Given data is
a) Wind load(WL) = 28lb/ft^2 = 28*10*12, b) Live load(LL) = 5000lb, c) Wall load(DL) = 120*10*12 ( since density of =3360lb =14400lb brick=120 lb/ft^2)
1) Governing ASD factored load combination equation = DL+0.75LL+0.75WL
2)Governing factored design load (k) = 14400 + 0.75*5000 + 0.75*3360 = 20670 lb
2. Calculate the factored design load (ASD) for 10 ft by 12 ft tributary area of...
CVE 313- 1- Design a square footing for the interior column shown in the figure. The column carries 2 k 't of uniform live load and 1.5 k/t of uniform dead load (including the self weiht of the structure), the base of the footing is 5 ft below grade, the soil weight is 100 lb/ft', and q," 5000 psf. and concrete unit weight is 150 lb/ (25 points) NOTE: Your design should only include the following (one-way shear, two-way shear,...
Homework Problem Solving TOPIC: Review of ASD and Load Resistance Factor Design A 2x6, #1 Grade, DF-L wood joists having a moisture content of 11% with a span of 13.5 ft. is shown below. The only combined load to be considered is (DL + RLL) of 34#/ft. DL-14 PSF & RLL 20 PSF with 12" O.C. Evaluate: Will the 2x6 section work for the given load? Show ASD and LRFD design method for Bending Only!! 34 13.5 Load: Wp 14...
Based on tributary load analysis, the dead and live loads, wd and wų, respectively, acting on a beam in a vertical load resisting system are shown below. The concrete is normalweight with compressive strength f=5000 psi. Note that the given dead load includes the self-weight of the beam and slab. Section wp=1.2 kip/ft, wu=1.5 kip/ft be-45 inch 5 inch 1 30 inch B A 10 inch In=10 ft 1. Design and detail the beam for positive flexure at section A....
2. (30 point) w 1.2 for dead load, and 1.6t is used the total factored lood and the reactions that a d 8 ft For the dead load, assume concrete and a slab that used to tors of l a thst ct o and as in a light storage the 4-inch con16or live load, determine and sketch r-dead load, ang 5 and ABC. The loads should be The steel framework is used to support warehouse. Using the load factors of...
A floor beam supports the following loads. Determine the load diagrams for the various (7) loac combinations DL= 1.15k/ft LL- 1.85 k/ft (horizontal) = 15k Eh-20 k Ev=.3k/ft CE Load Combinations Factored Loads from ASCE 7one in terms of pla (Strength Design Method) Ldesign 1.4D Ldesign 1.2D +1.6 L + 0.5 (Lroof or Snow) Ldesign 1.2D +1.6 (Lroof or Snow) + (.5L or .8 Wind) Ldesign = 1.2D + 1.6 Wind +0.5 L+ 1 2. 3. .5(Lroof or Snow) 4....
steel design
1. (20 points) A roof system with 16 x 40)sections spaced 10 ft on center is to be used to support a dead load O(30 psf (includling selfawoight-of_W_16x 40 sections); a roof live, snow, or rain load of 25 psf; and a wind load of 36 psf. Compute the factored load per linear foot for cach load combination (LRED from LCI to LC5). D 30 puf L,IR/S 25psf D-(lo t30 f)300/ft We t 36 L.IRIS(1G H 25 pif)....
The single-story unbraced frame shown below is subjected to dead load, roof live load, and wind load Figure 1 shows the results of a first-order analysis relative to the columns of the frame. The axial load and end moment (also equal to the maximum moment in the column) are given separately for the different load cases (i.e., dead load, roof live load, and lateral wind load). All vertical loads are symmetrically placed and contribute only to the Mnt moments (i.e.,...
An elevation of a concrete frame is shown below. A superimposed dead load of 200 lb/ft and a live load of 600 lb/ft are to be supported in addition to the beam self-weight. The beam's cross- section is shown as well. The weight density of reinforced concrete is 150 lbs/eu. ft. Use 1.2D+1.6L as your load combination. Use ACI moment coefficients and statics, as appropriate, to provide the Mu values for points A, B, C, and D. 48 5" 15"...
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Problem 2 Using the set of LRFD Load Combinations provided on the class sheet determine the controlling factored design strength, Mu, considering the following unfactored (actual) dead, roof live and wind loads. Find Mo. Mu and Mw considering each of the beam set-ups on their own and then determine the controlling factored combination of the loads, and therefore the moment carried by the beam. These loads are all acting on the same beam, but...
The T-beam shown in Figure 1 supports the un-factored dead load
of 1.4 kips/ft and live load of 1.5 kips/ft. The dead load does not
include the self-weight of the beam. The material properties are as
follows: fc’=3000 psi; fy=60,000 psi. Design the shear
reinforcement (stirrups). Plot the stirrups distribution along the
span of the beam.
DL= 1.4 kips/ft ; L2=1.5 kips/Ft * 75 Sz=7 X * b=3616. hr-6in k ) انا امه hw-lain + * bw=12 in