Design the section below as a doubly reinforced beam for resisting the ultimate positive moment of 425 kN.m. Use the given data. f = 21 MPa, fy = 420 MPa Assume d' = 85 mm. ● ♦ ● Use p25 mm for the tension reinforcement. For compression reinforcement, use only one bar.
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- W3D 25 Y= 155 %31 1oow= 2500 KN 27=310mm X= 475 Z= 60 %3D 102=600mm... Determine the nominal axial compressive strength, Pn for the following cases: a. L= 15 ft b. L = 20 ft Note: Pinned both end support W10 × 33 A992 steel r = 1.94 in E = 29000 ksi Fy = 50ksi L a) Blank 1 b) Blank 2 Blank 1 Add your answer Blank 2 Add your answerPERFORMED AS REQUIRED. SHOW YOUR COMPLETE SOLUTION. 4.3-1 Use AISC Equation E3-2 or E3-3 and determine the nominal axial compressive strength for the following cases: a. L = 10 ft b. L=30 ft L W10 x 100 A992 steel FIGURE P4.3-1
- Use 2015 NSCP. A compression member shown below with Fy = 50 ksi 30′ W12 x 87 A992 steel 1. Which of the following most nearly gives the value of critical buckling stress in ksi? 2. Which of the following most nearly gives the nominal strength in kips? 3. Considering ASD. Which of the following most nearly gives the allowable strength in kips? 4. Considering LRFD. Which of the following most nearly gives the design strength in kips?Bearing P plate Let the allowable stresses in the post be ost=120MPaand oco=6MPA. Compute the maximum safe axial load P that may be applied. The moduli of elasticity are 200 GPafor steel and 14 GPafor a concrete. Each steel bar has cross-sectional area of 900mm2. Steel - Concrete - 300 mm 300 mmDesign the reinforcements of the given T beam below. bf=900mm bw=420mm tf-120mm d=550mm d'=80mm fc'=34MPa fy=415MPa USE NSCP 2015 A.Mu = 1300kN-m, As = mm2 B.Mu = 1800kN-m, As = mm2, As' = mm2
- Answer the following for the section at Point D Only Calculate the distributed load "w" that: Will cause the section crack Will cause the reinforcement to yield. Material Properties: F'c = 5000 psi Fy = 60000 psi Es = 29000000 psi Ln = 27 ft L wl₂² 16 wl,² 14 CD L wl,2 vl₁² 10 11 win² 16 h: 28 in A=4 in² b=14 in n d: 25 inDetermine the nominal axial compressive strength, Pn for the following cases: a. L = 15 ft b. L = 20 ft Note: Pinned both end support W10× 33 r = 1.94 in E = 29000 ksi Fy = 50ksi A992 steel %3D Note: Ag = 9.71 in^2 a) Blank 1 b) Blank 2 Blank 1 Add your answer Blank 2 Add your answerH.W A Solid Steel bar of dlameter 6omm and length 350 mm is Placed inside an aluminum cylinder öf Inside diameter Femm and out side d lameter Ilomm, a comPressive load of looo lais applied on the assembly, find the stresses in the bar and the cylinder ? take Est = 200GPa and EAl = F0GPan %3D Cover Plate 0.25mm d- do 35omm A lum. cylinder Steel bar
- A traffic light signal pole is subjected to the weight of each traffic signal 45 lbs and the weight of the road lamp 55 lbs . The weight of lateral arms is included in the signal and lamp weights. The people is fixed and the base and is made of round structural steel section HSS 14.000 x 0.25. The details of the pole section may be obtained form the Manual of American society of steel Construction (AISC Steel Manual). Analyze the principle normal stresses and the maximum shear stresses on elements A and B located at the base in the pole as shown in the sketch below.4.9-7 Two plates %16 x 10 are welded to a W10 x 49 to form a built-up shape, as shown in Figure P4.9-7. Assume that the components are connected so that the cross section is fully effective. F, = 50 ksi, and K,L = KL = 25 ft. FIGURE P4.9-7 Problems 181 a. Compute the nominal axial compressive strength based on flexural buck- ling (do not consider torsional bucking). b. What is the percentage increase in strength from the unreinforced W10 x 49?4- The reduction factor (Ø) is equal to 0.9 if the: Es 2 Ety +0.003 Es S Ety +0.003 Es S Ety 5- The design method which use the factored loads is Working stress method Ultimate design Method - None of above 6- If p< Pb therefore the section is - Under reinforced - Over reinforced - None of above 7- For any given section, if c= 150mm , d = 450mm, the ɛs is equal to - 0.003 - 0.006 None of above 8- If area of steel is 2455 mm² and Ø25 mm the number of bars are -(4) - (5) - (6)