Question

Problem 1 Continuous Beams and One-Way Slabs (50 pts.) This is an Analysis Only Problem: a) For column line 2, use the ACI moment coefficients discussed in class to determine the maximum positive and negative factored moments at the support faces for columns A2 and B2, and at the midspans of an exterior span and the interior span. Also, determine the shears and plot the M and V diagrams. (25pts) b) For the one-way slab strip, use the ACI moment coefficients to determine the factored design moments at all of the points, "a" through "i", indicated in Fig. 1.1 below. Also, determine the shears and plot the M and V diagrams. (25pts) Tool 12 in. x 24 in. (all beams) 16 in x 16 in (all columns) Slab thickness - 6 in. SDL - 20 pst LL - 50 pst Figure 1.1 Floor Plan

Answer 1

Answer: Given that For Column line 2, use the Aco moment Coefficients discussed in class to determine the manimum positive and negative factored moment at the support tacts for Calamnya2 and B2 Doubling slaba ) selt weight = 150 x = 150X0.5-75 psf 2) - SDL 20psf 95PS 3) 22 = 50 psf wl = 1.2x95 +1.6 X 50 -1140+80 wl = 194 psf WI= 0.194 - 0.2 ksf a) The beam along column line 2 load in beam = 0.2% (6+55) + 150X72x24 144 X1000 -2.3403

*2.6 kift Column size -16"x16" Section Span beam = 30' end span as melempan where Ini= End span = 30 -1 = 30.1.33 [Ini = 2867) Interior span 2n2= 25-16 = 25-1.33 (Ing= 23.667) 12 End span The ACI moments Az = G) woln, a 16 = 2.6*(28.0 +)? = 2.68821.96 16 = 2137.119

Ag = 133. 569 133. 57 kft A2 = (-) 133.57 kft Mia Span = + Wulina = 2.8(28.67)2 14 = 2137.119 14 Mid span : 152.65 kft Then The supporyt foces for beam 2= (-) Wally subi the valuy in above egu =() 2.6% (28.67) = 2,68821.96 (0 10 - 21371119 = 213.711 k ft Interior span : 202=83-67 The support B=(-) Wulna = (-) 2.6 X (23.67)2 1

=6) 2.6X560.26 = (-) 1456 676 B - (-) 132.495 kft mid span= (+) ulna 26) 26(23.67)2 16 26X 560.26 = 16 - () 1456676 16 - (+) 91.04K47 The support (= (-) Woln ? The support (= 6) 2.6 X (23.67)? = (-) 2,6560.25 = 691456. 676 The support <= (-)132.42 kft The shear at suppost = wolna = 2.68 23.67 2 = 61.542 - 30.7718H.

M-Diagram (kH : 91kft 152.65 , 152.65KH 133.57 v-Diagrao kips) 30 hk. 42-86K, 37.276 30.776 37.276. b) For the one-way slab Strip use the ACI moment colsticents. Ini - 12-1=11' 2n2 = 11-1 = 10 End Span (12') 20= 11':- Ma> (-) Wulni? = -) 0.2 x (1) =62412 Ma = (-) 1.008 kat Mb = (-) wuln? =(-) 0.2(132 - 24 24 10

10 • (-3242 - (-)2.4.2 . The mid Point= (+) Walne 14 14 = 67) 0,28 (1)2 -ft) 042 = () 1.728 KW the six a) Wuly - 0.25 F2 .2.2 zl.lk b) wulu x 115= 1.181,15 1. = 1.27K. The interion span Ini =11':- Me= (-) w!», 2. Me= 7T = 6) 0.2 X112 3(-)-24:2 Me = (-) 2.2 Ma = (-) wouln² act) 0.2 x 1² = 67 242 =-2.2

1 76 Mid Point = (1) Wulm? = (+) 0.2X112 = (+) 24.2 16 - (1) 1.51 kft Sop = Wnln = 0.2X! 2,2 -1.lk, The Interior span 212 210' (e, f) (3,1);- Me = (-) Wolne ² M 3(-) 0.2X110)2 - (-) 20 =(-) 1.81kft Me = (-) Wolne? M = (-) 0.2x102 = (-) 20 (-) 1.82 KHC Mid Points w) wulp2 = 643 027102 - ( 1.95kfc The shear force = 0.2410 =1k. g&R are same as lng=10'

M-Diagram (kfe) 1-7KFC 1.51 1.25 1.25 B et o 2.2 of / g 1.01 dite 2.2 2.12 -1.62 2.4-2 lizy 1.82 v-Diagscenu(kips) :- 1.1K IR IT 122* TIKY 2m2 -10' 1n2 =10 291-11