  W12x14 compression fy= 36 [ksi] BE, EH  W10x33 DE, EF  W14x22 unbraced A B C 12 [ft] A) 1.00 B) 1.37 C) 1.43 D) 1.52 D E F 12 [ft] a 3-story structure expriences a compression load, of 180 kips, --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE PBE=180 [k] AB, BC  W12x14 compression fy= 36 [ksi] BE, EH  W10x33 DE, EF  W14x22 unbraced A B C 12 [ft] A) 1.00 B) 1.37 C) 1.43 D) 1.52 D E F 12 [ft] in member B E. The lengths of all columns and girders --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE PBE=180 [k] AB, BC  W12x14 compression fy= 36 [ksi] BE, EH  W10x33 DE, EF  W14x22 unbraced A B C 12 [ft] A) 1.00 B) 1.37 C) 1.43 D) 1.52 D E F 12 [ft] are provided, as well as the exact I beam type for 6 of those --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE PBE=180 [k] AB, BC  W12x14 compression fy= 36 [ksi] BE, EH  W10x33 DE, EF  W14x22 unbraced A B C 12 [ft] A) 1.00 B) 1.37 C) 1.43 D) 1.52 D E F 12 [ft] steel members. [pause] The effective length factor, as part of a --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE PBE=180 [k] AB, BC  W12x14 compression fy= 36 [ksi] BE, EH  W10x33 DE, EF  W14x22 unbraced A B C 12 [ft] A) 1.00 B) 1.37 C) 1.43 D) 1.52 D E F 12 [ft] structural frame, is determined by knowing the end condition --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE PBE=180 [k] compression KBE = f (GB, GE) fy= 36 [ksi] unbraced A B C 12 [ft] A) 1.00 B) 1.37 C) 1.43 D) 1.52 D E F 12 [ft] coefficients of the member. Which for this problem, refers to joints --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE PBE=180 [k] compression KBE = f (GB, GE) fy= 36 [ksi] unbraced A B C 12 [ft] A) 1.00 B) 1.37 C) 1.43 D) 1.52 D F E 12 [ft] B and E. [pause] The end condition coefficients, equal, the sum of the --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE λc>1.50 Σ Ec * Ic / Lc Σ Eg * Ig / Lg PBE=180 [k] columns KBE = f (GB, GE) Σ Ec * Ic / Lc λc>1.50 Gelastic= if Σ Eg * Ig / Lg girders A B C 12 [ft] D E F 12 [ft] rotational stiffness values in the columns, divided by, the sum of the --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE λc>1.50 Σ Ec * Ic / Lc Σ Eg * Ig / Lg PBE=180 [k] columns KBE = f (GB, GE) Σ Ec * Ic / Lc λc>1.50 Gelastic= if Σ Eg * Ig / Lg girders A B C 12 [ft] D E F 12 [ft] rotational stiffness values in the girders. [pause] Since it’s safe to assume the --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE λc>1.50 Σ Ec * Ic / Lc Σ Eg * Ig / Lg PBE=180 [k] columns KBE = f (GB, GE) Σ Ec * Ic / Lc λc>1.50 Gelastic= if Σ Eg * Ig / Lg girders A B C 12 [ft] Ec = Eg D E F 12 [ft] the moduli of elasticity is constant, then these terms cancel out, --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE λc>1.50 Σ Ec * Ic / Lc Σ Eg * Ig / Lg PBE=180 [k] columns KBE = f (GB, GE) Σ Ec * Ic / Lc λc>1.50 Gelastic= if Σ Eg * Ig / Lg girders A B C 12 [ft] Ec = Eg D E F 12 [ft] and we’re left with the area moment of inertia, ---- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE λc>1.50 Σ Ec * Ic / Lc Σ Eg * Ig / Lg area moment of inertia KBE = f (GB, GE) Σ Ec * Ic / Lc λc>1.50 Gelastic= if Σ Eg * Ig / Lg member A B C 12 [ft] length Ec = Eg D E F 12 [ft] I, and the length of the member, L. [pause] In the event member B E, is, --- G H I 12 [ft] 20 [ft] 18 [ft]

Find: KBE λc>1.50 λc≤1.50 Σ Ec * Ic / Lc Σ Eg * Ig / Lg area moment of inertia KBE = f (GB, GE) Σ Ec * Ic / Lc λc>1.50 Gelastic= if Σ Eg * Ig / Lg member length λc≤1.50 if (inelastic) inelastic, then, the end condition coefficient at a joint, equals, ---

Find: KBE λc>1.50 λc≤1.50 Σ Ic / Lc Σ Ig / Lg KBE = f (GB, GE) Gelastic= if Σ Ig / Lg Ginelastic= τ * Gelastic λc≤1.50 if (inelastic) Tau, times, the elastic end condition coefficient, for that joint. We’ll begin by evaluating, ---

Find: KBE λc>1.50 λc≤1.50 Σ Ic / Lc Σ Ig / Lg ? KBE = f (GB, GE) Gelastic= if Σ Ig / Lg ? Ginelastic= τ * Gelastic λc≤1.50 if (inelastic) lambda c, the slenderness parameter, which equals, ---

Find: KBE λc>1.50 λc≤1.50 Σ Ic / Lc Σ Ig / Lg ? λc = * KBE = f (GB, GE) Σ Ic / Lc λc>1.50 Gelastic= if Σ Ig / Lg ? Ginelastic= τ * Gelastic λc≤1.50 if (inelastic) K L, over r times Pi, times root f y over E. Since we’re interested in --- K * L fy λc = * r * π E

Find: KBE λc>1.50 λc≤1.50 Σ Ic / Lc Σ Ig / Lg ? λc = * KBE = f (GB, GE) Σ Ic / Lc λc>1.50 Gelastic= if Σ Ig / Lg ? Ginelastic= τ * Gelastic λc≤1.50 if (inelastic) member B E, we’ll add subscripts B E to the equation, --- K * L fy λc = * r * π E

Find: KBE λc>1.50 λc≤1.50 Σ Ic / Lc Σ Ig / Lg ? λc = * KBE = f (GB, GE) Σ Ic / Lc λc>1.50 Gelastic= if Σ Ig / Lg ? Ginelastic= τ * Gelastic λc≤1.50 if (inelastic) [pause] From the problem statement, we know the yield stress in the steel, --- KBE * LBE fy λc = * rBE * π E

Find: KBE λc>1.50 λc≤1.50 Σ Ic / Lc Σ Ig / Lg ? λc = * KBE = f (GB, GE) Σ Ic / Lc λc>1.50 Gelastic= if Σ Ig / Lg ? Ginelastic= τ * Gelastic λc≤1.50 if (inelastic) equals 36 ksi, which is the same as, --- KBE * LBE fy fy= 36 [ksi] λc = * rBE * π E

Find: KBE λc>1.50 λc≤1.50 Σ Ic / Lc Σ Ig / Lg ? λc = * KBE = f (GB, GE) Σ Ic / Lc λc>1.50 Gelastic= if Σ Ig / Lg ? Ginelastic= τ * Gelastic λc≤1.50 if (inelastic) 36,000 pounds per square inch. And it’s generally assumed the modulus of elasticity --- KBE * LBE fy fy= 36 [ksi] λc = * rBE * π E fy=36,000 [lb/in2]

Find: KBE λc>1.50 λc≤1.50 Σ Ic / Lc Σ Ig / Lg ? λc = * KBE = f (GB, GE) Σ Ic / Lc λc>1.50 Gelastic= if Σ Ig / Lg ? Ginelastic= τ * Gelastic λc≤1.50 if (inelastic) equals, 2.9*107 pound per inch squared. [pause] We can determine the length, ---- KBE * LBE fy fy= 36 [ksi] λc = * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2]

Find: KBE λc>1.50 λc≤1.50 Σ Ic / Lc Σ Ig / Lg ? λc = * KBE = f (GB, GE) Σ Ic / Lc λc>1.50 Gelastic= if Σ Ig / Lg ? Ginelastic= τ * Gelastic λc≤1.50 if (inelastic) L, by looking back the structure, --- KBE * LBE fy fy= 36 [ksi] λc = * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2]

Find: KBE λc>1.50 λc≤1.50 ? λc = * A B C 12 [ft] KBE = f (GB, GE) D if G H I 12 [ft] ? λc≤1.50 if 20 [ft] 18 [ft] (inelastic) locating column B E, and noting it’s length is --- KBE * LBE fy fy= 36 [ksi] λc = * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2]

Find: KBE λc>1.50 λc≤1.50 ? λc = * A B C 12 [ft] KBE = f (GB, GE) D if G H I 12 [ft] ? λc≤1.50 if 20 [ft] 18 [ft] (inelastic) 12 [in/ft] * 12 feet, or, 144 inches. For a W10 by 33 steel member, --- KBE * LBE fy λc = 144 [in] * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2]

Find: KBE λc>1.50 λc≤1.50 ? λc = * A B C 12 [ft] W10x33 D E F if G H I 12 [ft] ? λc≤1.50 if 20 [ft] 18 [ft] (inelastic) 12 [in/ft] * the radius of gyration about the strong axis, equals, --- KBE * LBE fy λc = 144 [in] * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2]

Find: KBE λc = * A B C 12 [ft] W10x33 D E F 12 [ft] rx=4.184 [in] G H 12 [in/ft] * 4.184 inches. [pause] The last unknown variable is --- KBE * LBE fy λc = 144 [in] * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2]

Find: KBE λc = * A B C 12 [ft] W10x33 D E F 12 [ft] G H I 12 [ft] 12 [in/ft] * the effective length factor, K, which is the variable --- KBE * LBE fy λc = 144 [in] * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2] r=4.184 [in]

Find: KBE λc = * A B C 12 [ft] W10x33 D E F 12 [ft] G H I 12 [ft] 12 [in/ft] * were solving for. So we’ll compute lamba c, as --- KBE * LBE fy λc = 144 [in] * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2] r=4.184 [in]

Find: KBE λc =0.3860 * KBE λc = * A B C 12 [ft] W10x33 D E F 12 [ft] G H I 12 [ft] λc =0.3860 * KBE 20 [ft] 18 [ft] 12 [in/ft] * 0.3860, times the effective length factor in columnd B E. Since this doesn’t answer our question regarding --- KBE * LBE fy λc = 144 [in] * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2] r=4.184 [in]

Find: KBE λc>1.50 λc≤1.50 λc =0.3860 * KBE Σ Ic / Lc Σ Ig / Lg ? Gelastic= if Σ Ig / Lg ? Ginelastic= τ * Gelastic λc≤1.50 if (inelastic) λc =0.3860 * KBE 12 [in/ft] * which equation to use, we’ll assume lambda c, is greater than 1.50, and compute --- KBE * LBE fy λc = 144 [in] * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2] r=4.184 [in]

Find: KBE λc>1.50 λc≤1.50 λc =0.3860 * KBE Σ Ic / Lc Σ Ig / Lg λc = Gelastic= if Σ Ig / Lg assume true Ginelastic= τ * Gelastic λc≤1.50 if (inelastic) λc =0.3860 * KBE 12 [in/ft] * the end condition coefficients using the top equation. Then, using those values, --- KBE * LBE fy λc = 144 [in] * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2] r=4.184 [in]

Find: KBE λc>1.50 λc =0.3860 * KBE Σ Ic / Lc Σ Ig / Lg λc = * Gelastic= if Σ Ig / Lg assume true GB KBE GE λc =0.3860 * KBE 12 [in/ft] * we’ll compute the effective length factor, K, for member B E, and then check our assumption, --- KBE * LBE fy λc = 144 [in] * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2] r=4.184 [in]

Find: KBE λc>1.50 λc =0.3860 * KBE Σ Ic / Lc Σ Ig / Lg λc = * Gelastic= if Σ Ig / Lg ? check GB assumption KBE GE λc =0.3860 * KBE 12 [in/ft] * that lambda c is greater than 1.50. [pause] The end condition coefficient for Joint B, --- KBE * LBE fy λc = 144 [in] * rBE * π E fy=36,000 [lb/in2] E=2.9*107 [lb/in2] r=4.184 [in]

Find: KBE λc>1.50 IBE / LBE GB= if IAB / LAB + IBC / LBC ? A B C 12 [ft] D E F is computed by plugging in the appropriate lengths, --- G H I 20 [ft] 18 [ft]

Find: KBE λc>1.50 IBE / LBE GB= if IAB / LAB + IBC / LBC ? A B C 12 [ft] D E F and plugging in the area moment of inertia values, --- G H I 20 [ft] 18 [ft]

Find: KBE λc>1.50 12 [ft] IBE / LBE GB= if IAB / LAB + IBC / LBC ? BE  W10x33  IBE=170 [in4] about the stronger axis. The end condition coefficient at joint B equals, --- AB  W12x14  IAB=88.6 [in4] G H I BC  W12x14  IBC=88.6 [in4] 20 [ft] 18 [ft]

Find: KBE λc>1.50 170 [in4] 12 [ft] IBE / LBE GB= if IAB / LAB + IBC / LBC ? 20 [ft] 18 [ft] 88.6 [in4] A B C GB=1.52 D E F 1.52. [pause] Joint E is computed --- G H I 20 [ft] 18 [ft]

Find: KBE λc>1.50 170 [in4] 12 [ft] IBE / LBE GB= if IAB / LAB + IBC / LBC ? 20 [ft] 18 [ft] 88.6 [in4] GB=1.52 IBE / LBE + IEH / LEH GE= in a similar manner. After plugging in the lengths and area moments of --- IDE / LDE + IEF / LEF

Find: KBE λc>1.50 170 [in4] 12 [ft] IBE / LBE GB= if IAB / LAB + IBC / LBC ? 20 [ft] 18 [ft] 88.6 [in4] GB=1.52 12 [ft] 170 [in4] IBE / LBE + IEH / LEH GE= inertia, the end condition coefficient, at Joint E, equals, --- IDE / LDE + IEF / LEF 20 [ft] 199 [in4] 18 [ft]

Find: KBE λc>1.50 170 [in4] 12 [ft] IBE / LBE GB= if IAB / LAB + IBC / LBC ? 20 [ft] 18 [ft] 88.6 [in4] GB=1.52 12 [ft] 170 [in4] IBE / LBE + IEH / LEH GE= 1.35. [pause] Keep in mind, these two end condition coefficients --- IDE / LDE + IEF / LEF 20 [ft] 199 [in4] 18 [ft] GE=1.35

Find: KBE λc>1.50 170 [in4] 12 [ft] IBE / LBE GB= if IAB / LAB + IBC / LBC ? 20 [ft] 18 [ft] 88.6 [in4] assumed elastic GB=1.52 12 [ft] 170 [in4] IBE / LBE + IEH / LEH GE= refer to an elastic condition in the I beam. Therefore, we can --- IDE / LDE + IEF / LEF 20 [ft] 199 [in4] 18 [ft] assumed elastic GE=1.35

Find: KBE λc>1.50 GB=1.52 if GE=1.35 ? effective length alignment chart determine the effective length factor, if lambda c is greater than 1.50, by marking --- Gtop K Gbottom

Find: KBE λc>1.50 GB=1.52 if GE=1.35 ? effective length alignment chart GE GB G B and G E on the effective length alignment chart, and the assumed effective length factor --- Gtop K Gbottom

Find: KBE λc>1.50 GB=1.52 if GE=1.35 ? effective length alignment chart GE GB equals 1.43. [pause] Now we can check our assuptions regarding lambda c, --- KBE=1.43 Gtop K Gbottom

Find: KBE λc>1.50 λc =0.3860 * KBE GB=1.52 if GE=1.35 ? effective length alignment chart GE GB which equals, 0.3860 times 1.43, or -- λc =0.3860 * KBE KBE=1.43 Gtop K Gbottom

Find: KBE λc>1.50 λc =0.3860 * KBE λc =0.552 GB=1.52 if GE=1.35 effective length alignment chart GE GB 0.552, which is less than 1.50. Therefore, we know the end conditions coefficients --- λc =0.3860 * KBE KBE=1.43 λc =0.552 Gtop K Gbottom

Find: KBE λc≤1.50 λc =0.3860 * KBE λc =0.552 GB,inelastic= τ * GB,elastic λc≤1.50 if GE,inelastic= τ * GE,elastic (inelastic) effective length alignment chart GE GB we calcaulted need to be multiplied by Tau, so we can find the inelastic end condition coefficients. λc =0.3860 * KBE KBE=1.43 λc =0.552 Gtop K Gbottom

Find: KBE λc≤1.50 λc =0.3860 * KBE λc =0.552 1.52 1.35 GB,inelastic= τ * GB,elastic λc≤1.50 if GE,inelastic= τ * GE,elastic 1.35 (inelastic) effective length alignment chart GE GB From before, we know G B and G E, for the elastic case. And Tau, represents the stiffness --- λc =0.3860 * KBE KBE=1.43 λc =0.552 Gtop K Gbottom

Find: KBE λc≤1.50 λc =0.3860 * KBE λc =0.552 1.52 1.35 stiffness GB,inelastic= τ * GB,elastic λc≤1.50 if GE,inelastic= τ * GE,elastic 1.35 (inelastic) effective length stiffness alignment chart reduction factor GE GB reduction factor, which can be determined using the stiffness reduction --- λc =0.3860 * KBE KBE=1.43 λc =0.552 Gtop K Gbottom

τ Find: KBE λc≤1.50 λc =0.3860 * KBE λc =0.552 1.52 1.35 stiffness GB,inelastic= τ * GB,elastic λc≤1.50 if GE,inelastic= τ * GE,elastic 1.35 (inelastic) stiffness reduction stiffness alignment chart reduction factor alignment chart. Where the elastic end condition coefficients --- λc =0.3860 * KBE λc =0.552 τ Gtop Gbottom

τ Find: KBE λc≤1.50 λc =0.3860 * KBE λc =0.552 1.52 1.35 stiffness GB,inelastic= τ * GB,elastic λc≤1.50 if GE,inelastic= τ * GE,elastic 1.35 (inelastic) stiffness reduction stiffness alignment chart reduction factor GE GB are plotted out as before, and Tau B E, equals, --- λc =0.3860 * KBE λc =0.552 τ elastic Gtop Gbottom

τBE=0.83 τ Find: KBE λc≤1.50 λc =0.3860 * KBE λc =0.552 1.52 1.35 GB,inelastic= τ * GB,elastic λc≤1.50 if GE,inelastic= τ * GE,elastic 1.35 (inelastic) stiffness reduction stiffness alignment chart reduction factor GE GB 0.83. [pause] Plugging in 0.83, for Tau, --- τBE=0.83 λc =0.3860 * KBE λc =0.552 τ elastic Gtop Gbottom

τBE=0.83 τ Find: KBE λc≤1.50 λc =0.3860 * KBE λc =0.552 1.52 1.35 GB,inelastic= τ * GB,elastic λc≤1.50 if GE,inelastic= τ * GE,elastic 1.35 (inelastic) stiffness reduction stiffness alignment chart reduction factor GE GB the inelastic end condition coefficients for joints B and E, equal, --- τBE=0.83 λc =0.3860 * KBE λc =0.552 τ elastic Gtop Gbottom

τBE=0.83 τ Find: KBE λc≤1.50 λc =0.3860 * KBE λc =0.552 0.83 1.52 1.35 GB,inelastic= τ * GB,elastic λc≤1.50 if GE,inelastic= τ * GE,elastic 1.35 (inelastic) GB,inelastic= 1.26 GE,inelastic= 1.12 GE GB 1.26 and 1.12, respectively. [pause] Now we can return to the --- τBE=0.83 λc =0.3860 * KBE λc =0.552 τ elastic Gtop Gbottom

Find: KBE λc≤1.50 λc =0.3860 * KBE λc =0.552 GB,inelastic= 1.26 if GE,inelastic= 1.12 effective length alignment chart effective length alignment chart. Plug in the 2 inelastic end condition --- λc =0.3860 * KBE λc =0.552 Gtop K Gbottom

Find: KBE λc≤1.50 λc =0.3860 * KBE λc =0.552 GB,inelastic= 1.26 if GE,inelastic= 1.12 effective length alignment chart GE coefficients, and the effective length factor, for member B E, --- GB λc =0.3860 * KBE λc =0.552 Gtop K Gbottom

Find: KBE λc≤1.50 λc =0.3860 * KBE λc =0.552 GB,inelastic= 1.26 if GE,inelastic= 1.12 effective length alignment chart GE equals, 1.37. [pause] GB λc =0.3860 * KBE KBE=1.37 λc =0.552 Gtop K Gbottom

Find: KBE λc≤1.50 GB,inelastic= 1.26 if GE,inelastic= 1.12 effective length A) 1.00 B) 1.37 C) 1.43 D) 1.52 alignment chart GE When reviewing the possible solutions, --- GB KBE=1.37 Gtop K Gbottom

Find: KBE λc≤1.50 GB,inelastic= 1.26 if GE,inelastic= 1.12 effective length A) 1.00 B) 1.37 C) 1.43 D) 1.52 alignment chart GE the answer is B. GB answerB KBE=1.37 Gtop K Gbottom