A uniformly distributed load of q = 44 kN/m is applied across the entire length of an L supported beam. The beam is composed of an $381 x 64 American standard section. A course textbook provides geometric properties of rolled steel shapes such as the S381 x E section. Determine the maximum normal stress (max), the maximum shear stress (Tmar), and the shear stress at the junction between the web and the flange (7₁). Omar = 237 Tmax= 41 MPa MPa ? 100% ? X0%

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Chapter11: Columns
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Problem 11.6.1P: A steel bar has a square cross section of width b = 2.0 in. (sec figure). The bar has pinned...
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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
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MPa
MPa
MPa
The
✓100%
x 0%
x0%
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Transcribed Image Text: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
Transcribed Image Text: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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