Question

-12 13 14 15 16 17 3. Calculation questions (20 scores for each question, 60 scores in total) (1) The I-section beam about the strong axis is subjected to static load. The section size is shown in the figure without any weakening. The designed bending moment value M-400kN.m and the designed shear value V=550kN. Check whether the strength of the beam at this section meet the requirements. The steel material is Q235B. (Full score is 20) Table 1. Strength design value of steel (N/mm²) 200 Steel Tensile, Face Shear Thickness Grade 20 compressive and pressure (mm) resistance bending resistance 516 215 125 500 > 16 -40 205 Q235B 120 325 > 40-60 200 H10 >60-100 190 110 20 115 I

Answer 1

bunen i 400 KN-m D M max. limax, - 550 kn 1 1£ * 미 % 500mm h. H lomm. 120mm 200 mm for given and the and bending gerade thickness tensile el comparessive reesistance (A) 205 N/mm² resistance (fel = 120 N/mn and shear Now first required of have section to Calculate modulus the Mman 400 x 10 Sreq'd f 205 3 1.951m

Now pete we have to section modules calculate the section o of Dt is given by s = I ग I is the moment where inertica from of is distance and the ce the noaudral axis :- so, from the modulers calculation for I- section section is gonen S = BH? 3. bhi 6H 6 Sol from the diagram : 2 s = 2000 540 3 190 X 500 6x 540 6 3. 2.389 m. so oreg then urred rection modulus postovided section section. the less modules is of

hence it is Now ginen maka 550 kn una allowable shear eusistance 120N lmm 2 Calculate e Aug Vmax Aw. where Aw Ć web. oy soj Aw - 540X 10 = 5400 mm ². Now 550 X10 yang 101.85 N/mm 5400 here e Aug en max (shear resistance) at so the this strength of the meet beam the section requirements.
Here when i calculating the section modulus of the section (provided) i am using a formula which is determined by the calculation. you can find out by the section modulus by the given formula which is I/ Y.

Thank you so much .