A bolted connection shown in the figure is bolted with a A325 bolts with an allowable shearing stress of 207 MPa
when
threads are excluded from shear plane. A36 steel is used with an Fy= 248MPa, Fu= 400 MPa. Assume bolt hole diameter
to be 2mm bigger than bolt diameter. Diameter of bolt is 20mm.
Use bearing type connection.
a.) Compute the tensile capacity of the connection if shearing stress governs.
b.) Compute the tensile capacity of the connection if bearing stress governs.
SOLUTION:
a.) Tensile capacity if shearing stress governs. (Double Shear)
T=Asfvn (2)
𝜋
𝑇 = (20)2 (207)(9)(2)
4
𝑻 = 𝟏𝟏𝟕𝟎. 𝟓𝟔 𝒌𝑵
b.) Tensile capacity if bearing stress governs.
T=Apfpn
𝑇 = 20(20)(1.2𝐹𝑢 )(9)
𝑇 = 20(20)(1.2)(400)(9)
𝑻 = 𝟏𝟕𝟐𝟖𝒌𝑵
The cross section shown in figure is that of a structural member known as a Z section. Determine the values of I x and Iy.
Determine the moment of inertia of the area shown in figure with respect to its centroidal axes.
A hollow square cross-section consists of an 8in square from which is subtracted a concentrically placed squarer
4in by 4in. Find the polar moment of inertia & the polar radius of gyration with respect to a z axis passing
through one of the outside corners.
Solution:
Determine the values of Ix and Iy.
For the given riveted connection shown, having the allowable stresses :
Fy = 248 MPa
Fv = 150 MPa (shear stress)
Tensile Stress = 0.60 Fy
Bearing Stress = 1.35 Fy
Rivet hole has a diameter 2 mm greater than the rivet diameter.
The main plate has a width of 300 mm and a thickness of 20 mm.
1. Compute the dia. of rivets such the shear capacity of the rivets is equal to its bearing capacity.
2. Compute the max. value of P so as not to exceed the allowable shearing stress.
3. Compute the max. value of P so as not to exceed the allowable tensile stress of plates.
Solution :
1. Diameter of rivets :
𝐴𝑏 𝑆𝑏 = 𝐴𝑣 𝐹𝑣
Double shear :
𝜋
d(20)(1.35)Fy = 4 (𝑑)2(150)(2)
𝜋150𝑑
20(1.35)(248) = (2)
4
d = 28.42
Use d = 28 mm 𝜃
2. Capacity P if shear governs :
P = 𝐴𝑣 𝐹𝑣
𝜋
P = 4 (28)2 (4)(150)(2)
P = 738903 N
P = 739 kN
3. Capacity P if tensile stress governs :
Dia. of hole = 28+2
Dia. of hole = 30 mm
P= 𝐴𝑛 0.60 Fy
P = [ 300 – 2(30) ] 20 (0.60) (248)
P = 714240 N
P = 714 kN
