1.050 Solid Mechanics, Fall 2004
Problem Set 9 Solution
Problem 9.1
10.086 in
10.244 in
w0 = 1000 lb/ft
A
B
C
L = 10 ft
RA
RB
SFD
Mb(max)
BMD
Figure 1.1
∑F
Y
= 0; R A + RB − 1000(10) = 0
R A = RB
1000(10)
= RB = 5000 lbs
2
M b(max) = 5000(5) − 5000(2.5) = 12500
lbs - ft
RA =
σx =
Mb ⋅ y
12500 ⋅ 10.244 ⋅ 12
lbs
= 2229.5 2
=
I
2 ⋅ 344.6
in
For the shear stress at the NA (at the left end of the beam), we get that
( )
V
V
ydA =
y A'
∫
Ib
Ib
5000
[(10.086)(0.682)(4.781) + (4.494)(0.422)(2.247 )]
σ xy @ NA =
(344.6)(0.422)
lbs
σ xy @ NA = 1277.3 2
in
σ xy @ NA =
From the stress transformation equations and the σ’xy must be zero on the plane that have
maximum and minimum normal stresses, we get that
⎡σ x − σ y ⎤
⎥ (sin 2φ ) + σ xy (cos 2φ ) = 0
⎣ 2 ⎦
σ xy' = − ⎢
1277.3(
cos 2φ ) = 0
cos 2φ = 0
2φ = 90 o
φ = 45 o
At the left end of the beam, the normal stress caused by the bending moment (σx) is zero.
Therefore, we get that
⎡σ x + σ y ⎤ ⎡σ x − σ y ⎤
⎥+⎢
⎥ (cos 2φ ) + σ xy (sin 2φ )
⎣ 2 ⎦ ⎣ 2 ⎦
σ x' = ⎢
σ x' = 1277.3(sin 90 o )
lbs
in 2
⎡σ x + σ y ⎤ ⎡σ x − σ y ⎤
= ⎢
⎥−⎢
⎥ (cos 2φ ) − σ xy (sin 2φ )
⎣ 2 ⎦ ⎣ 2 ⎦
σ x' = 1277.3
σ y'
σ y' = − 1277.3(sin 90 o )
σ y' = − 1277.3
lbs
in 2
For the shear stress at the point where the web meets the flange (at the left end of the
beam), we get that
( )
V
V
ydA =
y A'
∫
Ib
Ib
5000
[(10.086)(0.682)(4.781)]
=
(344.6)(0.422)
lbs
= 1130.7 2
in
σ xy =
σ xy
σ xy
Note that the maximum normal stress caused by the shear force at the left end of the
beam is much less than the maximum normal stress caused by the beam bending at the
middle (pt.C) of the beam (1277.3 psi vs. 2229.5 psi). Therefore, the maximum normal
stress that you should use in the design of the beam is still that of the flexure formula at
the maximum bending moment (2229.5 psi). Also note that the shear force at the middle
of the beam is zero and hence have no effect on the maximum normal stress at that point.
Problem 9.2
The main differences between the σx from the three section are from the y/I factor. Let’s
look at the y/I factor of each case. Case 1, 2, and 3 represent the I section wood chip,
2x10, and 4x10 section, respectively. Also note that the moment of inertia, I, for a
rectangle section is bh3/12.
Case 1:
7
y
=
= 0.0219 in - 3
320
I1
Case 2:
y
I
2
y
I
2
Case 3:
5 ⋅ 12
= 0.030 in - 3
3
2 ⋅ 10
5 ⋅ 12
=
= 0.015 in - 3
3
4 ⋅ 10
=
Therefore, we get that σx3 < σx1 < σx2 and
σ x1
σ
= 0.73 and x1 = 0.1.46 .
σ x2
σ x3
Problem 9.3
In this solution, I show you how to derive the moment of inertia, I, using the direct
integration method which is different from the method used in the class (in the class you
use J to calculate I). Figure 3.1 shows the sketch of the solid and hollow circular section
with relevant dimensions for the calculation of the moment of inertia.
y
dA = ρdθfρ
dρ
ρ
dθ
y = ρsinθ
θ
x
Solid Circular Section
y
dA = ρdθfρ
dρ
t
ρ
R
dθ
y = ρsinθ
θ
x
Hollow Circular Section
Figure 3.1
For a solid circular section with radius of r:
IX =
IX =
IX =
r 2π
∫ ∫ ρ (sin θ )dθdρ
2
∫ y dA =
4 2π
r
4
∫ sin
2
2
2
0 0
θ dθ
0
πr
4
4
For a hollow circular section with radius of R and thickness of t:
IX =
R+ t
⎡ρ4 ⎤
IX = ⎢ ⎥
⎣ 4 ⎦R
IX =
IX =
IX =
R + t 2π
∫ ∫ ρ (sin θ )dθdρ
2
∫ y dA =
2
2
0
R
2π
∫ sin
2
θ dθ
0
(R + t )4 − R 4 (π )
π
4
π
4
4
(R
4
+ 4 R 3t + 6 R 2 t 2 + 4Rt 3 + t 4 − R 4
(4R t + 6 R t
3
2 2
+ 4Rt 3 + t 4
)
)
For a hollow circular section with small t, we get that
IX =
π
4
(4 R t + 6R t )
3
2 2
From 2πRt = πr 2 , we get that
I x ( solid ) =
π (2Rt )2
=
π
(4 R t )
4
2 2
4
Therefore, for the same area, the Ix(hollow) is larger than Ix(solid).
I X ( hollow) =
π
4
(
)
4 R 3t + 6 R 2 t 2 = πR 3t +
πR 2 t 2
2
+ I X ( solid )