AW14 x 68 is used as a beam having a simple span of 6 m.Properties of WF Section.d = 356.62 mmbf = 254.89 mmtf = 18.29 mmtw = 10.54 mmSx = 1688 x 10° mmRy = 62.48 mmZx = 1885 x 10 mm3Fy = 345.6 MPaEl 150 000 kN-m2Determine the safe uniform live load it could carry so that it will not exceed theallowable live load deflection of 1/360 of the span. (kN/m)

Answer We know that deflection for a simply supported beam carrying uniform load is given by…

AW14 x 68 is used as a beam having a simple span of 6 m. Properties of WF Section. d = 356.62 mm %3D bf = 254.89 mm tf = 18.29 mm %3D tw 10.54 mm %3D Sx 1688 x 10° mm3 Ry = 62.48 mm !3! Zx = 1885 x 103 mm3 Fy = 345.6 MPa El 150 000 kN-m2 %3D Determine the safe uniform live load it could carry so that it will not exceed the allowable live load deflection of 1/360 of the span. (kN/m)

AW14 x 68 is used as a beam having a simple span of 6 m. Properties of WF Section. d = 356.62 mm %3D bf = 254.89 mm tf = 18.29 mm %3D tw 10.54 mm %3D Sx 1688 x 10° mm3 Ry = 62.48 mm !3! Zx = 1885 x 103 mm3 Fy = 345.6 MPa El 150 000 kN-m2 %3D Determine the safe uniform live load it could carry so that it will not exceed the allowable live load deflection of 1/360 of the span. (kN/m)

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter5: Beams

Section: Chapter Questions

Problem 5.5.12P

See similar textbooks

Similar questions

EYA 98 B/s docs.google.com/forms 100 Papertine J MA Papertine Papertine 45 - ls- das #2: For the Simply supa ted beam having the Sechin Shown and londed partially, Find: 1. Draw the bending moment diagram ? 2. Find the value of the maximum bendig moment Mmak 3. Find the value f the maximum bending stress Vmax ال لعة البطم رات المطع الو ضع المكل والحملة ۱. أسم الر البان لعن اوكماو : Mmax sls! e de i ană fic أوجه تعة أ على إ دإما کے ļ Umax KN/m - 20 Cm 4 1 m. 2 m. 1 m. Neutral 40 cm Axis Beam (AB) Beam Cross- Section ت إضافة ملف
8. H / The beam shown below has across seclion of channel Shape with width 360mm.and haghs h= loomm, the web thickness ist= 14mm.Deler- Mine ihe Max imum tensile aud Compressive stressin the beam due to unifom Load? 360Mm 3.6KN/M 14m tiom Please Sir 5T CASIC f-991ES PL DATURAL u CALC (-)
How much service live load, in kN/m, can be supported? The member weight is the only dead load. The axial compressive load consists of a service dead load of 45 kN and a service live load of 90 kN. Do not consider moment amplification. Use NSCP 2015. a. b. LRFD. ASD. wt A kN/m mm² W460x144 1.41 18400 472 13.6 283 SECTION tw bf tf Ix Sx mm mm mm mm³ 22.2 726 3080 mm mm mm 199 6.2m W460x144 Sy Ty Zx ly mm mm³ mm 83.6 591 Zy J Cw mmº mm³ mm³ 67.4 3450 906 2440 4230 mmª
A built-up section shown, 30 feet long, is fixed at both ends and braced in the weak direction at the midheight. A992 steel is used. Determine the following: 11. Ix 12. Ky 13. effective slenderness ratio 14. Fe 15. LRFD design strength 16. ASD allowable strength 18 4 W10 × 49 A 14.4 in² bf 10 in d. 10 in tw 0.340 in tf 0.560 in Ix= 272 in¹ Iy=93 4in¹ W10×54 A 15-3 in² bf 10 in d = 10 lin tw0-370 in tf 0.615 in Ix= 303 in 4 Ty = 103 in 1
DETERMINE THE THE IS MAXIMUM SAFE SERVICE LIVE LOAD THAT THE BEAM CAN CAN CARRY IF SERVICE DEAD L O A D THE BEAM WEIGHT OF THE BEAM SECTION fy = 415 MPa. TOP BARS BOTTOM BARS ARE CENTER DISTANCE 12 KM/M INCLUDING AND BASED ON THE CAPACITY SHOWN. fc : 40 MPa AND ARE 6-32 mm. AND ARE - 3 - 28 mm. CENTER TO 3 BETWEEN THE USE 2015 NSCP Is 108 mm 00 0 0 300 G ט1 108 352 70. A (1 7.2m B TOP BAR LAYERS ↓ 6.7M C OM
A built up section of A992 steel, F, = 345 mPa is %3D made from plates fully welded together. The flanges consist of PL16×380 and PL12×500 for the web. Use the NSCP 2015 specifications. PL16×380 W̟ = 1.5wp A Wu = 3.6wp 10m Which of the following best gives the maximum service dead load, wp? O 40.2 kN/m 26.8 kN/m O 10.8 kN/m O 96.6 kN/m B. PL12x500
For the given properties of the beam below, determine the safest wy that the beam will carry. bf = 850mm bw = 400mm tf = 100mm d = 400mm d' = 100mm fc' = 32MPa fy: 400MPa As = 8-32mm bars The answer is kN/m = 1m A Pu = 20 KN 3Pu Wu 5m 0.50 Pu ff ym 1B3 d bf As + bw-t + 1 tf 1
A simply supported pre-tensioned T-beam, as shown in Figure (2) has: f'ci 25MPa; f'-40 MPa; span length is 12m; Total Aps-800mm² for tendons; initially stress in the tendons is 1000 MPa. Consider 20% the total losses of pre stressed force then determine the allowable service live load can be carrying. 150mm 600mm T 500mm T Aps-800 mm² 50mm 150mm Typical Cross section Figure (2)
Determine the design moment capacity of the beam with given properties below. bf = 650mm bw = 350mm tf = 120mm d = 370mm d' = 100mm fc' = 34.5 MPa fy = 400 MPa As = 8 - 28mm bars USE NSCP 2015, Mu = kN-m
E CH2- Resultant paf - Adobe Acrobat Pro Estended Fle Edit View Document Comments Forms Tools Advanced Windew Help 2/87 If the resultant of the two forces and couple M passes through point O, determine M. 150 mm 60° mm 30 M, = M-400x 0.15cos30-320x0.3 0 M = 148 Nm CCW 320 N 160 400 N mm Example : Replace the three forces acting on the bent beam by a single equivalent force R. Specify the distance x from the point O in which the line of action of R passes. 200 N 160 N ahid hamoodi •..
Please answer the question with complete solution. Thank you
Pls answer asap