Prob. 16.30 Rivetted Joint in Torsion
Since there are rivets at two different radii from the centroid 
(the middle rivet), it is best to follow the method of Example 
16.9 and compute the moment of inertia J = Sum[ (Rj)^2 Aj ], 
where Aj is the area of the rivet stem, and Rj is the radius 
from the centroid to each rivet.  Here, there are 4 rivets at 
R = 120 mm, and 4 at R = (120)(1.414) mm , as well as the central 
rivet at R = 0.  Since A = ( pi d^2)/4 , J = 135,717 d^2 , 
where d is in mm.  We have to carry the "d" around as a variable, 
since we don't know what it is.  Then TauTorsion = T r / J, 
and you pick the biggest r for the biggest Tau = 15,000 / (d^2).
The TauDirect is the average direct shear = P / (9A) = 3395 / (d^2).  
We vectorially combine these.  Since TauTorsion is at a 45° angle, 
it has horizontal and vertical components of (0.707)(15,000)/d^2 = 
10,607/d^2.  The vector total is 17,566/d^2.  This gets set 
to 1/Ns of the allowable shear yield = (0.4)(600)/3 = 80 MPa.  
So d = 14.8 mm.

Problem 16.31 Rivets With Axial Load and Shear
Pivoting around the lower corner of the block makes the extension 
of the top rivet 140/20 = 7 times the extension of the bottom rivet, 
so Ft (Ftop) = 7 Fb (Fbottom).  Taking the moment about the lower 
corner, and remembering that there are 2 top and 2 bottom rivets, 
2(20 Fb + 140 (7 Fb)) = (250)(24,000), so Fb = 3000 N.  
Then Ft = 21,000 N.  Tensile stress = Ft / A or 26,738/ d^2.  
Average direct shear with 4 rivets = P/4A = 7639/ d^2.  
VonMises = 29,833/ d^2.  Then d = 12.2 mm.

Prob. 16.32 Welded Joint in Torsion
Centroid is (0.8, 1.8) from upper left corner.  By the Table, 
Ju = 54.53 in3.; J = 38.56h in4.  We have to carry the "h" around 
as a variable, since we don't know what it is.
TauDirect = W/WeldArea = (4000)/((4+6)(0.707)h) = 565.8/h psi.
TauTors = Tr/J.  Calculate as 2284/h for point B where r = 3.67in 
and 2664/h for point C where r = 4.28in.  Next break TauTors' 
into vertical and horizontal components so can combine vertical 
component with TauDirect and compute total component.  Totals 
come out to be 2792/h at B and 2664/h at C.  B is bigger, so 
select B.  Again set TauTotal to 1/Ns of the allowable shear 
yield = (0.4)(67000)/3 = 8933 psi.  Then h = 0.3125 in = 5/16.  
The throat length is (0.707)(0.3125) = 0.221 in.

Problem 16.36 Welded Joint in Bending
Redrawing the picture shows that b = 50mm and d = 20mm.  Z centroid
 is 17.86mm from corner; Y bar is 2.86 mm from top.  We'll analyze 
the top because it is in tension.  ZuTop = (4bd + d^2)/6 = 733.3 mm^2.  
Ztop = 3628 mm^3, and SigmaTop = M/Z = 241 MPa tensile.  
(SigmaBot = 1447 MPa compressive.)  Weld area is 346.4 mm^2, so 
average direct shear is P/A = 10.1 MPa.  The VonMises is 241.6 MPa.  
The E60XX electrode has a yield strength of 345 MPa, so the factor 
of safety is a puny 1.43.