Question
How do I solve this?

Problem 2(50 points) The roof of the Edwin A. Stevens building is to be redesigned to accommodate a billboard for Stevens Institute of Technology. The new billboard will weigh 40 kips and be supported equally by two columns located 9 ft from the rear of the building. The beams supporting the new roof will be simply supported as shown. (Note there is a 5 ft overhang at the front of the building). For the grade of steel to be used ơallow-24 ksi and Talk-14.5 ksi. Design the beams for the roof using w.(wide flange) section [Properties of W-Section are attached; from appendix C of the textbook] Details of 2 ft
Hints: (i) First calculate the loads on the beam to be designed. To that end- convert the load of 1,600 1b/ft to a uniformly distributed load of w Ib/ft (i.e., linear length) along the span of the beam (and add any concentrated load, İf required). To help in your calculation for w- the simple concept of tributary area for loading is described in an attachment. (ii) Then, using the Bending Moment Diagram check for Maximum Bending Stress to find the required x-section. A worked out Problem (Example 15.1 - Textbook) is attached to show a simple use of the Charts in the Appendix; finally check for Shear stress. Note: Example 15.1 uses the notation S INymx (S is called Section Modulus), for Bending Stress calculation. Check Art 15.2 (page 696 of the Textbook) for some details.
Tributary Loadings. When lat surfaces such as walls, foors or roos are supparted by a structural frame, it is necessary to determine how the load on these surfaces is transmitted to the various structural elements used for their support One-Way System. A sab or deck that is supported such that it delivers its load to the supporting members by one-way action, is often referred to as a one-way siab.To illustrate the metbod of loed transmission,consider tbe framing system shown in Fig where the beams AB.CD, and E rest on the girders AE and BF.IH a uniform load of 100 Ib/t is placed o the slab, then the center beam CD is assumed to support the load acting on the tributary area shown dark shaded on the structural traming plan in Fg Member CD is therefore subjected to linear distribution of loed ot (100 b/t)(s ft)-s00 b/ft, showa oa the idealised beam in Fie. 100 / 258 LC 56 dealized framing plan 10 t 2500 t 2500 Ib idealized beam
EXAMPLE 15.1 2 A beam is to be made of steel that has an allowable bending stress of οι .-24 ksi and an Iowable shear stress of ace-145 ksi. Select n appropriate W shape that will carry the loading sbowa in Fig.15-7a. SOLUTION Shear and Moment Diagrams. The support reactions have been calculated, and the sbear and moment diagrams are shown in Fie 15-7b. 68-6-6 Prom these diagrams, V-30 kip and Ma -120 kip ft Bending Stress. The required section modulus for the beam is determined from the lexure foníula, 15 40 ldip 20 Hp Using the table in Appendix C, the folowing beams are adoquate W18 x 40S 68.4 in W14 x 43$-62.7 i W10 X S4 S-60.0 in WS X 67S-60.4 in 6 D e n 50 kp 20 -30 The beam haviog the least weight per foot is chosen, ie. wis x 40 The actual maximu oment Max, which inctudes the weight of the beam, can be calculated and the adequacy of the selectied beam cas be chocked. In comparison with the applied loads, however, the beams weight, (0.040 ip/h)(18 D)-0.720kp. will only slightly -120 ncrease Se In spite of hi oK Fig.15-7 Shear Stress. Since the beam is a wide flange section, the average shear acress within the web will be coesidered. (Sea Eample 123) Here the wob is assumed to extend from the very top to the very bottom of the beacn From Appendix C, for a W18 x 40, d-1790 in. 0315 in Thus Use& W18 x 40 Ans
Geometric Properties of Wide-Flange Sections APPENDIX Wide Flange Sections ar W Shapes FPS Units Range Web Area Dapth hickness width thickness in. W24× 104 | 306 | 2406 0500 | i250 | 0.230 | 3100 258 | 101 | 259 | 40,7 291 W24 X 84 247 2410 0470 900 0.770 2370 196 9.79 944 20.9 195 4x68 201 23.73 0415 W4 x 5 62 257 130114 911 291 8.30 134 wi8x 65 | 29.1 | 1835 | 0.450 | 7590 | 0.750 | 1070 | 1171 749 | SASİKA W18 X60 176 1824 041S 7555 0695984 108 747 50.1 13.3 169 W18 x 35162 18110390 7530 0.630 890 983 7.41449 119 1.67 W18 x 40 18 1790 0315 w16 x57 168 1643 04307120 0.715758 92.262 431 121 160 W16 x 50 14.71626 0380 700 0.630 659 10668 372 10 1.59 WI6 X 45 133 1613 0345 7.035 0.55 86 72.7 66 32893 157 w16 x 26 768 15.69020500 034530184 626 99 5525 0.440 375 472 641 1244 117 112 W14 x38112 1410 0310 WI4 x 3410.0 1398 0.285 x 25.69 13910255 5.005 0420 w14x 22 649 134 0.2305000 033 199054 7.00 2 LD
Wice-FLANGE SECTIONS OR W SHAPES FPS UNIm 811 Wid Flange Sections or W Shapes FRs Units Fange Wob Area | Depth | thickness width | thickness W12 x 87 25.6 1253 ss 12125 0.810 740 118 241 39,7307 W12 x 45 132 1203358045 0575 350 81 .15 00 124 194 w12 x 22 6.48 12310.260 4.0300425 156 254 491 466 231 0.847 w12 x 16 4.11990.220 9900.265 10 17.1 467 282 1A 6 wi2x14 416 | 11.91 w10 x 100 29A 1140 W10 x 54 15.8 10.09 0370 100so 0615 0200 | 3 70| 0225 | 88.6 | 14.9 | 4S2 | 236| 1.19|0753 0680 10:340 | 1.120 623 | 112 460 | 207 400 | 2S 303 60.0 437103 20.6 256 0620 248 49.1.1 432 | 534 193 | 201 050 21 $50 113 198 0.300 510 0.510 170 324 438 67 5.75 137 5.62 20.24 w0x15 441| 999 | .0250 0.270 69 13.8 3.95 289 145 0810 wi0× 12 | 3SAİ 9871 0.190 | 3860| 0.220 153.8 | 10.913.00 | 21s| 1.101 ans ws x 67 197 900 070 W8 x 48 141 8.500400 w8 x 40 11.78.25 ws x 24 we x 25134 638 320 6.080 0.455 534 167 2.70 171 5.613 w6 x 20 5.87620 0.26060000365 414 134 2.56 133 441 w6 x 15 443 599 023059900260 2919.72 256 932 3.11 WE×12 0.935 272 604 372886 214 2.12 220.810 228 52.0 365751 18. 2.10 tol 5 1184 | 433 | 361 | 609 | 15.0.1 248 070560 146 35S 353 491 122 204 ias | 110 | 275 | 347 | 371 | 927| 2ng 50.400 88 209 342 183 5.63 16 9.13 8.00 7.08 793 0.245 0.31540 118 329 34 170 0s76 4.74 628 4.030 0405321102 2.60443 2.20 3.551 6D3 | 0.230 |dool 0.280 122.1 | 731 | 249 | 299| 268 590 01030215 14 56 247 219 :の
0 0
Add a comment Improve this question Transcribed image text
Know the answer?
Add Answer to:
How do I solve this? Problem 2(50 points) The roof of the Edwin A. Stevens building...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT