Find: The Lightest I Beam

Find: The Lightest I Beam
P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] LL = 80,000 [lb] pin DL= 20,000 [lb] φc = 0.85 L L= 14 [ft] A) W8x18 B) W8x24 C) W10x26 D) W10x33 fixed K * L Find the lightest I beam. [pause] In this problem, --- ≤ 200 r P

P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] LL = 80,000 [lb] pin DL= 20,000 [lb] φc = 0.85 L L= 14 [ft] A) W8x18 B) W8x24 C) W10x26 D) W10x33 fixed K * L a 14-foot long I beam is subjected to a compresssion load, P. The unfactored --- ≤ 200 r P

P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] LL = 80,000 [lb] pin DL= 20,000 [lb] φc = 0.85 L L= 14 [ft] A) W8x18 B) W8x24 C) W10x26 D) W10x33 fixed K * L live load and dead load values are provided. The beam is connected --- ≤ 200 r P

P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] LL = 80,000 [lb] pin DL= 20,000 [lb] φc = 0.85 L L= 14 [ft] A) W8x18 B) W8x24 C) W10x26 D) W10x33 fixed K * L with a fixed-pin connection, and the maximum slenderness ratio of the I beam, is, --- ≤ 200 r P

P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] LL = 80,000 [lb] pin DL= 20,000 [lb] φc = 0.85 L L= 14 [ft] A) W8x18 B) W8x24 C) W10x26 D) W10x33 fixed K * L 200. [pause] Of the 4 possible I beams, --- ≤ 200 r P

P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] LL = 80,000 [lb] pin DL= 20,000 [lb] φc = 0.85 L L= 14 [ft] A) W8x18 B) W8x24 C) W10x26 D) W10x33 fixed K * L we want to choose the beam with the lightest linear weight, --- ≤ 200 r P

P E = 2.9 * 107 [lb/in2] fy = 60,000 [lb/in2] linear LL = 80,000 [lb] pin weight DL= 20,000 [lb] [lb/ft] φc = 0.85 L L= 14 [ft] A) W8x18 B) W8x24 C) W10x26 D) W10x33 fixed K * L which will still satisfy the strength requirement, and, the --- ≤ 200 r P

P strength Σ γ * Q ≤ φc * Rn requirement LL = 80,000 [lb] linear pin weight DL= 20,000 [lb] [lb/ft] φc = 0.85 L L= 14 [ft] A) W8x18 B) W8x24 C) W10x26 D) W10x33 fixed K * L slenderness requirement. [pause] We’ll first look at the slenderness requirement, --- ≤ 200 r slenderness requirement P

P strength Σ γ * Q ≤ φc * Rn requirement LL = 80,000 [lb] pin DL= 20,000 [lb] φc = 0.85 L L= 14 [ft] slenderness requirement fixed K * L where the radius of gyration of the beam, r, must be at least --- ≤ 200 r K * L r ≥ P 200

P strength Σ γ * Q ≤ φc * Rn requirement LL = 80,000 [lb] pin DL= 20,000 [lb] φc = 0.85 L L= 14 [ft] slenderness requirement fixed K * L the effective length of the beam, divided by The effective length equals, --- ≤ 200 r effective length K * L r ≥ P 200 radius of gyration

P strength Σ γ * Q ≤ φc * Rn requirement LL = 80,000 [lb] pin DL= 20,000 [lb] φc = 0.85 L L= 14 [ft] slenderness requirement fixed K * L the length, L, which is 14 feet, times the effective length factor, --- ≤ 200 r length K * L r ≥ P 200 radius of gyration

P strength Σ γ * Q ≤ φc * Rn requirement LL = 80,000 [lb] pin fixed-pin DL= 20,000 [lb] connection φc = 0.85 L K = 0.80 L= 14 [ft] effective fixed K, which equals, 0.80, for a beam having a fixed-pin connection. After converting the length to inches, --- slenderness length requirement factor length K * L r ≥ P 200 radius of gyration

P strength Σ γ * Q ≤ φc * Rn requirement LL = 80,000 [lb] pin fixed-pin DL= 20,000 [lb] connection φc = 0.85 L 12[in/ft] * K = 0.80 L= 14 [ft] 168 [in] effective fixed the radius of gyration of the beam, much be at least, --- length slenderness factor requirement length K * L r ≥ P 200 radius of gyration

P strength Σ γ * Q ≤ φc * Rn requirement LL = 80,000 [lb] pin fixed-pin DL= 20,000 [lb] connection φc = 0.85 L 12[in/ft] * K = 0.80 L= 14 [ft] 168 [in] effective fixed 0.672 inches. [pause] We’ll return to this slenerness requirement later in the problem. But next, --- length slenderness factor requirement K * L r ≥ r ≥ [in] P 200

P strength Σ γ * Q ≤ φc * Rn requirement LL = 80,000 [lb] pin fixed-pin DL= 20,000 [lb] connection φc = 0.85 L 12[in/ft] * K = 0.80 L= 14 [ft] 168 [in] effective fixed we’ll look at the strength requirement, which states, --- length slenderness factor requirement K * L r ≥ r ≥ [in] P 200

P strength Σ γ * Q ≤ φc * Rn requirement sum load overload factor L the sum of the product of the overload factor and loads, shall not exceed, --- LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P nominal Σ γ * Q ≤ φc * Rn resistance sum load resistance factor overload factor L the resistance factor, times the nominal resistance, of the steel member. We can also describe the --- LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P nominal Σ γ * Q ≤ φc * Rn resistance resistance load factor demand L left hand side as the load demand, and the right hand side, as, --- LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn load design demand strength L the design strength. [pause] The load demand equals, --- LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn load demand design strength various load L Σ γ * Q = max combination equations the largest load combination resulting from various load combination equations. The equations --- LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn load demand design strength L 1.4 * DL Σ γ * Q = max 1.2 * DL+1.6 * LL we need for dead load and live loads are shown here. After plugging in --- LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn load demand design strength L 1.4 * DL Σ γ * Q = max 1.2 * DL+1.6 * LL the given live load and dead load, the load demand for this problem, --- LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn load demand design strength L 1.4 * DL Σ γ * Q = max 1.2 * DL+1.6 * LL Σ γ * Q = max (28,000 [lb], 152,000 [lb]) equals, 152,000 pounds. [pause] The design strength of the I beam, --- LL = 80,000 [lb] 152,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn load design demand strength L φc * Rn = φc * Fcr * Ag Σ γ * Q = max (28,000 [lb], 152,000 [lb]) is a function of the critical stress, F cr, and the gross area, --- LL = 80,000 [lb] 152,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn 152,000 [lb] critical stress φc * Rn = φc * Fcr * Ag gross area L A g. [pause] After plugging in the given LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn 152,000 [lb] critical stress φc * Rn = φc * Fcr * Ag gross area L resistance factor for a column, we can rewrite our strength requirement, --- LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn 0.85 * Fcr*Ag 152,000 [lb] 152,000 [lb] ≤ 0.85 * Fcr*Ag L and isolate the critical stress, where the critical stress --- LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn 0.85 * Fcr*Ag 152,000 [lb] 152,000 [lb] ≤ 0.85 * Fcr*Ag L 1.788*105 [lb] Fcr≥ Ag must be at least 1.788*105 pounds, divided by --- LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn 0.85 * Fcr*Ag 152,000 [lb] 152,000 [lb] ≤ 0.85 * Fcr*Ag L 1.788*105 [lb] strength Fcr≥ Ag requirement the gross are of the beam, A g. [pause] Now we’ll return ---- LL = 80,000 [lb] DL= 20,000 [lb] P φc = 0.85

P Σ γ * Q ≤ φc * Rn 0.85 * Fcr*Ag 152,000 [lb] 152,000 [lb] ≤ 0.85 * Fcr*Ag L 1.788*105 [lb] strength Fcr≥ Ag requirement to our 4 possible I beams, and create a table, --- A) W8x18 B) W8x24 C) W10x26 D) W10x33 P

r ≥ [in] Fcr ≥ *105 [lb] / Ag Fcr [lb/in2] λc beam ry [in] required actual W8x18 where each beam has a row, and the columns will help us check --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength requirement slenderness requirement Fcr [lb/in2] λc beam ry [in] required actual W8x18 the slenderness requirement and the strength requirement. Since we’re asked to find --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength requirement slenderness requirement Fcr [lb/in2] λc beam ry [in] required actual W8x18 the lightest I beam, we’ll start with beam W --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength requirement slenderness requirement Fcr [lb/in2] λc beam ry [in] required actual W8x18 8 by 18, and if this beam doesn’t pass the requirements, we’ll check ---- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength requirement slenderness requirement Fcr [lb/in2] λc beam ry [in] required actual W8x18 the next lightest beam, W 8 by 24, and so on, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength requirement slenderness requirement Fcr [lb/in2] λc beam ry [in] required actual W8x18 until we find a beam which meets --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength requirement slenderness requirement Fcr [lb/in2] λc beam ry [in] required actual W8x18 both requirements. For beam W 8 by --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength requirement slenderness requirement Fcr [lb/in2] λc beam ry [in] required actual W8x18 18, the radius of gyration, r, equals, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] y Fcr ≥ *105 [lb] / Ag slenderness x requirement λc beam ry [in] rx= [in] W8x18 3.430 inches about the x axis, and inches about the y axis. Since we’re checking for --- 1.231 1.581 ry= [in] W8x24 1.607 1.211 W10x26 W10x33

r ≥ [in] y Fcr ≥ *105 [lb] / Ag slenderness x requirement λc beam ry [in] rx= [in] W8x18 a minimum, r, value, we’ll always use the smaller value, which will be --- 1.231 1.581 ry= [in] W8x24 1.607 1.211 W10x26 W10x33

r ≥ [in] y Fcr ≥ *105 [lb] / Ag slenderness x requirement λc beam ry [in] rx= [in] W8x18 the radius of gyration about the y axis, r y. Since --- 1.231 1.581 ry= [in] W8x24 1.607 1.211 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag slenderness requirement Fcr [lb/in2] λc beam ry [in] required actual W8x18 1.231 inches is greater than inches, the slenderness requirement for beam ---- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag slenderness requirement Fcr [lb/in2] λc beam ry [in] required actual W8x18 W 8 by 18, is met. Next we’ll check --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength requirement slenderness requirement Fcr [lb/in2] λc beam ry [in] required actual W8x18 the strength requirement for this I beam. We can look up the gross area, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength requirement Ag= 5.26 [in2] Fcr [lb/in2] λc beam ry [in] required actual W8x18 which equals, 5.26 inches squared. If we plug this area into our strength requirement, for A g, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength requirement Ag= 5.26 [in2] Fcr [lb/in2] λc beam ry [in] required actual W8x18 the critical stress of the beam must be at least --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength Fcr ≥ 33,992 [lb/in2] requirement Ag= 5.26 [in2] Fcr [lb/in2] λc beam ry [in] required actual W8x18 33,992 pounds per square inch. [pause] To compute the actual, critical stress, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength Fcr ≥ 33,992 [lb/in2] requirement Ag= 5.26 [in2] Fcr [lb/in2] λc beam ry [in] required actual W8x18 ? provided by the beam, we’ll first need to compute the unitless variable, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag strength Fcr ≥ 33,992 [lb/in2] requirement Fcr [lb/in2] λc beam ry [in] required actual W8x18 ? ? lambda c. Because, if lambda c is less than 1.5, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

if λc < 1.5 then λc 2 strength Fcr=0.658 fy * requirement if λc ≥ 1.5 then Fcr [lb/in2] fy Fcr=0.877 * required actual λc 2 33,992 21,052 ? we’ll calculate the critical stress using the top equation, and if lambda c is at least 1.5, --- 25,254 32,477

if λc < 1.5 then λc 2 strength Fcr=0.658 fy * requirement if λc ≥ 1.5 then Fcr [lb/in2] fy Fcr=0.877 * required actual λc 2 33,992 21,052 ? we’ll calculate the critical stress using a second equation. This threshold of 1.5 distiguishes between --- 25,254 32,477

if λc < 1.5 then inelastic buckling λc 2 strength Fcr=0.658 fy * requirement if λc ≥ 1.5 then Fcr [lb/in2] fy Fcr=0.877 * required actual λc 2 the inelastic buckling range and the elastic --- 33,992 21,052 ? 25,254 32,477

if λc < 1.5 then inelastic buckling λc 2 Fcr=0.658 fy * elastic buckling if λc ≥ 1.5 then Fcr [lb/in2] fy Fcr=0.877 * required actual λc 2 buckling range. [pause] Lambda c, equals, --- 33,992 21,052 ? 25,254 32,477

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 * λc 2 K * L fy λc = Fcr [lb/in2] * r * π E required actual 33,992 21,052 ? K L over r Pi, times root f y over E. In this equation, we already determined --- 25,254 32,477

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 * λc 2 K * L fy λc = Fcr [lb/in2] * r * π E required actual K = 0.80 the effective length factor, K, and the length, L, and the problem statement provided the --- 33,992 21,052 ? L=168 [in] 25,254 32,477

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 * λc 2 K * L fy fy= 60,000 [lb/in2] λc = * r * π E E = 2.9 * 107 [lb/in2] K = 0.80 the yield stress, f y, and the modulus of elasticity, E. If we simplify this equation, --- L=168 [in]

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 * λc 2 K * L fy fy= 60,000 [lb/in2] λc = * r * π E E = 2.9 * 107 [lb/in2] K = 0.80 we find that lambda c, is a function of, --- L=168 [in]

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 * λc 2 K * L fy fy= 60,000 [lb/in2] λc = * r * π E E = 2.9 * 107 [lb/in2] K = 0.80 the radius of gyration, r. If we plug in our value of --- L=168 [in] λc = [in] / r

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 * λc 2 K * L fy fy= 60,000 [lb/in2] λc = * r * π E E = 2.9 * 107 [lb/in2] K = 0.80 1.231 inches, for r, we compute lambda c, as, --- ry = [in] L=168 [in] λc = [in] / r W8x18

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 * λc 2 K * L fy fy= 60,000 [lb/in2] λc = * r * π E E = 2.9 * 107 [lb/in2] K = 0.80 [pause] is greater than 1.5, therefore, --- ry = [in] L=168 [in] λc = [in] / r W8x18 λc =1.581

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 * λc 2 K * L fy fy= 60,000 [lb/in2] λc = * r * π E E = 2.9 * 107 [lb/in2] K = 0.80 we’ll use the second equation to compute the critical stress, f cr. [pause] Plugging in the --- ry = [in] L=168 [in] λc = [in] / r W8x18 λc =1.581

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 * λc 2 fy= 60,000 [lb/in2] λc =1.581 Fcr [lb/in2] λc beam ry [in] required actual W8x18 ? yield stress and lambda c, we find the critical stress in the beam equals, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 =21,052 * λc 2 [lb/in2] fy= 60,000 [lb/in2] λc =1.581 Fcr [lb/in2] λc beam ry [in] required actual W8x18 21,052 pounds per square inch. Which turns out to be --- 1.231 1.581 33,992 21,052 ? W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 =21,052 * λc 2 [lb/in2] fy= 60,000 [lb/in2] λc =1.581 Fcr [lb/in2] λc beam ry [in] required actual W8x18 > less than the minimum required critical stress, therefore, beam W 8 by 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 =21,052 * λc 2 [lb/in2] fy= 60,000 [lb/in2] λc =1.581 Fcr [lb/in2] λc beam ry [in] required actual W8x18 does not meet the strength requirement. [pause] Next, we’ll evaluate --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

λc 2 if λc < 1.5 then Fcr=0.658 fy * fy if λc ≥ 1.5 then Fcr=0.877 =21,052 * λc 2 [lb/in2] fy= 60,000 [lb/in2] λc =1.581 Fcr [lb/in2] λc beam ry [in] required actual W8x18 the next lightest beam, W 8 by 24. [pause] The radius of gyration, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag Fcr [lb/in2] λc beam ry [in] required actual W8x18 about the y axis, is, inches, which means this I beam also satisfies, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag Fcr [lb/in2] λc beam ry [in] required actual W8x18 the slenderness requirement. [pause] The gross area of this beam equals, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag Ag= 7.08 [in2] Fcr [lb/in2] λc beam ry [in] required actual W8x18 7.08 inches squared, which makes the minimum critical stress in the I beam, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

r ≥ [in] Fcr ≥ *105 [lb] / Ag Fcr≥ 25,254 [lb/in2] Ag= 7.08 [in2] Fcr [lb/in2] λc beam ry [in] required actual W8x18 25,254 pounds per square inch. [pause] Next we’ll calculate lambda c, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

λc = [in] / r Fcr [lb/in2] λc beam ry [in] required actual W8x18 which equals inches divided by inches, or --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

λc = [in] / r λc =1.211 Fcr [lb/in2] λc beam ry [in] required actual W8x18 [pause] Since lambda c is less than 1.5, we’ll use --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

λc = [in] / r λc =1.211 if λc < 1.5 then λc 2 Fcr=0.658 fy * Fcr [lb/in2] λc beam ry [in] required actual W8x18 the inelastic bucking equation to find the critical stress, f cr. Which equals, --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

λc = [in] / r fy= 60,000 [lb/in2] λc =1.211 if λc < 1.5 then Fcr= 32,477 [lb/in2] λc 2 Fcr=0.658 fy * Fcr [lb/in2] λc beam ry [in] required actual W8x18 32,477 pounds per square inch, which is greater than the --- 1.231 1.581 33,992 21,052 W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

λc = [in] / r fy= 60,000 [lb/in2] λc =1.211 if λc < 1.5 then Fcr= 32,477 [lb/in2] λc 2 Fcr=0.658 fy * Fcr [lb/in2] λc beam ry [in] required actual W8x18 minimum required value, and the strength requirement, is met. Which means beam W 8 by --- 1.231 1.581 33,992 21,052 < W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

λc = [in] / r fy= 60,000 [lb/in2] λc =1.211 if λc < 1.5 then Fcr= 32,477 [lb/in2] λc 2 Fcr=0.658 fy * Fcr [lb/in2] λc beam ry [in] required actual W8x18 24 is the lightest beam which satisifies both requirements. [pause] 1.231 1.581 33,992 21,052 < W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

A) W8x18 B) W8x24 C) W10x26 D) W10x33 Fcr= 32,477 [lb/in2] Fcr [lb/in2] λc beam ry [in] required actual W8x18 When reviewing the possible solutions, --- 1.231 1.581 33,992 21,052 < W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

A) W8x18 B) W8x24 C) W10x26 D) W10x33 answerB Fcr= 32,477 [lb/in2] Fcr [lb/in2] λc beam ry [in] required actual W8x18 the answer is B. 1.231 1.581 33,992 21,052 < W8x24 1.607 1.211 25,254 32,477 W10x26 W10x33

Find: The Lightest I Beam

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