Please help with problem 2 Given: Following Bars/Beams.Required: Construct displacement, shear, and moment diagrams for each case.Problem 1Problem 2min1.2mSizeWWx 0.6m4 x5 xX6mp=-k sin12 kN/mCarbon - Stainless - Alloy- Aluminumtexasironandmetal.com6 kN/mFor Each beam, assume an allowable stress of 440 MPa. And, assume a cross sectional area asshown below:(2πχsin (²x)L1.2mWIDE FLANGE BEAMIWide Flange Beam . H Beam(A)(B)Weight Section WebPer Ft. Depth In. Thk.13.000 4.160 0.28016.00019.0005.0100.2605.1500.270(C) (D)Flange FlangeWidth Thk.4.060 0.3755.000 0.3605.030 0.430TEXAS IRON& METALSUPPLY ON DEMAND713.672.7595800.839.4766Fax 713.672.0653texasironandmetal.com4.16 in4.16-2*0.375-3.41 in.4.060 inthickness0.375 in.thickness=0.280 in.thickness= 0.375 in.

Given data,simply supported beam is givenFINDsshear, moment and diplacement diagram

Problem 2 1.2 m Size W 4 x W 5 x x 0.6 m X For Each beam, assume an allowable stress of 440 MPa. And, assume a cross sectional area as shown below: 12 kN/m Weight Per Ft. 6 kN/m 13.000 16.000 19.000 1.2 m WIDE FLANGE BEAM Carbon - Stainless - Alloy-Aluminum texasironandmetal.com Do Wide Flange Beam . H Beam (A) (B) Section Web Depth In. Thk. 0.280 4.160 5.010 0.260 5.150 0.270 I T (C) (D) Flange Flange Width Thk. 4.060 0.375 5.000 0.360 5.030 0.430 TEXAS IRON & METAL SUPPLY ON DEMAND 713.672.7595 800.839.4766 Fax 713.672.0653 texasironandmetal.com 4.16 in = 416 2*0.375 -3.4in. O " 4.060 in thickness= 0.375 n. thickness=0.280 in. thickness = 0.375 in.

Problem 2 1.2 m Size W 4 x W 5 x x 0.6 m X For Each beam, assume an allowable stress of 440 MPa. And, assume a cross sectional area as shown below: 12 kN/m Weight Per Ft. 6 kN/m 13.000 16.000 19.000 1.2 m WIDE FLANGE BEAM Carbon - Stainless - Alloy-Aluminum texasironandmetal.com Do Wide Flange Beam . H Beam (A) (B) Section Web Depth In. Thk. 0.280 4.160 5.010 0.260 5.150 0.270 I T (C) (D) Flange Flange Width Thk. 4.060 0.375 5.000 0.360 5.030 0.430 TEXAS IRON & METAL SUPPLY ON DEMAND 713.672.7595 800.839.4766 Fax 713.672.0653 texasironandmetal.com 4.16 in = 416 2*0.375 -3.4in. O " 4.060 in thickness= 0.375 n. thickness=0.280 in. thickness = 0.375 in.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Types and properties of steel

Steel is an alloy of iron. It is formed by heating iron and combining carbon (0.1-1.7%) with it and minor amounts of other elements such sulfur, chromium, manganese, nickel, etc. Thousands of steel grades have been published, registered, or regulated around the world, each with its chemical composition, and special numbering systems have been devised in various countries to classify the vast number of alloys.

Question

Please help with problem 2

Given: Following Bars/Beams.
Required: Construct displacement, shear, and moment diagrams for each case.
Problem 1
Problem 2
min
1.2
m
Size
W
W
x 0.6
m
4 x
5 x
X
6m
p=-k sin
12 kN/m
Carbon - Stainless - Alloy- Aluminum
texasironandmetal.com
6 kN/m
For Each beam, assume an allowable stress of 440 MPa. And, assume a cross sectional area as
shown below:
(2πχ
sin (²x)
L
1.2
m
WIDE FLANGE BEAM
I
Wide Flange Beam . H Beam
(A)
(B)
Weight Section Web
Per Ft. Depth In. Thk.
13.000 4.160 0.280
16.000
19.000
5.010
0.260
5.150
0.270
(C) (D)
Flange Flange
Width Thk.
4.060 0.375
5.000 0.360
5.030 0.430
TEXAS IRON
& METAL
SUPPLY ON DEMAND
713.672.7595
800.839.4766
Fax 713.672.0653
texasironandmetal.com
4.16 in
4.16-
2*0.375
-3.41 in.
4.060 in
thickness
0.375 in.
thickness=0.280 in.
thickness
= 0.375 in.

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