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

225 P= 8 P= 10 9,- 18 H3 2 (m) 1 (m) . 4.50 (m) 1.50 (m)
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
A steel pipe (D = 3.500 in.; d = 3.068 in.) supports a concentrated load of P = 617 lb as shown. The span length of the cantilever beam is L = 4 ft. Determine the magnitude of the maximum bending stress in the pipe. O 21.55 ksi O 13.81 ksi O 15.73 ksi O 24.04 ksi O 17.18 ksi L B P d Pipe cross section. D
The beam shown in figure carries a uniformly distributed load of w on overhang BC. The beam is constructed of a wood beam [E,=12 GPa] that is reinforced on its upper surface by a steel plate [E,-200 GPa] plate. Assume that the allowable bending stresses of the wood and the steel are o. = 9 MPa and o = 165 MPa, respectively. Determine the largest acceptable magnitude for distributed load w. [L = 4m, a = 1.25 m, b. =150 mm, d, =280 mm, b. = 230 mm, t, = 6 mm] IC
The steel beam has the cross section shown. The beam length is L = 5.6 m, and the cross-sectional dimensions are d = 335 mm, bf = 195 mm, tf = 12 mm, and tw = 10 mm. Calculate the largest intensity of distributed load wo that can be supported by this beam if the allowable bending stress is 165 MPa. Wo X Answer: Wo= 12 B 72 C kN/m
2. The beam with cross-section shown carries a uniformly distributed load of intensity w over its entire length. Determine w if the allowable stresses are g = 40 MPa, g = 80 MPa, and t = | T 12 MPa. w (kN/m) 75 mm 10 mm 0.5 m : 10 mm 3m 10 mm- 150 mm 150 mm
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).
The steel beam has the cross section shown. The beam length is L = 7.5 m, and the cross-sectional dimensions are d = 385 mm, b = 190 mm, t+= 12 mm, and tw = 10 mm. Calculate the largest intensity of distributed load wo that can be supported by this beam if the allowable bending stress is 155 MPa. Wo -1/2 Answer: Wo= B sla -1/12 kN/m
.: 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
A balcony is supported outside a building by steel cantilever. The balcony is a m wide (length of the cantilever beam) and the total load carried is y kN/m² of floor area. The cantilever cros s- section is b x d and the floor thickness is c. Note that y kN/m² varies between 11 and 19 kN/m2 (each group to select their own value) Figure 1 a) Design your section if d = a/10 b = 3d/10 c = d/4 b) Calculate the maximum permissible spacing l if the allowable stress is 12 MPa. p
Consider a 3 metre long uniform beam with flexural rigidity of 100 kNm². The beam is supported with a fixed (clamped) support on the left, and a roller support 1 metre from the right hand side. 3m 2m A 1kN/m point load is applied 1 metre from the left hand side. If we find the deflection for this beam, using the boundary conditions for the fixed (clamped) end and for the roller support, we get the equation. y(z) 9600 ((16³-48x² +48r - 16) H(r− 1) - 112³ +18r²) (you do not need to calculate this yourself, but you may verify it if you wish) Have Desmos (or any other computer graphing system you like) draw a picture of the beam under load. Make sure the picture is accurate, and is zoomed appropriately to see the actual bends in the curve. What is wrong with the deflection function to the right of the roller support? b) Solve the deflection for this beam in two parts, as if there were two separate beams: 3m Om 2m 2m For the roller support use the following boundary conditions: y (2)= 0 and…