Unusual problem for my course and struggling

  • Thread starter Mechanical94
  • Start date
  • Tags Tags
    Course
6 replies · 3K views · last reply
Mechanical94
Messages
7
Reaction score
0
A boom AC is supported by a rectangular steel bar BD, and it is attached to a bracket at C by a high-strength steel pin. Assume that the pin at B is adequate to sustain the loading applied to it, and that the design critical components are the bar BD and the pin at C. The factor of safety against the failure of BD by yielding in uniaxial tension is 3.0. The factor of safety against a shear failure of the pin at C is 3.3. Determine the required:


a) Thickness, t of the rectangular bar BD, whose width is b; and, b) Diameter, d, of the pin at C.

known data

P = 10 kN,
L = 3 m,
h = 2 m
Bar BD: b = 25 mm,
shear stress = 250 MPa
Pin C: = 400 MPa
 
Attachments
  • Diagram of a boom AC held horizontally at a wall by pin C, with a diagonal bar BD connecting point B on the boom to a pin D higher on the wall. A downward load P acts at end A, with dimensions L/2 and L/2 along the boom and h between the pins; labels b and t = thickness mark the bar, and View a-a shows a pin with diameter d.
    mfskf.jpg
    20.7 KB · Views: 804
In case you missed it when you read the Terms & Conditions, PF is not a homework service. You must put forth some effort in order to receive help.
 
this is my efforts on this question i believe i have part a correct in this however i can't piece together how to do part b in it
 
Attachments
  • Handwritten notes on spiral-bound paper: stress = F/A, then 250N/A = 3(25000)/A, A = 3x10^-4, b = 25x10^-3, 3x10^-4 = 1x25x10^-3, 3x10^-4 / 25000 = l, l = 0.0012 m, l = 12 mm.
    picture025.jpg
    40.3 KB · Views: 589
  • Handwritten statics working in a spiral notebook held up to a webcam, showing a beam diagram and moment equations about points, with a horizontal force balance below.
    picture024.jpg
    43.6 KB · Views: 586
Mechanical94 said:
this is my efforts on this question i believe i have part a correct in this however i can't piece together how to do part b in it

That's fairly hard to read. We prefer that you type your work into the forum posts (preferably with Latex, but at least type in the equations). If you could take a closer picture of your figure (or scan it), that would be a help too. Thanks.
 
hopefully these photos are clearer, i got an answer of 12mm for the thickness and cannot get a solution to part b
 
Attachments
  • Handwritten free-body diagram of a triangular frame with a 2 m vertical side, 1.5 m base and a 3 m sloped member at 53 degrees, with 1.5 kN load and reaction forces Cx and Cy, followed by moment equations about D and C.
    picture027.jpg
    25.9 KB · Views: 508
  • Handwritten working on lined paper in a spiral notebook: moment about C = 0 with terms (6/5 T)(15), -(Dx)(2), -(10kN)(3), giving 12 T - 2Dx - 30kN = 0, then Dx = 0.6T - 15 kN; horizontal sum of forces = 0 gives 3/5 T - 5 kN + Dx = 0, then 6/5 T = 30 kN and T = 25 kN.
    picture028.jpg
    26.4 KB · Views: 565
  • Handwritten working on a spiral notebook page: stress = F/A, then 250 MPa = 3(25000)/A, A = 3x10^-4, b = 25x10^-3, 3x10^-4 = 1x25x10^-3, 3x10^-4 / 25x10^-3 = t, = 0.0012 m, = 12 mm.
    picture029.jpg
    38 KB · Views: 549
The dimensions used in your answer don't match with the symbolic notation used in the question. You could try using the property that when a body is acted upon by three forces that are in equilibrium, then they must all meet at one point (otherwise there is a turning moment). You could use that property to determine the force on the hinge at C. Even if you don't do it that way, the method I suggest acts as a check on the method you do use.