Problem Statement from book A W530 X 92 structural steel wid

Problem Statement from book:

A W530 X 92 structural steel wide-flange shape [E = 200 GPa; I = 554x10⁶ mm⁴] is loaded and supported as shown below.

Determine:

a. the force and moment reactions at supports A and C

b. the maximum bending stress in the beam

c. the deflection of the beam at B

For the picture: the support on the left is point A and the origin, at x = 4.5 m, where the load ends is point B and the second support located at x = 9 m is point C

A W530 times 92 standard steel wide. Flays shape [E = 200 GR, I = 554 times 10^6 m^4] is loaded and supported as shown in the figure. Determine the free and moment release at suppose A and C Determine the moment find stress in the Determine the deflection of the Gram at B

Solution

solution:

1)here wide flange beam W530*92 has depthh= 533 mm and w1=92 kg/m or 902.52 N/m

hence total load acting at point B

W1=w1*l=902.52*9=8.1226 kN

2)where UDL is acting between 0 to B, w2=80000 N/m

hence W2=w2*l1=80*4.5=360 KN it is acting at cg of udl at 2.25 m from point A

3)hence support reaction at A and C are

Ra+Rc=360+8.1226=368.1226

taking moment about A is

Rc*9=360*2.25+8.1226*4.5

Rc=94.0613 kN

Ra=274.0613 KN

4)hence shear forcea at point A,A\',B,C are as follows

Fla=0 KN

Fra=274.0613 kn

Fla\'=274.0613 KN

Fra\'=-85.9387 KN

Flb=-85.93 KN

Frb=-94.0613 KN

Flc=-94.0613 KN

Frc=0 KN

here line connecting all point is straigh between points and wherever it cuts axis is point of shaer force at which bending moment is maximum hence for that point as follows

274.0613/x=85.9387/2.25-x

x=1.7128 m

4)hence at x=1.7128 m from A will have point of maximum bending moment hence it will be point of maximum bending stresses

Mx=1.7128=Ra*x-w1*x^2/2-w2*x^2/2=350.74*10^6 N mm

5)hence bending stresses are

sb=Mx*y/I=350.74*10^6*533/554*2*10^6=168.722 MPa

6)deflection of fixed beam due to self wieght is given at B is

d1=8.1226*1000*9^4/200*10^3*554*10^6=1.2525*10^-6 mm

deflection due f=360KN

d2=Wa^3b^3/3*EIL^3=360*1000*2.25^3*6.75^3/3*200*554*10^9=5.204*10^-6 mmhence total deflection at B is

d=d1+d2=6.4569*10^-6 mm