Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Note: The resultant force should always be used in computing the shear stress in a pin
or bolt.
(a) The cross-sectional area of a 10 mm diameter pin is A pin = 78.540 mm 2 . Since pin A
is a single-shear connection, its shear area A V is equal to the pin cross-sectional area
A pin . The average shear stress in pin A is found from the shear force V A that acts on
the pin (i.e., the 6.448 kN reaction force) and A V :
VA
(6.448 kN)(1, 000 N/kN)
τ A = = = 82.1 MPa
2
A 78.540 mm
V
Pin B is a double-shear connection; therefore, the pin area subjected to shear stress A V
is equal to twice the pin cross-sectional area A pin . The shear force V B that acts on the
pin equals the resultant force at B. Thus, the average shear stress in pin B is
VB
(14.210 kN)(1, 000 N/kN)
τ B = = = 90.5 MPa
2
A 2(78.540 mm )
V
By Equation (4.2), the pin factors of safety with respect to the 280 MPa ultimate shear
strength are
FS
A
τ failure 280 MPa
τ failure 280 MPa
= = = 3.41 and FSB
= = = 3.09
τ 82.1 MPa
τ 90.5 MPa
actual
(b) The bearing stress at B occurs on the contact surface between the 10 mm diameter
pin and the 6 mm thick steel plate. Although the actual stress distribution in the
steel plate at this contact point is quite complex, the average bearing stress is
customarily computed from the contact force and a projected area equal to the
product of the pin diameter and the plate thickness. Therefore, the average bearing
stress in the steel plate at pin B is computed as
actual
RB
(14.210 kN)(1, 000 N/kN)
σ b = = = 236.8 MPa
d t (10mm)(6 mm)
B
The factor of safety of the plate with respect to the 530 MPa ultimate bearing
strength is
FS
530 MPa
= = 2.24
236.8 MPa
Ans.
bearing Ans.
ExAMpLE 4.2
y
(1) (2)
45° 30°
P
x
A truss joint is shown in the sketch. Member (1) has a cross-sectional area of
7.22 in. 2 , and member (2) has a cross-sectional area of 3.88 in. 2 . Both members
are A36 steel with a yield strength of 36 ksi. If a factor of safety of 1.5
is required, determine the maximum load P that may be applied to the joint.
Plan the Solution
Since truss members are two-force members, two equilibrium equations can
be written for the concurrent force system. From these equations, the unknown
load P can be expressed in terms of member forces F 1 and F 2 . An
allowable stress can then be determined from the yield strength of the steel
and the specified factor of safety. With the allowable stress and the cross-sectional area,
the maximum allowable member force can be determined. However, it is not likely that
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