k What is the maximum uniform distributed load w- ft on a simply supported steel beam with size W21x55 A992 and length 10 ft.
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- Where is/are the location(s) of the maximum compressive flexural stress? A simple I-beam is loaded as shown. 20 mm P KN PKN P KN B 20 mm- B с D L/4 m L/4 m L/4 m Pin support at the NA Midspan at point B Mid span at point D Midspan at the top fiber Roller support at top fiber Section D at the top fiber Section C at +170 mm from the NA L/4 m 20 mm C 250 mm 150 mm 150 mm D AUniversity of Al-Qadisiyah- Civil Eng. Dept. Lecture 11; Shear design.mp4 Example 1 GIF the rectangular beam given in figure below has been designed for flexure; design the necessary shear reinforcement given that that beam is loaded by a uniformly distributed live load of 25 kN/m and a dead load (including self-weight) of 40 kN/m, take f. = 28 MPа, fyt 3 420 MPа, о 542 mm 600 mm 5020- +300 mm Face of Support 300 mm 7.0 m 44:רן •..Q2 a) This is a concrete beam (the interior lines represent reinforcing bars), If P is live load, draw the BMD and the SFD as a result of ULTIMATE LOAD. P= 50 kips 20 ft. 20 ft. 40 ft. Own weight of the beam = 1.0 kip/ft. (Uniform dead load) b) This is a concrete frame (the interior lines represent reinforcing bars), If P is dead load, draw the BMD and the SFD as a result of ULTIMATE LOAD. 6.0 ft. P= 1.0 Kip Own weight of the beam = 1.0 kip/ft. (Uniform dead load)
- USE NSCP 2010 A rectangular beam reinforced for both tension and compression barshas an area of 1450 mm² for compression bars and 4350 mm² fortension bars. The tension bars and compression bars are placed at a distance of 600 mm and 62.5 mm respectively from the top of thebeam. The beam width 300 mm, fc’ = 21 MPa, fy = 415 MPa andtension steel covering is 60 mm.If it is 8 m-simply-supported beam that carries three concentratedservice live loads P applied at three quarter points of the beam (exceptat the supports), neglecting the self weight of the beam, determine themaximum value of service load P in kiloNewtons.PROBLEM 2: The steel beam loaded below has the built-up cross section as shown. Determine the maximum permissible value of the load w so as not to exceed allowable bending stresses of 110 MPa. -125 mm- w (including beam weight) 20 mm 20 mm 100 mm .-- NA A B - 2 m - 6 m 20 mm FINAL ANSWER: Max. allowable w kN/mA 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
- Which of the following best gives the nominal shear strength of a beam, Vn? O FyAw O FySx O 1.0Fydtw O 0.6FydtwTutorial 7 Вeams 1. A simply supported beam must carry the factored ultimate loads (two 15kN point loads) shown in the figure. Do not consider the self-weight of the beam. The beam is laterally and torsional supported only at A, B, C and D. For a 203 x 133 x 25 I- section: (Grade 350 W steel) (a) Determine the class of the section. (b) Determine whether the section has sufficient flexural resistance. 15 kN 15 kN 4,0 m 2,0 m 4,0 mThe given girder has beams framing into it at the ends and at every L/3 point. The beam carries a service live load of 20 kips as shown and superimposed uniformly distributed service dead load of 10 kip/ft. Select the lightest A992 W-section that can carry the load. Do not check for deflection. P, = 20 kips LL PLL W. = 10 kip/ft DL = 20 ft WDL L/3 BEAMS FRAMING INTO GIRDER L
- A built up section of A992 steel, F, = 345 mPa is made from plates fully welded together. The flanges consist of PL16X380 and PL12X500 for the web. Use the NSCP 2015 specifications. PL16x380 WL = 1.5wp %3D A Wu = 3.6wp 10m Which of the following best gives the maximum service dead load, wp? 40.2 kN/m 26.8 kN/m O 10.8 kN/m O 96.6 kN/m PL12x5005. The beam-column has a height of 7.5m and is a member of a braced frame. It is subjected to an axial load of DL-65KN and LL= 1OOKN. It is pinned at both ends. A W12X35 section was used and was made up of A-36 steel with Fy-250MPA. Compute the interaction value for beam- column based on NSCP 2015 code provisions. Ngelect the weight of the section. A = 6645 mm2 Zx- 839x10^3 mm3 d- 317.50mm Sx- 747x10^3 mm3 tw= 7.62 Rx= 133.35 bf= 166.62 ly= 10x10^6 mm4 tf- 13.21 Sy= 122x10^3 mm3 Ix= 119x10^6 mm4 Ry3 39.12mm 3.5 Note: Use C=D1.00%--0.2Situation 3: A W12x22 simply supported beam carries a uniformly distributed load. The properties of the section relevant to the problem are as follow: Use A50 steel and Cb - 1.14 d-312 mm bf - 102 mm tf = 10.8 mm tw = 6.60 mm kdes - 18.4 mm ry-21.5 mm Ix - 64.5 x 10^6 mm^4 ly-1.94 x 10^6 mm^4 Sx - 416 x 10^3 mm^3 Zx 480 x 10^3 mm^3 J-122 x 10^3 mm^3 Cw-44.0 x 10^9 mm^6 1. Which of the following nearly gives the maximum safe uniform load (Wu) that the beam could if the span of the beam is 0.800m? Neglect weight of the beam. 960 KN/m 933 KN/m 922 KN/m 975 KN/m