Find: Fcr [lb/in2] Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] fixed-fixed connection W 10 x 22 L=15 [ft] A) 24,000 B) 28,000 C) 32,000 D) 36,000 Find the critical stress, in pounds per square inch. [pause] In this problem, ---- Fcr
Find: Fcr [lb/in2] Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] fixed-fixed connection W 10 x 22 L=15 [ft] A) 24,000 B) 28,000 C) 32,000 D) 36,000 a 15 foot long I beam is connected in a fixed-fixed orientation. The Young’s Modulus and --- Fcr
Find: Fcr [lb/in2] Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] fixed-fixed connection W 10 x 22 L=15 [ft] A) 24,000 B) 28,000 C) 32,000 D) 36,000 yield stress of the steel are provided. [pause] The critical stress of a steel member in compression --- Fcr
Find: Fcr [lb/in2] ≤ 4.71 > 4.71 Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] (1) Fcr=0.658 fy/fe * fy K * L E ≤ 4.71 if * L=15 [ft] r fy (2) Fcr=0.877 * fe is computed using 1 of 2 equations. The particular equation to use, depends on, the terms --- K * L E > 4.71 if * W10x22 r fy Fcr
Find: Fcr [lb/in2] ≤ 4.71 > 4.71 Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] (1) Fcr=0.658 fy/fe * fy K * L E ≤ 4.71 if * L=15 [ft] r fy (2) Fcr=0.877 * fe K L over r and 4.71 times root E over f y. Therefore, we’ll first evaluate these two terms. The former term, --- K * L E > 4.71 if * W10x22 r fy Fcr
Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy L=15 [ft] K * L r K L over r, is the effective length factor, K, --- W10x22 Fcr
Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy L=15 [ft] length K * L r times the length of the member, L, divided by the radius of gyration, --- effective W10x22 length Fcr factor
Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy L=15 [ft] length K * L r radius of r. [pause] This term, as a whole, --- gyration effective W10x22 length Fcr factor
Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy L=15 [ft] length K * L r radius of is referred to as the slenderness ratio. [pause] From the problem statement, --- gyration effective W10x22 slenderness length Fcr ratio factor
Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy L=15 [ft] length K * L r radius of we know the length is 15 feet, which is the same as, --- gyration effective W10x22 slenderness length Fcr ratio factor
Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy L=15 [ft] length 12[in/ft] K * L L = 180 [in] * r radius of 180 inches. [pause] The effective length factor, for fixed-fixed --- gyration effective W10x22 slenderness length Fcr ratio factor
Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * fixed r fy K=0.65 L=15 [ft] L = 180 [in] K * L fixed r radius of end connections, equals 0.65. And the radius of gyration, r, --- gyration effective W10x22 slenderness length Fcr ratio factor
Find: Fcr [lb/in2] ≤ or > 4.71 Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * fixed r fy K=0.65 L=15 [ft] L = 180 [in] K * L fixed r radius of equals the smaller of the radii of gyration, in the x and y directions. gyration r = min (rx, ry) W10x22 Fcr
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ K=0.65 L=15 [ft] L = 180 [in] K * L fixed r radius of If we take a cross section of the beam, A A prime, --- gyration r = min (rx, ry) W10x22 Fcr
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ K=0.65 y L=180 [in] K * L r we can look up the radii of gyration values as being, --- W10x22 x r = min (rx, ry) section A-A’
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ K=0.65 y L=180 [in] K * L r 4.27 inches and 1.33 inches, for a W 10 by 22 I beam. Therefore, --- W10x22 x rx= 4.27 [in] r = min (rx, ry) ry= 1.33 [in] section A-A’
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ K=0.65 y L=180 [in] K * L r we’ll use 1.33 inches for the radius of gyration, r. And our slenderness ratio, --- x rx= 4.27 [in] r = min (rx, ry) ry= 1.33 [in] r = 1.33 [in] section A-A’
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 88.0 K=0.65 y L=180 [in] K * L r computes to 88.0. [pause] To solve the right hand side of the equation, --- x rx= 4.27 [in] r = min (rx, ry) ry= 1.33 [in] r = 1.33 [in] section A-A’
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 88.0 K=0.65 y L=180 [in] K * L r we can plug in the given Young’s Modulus and yield stress of the steel, and the right hand side, --- x rx= 4.27 [in] r = min (rx, ry) ry= 1.33 [in] r = 1.33 [in] section A-A’
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 88.0 K=0.65 133.7 L=180 [in] K * L r equals, 133.7. [pause] Since 88.0 is less than 133.7, ---- x rx= 4.27 [in] r = min (rx, ry) ry= 1.33 [in] r = 1.33 [in] section A-A’
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 88.0 133.7 (1) Fcr=0.658 fy/fe * fy we’ll use the first equation to compute the critical stress, F cr. [pause] In this equation, --- x section A-A’
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 88.0 133.7 (1) Fcr=0.658 fy/fe * fy we already know the yield stress, f y. [pause] And the variable f e, --- x section A-A’
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 88.0 133.7 (1) Fcr=0.658 fy/fe * fy represents the buckling stress, which equals, PI squared, --- buckling x stress section A-A’
Find: Fcr [lb/in2] π2 * E ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 88.0 133.7 (1) Fcr=0.658 fy/fe * fy times the Young’s Modulus, divided by the slenderness ratio, squared. [pause] buckling π2 * E x Fe= stress K * L 2 r section A-A’
Find: Fcr [lb/in2] π2 * E ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 88.0 133.7 (1) Fcr=0.658 fy/fe * fy Since we’ve be given the Young’s modulus, and already solved for --- buckling π2 * E x Fe= stress K * L 2 r section A-A’
Find: Fcr [lb/in2] π2 * E ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 88.0 133.7 (1) Fcr=0.658 fy/fe * fy the slenderness ratio, we compute the buckling stress as --- buckling π2 * E x Fe= stress K * L 2 r section A-A’
Find: Fcr [lb/in2] π2 * E ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 88.0 133.7 (1) Fcr=0.658 fy/fe * fy 36,960 pounds per square inch. [pause] After plugging in the yield stress, and, --- buckling π2 * E x Fe= stress K * L 2 r Fe= 36,960 [lb/in2] section A-A’
Find: Fcr [lb/in2] π2 * E ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 88.0 133.7 (1) Fcr=0.658 fy/fe * fy buckling stress, the critical stress on the steel member, equals, --- π2 * E x Fe= K * L 2 r Fe= 36,960 [lb/in2] section A-A’
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 133.7 (1) Fcr=0.658 fy/fe * fy 23,947 pounds per square inch. [pause] Fe= 36,960 [lb/in2] x Fcr = 23,947 [lb/in2] section A-A’
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 133.7 (1) Fcr=0.658 fy/fe * fy A) 24,000 B) 28,000 C) 32,000 D) 36,000 When reviewing the possible solutions, --- Fe= 36,960 [lb/in2] Fcr = 23,947 [lb/in2]
Find: Fcr [lb/in2] ≤ or > 4.71 A A’ Fcr E = 2.9 * 107 [lb/in2] fy = 36,000 [lb/in2] K * L E ≤ or > 4.71 * r fy A A’ 133.7 (1) Fcr=0.658 fy/fe * fy A) 24,000 B) 28,000 C) 32,000 D) 36,000 the answer is A. Fe= 36,960 [lb/in2] Fcr = 23,947 [lb/in2] answerA