Question

'BACK STANDARD VTEW PRİNTER VERSİON P9.029 G0 Multipart Part 1 The cantilever beam shown is subjected to a concentrated load of P- 125 kN. The cross-sectional dimensions of the t ectangular tube shape are shewn, where b- 150 mm, d 250 mm -8 mm, and 60 mm. Determine (a) the stear stress at point K, located d- b) the maximum horizontal shear stress in the rectangular tube shape. which s loceted d - 60 mm below the cent of the rectanguilar tube shspe. d of the rectangular tube shape. УК : (typ.) Determine the moment of inertia I, for the cross-section abou: the z centroidal axis Answer: the tolerance is +/-2% Click if you would like to Show work for this question: asnstenwoch Attempts: Unlimited SAVE FOR LATER SUBHIT ANEIWER Part 2 connpute the val of of the first r ont of area Q t is asociated with Pol t κ which s located d 60 mm be ow the cent old of the rectangular tube shape. Answer: Q the tolerance is +/-2% Click if you would like to Show Work for this ouestinen

Answer 1

Q1 part (a)

$\tau=\frac{F*A*\widehat{y}}{I*b}$ (eq. 1)

where

$\tau$ = Shear stress

F= Applied load

$\widehat{y}$ = mid point of area of interest

I = Moment of inertia

A= area

b = width at point of interest

In our case

F= 125 kN

$A(flange)*\widehat{y}$ $=B(\frac{d-(d-2*t)}{2})*(\frac{d-(d-2*t)}{4})$ (eq. 2)

$A(web )*\widehat{y}$ $=2*t(\frac{(d-2*t)}{2}-y)*\frac{1}{2}(\frac{(d-2*t)}{2}+y)$ (eq. 3)

$A*\widehat{y} (total) =$ $A(flange)*\widehat{y}$ + $A(web )*\widehat{y}$

= 85511.99 mm^3

   $I =\frac{bd^3-hk^3}{12}$ (eq.4)

here h=b - 2*t=150-2*8 =134

k= d - 2*t=250-2*8 = 234

I = 52235072 $mm^4$

A= (b*d)-(h*k)/2-(2*t*y) = 2112 $mm^2$ (here y is 60 mm that is given in question.)

b = 2*t= 2*8= 16 mm

Putting these values in eq. 1

$\tau$ = 12.789539   $mm^2$ at point 60 mm below centroid.

Q1 part(b)

In eq. 2 and eq . 3

take y= 0

as maximum stress occurs at centroid

then

$\widehat{y}$ = 62.5 mm

A= 3072 $mm^2$

$\tau$ = 28.71633   $mm^2$

Part 2

$\widehat{y} (total) =$ d/2 -   $A(flange)*\widehat{y}$ + $A(web )*\widehat{y}$

=250/2- 85511.99/3872

= 102.91 mm

Part 3

answered above at point k stress

Part 4

area calculation in part 1