A uniformly distributed load of q = 44 kN/m is applied across the entire length of an L = 6.5 m long simply-supported beam. The beam is composed of an S381 x 64 American standard section. Appendix C of yourcourse textbook provides geometric properties of rolled steel shapes such as the S381 x 64 American standardsection.Determine the maximum normal stress (max), the maximum shear stress (Tmaz), and the magnitude of theshear stress at the junction between the web and the flange (7₁)JmarTmarTj=2237Save & Grade 3 attempts leftSave onlyMPaMPaMPaThe✓100%x 0%x0%Additional attempts available with new variants Designation*S610 x 180x 158x 149x 134x 119S508 × 143x 128x 112x 98S457 × 104x 81S381 x 74× 64Area(mm²)Depth(mm)22970 622.320130622.318900171001516018190163201419012520132901039094858130609.6609.6609.6515.6515.6508.0508.0457.2457.2381.0381.0FlangeWidth Thickness(mm) (mm)204.5199.9184.0181.0177.8182.9179.3162.2158.9158.8152.4143.3139.727.727.722.122.122.123.423.420.220.217.617615.815.8WebAxis Z-ZZThicknessI(mm) (106 mm) (10³ mm³)20.315.718.915.912.720.316.816.112.818.111.714.010.413151225995937874695658533495358335202186422539353260306528702705254021001950169014651060977T(mm)240247229234241196200194199170180146151Axis Y-YZI(106 mm) (10³ mm³)34.732.119.918.717.620.919.512.411.510.08.666.535.9933932121620619822821815314512711491.385.7

The uniformly distributed load .The length of the beam is .The beam is simply supported.For S381*64

Answered: A uniformly distributed load of q = 44…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

A uniformly distributed load of q = 44 kN/m is applied across the entire length of an L = 6.5 m long simply-
supported beam. The beam is composed of an S381 x 64 American standard section. Appendix C of your
course textbook provides geometric properties of rolled steel shapes such as the S381 x 64 American standard
section.
Determine the maximum normal stress (max), the maximum shear stress (Tmaz), and the magnitude of the
shear stress at the junction between the web and the flange (7₁)
Jmar
Tmar
Tj=
2
237
Save & Grade 3 attempts left
Save only
MPa
MPa
MPa
The
✓100%
x 0%
x0%
Additional attempts available with new variants

Designation*
S610 x 180
x 158
x 149
x 134
x 119
S508 × 143
x 128
x 112
x 98
S457 × 104
x 81
S381 x 74
× 64
Area
(mm²)
Depth
(mm)
22970 622.3
20130
622.3
18900
17100
15160
18190
16320
14190
12520
13290
10390
9485
8130
609.6
609.6
609.6
515.6
515.6
508.0
508.0
457.2
457.2
381.0
381.0
Flange
Width Thickness
(mm) (mm)
204.5
199.9
184.0
181.0
177.8
182.9
179.3
162.2
158.9
158.8
152.4
143.3
139.7
27.7
27.7
22.1
22.1
22.1
23.4
23.4
20.2
20.2
17.6
176
15.8
15.8
Web
Axis Z-Z
Z
Thickness
I
(mm) (106 mm) (10³ mm³)
20.3
15.7
18.9
15.9
12.7
20.3
16.8
16.1
12.8
18.1
11.7
14.0
10.4
1315
1225
995
937
874
695
658
533
495
358
335
202
186
4225
3935
3260
3065
2870
2705
2540
2100
1950
1690
1465
1060
977
T
(mm)
240
247
229
234
241
196
200
194
199
170
180
146
151
Axis Y-Y
Z
I
(106 mm) (10³ mm³)
34.7
32.1
19.9
18.7
17.6
20.9
19.5
12.4
11.5
10.0
8.66
6.53
5.99
339
321
216
206
198
228
218
153
145
127
114
91.3
85.7

Expert Solution

Step 1: Step 1 Given

The uniformly distributed load $q equals 44 space k N divided by m$ .

The length of the beam is $L equals 6.5 space m$ .

The beam is simply supported.

For S381*64

$Z left parenthesis z minus z space a x i s right parenthesis equals 977 asterisk times 10 cubed space m m cubed t subscript w equals 10.4 space m m t subscript f equals 15.8 space m m w subscript f equals 139.7 space m m A equals 8130 space m m squared I equals 186 asterisk times 10 to the power of 6 space m m to the power of 4 d equals 381 space m m$

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