5 Decimal Places.. External support reactions:RA=RC =RF =KNKNKNThe shear and moment at point B areThe moment at point C isThe moment at point D isThe maximum positive moment at segment AC isThe maximum positive moment at segment CF isThe maximum shear along the entire span of the beam isThe maximum positive moment along the entire span of the beam isThe maximum positive moment along the entire span of the beam iskN-m.kN-m.kN andkN-m. It is located atkN-m.KN.kN-m.kN-m.kN-m, respectively.m from point A.Express all your answers in three decimal places.For the reactions, there is NO NEED to put the sign.For V and M values, ALWAYS include the sign (+ or -) to denote the nature of shear and bending. Not indicating the sign will mark your answer wrong. Noconsiderations. L1ANCW1BL2TOCDAL32 mEW2L42Consider the following values in the given beam above:L1 = 6 mL2 = 2 mL3 = 4 mL4 = 3 mL5 = 2 mL6 = 1 mW1 = 90 kN/mW2 = 30 kN/mP = 50 kNM = 60 kN-mL5MGL6H

Answered: Image /qna-images/answer/e00bd07f-ae53-4235-a4c6-1af651dc704d.jpg

External support reactions: RA= RC = RF = KN KN KN The shear and moment at point B are The moment at point C is The moment at point D is The maximum positive moment at segment AC is The maximum positive moment at segment CF is The maximum shear along the entire span of the beam is The maximum positive moment along the entire span of the beam is kN-m. kN and kN-m. kN-m. It is located at kN-m. KN. kN-m. KN-m, respectively. m from point A.

External support reactions: RA= RC = RF = KN KN KN The shear and moment at point B are The moment at point C is The moment at point D is The maximum positive moment at segment AC is The maximum positive moment at segment CF is The maximum shear along the entire span of the beam is The maximum positive moment along the entire span of the beam is kN-m. kN and kN-m. kN-m. It is located at kN-m. KN. kN-m. KN-m, respectively. m from point A.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

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
1K 2K 1K 60° 60° 60° 60° 10 60° 60° 10 OE 10' G. 5 K Solve for member FG.
For the given properties of the beam below, determine the safest wu that the beam will carry. bf = 850mm bw = 400mm tf = 100mm %3D d = 400mm d' = 100mm fc' = 34MPA fy = 400MPA As = 8-32mm bars %3D The answer is kN/m Pu bf キ 50 KN (Factore d) load Wu As t. A 5m + bw +
concentrated load W and a total uniformly distributed load of 2W. 587. The wide flange beam shown in Fig. P-587 supports the Determine the maximum safe value of W if o, < 10 MPa and r < 1.4 Ans. W = 2.62 kN %3D MPa. 20 mm -200 mm- W 2 m 2 m 2W 1 m 180 mm 20 mm R2 R1 20 mm Flgure P-587.
Q5) 1- Max shearing stress? 2- The necessary distance between pins and number of these pins of diameter 10mm can be used in longitudinal direction of beam? Knowing that the allowable shear stress of bolt is 25 MPa. 20mm 120mm re) For the beam shown in Figure (4), compute: 180mm H H 20mm 20mm 5 kN 1m m Figure (4) 2m 10 kN/m +4 2m 1m 5 KN
V M R = R R = D RA M 40 kN = VD₁² M VD2= MB1 B2 D = 2 m 6 m MB2 MB1 hinge 4m-4m VB Vpz KN KN 2 kN/m KN KN KN KN kNm kNm kNm D RD VD1 MD 8 m 2° 30 kN-m RE
A cantilever beam shown in Fig , with rectangular section shown. Its carry uniform load wand concentrated load Pa =12 KN, F-0.5. a- Calculate the instantaneous and 6- years long-term deflections at free end b- Check the deflections with ACI-Code limitations. 012 stirrups 60 29 h 2429 K-bw fy bw, W, kN/m² h, mm Wa mm 20 350 155 1.75 | 8.5 | 1.6 500
compute the safe factored uniformly distributed deadload that will be added to the weight of the beam. the beam is simply supported with an span of 6m and carries a uniformly distributed liveload of 20kN/m. use ycon=2 4KN/m³ b=280mm d=415mm fc'=25MPa As=6-20mm diameter bar the answer is_____KN/m
W₂ = 30 kN/m W₁ = 10 kN/m B A L=12 m Given: E-200x10³ MPa, I-120x106 mm² For the beam shown, you are at liberty to choose any of the two methods (DIM or MAM) to determine the following: Upward reaction at support B. Re kN Upward reaction at support A RA KN 8m C
A 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 strength
Situation 1: A W8x67 has a length of 7.50 m. Fy- 248 MPa. The properties are given below. Properties of W shape A = 12,700 mm^2 d=229 mm bf = 210 mm tf-23.7 mm kdes - 33.8 mm tw - 14.5 mm rx - 94.5 mm ry-53.80 mm 1. Which of the following most nearly gives the axial capacity (Pu) of the column if both are pinned? 1019 KN 1001 KN 1111 KN 979 KN
4.5.12 OIC Section Properties Write functions for A, I, c, radius of gyration. Each function works for the three sections below. Example Function name: OICSEC_A() • OICSec_() Etc. 을 + 12 BIP - bhs V, 12(BII-bh) BII -- bha 6/1 - 8