Find: λc P P 1[in] fixed y 8 [in] A A’ 14[in] x 1[in] L z z fixed L/2 section A-A’ Find the slenderness parameter, lambda c. [pause] In this problem, --- P P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc P P 1[in] fixed y 8 [in] A A’ 14[in] x 1[in] L z z fixed L/2 section A-A’ an 12 foot long I beam is subjected to a compression load, P. Note that the beam is --- P P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc P P 1[in] fixed y 8 [in] A A’ 14[in] x 1[in] L z z fixed L/2 section A-A’ supported half way along it’s length, in one direction. A cross section of the beam --- P P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc P P 1[in] fixed y 8 [in] A A’ 14[in] x 1[in] L z z fixed L/2 section A-A’ is provided, and we notice the end connections are --- P P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc P P 1[in] fixed y 8 [in] A A’ 14[in] x 1[in] L z z fixed L/2 section A-A’ fixed. The modulus of elasticity and yield stress --- P P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc P P 1[in] fixed y 8 [in] A A’ 14[in] x 1[in] L z z fixed L/2 section A-A’ of the steel are also given. [pause] Keep in mind, there is only 1 beam in this problem, --- P P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc P P 1[in] fixed y 8 [in] A A’ 14[in] x 1[in] L z z fixed L/2 section A-A’ viewed from 2 different directions. [pause] The slenderness parameter, --- P P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc λc = * P P K * L fy A A’ r * π E L z z L/2 y x P P lambda c, equals, K L, over r times Pi, multiplied by root --- P P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc λc = * P P slenderness parameter K * L fy A A’ r * π E L z z x f y over E. In this equation we already know --- P P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc λc = * P P slenderness parameter K * L fy A A’ r * π E L z z x the yield stress of the steel, fy, and the modulus of elasticity, --- P P E = 2.9 * 107 [lb/in2] fy= 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc λc = * P P slenderness parameter K * L fy A A’ r * π E L z z x E. [pause] If we simplify our equation, lamba c, equals, --- P P E = 2.9 * 107 [lb/in2] fy= 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc λc = λc = 0.01448 * * P P slenderness parameter K * L fy A A’ z z K * L λc = 0.01448 * L/2 r y x 0.01448 times K L over r. In this equation, --- P P E = 2.9 * 107 [lb/in2] fy= 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc λc = 0.01448 * P P slenderness parameter effective A A’ length factor L z z K * L λc = 0.01448 * L/2 r y x K represents the effective length factor, L refers to --- P P E = 2.9 * 107 [lb/in2] fy= 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc λc = 0.01448 * P P slenderness parameter effective length A factor L z z K * L λc = 0.01448 * L/2 r y x the length, and r is the radius of gyration --- P P E = 2.9 * 107 [lb/in2] fy= 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc λc = 0.01448 * P P slenderness parameter effective length A factor L z z K * L λc = 0.01448 * L/2 r y x radius of gyration [pause] Since the I beam can globally buckle --- P P E = 2.9 * 107 [lb/in2] fy= 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc λc = 0.01448 * P P slenderness parameter effective length A factor L z z K * L λc = 0.01448 * L/2 r y x radius of gyration about the x or y axis, our value for K L over r will be the larger between ---- P P E = 2.9 * 107 [lb/in2] fy= 60,000 [lb/in2] A) 0.24 B) 0.39 C) 0.48 D) 0.78 L = 12[ft]

Find: λc λc = 0.01448 , =max * x-dir y-dir slenderness parameter L = 12[ft] A A’ L z z K * L λc = 0.01448 * L/2 r y x radius of gyration K L over r in the x direction and K L over r in the y direction. The length in the x direction, --- K * L Kx*Lx Ky*Ly =max r rx , ry

Find: λc λc = 0.01448 , =max * x-dir y-dir slenderness parameter K * L L = 12[ft] L z z Lx = L= 12 [ft] L/2 y x 12[in/ft] * equals, 12 feet, or 144 inches. [pause] And the effective length factor, K x, --- Lx=144 [in] K * L Kx*Lx Ky*Ly =max r rx , ry

Find: λc λc = 0.01448 , =max * x-dir y-dir slenderness parameter K * L L = 12[ft] L z z L/2 y x Lx=144 [in] equals, 0.65, because of the fixed-fixed end connections, in the x direction. [pause] The y direction --- Kx=0.65 K * L Kx*Lx Ky*Ly =max r rx , ry

Find: λc λc = 0.01448 , =max * x-dir y-dir slenderness parameter K * L L = 12[ft] L z z L/2 y x Lx=144 [in] is supported at L over 2, with a roller, or pin, connection, therefore L y, equals, --- Kx=0.65 K * L Kx*Lx Ky*Ly =max r rx , ry

Find: λc λc = 0.01448 , =max * x-dir y-dir slenderness parameter K * L L = 12[ft] L z z Ly = L/2 12 [in/ft] * L/2 y x Lx=144 [in] L over 2, or 72 inches. [pause] And a fixed-pin end connections have an effective length factor of --- Ly=72 [in] Kx=0.65 K * L Kx*Lx Ky*Ly =max r rx , ry

Find: λc λc = 0.01448 , =max * x-dir y-dir slenderness parameter K * L L = 12[ft] L z z Ly = L/2 12 [in/ft] * L/2 y x Lx=144 [in] 0.8. [pause] After simplifying these quotients, --- Ly=72 [in] Kx=0.65 K * L Kx*Lx Ky*Ly =max Ky=0.8 r rx , ry

Find: λc λc = 0.01448 , =max * x-dir y-dir slenderness parameter K * L L = 12[ft] L z z L/2 y x the last variable we need to solve for is ---- K * L 93.6 [in] 57.6 [in] =max r rx , ry

Find: λc λc = 0.01448 , =max * x-dir y-dir slenderness parameter K * L L = 12[ft] L z z L/2 y x the radii of gyration, r x and r y. [pause] Where the radius of gyration, equals, --- K * L 93.6 [in] 57.6 [in] =max r rx , ry

Find: λc λc = 0.01448 , =max * x-dir y-dir slenderness parameter K * L L = 12[ft] L z z L/2 y x the square root of the area moment of inertia, divided by the area. [pause] Next we’ll return to --- Ix Iy rx= ry= A A K * L 93.6 [in] 57.6 [in] =max r rx , ry

Find: λc , =max x-dir y-dir 1[in] y 8 [in] A A’ 14[in] x 1[in] L z z section A-A’ 1[in] the given cross section of the I beam, A A prime, and increase it’s size, --- Ix Iy rx= ry= A A K * L 93.6 [in] 57.6 [in] =max r rx , ry

Find: λc y-dir y 1 [in] A A’ 8 [in] 14[in] x z 1 [in] x 1 [in] [pause] We’ll divide the cross section into 3 parts, where parts 1 and 3 --- 1 [in] section A-A’ Ix Iy rx= ry= A A

Find: λc part 1 y-dir y 1 [in] A A’ 8 [in] 14[in] x z 1 [in] x part 3 refer to the flanges, and part 2, will refer to --- 1 [in] section A-A’ Ix Iy rx= ry= A A

Find: λc part 1 y-dir y 1 [in] A A’ 8 [in] 14[in] part 2 x z 1 [in] x the web. [pause] Now we can calculate --- 1 [in] section A-A’ Ix Iy rx= ry= A A

Find: λc part 1 y-dir y 1 [in] A A’ 8 [in] 14[in] part 2 x z 1 [in] x the cross sectional area of the beam, A, by adding --- 1 [in] section A-A’ Ix Iy rx= ry= area A A

Find: λc part 1 y-dir y 1 [in] A A’ 8 [in] 14[in] part 2 x z 1 [in] x the area from each of the 3 parts. If we add 8 inches squared plus --- 1 [in] A = A1+A2+A3 Ix Iy rx= ry= area A A

Find: λc A1= 8 [in2] y-dir y 1 [in] A A’ 8 [in] 14[in] A2=12 [in2] x z 12 inches squared plus 8 inches squared, the total area equals, --- 1 [in] A = A1+A2+A3 Ix Iy rx= ry= area A A

Find: λc A1= 8 [in2] y-dir y 1 [in] A A’ 8 [in] 14[in] A2=12 [in2] x z 28 inches squared. [pause] The area moment of inertia, --- 1 [in] A = A1+A2+A3 Ix Iy rx= ry= 28 [in2] A A

Find: λc part 1 y 1 [in] Ix = Σ Ic,i,x+Ai * dy,i 8 [in] 2 i n 8 [in] Iy = Σ Ic,i,y+Ai * dx,i 2 14[in] i x part 2 1 [in] part 3 I, is computed about both the x and y axis. If we write out the equation for --- area moment 1 [in] of inertia Ix Iy rx= ry= 28 [in2] A A

Find: λc +8 [in2]*(6.5 [in])2 +1[in]*(12[in])3/12 +12 [in2]*(0 [in])2 part 1 y n 1 [in] Ix = Σ Ic,i,x+Ai * dy,i 2 i Ix = 8 [in]*(1 [in])3/12 8 [in] 14[in] +8 [in2]*(6.5 [in])2 x +1[in]*(12[in])3/12 part 2 +12 [in2]*(0 [in])2 1 [in] +8 [in]*(1 [in])3/12 I x, we find the area moment of inertia about the x axis equals, --- +8 [in2]*(6.5 [in])2 1 [in] part 3 Ix Iy rx= ry= 28 [in2] A A

Find: λc +8 [in2]*(6.5 [in])2 +1[in]*(12[in])3/12 +12 [in2]*(0 [in])2 part 1 y n 1 [in] Ix = Σ Ic,i,x+Ai * dy,i 2 i Ix = 8 [in]*(1 [in])3/12 8 [in] 14[in] +8 [in2]*(6.5 [in])2 x +1[in]*(12[in])3/12 part 2 +12 [in2]*(0 [in])2 1 [in] +8 [in]*(1 [in])3/12 821.4 inches to the fourth power. Next, if we calculate --- +8 [in2]*(6.5 [in])2 1 [in] part 3 Ix = 821.4 [in4] Ix Iy rx= ry= 28 [in2] A A

Find: λc part 1 y 1 [in] Iy = Σ Ic,i,y+Ai * dx,i 8 [in] 14[in] x 2 i 8 [in] 14[in] x part 2 1 [in] the area moment of inertia about the y axis, --- 1 [in] part 3 Ix = 821.4 [in4] Ix Iy rx= ry= 28 [in2] A A

Find: λc +8 [in2]*(0 [in])2 +12[in]*(1[in])3/12 +12 [in2]*(0 [in])2 part 1 y n 1 [in] Iy = Σ Ic,i,y+Ai * dx,i 2 i Iy = 1 [in]*(8 [in])3/12 8 [in] 14[in] +8 [in2]*(0 [in])2 x +12[in]*(1[in])3/12 part 2 +12 [in2]*(0 [in])2 1 [in] +1 [in]*(8 [in])3/12 we find the value of I y, equals, --- +8 [in2]*(0 [in])2 1 [in] part 3 Ix Iy rx= ry= 28 [in2] A A

Find: λc +8 [in2]*(0 [in])2 +12[in]*(1[in])3/12 +12 [in2]*(0 [in])2 part 1 y n 1 [in] Iy = Σ Ic,i,y+Ai * dx,i 2 i Iy = 1 [in]*(8 [in])3/12 8 [in] 14[in] +8 [in2]*(0 [in])2 x +12[in]*(1[in])3/12 part 2 +12 [in2]*(0 [in])2 1 [in] +1 [in]*(8 [in])3/12 86.3, inches to the fourth power. [pause] Now we can --- +8 [in2]*(0 [in])2 1 [in] part 3 Iy = 86.3 [in4] Ix Iy rx= ry= 28 [in2] A A

Find: λc +8 [in2]*(0 [in])2 +12[in]*(1[in])3/12 +12 [in2]*(0 [in])2 part 1 y n 1 [in] Iy = Σ Ic,i,y+Ai * dx,i 2 i Iy = 1 [in]*(8 [in])3/12 8 [in] 14[in] +8 [in2]*(0 [in])2 x +12[in]*(1[in])3/12 part 2 +12 [in2]*(0 [in])2 1 [in] +1 [in]*(8 [in])3/12 compute the radii of gyration as --- +8 [in2]*(0 [in])2 Ix = 821.4 [in4] Iy = 86.3 [in4] Ix Iy rx= ry= A A A=28 [in2]

Find: λc +8 [in2]*(0 [in])2 +12[in]*(1[in])3/12 +12 [in2]*(0 [in])2 part 1 y n 1 [in] Iy = Σ Ic,i,y+Ai * dx,i 2 i Iy = 1 [in]*(8 [in])3/12 8 [in] 14[in] +8 [in2]*(0 [in])2 x +12[in]*(1[in])3/12 part 2 +12 [in2]*(0 [in])2 1 [in] +1 [in]*(8 [in])3/12 5.416 inches for r x, and 1.756 inches for r y. [pause] Returning to ---- +8 [in2]*(0 [in])2 Ix = 821.4 [in4] Iy = 86.3 [in4] Ix Iy =5.416 [in] rx= ry= =1.756 [in] A A A=28 [in2]

Find: λc λc = 0.01448 , =max rx= 5.416 [in] ry=1.756 [in] * y-dir slenderness parameter A A’ K * L λc = 0.01448 * r z L/2 x our equation for K L over r, K L over r equals the larger value between --- K * L 93.6 [in] 57.6 [in] =max r rx , ry rx= 5.416 [in] ry=1.756 [in]

Find: λc λc = 0.01448 , =max rx= 5.416 [in] = max (17.28, 32.80) y-dir slenderness parameter A A’ K * L λc = 0.01448 * r z L/2 x 17.28 and 32.80, which is --- K * L 93.6 [in] 57.6 [in] =max r rx , ry rx= 5.416 [in] = max (17.28, 32.80) ry=1.756 [in]

Find: λc λc = 0.01448 , =max rx= 5.416 [in] = max (17.28, 32.80) y-dir slenderness parameter A A’ K * L λc = 0.01448 * r z L/2 x 32.80. [pause] Lastly, we can solve for lambda c by --- K * L 93.6 [in] 57.6 [in] =max r rx , ry rx= 5.416 [in] = max (17.28, 32.80) ry=1.756 [in] = 32.80

Find: λc λc = 0.01448 , =max rx= 5.416 [in] = max (17.28, 32.80) y-dir slenderness parameter A A’ K * L λc = 0.01448 * r z L/2 x plugging in 32.80 for K L over r, and lambda c, --- K * L 93.6 [in] 57.6 [in] =max r rx , ry rx= 5.416 [in] = max (17.28, 32.80) ry=1.756 [in] = 32.80

Find: λc λc = 0.01448 λc = 0.475 , =max rx= 5.416 [in] y-dir slenderness parameter A A’ K * L λc = 0.01448 * r z λc = 0.475 L/2 x equals, 0.475. [pause] K * L 93.6 [in] 57.6 [in] =max r rx , ry rx= 5.416 [in] = max (17.28, 32.80) ry=1.756 [in] = 32.80

Find: λc λc = 0.01448 λc = 0.475 , =max rx= 5.416 [in] y-dir slenderness A) 0.24 B) 0.39 C) 0.48 D) 0.74 parameter A A’ K * L λc = 0.01448 * r z λc = 0.475 L/2 x When looking over the possible solutions, --- K * L 93.6 [in] 57.6 [in] =max r rx , ry rx= 5.416 [in] = max (17.28, 32.80) ry=1.756 [in] = 32.80

Find: λc λc = 0.01448 λc = 0.475 , =max rx= 5.416 [in] y-dir slenderness A) 0.24 B) 0.39 C) 0.48 D) 0.74 parameter A A’ K * L λc = 0.01448 * r z λc = 0.475 L/2 x answerC the answer is C. K * L 93.6 [in] 57.6 [in] =max r rx , ry rx= 5.416 [in] = max (17.28, 32.80) ry=1.756 [in] = 32.80