i need the answer for question 2 and when i post this question again thats mean

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i need the answer for question 2 and when i post this question again thats mean the answer that u give me before it is wrong so please do not send same answer and read the question carfuly ... thanks

Problem 1) Using tabulated information from the AISC Manual, find My and Mp (A992, Gr. 50 steel) for strong axis bending of a a) W18X3:5 b) W8X35 (note that both sections have the same amount of steel and weight) c) Are these sections "compact" for local flange and local web buckling considerations? Show calculations to justify your answer. If not compact for FLB calculate the moment capacity of the section. If not compact for WLB just state this with no further calculations Problem 2) a) Use hand calculations to find My and Mp for strong axis bending of the W18X35 section in Problem #1 (provide hand calculations for Sx and Z). Ignore all fillet dimensions (assume the shape is made up of rectangles as shown below) Re-work part a, but assume that the bottom flange is removed to result in a t-shape beam as shown below. Note that this section is no longer symmetric. b) bf bf d-tf A) B)

Explanation / Answer

a) bf = 6in

tf = 0.425in

tw=0.3in

d = 17.7 in

Due to symmetry, centroid of section is at mid-depth

moment of inertia of section about centroid = 2*[(6*0.4253/12)+(6*0.425*8.63752)] + (0.3*16.853/12)=500.17in4

elastic section modulus,S = 500.17/(17.7/2)=56.52 in3

My = 50*56.52/12=235.5 kip-ft

Plastic section modulus of section about neutral axis = Z=2*[(6*0.425*8.6375)+(0.3*8.425*4.2125)]=65.34in3

Mp = 50*65.34/12=272.3 kip-ft

b) Let us determine the depth of neutral axis from the topmost face of the section

Depth of elastic neutral axis = [(6*0.425*0.2125)+(16.85*0.3*8.85)]/[(6*0.425)+(16.85*0.3)] = 5.95in

moment of inertia of the section about neutral axis = (6*0.4253/12)+(6*0.425*5.73752)+(0.3*16.853/12)+(0.3*16.85*2.92)=246.1in4

Elastic section modulus,S=246.1/(11.325)=21.73 in3

My=50*21.73/12=90.54 kip-ft

Since the section is not symmetric,the plastic neutral axis will not be at centroid

Let depth of plastic neutral axis be d from bottom of flange

6*0.425+0.3d = 0.3(16.85-d)

d=4.175 in

Depth of plastic neutral axis from topmost face=4.175+0.425=4.6in

Plastic section modulus about plastic neutral axis = 6*0.425*(4.6-0.2125)+(0.3*4.175*2.0875)+(0.3*12.675*6.3375)=37.9 in3

plastic section capapcity = 50*37.9/12=157.9 kip-ft

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