design the beam for shear
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- University 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 •..H.W-4: For the beam of sections and detail shown in Fig. find the maximum live load (PL) that the beam can be carry it. Neglect self-weight? Ans. PL=195.74 kN Use f 35 = - MPa fy 420 -2.0m- 70mm 2PL 2020mm 5025mm 350 mm sct.a-a 600 mm- a a -2.0m- -LHHH] .... .... 8025mm Soemy 3025mm 350 mm sct.b-b b b -2.0 m- 600 mm PLThe cross-sectison of a simply supported plate girder is shown in figure. The loading on the grider is symmetrical. The bearing stiffeners at supports are the sole means of providing restraint against torsion. Design the bearing stiffeners at supports. with minimum moment of inertia about the center line of web palte only as the sole design criterion. The flat section available are: 250 x 25, 250 x 32, 200 x 28, and 200 x 32 mm. Draw a sketch 500 25 8 .1445 Dimensions in mm 20 +425
- Task 12 Z A C B 2a a a The figure above shows a static indeterminate beam ABCD. The beam has a fixed joint bearing in A and a sliding bearing in B and C. Note that the beam is continuous at point B and that the joint in the bearing is below the beam. The beam is subjected to an evenly distributed load q between A and C, while there is a point load F at point D. The beam is made of steel with an E-module E = 210 000 MPa and has a cross section with a second area moment ly = 12x106 mm4. The value of the evenly distributed load q and the length a are determined from: 9= ID+25 20 kN/m ID+40 a= m 50 ID=40 a) Determine how much the maximum force can be when the bearing in point B can withstand a maximum vertical force of 50 kN (Bz< 50 kN). Write your candidate number on all the sheets you upload. b) Now set the force F to the value you found in question a) and decide how big the deflection of the beam will be in point D, under the point load F. Write your candidate number on all the sheets…A cantilever beam AB of length L = 6.5 ft supportsa trapezoidal distributed load of peak intensity q,and minimum intensity q/2, that includes the weight ofthe beam (see figure). The beam is a steel W12 X14wide-flange shape (see Table F-1(a), Appendix F).Calculate the maximum permissible load q basedupon (a) an allowable bending stress σallow =18 ksiand (b) an allowable shear stress τallow = 7.5 ksi.Note: Obtain the moment of inertia and section modulusof the beam from Table F-1(a).1%9. l. In. • f 10 Y:- quiz2.pdf > Quiz: Design the reinforcement steel that necessary for the shcar and torsion for the rectangular cantilever beam shown in figure. Use for stirrups (o12) and let f.=25 MPa, f-400 MPa, dead load (5 kN/m?) and live load (10 kN/m), take into account the self-weight of beam. All dimensions in (mm) 1000 800 3000 300 Cnand TEXTEDT II
- Current Attempt in Progress A pin-connected structure is supported and loaded as shown. Member ABCD is rigid and is horizontal before the load Pis applied. Bars (1) and (2) are both made from steel (E - 30600 ksi) and both have a cross-sectional area of 0.87 in.?. Assume L;-87 in, Lg-111 in., a-50 in., b-93 in, and c-39 in. If the normal stress in each steel bar must be limited to 19.4 ksi, determine the maximum load Pthat may be applied to the rigid bar. (2) (1) D Answer: P- kips.H.W Draw S.f,B.M and elastic dliagrams for the beam Shown below by method of equntions ? 6W/m 20 KN 6m 2m 2mA bracket of length 1.5m is rigidly attached to the simply-supported beam at its end. The beam is subjected to the combined effect of M,, M, and compressive force P caused by the factored loads as shown. There are lateral supports only at the ends. The member is made of HE300A section and steel grade is S275 (Fy-275MPA, Fu= 430 MPa). The loads acting are determined from LRFD combinations. Determine whether the member can safely carry the applied loads. Multiply first-order moment diagrams by a factor of 1.2 to account for second-order effects. No need to check shear limit state. You do not need to additionally consider the beam self-weight. E Use the following simple expression to calculate the value of L:: L, = T*rts 0.7*Fy y P = 400kN 1.5 m F = 150kN 2 т 4 т X: Lateral support НЕЗ00А
- 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 ملفA steel beam having a span of 14 m. carries a dead load concentrated load Pp = 60 kN at its midspan. The beam is laterally supported at its span. Properties of the wide flange section. rts - 104.39 mm ho - 338.08 mm Jo -1.69 x 10^6 Cw - 4.3 x 10^12 d=356.11 mm tf = 18.03 mm tw - 11.18 mm Sx - 2343 x 103 mm rx = 155.96 mm ry - 93.98 mm Sy = 818 x 10^3 mm Zx = 2573 x 10^3 mm Zy= 1239 x 10^3 mm Fy = 345 MPa Iy = 151 x 10 mm Lp=3.98 Lb=14m Lr=12.98 Cb=1.14 Determine the safe concentrated live load that the beam could support at the center based on its design strengthFor the beam shown pind IL. Por RA dueto uniferm load (sENim) and concentrated load (lokn) only positive RA J 2m Im, Lm, 1m IL RA