Tensile stress in perforated steel plate

[DevonZA]

Homework Statement

Two lengths of steel plate 100mm wide and 15mm thick are riveted together by a single 18mm diameter rivet. If the joint carries a tensile load of 8kN calculate:
a) The shear stress in the rivet
b) The tensile stress in the perforated plate

Homework Equations

stress = force / cross sectional area

The Attempt at a Solution

a) stress = force / cross sectional area
= 8x10^3 / pi/4(0.018)^2
= 31.43MPa

This is the answer given in the textbook so I am satisfied it is correct.

b) This is where I am stuck.
The given answer is 6.51MPa.

I thought maybe if I use 100mmX15mm (1500mm) as the area for the plate and subtract the area of the 18mm diameter rivet (pi/4(0.018)^2) I could divide the 8kN force (although this is the tensile load of the joint) by this area and get the stress for the plate.

However if I do the above:

8x10^3 / [(1.5x10^-3)-pi/4(0.018)^2]
= 6.42MPa

Where am I going wrong?

 

[gneill]

I thought maybe if I use 100mmX15mm (1500mm) as the area for the plate and subtract the area of the 18mm diameter rivet (pi/4(0.018)^2)...

Those areas are not oriented in the same plane. One is perpendicular to the plate surface, the other parallel to the plate surface.

What's the minimum cross sectional area of the plate? Note that would be the area of the section (so a slice through contiguous material).

 

[DevonZA]

Those areas are not oriented in the same plane. One is perpendicular to the plate surface, the other parallel to the plate surface.

What's the minimum cross sectional area of the plate? Note that would be the area of the section (so a slice through contiguous material).

100mmX15mm?

 

[gneill]

100mmX15mm?

Nope. Suppose you make your cross section through where the rivet is. How much plate material is sliced?

 

[DevonZA]

Nope. Suppose you make your cross section through where the rivet is. How much plate material is sliced?

Um if each plate is 100mm wide and 15mm thick then I would say 100mmX15mm - 2X15mmX18mm for the hole
2(100x15)-(2)(15)(18) = 2460mm^2

Not sure if I am visualizing this correctly..

 

[gneill]

Not sure why you're multiplying by two. You want to deal with only one plate.

 

[DevonZA]

Not sure why you're multiplying by two. You want to deal with only one plate.

Okay so 100x15mm - 15x18mm = 1230mm^2

Then stress = force / cross sectional area
= 8x10^3 / 1.23x10^-3
= 6.5MPa

thank you