Beam and Girder Design Calculations Using LRFD Methods
School
University of Dayton**We aren't endorsed by this school
Course
CEE 411
Subject
Mechanical Engineering
Date
Dec 10, 2024
Pages
3
Uploaded by GrandFlowerJackal15
Problem 9-6 Neglect beam self-weight—CK after selection Problem 9-4 LRED w, = 1.2(0.75)+ 1.6¢2.0) = 4. 10 k/ft P =16(15)=24k 133P =32k LRFD e Neglect beam self-weight—CK after selection 24k w =1.2(1.0) = 1.20 k/ft o % R 1 p 7?{@..::1 | ’;"'I P, =1.6(16)+25.6k —~=12.8k U S S i 2 [q132] : 4361 M, - "2(’;'4) £25.6(7)+12.8(14) = 476 fi-k l\ From AISC Table 3-2 v, W24x55, ¢M, =503 fik " w, =1.2(1.055) = 1.266 k/ft 2 M, = 1'2662(14) +25.6(7)+12.8(14) = 482.5 fi-k M, <gM, =503 ft-k OK - Use W24 x 55| LRFD M, ... =2178ft-k WIS x 35 gM_ =249 fi-k (w/self-weight) w, =1.2(0.785)+1.6(2.0) =414 k'ft M_ . =2185f-k < ¢M, =249 fi-k Use W18 x 35| LRFD
LRFD w, =1.2(0.50)+1.6(0.64) = 1.624 k/ft =182.7 ft-k From AISC Table 3-2 uw (a) Design of beams Neglect beam self-weight—CK after selection WI16x31, M =203 fik Concrete slab = %(l 50)(8) = 500 Ib/ft = w,, w, =1.2(0.53) +1.6(0.64) = 1.66 k/ft Live Load =80(8) = 640 Ib/ft = w, M = 1.66(30) i <M, =203 fik OK =186.8 ft-k Beam, W16x31-LRFD (b) Design of girders LRFD P = 1.66(370)(2) =49.80 k M, =49.80(8) =398.4 fi-k W21x50, gM =413 ft-k w, =1.2(0.05) = 0.06 k/ft M, = 0‘06;24)“ +49.80(8) M. =402.7 ft-k <¢M_=413fik OK Girder, W21x50 LRFD
Problem 9-10 - 10" ~ 3" J A= 10(1]+8[1J+9(l) =13.5in’ 2 2 2 Plastic N.A. located an area % =6.75in" From top or bottom 13.5(490) self-weight = =45.9 Ib/ft A36 Steel Z =5(3.75)+1 .75(?} + 2.75(5—:)+ 4(5.75) =52.38in’ _F,Z 36(52.38) ) =157.14 fi-k 12 M LRFD oM =0.9(157.14) = 141.4 f-k ACUNSYY w, = 2.83 k/ft 1.2(0.80+0.046) +1.6(w, ) = 2.83 w, =1.13 k/ft| LRFD