A simply supported beam, having thefollowing properties: 5.5 m length, Mcr= 30 kN-m, Ig = 7000x10^6 mm^4, Icr =3000x10^6 mm^4, uniformly distributeddead load = 5 kN/m, uniformly%3Ddistributed live load = 7 kN/m, Ec =21,000 MPa.a) Determine the value of Ma (DL+LL).None of the Choices37.500 kN-m73.500 kN-m45.375 kN-m54.000 kN-m63.375 kN-m

Answered: Image /qna-images/answer/a031dccc-a738-464e-809e-1737a6ec7518.jpg

A simply supported beam, having the following properties: 5.5 m length, Mcr = 30 kN-m, Ig = 7000x10^6 mm^4, Icr = 3000x10^6 mm^4, uniformly distributed dead load = 5 kN/m, uniformly distributed live load = 7 kN/m, Ec = 21,000 MPa. a) Determine the value of Ma (DL+LL). None of the Choices O 37.500 kN-m 73.500 kN-m 45.375 kN-m 54.000 kN-m 63.375 kN-m

A simply supported beam, having the following properties: 5.5 m length, Mcr = 30 kN-m, Ig = 7000x10^6 mm^4, Icr = 3000x10^6 mm^4, uniformly distributed dead load = 5 kN/m, uniformly distributed live load = 7 kN/m, Ec = 21,000 MPa. a) Determine the value of Ma (DL+LL). None of the Choices O 37.500 kN-m 73.500 kN-m 45.375 kN-m 54.000 kN-m 63.375 kN-m

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

Calculate the Bending Stress using the equation ?? = ?.?max /? , where M is the Bending Moment for each of your two beams. Also calculate the longitudinal shear ? = ?.? /?.? for each beam. Compare and discuss the results for 3 point load and 4 point load. Draw Bending moment diagram and shear force diagram for each test.
1. PLEASE I NEED IT ASAP.
Use transformed-area method. The cross section shown consists of a hollow concrete beam with an effective depth of 400 mm. The beam is reinforced with 4 - 16mm diameter bars placed 50 mm from the bottom of the beam. If the beam has a simple span of 6 m, what total uniform load (including its own weight) could the beam carry besides a concentrated load of 8 kN at midspan. Fc' = 7 MPa, fs = 124 MPa, n=12. 75mm 100mm & 75mm 4 D 4 150mm 200mm 50mm 50mm
2.1.Write the Equation and describe the style in the latitude of the structure
A simply supported beam, having the following properties: 5.5 m length, Mcr = 30 kN-m, Ig = 7000x10^6 mm^4, Icr = 3000x10^6 mm^4, uniformly distributed dead load = 5 kN/m, uniformly distributed live load = 7 kN/m, Ec = 21,000 MPa. a) Determine the value of Ma (DL+LL).
Rectangular reinforced concrete beam section b = 40 cm, h = 60 cm, d = 53 cm, material strength fy = 4200 = 4200 kgf / cm^2, fc ' = 280 kgf/cm^2, beam section configuration #4 For closed stirrups @15 cm, the net cover is 4 cm, and the beam section is subjected to a torque of Tu = Tcr. Q: What is the ultimate shear load value Vu that the beam section can bear. Among them, Tcr is the cracking torque of the beam section, #4 steel bar (db=1.27 cm, Ab=1.267 cm^2).
In the following beam with the working section shown is subjected to the indicated loads and supported as shown, determine the magnitude of the twist BA and the deflection YC in the middle. Consider E = 70 GPa, A = 7.5 x 10-3 m2, IZ = 61.3 x 10-6 m4, IY = 23.2 x 10-6 m4 Develop by: 1) Double integration method 10 kN 2) Overlay Method 3) Moment area method 4m 4 m develop: a) Problem approach and solution sequence b) Free body diagram c) Calculation of intermediate and constant forces d) Bending moment diagram e) Application of equations and associated tables Caniatidado AM 141do earde tier ade
Question: Calculate for Mmax of the beam
Strength of Materials, i need solutions in 15 minutes please. Q/ The Inverted T section of a L - m simply supported beam has the properties shown in Fig. (4). Find the flange width (b) so that the allowable stresses( σ tension max = 30 Mpa) and (σ compression max. = 90 Mpa.)
ASAP
.: Find the deflection of the simply supported beam of 12 m span, and then compare with ACI Code limit. The beam width (500 mm) and high (800 mm). f 'e = 21 MPa, live load= 8 k/m², surfacing load=1.0 kN/m²). Solution:- Beam weight=0.5(0.8) (24)= 9.6 kN/m WD 9.6+(1.0) (0.5)= 10.1 kN/m WL-8 (0.5) = 4 kN/m 5 Ws.14 5 384 Eclg (10.1+4). (12000)* 384 4700 √21 (500) (800)³/12 (Table 6.2) ACI Code limit A 1 360 12000 360 8.3 mm check A limit = 33.3 mm > 8.3 mm (O.K.) 1 360
Question A1. The doubly reinforced rectangular concrete section shown in Figure A-1 is designed for the hogging moment of a continuous beam. The section has a total depth of D mm, width B= 300 mm and is reinforced with 4N28 bars at the top and 2N28 bars at the bottom. Given: f. = 40 MPa, fay = 500 MPa. Find other required data from AS3600-2018. a) Calculate the beam depth, D where D = 500+20n mm. Take n = 3rd digit of your student number. For example: your student number = 18505679, n = 5; D = (500+ 20*5)=600 mm) Note: Use the calculated D from (a) for the further calculation in (b-c). b) Calculate ultimate moment capacity (Mu) of the section. c) Check ductility of the section according to AS3600-2018. 300 4N28 2N28 Figure A-1