How do you find the stress at the M-max of a cantilever beam with a "z" cross section + an end load?

Question

M-max being max moment.

like i was given this problem and I have literally no idea how to solve it because the cross section is a Z shape and i don't know where the max moment is....

if you have any idea i would really appreciate help

Answer 1

Use what you know of statics to give you the maximum moment - draw the moment diagram for the beam. The cross section doesn't have any effect on this.



The maximum stress on the cross section in bending will always be farthest from the centroid. Find the centroid, with the distance from the top and the bottom of the Z.



Then use familiar equations which you have been taught



stress = M * Y / I



where I (moment of inertia) for a Z shaped section should either be given, or be something you can calculate using methods you already know and tables from the back of your textbook.

Answer 2

The source site has a lot of references and even spreadsheets for beam calculations. The first source is a page with links to engineering spreadsheets, just go to Abbott Aerospace and you will find a spreadsheet that calculates the moments and indicates the maximum. Also there you will find spreadsheets for section properties calculation (any type of section, on the "arbitrary" section).

On the source 2 you can download this report: BEAM DESIGN FORMULAS WITH SHEAR AND MOMENT DIAGRAMS.

On the source 3 you find the most used books for structures calculations and the ones for stress analysis have all you may need for beam calculation.

Answer 3

Since you have none symmetry axis , S= (M/I )Y max is not valid , unless I is the principal moment of inertia .- You must turn the axis passing by center of gravity in an angle Tan 2T = 2 Ixy / (Ix-Iy) , and from this axis , measure Ymax .- Ixy is the product of Inertia .-

Ymax= (h/2cosT)

I = (Ix+Iy)/2 +( (Ix-Iy)/2 ) cos2T -Ixy sin2T



Example b= 20 cm , H= 60 cm , t=1 cm

Ix= (1/12) 60^3 *1 + 2 (20*1) (60/2)^2 =54000cm^4

Iy= 2* (20/2) ^2 * (20*1) =4000 cm4

Ixy= 2* ( ( 20/2)*30) * (20*1) =12000 cm^4



Tan 2T= 2*12000/ ( 54000-4000) = 0.48

T=12.82°

I principal = (58000/2) +(50000/2) *.0.90 - 12.000* 0.4327 =46308 cm4

Y max= 30/0.975 =31 cm

Smax = Smin = (M/46308 ) * 31 =M/1494

If M= 10 t-m = 1000000 kg-cm , and Smax = 1000000/1494 = 670 (kg/cm2)