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

The following equilibrium equations can be developed from the FBD:
∑ Fx = Ax − F1
cos(33.69 ° ) = 0
(a)
∑ Fy = Ay + F1 sin( 33.69° ) − 20 kips = 0
(b)
∑ MA = F1sin(33.69 ° )(72 in.) + F1cos(33.69 ° )(8 in.)
− (20 kips)(72 in. + 36 in.) = 0
(c)
From Equation (c), the internal axial force in rod (1) is found to be F 1 = 46.4 kips. This
result can be substituted into Equations (a) and (b) to obtain the reactions at pin A: A x =
38.6 kips and A y = −5.71 kips. Since the value computed for A y is negative, this reaction
force actually acts opposite to the direction assumed initially.
FBD Exposing internal Forces at H
An FBD that shows the external reaction forces at pin A is cut through the
section containing point H. The internal forces acting at the section of
interest can be calculated from this FBD.
The internal axial force is F = 38.6 kips, acting in compression. The
internal shear force is V = 5.71 kips, acting upward on the exposed right
face (i.e., the +x face) shown in the FBD. The internal bending moment
can be calculated by summing moments about the centerline of the HSS
at the section containing point H:
38.6 kips
A
H
27 in.
5.71 kips
FBD at H.
5.71 kips
38.6 kips
154.2 kip·in.
∑ M = (5.71 kips)(27 in.) − M = 0 ∴ M = 154.2 kip⋅ in. (d)
H
Section Properties
The cross-sectional area of the HSS is
6 in.
A = (6 in.)(10in.) − (5.5 in.)(9.5 in.) = 7.75 in. 2
The moment of inertia of the cross-sectional area about the z centroidal
axis is
I
z
3 3
(6 in.)(10in.) (5.5 in.)(9.5 in.)
= − = 107.04 in.
12
12
The first moment of area corresponding to point H is calculated for the
highlighted area as
4
2.5 in.
H
z
y
0.25 in.
10 in.
Q = 2(0.25 in.)(2.5 in.)(3.75 in.) + (5.5 in.)(0.25 in.)(4.875 in.)
H
= 11.391 in.
3
Stress Calculations
Axial stress due to F: The internal axial force F = 38.6 kips creates a uniformly
distributed compressive normal stress that acts in the x direction.
The stress magnitude is
F 38.6 kips
σ axial = = = 4.98 ksi(C)
2
A 7.75 in.
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