Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2]

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Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] (2) C 12 x 25 φc = 0.85 x A) 5 * 103 B) 5 * 104 C) 5 * 105 D) 5 * 106 Find the design compressive strength if a column, in pounds. [pause] In this problem, --- 10 [in]

Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] (2) C 12 x 25 φc = 0.85 x A) 5 * 103 B) 5 * 104 C) 5 * 105 D) 5 * 106 2 C 12 by 25 members are positioned back to back, with a separation of 10 inches. The effective length 10 [in]

Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] (2) C 12 x 25 φc = 0.85 x A) 5 * 103 B) 5 * 104 C) 5 * 105 D) 5 * 106 of the column, in both the x and y directions, equal 20 feet, and the Young’s Modulus and --- 10 [in]

Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] (2) C 12 x 25 φc = 0.85 x A) 5 * 103 B) 5 * 104 C) 5 * 105 D) 5 * 106 yield stress of the steel are provided. [pause] The design strength of a column equals, --- 10 [in]

Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] x φc Pn = φc * Fcr * Ag phi c, times F cr, times A g, where phi c equals, --- 10 [in] (2) C 12 x 25

Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] resistance factor x φc Pn = φc * Fcr * Ag the resistance factor for a column, F cr is the --- 10 [in] (2) C 12 x 25

Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] resistance critical stress factor x φc Pn = φc * Fcr * Ag critical stress, and A g repressents --- 10 [in] (2) C 12 x 25

Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] resistance critical stress factor x φc Pn = φc * Fcr * Ag the gross area of the column. From the problem statement, --- gross area 10 [in] (2) C 12 x 25

Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] resistance critical stress factor x φc Pn = φc * Fcr * Ag we know phi c equals, 0.85. The gross area ---- gross φc = 0.85 area 10 [in] (2) C 12 x 25

Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] resistance critical stress factor x φc Pn = φc * Fcr * Ag can be determined by looking up the area of a 12 by 25 ---- gross φc = 0.85 area 10 [in] (2) C 12 x 25

Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] resistance critical stress factor x φc Pn = φc * Fcr * Ag channel member, and multiplying that area by 2 members, which equals, --- gross φc = 0.85 area 10 [in] 7.34 [in2/member] 2 [member] (2) C 12 x 25 *

Find: φcPn [lb] column in compression E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] y Kx * Lx=Ky * Ly = 20 [ft] resistance critical stress factor x φc Pn = φc * Fcr * Ag 14.68 inches squared. [pause] The critical stress in the column, --- φc = 0.85 Ag=14.68 [in2] 10 [in] 7.34 [in2/member] 2 [member] (2) C 12 x 25 *

Find: φcPn [lb] ≤ 4.71 > 4.71 E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] φc Pn = φc * Fcr * Ag F cr, is computed using 1 of 2 equations, which depends on ---- K * L E ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ? ≤ 4.71 > 4.71 E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] φc Pn = φc * Fcr * Ag the slenderness ratio and the value of 4.71 times root E over f y. We’ll first determine --- K * L E ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy ? or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ? ≤ 4.71 > 4.71 E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] φc Pn = φc * Fcr * Ag K * L E these two values, so we know how to compute the critical stress. We’ll plug in --- ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy ? or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ≤ 4.71 > 4.71 E 4.71 E = 2.9 * 107 [lb/in2] * fy fy = 50,000 [lb/in2] Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] RHS φc Pn = φc * Fcr * Ag the Young’s Modulus and yield stress into the right hand side, and find ---- K * L E ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ≤ 4.71 > 4.71 E 4.71 E = 2.9 * 107 [lb/in2] * fy fy = 50,000 [lb/in2] Kx * Lx=Ky * Ly = 20 [ft] 113.43 φc = 0.85 Ag=14.68 [in2] RHS φc Pn = φc * Fcr * Ag E the right hand side equals, 113.43. [pause] Next we’ll solve for the --- K * L ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ≤ 4.71 > 4.71 E 4.71 E = 2.9 * 107 [lb/in2] * fy fy = 50,000 [lb/in2] Kx * Lx=Ky * Ly = 20 [ft] 113.43 φc = 0.85 Ag=14.68 [in2] LHS RHS φc Pn = φc * Fcr * Ag the slenderness ratio, K L over r. The most conservative slenderness ratio --- K * L E ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ≤ 4.71 > 4.71 = max , K * L Kx * Lx Ky * Ly r rx ry Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] 113.43 LHS RHS φc Pn = φc * Fcr * Ag E is the larger slenderness ratio, about the x and y axis. From the problem statement, --- K * L ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ≤ 4.71 > 4.71 = max , K * L Kx * Lx Ky * Ly r rx ry Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] 113.43 LHS RHS φc Pn = φc * Fcr * Ag E the effective length about both axis equals, K * L ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ≤ 4.71 > 4.71 = max , 240 [in] K * L Kx * Lx Ky * Ly = max r rx , ry 12[in/ft] * Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] 113.43 LHS RHS φc Pn = φc * Fcr * Ag 240 inches. [pause] The radius of gyration, r, --- K * L E ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ≤ 4.71 > 4.71 = max , 240 [in] K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] 113.43 LHS RHS φc Pn = φc * Fcr * Ag equals the square root of the area moment of inertia, divided by --- K * L E ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ≤ 4.71 > 4.71 = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A area Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] 113.43 LHS RHS φc Pn = φc * Fcr * Ag the gross area of the column. We’ve already computed the gross area as, --- K * L E ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ≤ 4.71 > 4.71 = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A area Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] 113.43 LHS RHS φc Pn = φc * Fcr * Ag 14.68 inches squared. The area moment of inertia, equals, --- K * L E ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A area Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] 10 [in] φc Pn = φc * Fcr * Ag the sum of the centoidal area moment of inertia, plus, the area times the distance from the centroidal axis, for each channel member. Since both --- x Ix =Σ (Ic,x + A * dx2) Iy =Σ (Ic,y + A * dy2) y

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A area Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] 10 [in] φc Pn = φc * Fcr * Ag channel members are symmetric about both axis, we can simplify the equations. --- x Ix =Σ (Ic,x + A * dx2) Iy =Σ (Ic,y + A * dy2) y C 12 x 25

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A area Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] 10 [in] φc Pn = φc * Fcr * Ag [puause] Next, we’ll look up the centroidal area moments of intertia --- x Ix =2 * (Ic,x + A * dx2) Iy =2 * (Ic,y + A * dy2) y C 12 x 25

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A area Kx * Lx=Ky * Ly = 20 [ft] φc = 0.85 Ag=14.68 [in2] 10 [in] φc Pn = φc * Fcr * Ag Ix =2 * (Ic,x + A * dx2) for both the x and y axis, for a C 12 by 25 channel member. From earlier, we already know --- x Ic,x=144 [in4] Iy =2 * (Ic,y + A * dy2) y Ic,y=4.45 [in4] C 12 x 25

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] A=14.68 [in2] 10 [in] A=7.34 [in2/member] Ix =2 * (Ic,x + A * dx2) the area of a single member. Next we’ll focus in on --- x Ic,x=144 [in4] Iy =2 * (Ic,y + A * dy2) y Ic,y=4.45 [in4] C 12 x 25

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] A=14.68 [in2] 10 [in] A=7.34 [in2/member] Ix =2 * (Ic,x + A * dx2) one of the two steel members. Here we notice --- x Ic,x=144 [in4] Iy =2 * (Ic,y + A * dy2) y Ic,y=4.45 [in4] C 12 x 25

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] A=14.68 [in2] 5 [in] C 12 x 25 A=7.34 [in2/member] Ix =2 * (Ic,x + A * dx2) the centroid of the member, lies on the x axis, therefore, --- x 144 [in4] centroid Iy =2 * (Ic,y + A * dy2) 4.45 [in4] y

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] A=14.68 [in2] 5 [in] C 12 x 25 A=7.34 [in2/member] dx=0 Ix =2 * (Ic,x + A * dx2) the distance, d x equals, 0. [pause] To find d y, we’ll add 5 inches to --- x 144 [in4] centroid Iy =2 * (Ic,y + A * dy2) 4.45 [in4] y

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] A=14.68 [in2] 5 [in] C 12 x 25 A=7.34 [in2/member] dx=0 centroid Ix =2 * (Ic,x + A * dx2) the length, y bar, which, for a C 12 by 25 member, equals, --- x 144 [in4] y Iy =2 * (Ic,y + A * dy2) y 4.45 [in4] dy = 5 [in] + y

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] A=14.68 [in2] 5 [in] C 12 x 25 A=7.34 [in2/member] dx=0 centroid Ix =2 * (Ic,x + A * dx2) y bar equals, 0.674 inches. Which makes d y, equal to, --- x 144 [in4] y=0.674[in] Iy =2 * (Ic,y + A * dy2) y 4.45 [in4] dy = 5 [in] + y

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] A=14.68 [in2] 5 [in] C 12 x 25 A=7.34 [in2/member] dx=0 centroid Ix =2 * (Ic,x + A * dx2) 5.674 inches. [pause] After plugging in the known variables, --- x 144 [in4] y=0.674[in] Iy =2 * (Ic,y + A * dy2) y 4.45 [in4] dy = 5 [in] + y= 5.674 [in]

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] A=14.68 [in2] 5 [in] C 12 x 25 A=7.34 [in2] dx=0 centroid Ix =2 * (Ic,x + A * dx2) the area moment of inertia equal, --- x 144 [in4] y=0.674[in] Iy =2 * (Ic,y + A * dy2) y 4.45 [in4] dy = 5 [in] + y= 5.674 [in]

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] A=14.68 [in2] A=7.34 [in2] dx=0 Ix =2 * (Ic,x + A * dx2) 288 inches to the 4th power and 482 inches to the 4th power, for I x and I y, repsectively. Ix = 288 [in4] 144 [in4] Iy = 482 [in4] Iy =2 * (Ic,y + A * dy2) 4.45 [in4] dy = 5 [in] + y= 5.674 [in]

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] A=14.68 [in2] A=7.34 [in2] dx=0 Ix =2 * (Ic,x + A * dx2) Next we’ll solve for the radii of gyration about both axis, where r x --- Ix = 288 [in4] 144 [in4] Iy = 482 [in4] Iy =2 * (Ic,y + A * dy2) 4.45 [in4] dy = 5 [in] + y= 5.674 [in]

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] 14.68 [in2] rx= 4.429 [in] ry= 5.730 [in] A=7.34 [in2] dx=0 Ix =2 * (Ic,x + A * dx2) equals 4.429 inches, and r y equals, 5.730 inches. [pause] Now we can solve for the slenderness ratio --- Ix = 288 [in4] 144 [in4] Iy = 482 [in4] Iy =2 * (Ic,y + A * dy2) 4.45 [in4] dy = 5 [in] + y= 5.674 [in]

Find: φcPn [lb] = max , area moment 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A Kx * Lx=Ky * Ly = 20 [ft] 14.68 [in2] rx= 4.429 [in] ry= 5.730 [in] A=7.34 [in2] dx=0 Ix =2 * (Ic,x + A * dx2) by dividing the effecitve length, by, the radii of gyration, the slenderness ratio, --- Ix = 288 [in4] 144 [in4] Iy = 482 [in4] Iy =2 * (Ic,y + A * dy2) 4.45 [in4] dy = 5 [in] + y= 5.674 [in]

Find: φcPn [lb] = max , = max (54.19, 41.88) = 54.19 area moment of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A K * L = max (54.19, 41.88) 14.68 [in2] r rx= 4.429 [in] K * L ry= 5.730 [in] = 54.19 r dx=0 Ix =2 * (Ic,x + A * dx2) equals 54.19. [pause] Since we know the left hand side of the equation, --- Ix = 288 [in4] 144 [in4] Iy = 482 [in4] Iy =2 * (Ic,y + A * dy2) 4.45 [in4] dy = 5 [in] + y= 5.674 [in]

Find: φcPn [lb] ≤ 4.71 > 4.71 = max , = max (54.19, 41.88) = 54.19 area moment Find: φcPn [lb] 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A K * L = max (54.19, 41.88) 14.68 [in2] r 113.43 K * L = 54.19 LHS RHS r E is less than the right hand side of the equation, --- K * L ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] ≤ 4.71 > 4.71 = max , = max (54.19, 41.88) = 54.19 area moment Find: φcPn [lb] 240 [in] of inertia K * L Kx * Lx Ky * Ly I = max r= r rx , ry A K * L = max (54.19, 41.88) 14.68 [in2] r 113.43 K * L = 54.19 r we’ll use the first equation to solve for the critical stress in the column, which is the --- K * L E ≤ 4.71 if * (1) Fcr=0.658 fy/fe * fy r fy or K * L E (2) Fcr=0.877 * fe > 4.71 if * r fy

Find: φcPn [lb] = 54.19 area moment of inertia E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] r= A φc = 0.85 Ag=14.68 [in2] 14.68 [in2] φc Pn = φc * Fcr * Ag K * L = 54.19 r Fcr=0.658 fy/fe * fy last unkown variable we need to solve for the design strength of the column. [pause] From the problem statement, ---

Find: φcPn [lb] = 54.19 area moment of inertia E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] r= A φc = 0.85 Ag=14.68 [in2] 14.68 [in2] φc Pn = φc * Fcr * Ag K * L = 54.19 r Fcr=0.658 fy/fe * fy we know the yield stress of the steel, f y. The buckling stress, f e, ---

Find: φcPn [lb] π2 * E = 54.19 area moment of inertia E = 2.9 * 107 [lb/in2] I fy = 50,000 [lb/in2] r= A φc = 0.85 Ag=14.68 [in2] 14.68 [in2] φc Pn = φc * Fcr * Ag K * L = 54.19 r Fcr=0.658 fy/fe * fy equals, Pi squared times the Young’s Modulus, divided by the slenderness ratio squared. We know --- buckling π2 * E Fe= stress K * L 2 r

Find: φcPn [lb] π2 * E = 54.19 area moment of inertia E = 2.9 * 107 [lb/in2] I fy = 50,000 [lb/in2] r= A φc = 0.85 Ag=14.68 [in2] 14.68 [in2] φc Pn = φc * Fcr * Ag K * L = 54.19 r Fcr=0.658 fy/fe * fy the Young’s Modulus, E, and we already computed the slenderness ratio, --- π2 * E Fe= K * L 2 r

Find: φcPn [lb] π2 * E = 54.19 area moment of inertia E = 2.9 * 107 [lb/in2] I fy = 50,000 [lb/in2] r= A φc = 0.85 Ag=14.68 [in2] 14.68 [in2] φc Pn = φc * Fcr * Ag K * L = 54.19 r Fcr=0.658 fy/fe * fy K L over r. [pause] This makes the buckling stress equal to, --- π2 * E Fe= K * L 2 r

Find: φcPn [lb] π2 * E = 54.19 area moment of inertia E = 2.9 * 107 [lb/in2] I fy = 50,000 [lb/in2] r= A φc = 0.85 Ag=14.68 [in2] 14.68 [in2] φc Pn = φc * Fcr * Ag K * L = 54.19 r Fcr=0.658 fy/fe * fy 97,467 pound per square inch. [pause] And subsequently we compute the critical stress to equal, --- π2 * E Fe= Fe= 97,467 [lb/in2] K * L 2 r

Find: φcPn [lb] = 54.19 area moment of inertia E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] r= A φc = 0.85 Ag=14.68 [in2] 14.68 [in2] φc Pn = φc * Fcr * Ag K * L = 54.19 r Fcr=0.658 fy/fe * fy Fe= 97,467 [lb/in2] 40,339 pounds per square inch. [pause] And finally --- Fcr= 40,339 [lb/in2]

Find: φcPn [lb] = 54.19 area moment of inertia E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] r= A φc = 0.85 Ag=14.68 [in2] 14.68 [in2] φc Pn = φc * Fcr * Ag K * L = 54.19 r Fcr=0.658 fy/fe * fy Fe= 97,467 [lb/in2] the design compressive strength of the column, equals, --- Fcr= 40,339 [lb/in2]

Find: φcPn [lb] = 54.19 area moment of inertia E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] r= A φc = 0.85 Ag=14.68 [in2] 14.68 [in2] φc Pn = φc * Fcr * Ag K * L = 54.19 r Fcr=0.658 fy/fe * fy Fe= 97,467 [lb/in2] 5.03*105 pounds. [pause] Fcr= 40,339 [lb/in2] φc Pn = 5.03 * 105 [lb]

Find: φcPn [lb] = 54.19 area moment of inertia E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] r= A φc = 0.85 Ag=14.68 [in2] 14.68 [in2] φc Pn = φc * Fcr * Ag K * L = 54.19 r Fcr=0.658 fy/fe * fy A) 5 * 103 B) 5 * 104 C) 5 * 105 D) 5 * 106 When reviewing the possible solutions, --- Fcr= 40,339 [lb/in2] φc Pn = 5.03 * 105 [lb]

Find: φcPn [lb] = 54.19 area moment of inertia E = 2.9 * 107 [lb/in2] fy = 50,000 [lb/in2] r= A φc = 0.85 Ag=14.68 [in2] 14.68 [in2] φc Pn = φc * Fcr * Ag K * L = 54.19 r Fcr=0.658 fy/fe * fy A) 5 * 103 B) 5 * 104 C) 5 * 105 D) 5 * 106 the answer is C. Fcr= 40,339 [lb/in2] φc Pn = 5.03 * 105 [lb] answerC