PROBLEMS (A.1 through A.4) In terms of dimensions b, h, and t, derive expressions for
distances z and y, as needed, to the centroids C of the areas shown in Figs. PA.1 through PA.4. Calculate
the values of x and y for b = 3 in., h = 4 in., and t = 3/8 in.
SOLUTION (A.1)
b
A1
y
t
x
C
y1
A2
h
y
y2
x’
2t
Due to symmetry about y axis:
x =0
We have
(PA.1a)
t
h
bt (h + ) + 2ht ( )
A y + A2 y2
2
2
y= 1 1
=
A1 + A2
bt + 2ht
or
b(2h + t ) + 2h 2
2(b + 2h)
Substituting the given data
3(2 × 4 + 0.375) + 2(4) 2
y=
= 2.597 in.
2(3 + 2 × 4)
y=
(PA.1b)
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SOLUTION (A.2)
b
A2
y
2t
x
C
h
A1
A1
y1 y
y2
x’
t
t
Due to symmetry about y axis:
x =0
We write
2 A y + A2 y2
y= 1 1
2 A1 + A2
2ht (h 2) + (b − 2t )(2t )(h − t )
=
2ht + (b − 2t )2t
or
1 h 2 + 2(b − 2t )(h − t )
y=
2
b + h − 2t
Substituting the given numerical values:
1 (4) 2 + 2(3 − 2 × 0.375)(4 − 0.375)
y=
= 2.585 in.
2
3 + 4 − 2 × 0.375
(PA.2a)
(PA.2b)
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SOLUTION (A.3)
y’
y’
y’
x1
2b/3
A2
x +
A1
h
h
2
x
A3
−
x2
x
x3
b/6
b/3
Due to symmetry about x axis:
y =0
We obtain
A x + A2 x2 + A3 x3
x= 1 1
A1 + A2 − A3
bh b 1 h 2b b 2b
b b
( + ) − π ( )2
+
6 3
= 3 6 22 3 3 9 2
bh bh π b
+ −
3
6
36
or
1 16bh − π b 2
x=
3 18 − π b
(PA.3a)
(PA.3b)
Introducing the given data:
1 16(3 × 4) − π (3) 2
= 0.872 in.
x=
3 18(4) − 3π
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SOLUTION (A.4)
y
y’
A2
x
x2
x
x1
t
C
y1
y
h
y1
x’
A1
b
x=
or
A1 x1 + A2 x2
=
A1 + A2
b−t
t
) + ht (b − )
2
2
(b − t )t + ht
(b − t )t (
(b − t ) 2 + h(2b − t)
x=
2(b + h − t )
=
(PA.4a)
(3 − 0.375) 2 + 4(6 − 0.375)
= 2.218 in.
2(3 + 4 + 0.375)
Similarly,
t
h
t (b − t )( ) + ht ( )
A y + A2 y2
2
2
=
y= 1 1
A1 + A2
(b − t )t + ht
or
=
(b − t )t + h 2
2(b + h − t )
=
(3 − 0.375)(0.375) + (4) 2
= 1.282 in.
2(3 + 4 − 0.375)
(PA.4b)
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________________________________________________________________________
PROBLEMS (A.5 and A.6) In terms of dimensions b, h, and t, write expressions for distances z
and y, as needed, to the centroids C of the areas shown in Figs. PA. 5 and PA.6. What are the values of .x
and y for b = 3 in., h = 4 in., and t = 3/8 in.?
SOLUTION (A.5)
b/2
y’
y1
y
x
x1
y2
x
C
h
A2
A1
x2
y3
x3
A3
y
x’
b
A1 x1 + A2 x2 + A3 x3
x=
A1 + A2 + A3
bt b
t
b
+ (h − 2t )t ( ) + bt
2
2
= 2 4
bt
+ (h − 2t )t + bt
2
or
1 5b 2 + 4(h − 2t )t
x=
4 3b + 2h − 4t
1 5(3) 2 + 4(4 − 2 × 0.375)(0.375)
= 0.804 in.
=
4
3(3) + 2(4) − 4(0.375)
Similarly,
bt
t
h
t
(h − ) + (h − 2t )t ( ) + bt ( )
A1 y1 + A2 y2 + A3 y3
2
2
2
y=
= 2
bt
A1 + A2 + A3
+ (h − 2t )t + bt
2
or
1 b(2h + t ) + 2(h − 2t )h
y=
2
3b + 2h − 4t
1 3(2 × 4 + 0.375) + 2(4 − 2 × 0.375)4
=
= 1.649 in.
2
3(3) + 2(4) − 4(0.375)
(PA.5a)
(PA.5b)
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SOLUTION (A.6)
b
y’
y
x2
h
y3
x
C
A1
A2
y2 y
x3
x
y1
A3
x’
A1 x1 + A2 x2 + A3 x3
A1 + A2 + A3
b
b 1
b+t
bt ( ) + ht ( ) + (b + t )t (
)
2
2 2
4
=
1
bt + ht + (b + t )t
2
x=
or
1 5b 2 + 2b + (t + 2h) + t 2
4
3b + 2h + t
2
1 5(3) + 2(3)(0.375 + 2 × 4)(0.375) 2
=
= 1.373 in.
4
3(3) + 2(4) − 4(0.375)
x=
(PA.6a)
In a like manner
A y + A2 y2 + A3 y3
y= 1 1
A1 + A2 + A3
h
t
3t
1
bt (h + ) + ht ( + t ) + (b + t )t ( )
2
2
2
2
=
1
bt + ht + (b + t )t
2
1 2h(2b + h + 2t ) + t 2 + 7bt
or
y=
(PA.6b)
2
3b + 2h + t
1 2(4)(2 × 3 + 4 + 2 × 0.375 + (0.375) 2 + 7(3)0.375
=
= 2.705 in.
2
3(3) + 2(4) − 0.375
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________________________________________________________________________
PROBLEMS (A.7 through A.9) In terms of dimensions b, h, and t, derive expressions for
the centroidal moments of inertia and the products of inertia of the areas shown in Figs.PA.1
through PA. 3. Write the equations for the centroidal principal moments of inertia and their
orientations with respect to x in a form suitable for computer analysis; calculate the value of
these quantities for b = 50 mm. h = 75 mm, and t = 5 mm.
SOLUTION (A.7)
y
b
t
x
C
h
2t
y
1
I x = ∑ ( bh3 + Ay 2 )
12
1
t
1
h
or
I x = bt 3 + bt (h − y + ) 2 + (2t )h3 + 2th( y − ) 2
12
2
12
2
1 3
I y = ∑ ( bh + Ax 2 )
12
or
1
1
I y = tb3 + h(2h)3
12
12
Due to symmetry
I xy = 0
For b=50 mm, h=75 mm, and t=5 mm, Eqs. (PA.1) give
x =0
y = 47.5 mm
Thus, substituting the data:
I x = 65.2 × 104 mm 4
I y = 5.83 ×104 mm 4
I xy = 0
Equations (A.13) and (A.14):
I1 = I x
I2 = I y
θθ = 0
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SOLUTION (A.8)
b
A2
y
2t
x
C
h
y
A1
A1
t
t
1
I x = ∑ ( bh3 + Ay 2 )
12
1 3
h
1
= 2[ th + ht ( y − ) 2 ] + (b − 2t )2t (h − y − t ) 2 + (b − 2t )(2t )3
12
2
12
1 3
I y = ∑ ( bh + Ax 2 )
12
1
b t
2t
= 2[ ht 3 + th( − ) 2 ] + (b − 2t )3
12
2 2
12
Due to symmetry
I xy = 0
For b=50 mm, h=75 mm, and t=5 mm, Eqs.(PA.2): x = 0, y = 48.8 mm.
Then,
I x = 63 × 104 mm 4
I y = 43.5 × 104 mm 4
I xy = 0
Equations (A.13) and (A.14):
I1 = I x
I2 = I y
θp = 0
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SOLUTION (A.9)
y
1
I x = ∑ ( bh3 + Ay 2 )
12
x
or
A3
1 b 3 1 2b h 3 π b 4
Ix =
h +
( ) − ( )
h
12 3
48 3 2
4 6
c A2
1
A1
I y = ∑ ( hb3 + Ax 2 )
12
b/3 2b/3
or
1 b
b
b
I y = h( )3 + h( )( x − ) 2
12 3
3
6
1 h 2b 3 h 2b b
2b
π b
b
( ) +
( − x + )2 − [ ( )2 + ( − x )2 ]
+
36 2 3
2 3 3
9
4 6
3
Due to symmetry
x
I xy = 0
For b=50 mm, h=75 mm, and t=5 mm, Eqs. (PA.3) : y = 0, x = 14.57 mm.
Then,
I x = 61.9 × 104 mm 4
I x = 32.94 × 104 mm 4
I xy = 0
Equations (A.13) and (A.14):
I1 = I x
I2 = I y
θp = 0
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________________________________________________________________________
PROBLEMS (A.10 through A.12) In terms of dimensions b, h, and t, write expressions for the
centroidal moments of inertia and the products of inertia of the areas shown in Figs.PA.4 through PA. 6.
Derive the equations for the centroidal principal moments of inertia and their orientations with respect to x
in a form suitable for computer analysis; calculate the value of these quantities for b = 50 mm. h = 75 mm,
and t = 5 mm.
SOLUTION (A.10)
y
A2
x
h
C
A1
y
x
t
b
1
I x = ∑ ( bh3 + Ay 2 )
12
1
t
1
h
= (b − t )t 3 + t (b − t )( y − ) 2 + th3 + ht ( − y ) 2
12
2
12
2
1
I y = ∑ ( hb3 + Ax 2 )
12
1
b−t 2 1 3
t
) + ht + ht (b − x − ) 2
= t (b − t )3 + (b − t )t ( x −
12
2
12
2
b−t
t
t h
I xy = 0 + (b − t )t (
− x )( − y ) + 0 + ht (b − x − )( − y )
2
2
2 2
For b=50 mm, h=75 mm, and t=5 mm, Eqs.(PA.4): x = 38.125 mm, y = 24.375
Then, we have
I x = 34.9 × 104 mm 4
I y = 12.7 ×104 mm 4
I xy = 12.3 × 104 mm 4
Equations (A.13) and (A.14):
I1 = 40.4 × 104 mm 4
I1 = 7.2 × 104 mm 4
θ p ' = 24o
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SOLUTION (A.11)
b/2
y
A1
x
h
x
C
A2
y
t
A3
b
1
I x = ∑ ( bh3 + Ay 2 )
12
1 b 3 bt
t
1
h
1
t
=
t + (h − y − ) 2 + t (h − 2t )3 + t (h − 2t )( − y ) 2 + bt 3 + bt ( y − ) 2
12 2
2
2
12
2
12
2
1
I y = ∑ ( hb3 + Ax 2 )
12
1 b 3 bt
b
1
t
1
b
= t ( ) + ( x − ) 2 + (h − 2t )t 3 + (h − 2t )t ( x − ) + tb3 + bt ( − x ) 2
12 2
2
4
12
2 12
2
I xy = ∑ ( I x y + Ax y )
= 0+
bt b
t
t
h
b
t
( − x )(h − y − ) + 0 + (h − 2t )t ( − x )( − y ) + 0 + bt ( − x )(− y + )
2 4
2
2
2
2
2
For b=50 mm, h=75 mm, and t=5 mm, Eqs. (PA.5) give: x = 12.32 mm,
y = 31.25 mm
Then,
I x = 54.7 × 104 mm 4
I y = 13.1×104 mm 4
I xy = −11.0 × 104 mm 4
Equations (A.13) and (A.14) result in
I1 = 57.5 ×104 mm 4
I 2 = 10.3 × 104 mm 4
θ p ' = 13.9o
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SOLUTION (A.12)
y
b A
1
t
A2
h
x
C
A3
y
t
b/2
1
I x = ∑ ( bh3 + Ay 2 )
12
h
1 3
3t
1
= bt + bt (h + − y ) 2 + th3 + th( y − − t ) 2
12
2
12
2
b t
t
1 b t
+ ( + )t 3 + ( + )t ( y − ) 2
12 2 2
2 2
2
1
I y = ∑ ( hb3 + Ax 2 )
12
b
b
b t
b+t
1 3
1
1 b t
+ x )2
= tb + tb( − x ) 2 + ht 3 + ht ( − x ) 2 + ( + )t + ( + )t (−
12
2
12
2
12 2 2
2 2
4
I xy = ∑ ( I x y + Ax y )
b
h
b
3t
= 0 + bt ( − x )(h + − y ) + 0 + th(− y + + t )( − x )
2
2
2
2
b t
b t
t
+0 + ( + )t (− x + + )(− y + )
2 2
4 4
2
For b=50 mm, h=75 mm, and t=5 mm, Eqs. (PA.6) yield: x = 22.97 mm,
y = 48.8 mm
Then,
I x = 77.0 × 104 mm 4
I y = 7.6 × 104 mm 4
I xy = 7.1× 104 mm 4
Continued on next slide
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Equations (A.13) and (A.14) give
I1 = 77.7 × 104 mm 4
I 2 = 6.9 ×104 mm 4
θ p ' = 5.78o
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Continued on next slide
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Problem 6.70 continued
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PROBLEM (7.1) A copper wire of diameter d and modulus of elasticity E is coiled around
a spool of radius r . Compute the maximum stress and the bending moment in the wire.
Given: d = 1/32 in., r = 6 in., E = 17 x 10 6 psi
SOLUTION
d
ρ
r
C
r = 6 in.
ρ = 6.015625 in.
1
d=
= 0.03125 in.
32
Equation (7.4):
Ec 17 × 106 (0.015625)
=
σ max =
= 44.2 ksi
ρ
6.015625
Mc 32M
σ max =
=
πd3
I
or
πd3
M = σ max
32
44.2 ×103 (0.03125)3 π
=
= 0.138 lb ⋅ in.
32
________________________________________________________________________
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PROBLEM (7.2) An aluminum bar of modulus of elasticity E, width b, depth h , and length
L , is acted upon by bending moments M at its ends so that the midpoint deflection is δ (Fig.
P7.2). Determine the maximum longitudinal strain and the bending moment in the bar.
Given: E = 70 GPa, b = 10 mm, h = 30 mm, L = 2 m,
δ = 10 mm
Assumption: The deflection curve is nearly flat so that the distance between the ends of the bent
bar equals the length of the beam as indicated in the figure.
SOLUTION
0.01(0.03)3
12
= 22.5 ×10−9 m 4
I=
O
ρ
y
1m
ρ
10 mm
1m
y
M
M
x
z
h=30 mm
vmax
From geometry:
ρ 2 − 1 = ( ρ − vmax ) 2 = ( ρ − 0.01) 2 = ρ 2 − 0.02 ρ + 10−4
or
ρ = 50 m
Equation (7.2), for y = h 2 :
h
0.03
ε max =
=
= 300 μ
2 ρ 2(50)
Equation (7.6):
EI 70 ×109 (22.5 ×10−9 )
M=
=
= 31.5 N ⋅ m
ρ
50
________________________________________________________________________
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PROBLEM (7.3)
A steel beam having the modulus elasticity E and allowable normal stress
that may be
σ all experiences pure bending (Fig. P7.3). Calculate the maximum moment M
applied and the corresponding radius of curvature ρ for the two cases:
(a) The cross section is a circle of diameter d.
(b) The cross section is an equilateral triangle of sides b.
Given: d = 12 mm,
b = 15 mm,
E = 200 GPa, σ all = 160 MPa
SOLUTION
σ all =
(a)
32M
πd3
or
D=2c=12 mm
M=
σ allπ d 3
32
(160 ×106 )π (0.012)3
M=
32
= 27.14 N ⋅ m
Equation (7.4):
1 σ all
160 × 106
=
=
= 0.133 m −1
9
ρ Ec 200 ×10 (0.006)
ρ = 7.52 m
Alternatively, Eq. (7.6) yields the same result.
1 3 15(12.99)3
I = bh =
(b)
36
36
−12
= 913 × 10 m 4
2h
c=
σ all = Mc I
h=12.99 mm
3
C
o
or
60o 60
M = σ all I (2h 3)
B=15 mm
M=
160 × 106 (913 × 10−12 )
= 16.87 N ⋅ m
8.66 × 10−3
Equation (7.9):
1 σ all
160 × 106
=
=
= 0.0924 m −1
−3
9
ρ Ec 200 × 10 (8.66 × 10 )
or
ρ = 10.82 m
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________________________________________________________________________
PROBLEM (7.4) If a circular beam and equilateral triangular beam are to resist the same
maximum stress and bending moment (Fig. P7.3), what is the ratio of their cross-sectional areas?
SOLUTION
σ max =
y
32M
πd3
d
z
bh3 b b 3 3
= (
)
36 36 2
3 4
=
b
96
I=
y
h=b 3 2
z
b
b
σ max =
M (2h 3) 32 M
= 3
b
3b 4 96
Thus,
32 M 32M
= 3 ,
b
πd3
b = 1.46d
πd2
1
1.46d 3
At = (1.46d )
2
4
2
At Ac = 3.69 π = 1.19
Ac =
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________________________________________________________________________
PROBLEM (7.5) A steel rod of moludulus of elasticity E and semicircular cross section with radius
r is acted upon by a bending moment Mz (Fig. P7.5). Calculate:
(a) The maximum tensile and compressive strains in the rod.
(b) The radius of curvature of the rod.
Given: Mz = - 1.5 kip ⋅ in. ,
E = 30 x 106 psi,
r = 1 in.
SOLUTION
y
x
M
Mz=M
C
0.576 in.
r
2 in.
4r
=0.424 in.
r =1 in.
(a) I z = 0.11r = 0.11(1) = 0.11 in.
Mc 1500(0.576)
σt =
=
= 7.855 ksi
0.11
I
1500(0.424)
Mc
σc = −
=−
= −5.782 ksi
0.11
I
Then
σ
7855
εt = t =
= 262 μ
E 30 ×106
σ
5782
εc = c = −
= −193 μ
30 ×106
E
4
(b)
1
ρ
=
4
M
1500
=
,
EI 30 × 106 (0.11)
4
ρ = 2200 in. = 183.3 ft
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________________________________________________________________________
PROBLEM (7.6) A brass bar having a modulus of elasticity E and semicircular cross
section of radius r (Fig. P7.5) is bent over a cylinder of diameter D. What is the bending
moment M z in the bar?
Given: E = 15 x 10 6 psi, r = 1 in., D = 36 ft
SOLUTION
Refer to figure in
Solution of Prob. 7.5.
C
ρ
ρ = 18 ×12 + 0.424 = 216.424 in.
Mz
15 × 106 (0.576)
σ max =
=
= 39.92 ksi
ρ
216.424
Thus,
σ I 39.92 ×103 (0.11)
= 7.62 kip ⋅ in.
M z = max =
0.576
c
Ec
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________________________________________________________________________
PROBLEM (7.7) Moment M acts about the horizontal (x) axis of a square bar of sides b in
the two different positions shown in Fig. P7.7. Determine the maximum stress and the radius of
curvature of the bar, in terms of the modulus of elasticity E, b, and M, as required.
SOLUTION
Refer to Table B.6 and Appendix A.
M (b 2) 6 M
= 3
b 4 12
b
1 M
12M
M
=
=
=
4
ρ EI E (b 12) Eb 4
σ=
y
C
z
b
b
For a triangle:
bh3
bh3 bh h 2 bh3
Iz =
+ Ad 2 =
+ ( ) =
36
36 2 3
12
Hence, for square:
y
b
2
b
z
C
b
2
b 2
Therefore,
σ max =
6 2M
b3
1
b 3 b4
I z = 2[ (b 2)( ) ] =
12
12
2
Mc M (b 2) 6 2 M
=
=
Iz
b 4 12
b3
1 M 12M
=
=
ρ EI z Eb 4
σ=
ρ=
Eb 4
12 M
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________________________________________________________________________
PROBLEM (7.8) An aluminum bar of rectangular cross section is acted upon by a bending
moment M about the z axis as shown in Fig. P7.8. Calculate:
(a) The radius of curvature ρ of the beam.
(b) The radius of curvature ρ1 of a transverse cross section of the beam.
Given:
b = 25 mm,
h = 75 mm,
M = 5 kN ⋅ m,
E = 72 GPa,
ν = 0.29
SOLUTION
1 3 1
bh = (25)(75)3 = 878.91 (103 ) mm 4
12
12
1 M
5(103 )
=
=
= 0.07901
(a)
ρ EI 72(109 )(878.91×10−9 )
or
ρ = 12.657 m
I=
(b) Equation(7.3a)
ρ1 = −
ρ
12.657
=−
= −43.645 m
ν
0.29
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________________________________________________________________________
PROBLEM (7.9) An S 8 x 23 rolled-steel I beam (see Table B.7) carries a moment M
about the z axis, as shown in Fig. P7.9. Determine:
(a) The radius of curvature ρ of the beam.
(b) The radius of curvature ρ1 of a transverse cross section of the beam.
Given: E = 29 x 10 6 psi,
ν = 0.3,
M = 400 kip ⋅ in.
SOLUTION
For S150x26 rolled-steel section:
I = 64.4 in.4
(a)
400 × 103
M
=
= 2.142(10−4 )
6
ρ EI (29 × 10 )(64.4)
ρ = 4, 669 in. = 389 ft
1
=
(b) ρ1 = −
ρ
4, 669
=−
= −15,563 in. = −1, 297 ft
ν
0.3
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________________________________________________________________________
PROBLEM (7.10) A rectangular tube is made of an aluminum alloy with an allowable stress
of σ all and is acted upon by a bending moment M about the z axis (Fig. P7.10). Calculate:
(a) The bending moment.
(b) The radius of curvature ρ of th beam.
Given :
σ all = 120 MPa,
E=70 GPa
SOLUTION
y
bh
1
− (b − 2t )(h − 2t )3
12 12
1
1
= (100)(150)3 − (80)(130)3
12
12
6
4
= 13.478(10 ) mm
Iz =
t
3
t
M
C
z
t
(a)
Mc
,
σ all =
Iz
h
t
b
I
13.478(10−6 )
M = σ all =
120(106 ) = 21.565 kN ⋅ m
c
0.075
1 M
21,565
(b)
=
=
= 0.02286
9
ρ EI (70 ×10 )(13.478)(10−6 )
ρ = 43.74 m
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________________________________________________________________________
PROBLEM (7.11) An aluminum beam with an hollow circular cross section is acted upon by
a bending moment M about the z axis (Fig.P7.11). Calculate:
(a) The normal stress at point A.
(b) The normal stress at point B.
(c) The radius of curvature ρ1 of the beam of a transverse cross section.
Given: M = 400 N ⋅ m, D = 50 mm,
d = 30 mm,
E = 70 GPa,
ν = 0.29
SOLUTION
I=
π
1
( D 4 − d 4 ) = (504 − 304 ) = 267.035(103 ) mm 4
64
64
(a) y A = 25 mm = 0.025 m.
My A (400)(0.025)
=
= 37.45 MPa
σA =
I
267.035(10−9 )
(b) yB = 15 mm = 0.015 m.
MyB
(400)(0.015)
=
= 22.47 MPa
σB =
I
267.035(10−9 )
( c)
1
ρ
=
=
M
EI
z
400
70(10 )(267.035 ×10−9 )
9
= 0.0214
ρ = 46.73 m
y
M
C
B
A
d
D
Hence,
ρ1 = −
ρ
46.73
=−
= −161.14 m
ν
0.29
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________________________________________________________________________
PROBLEM (7.12) An aluminum, alloy 6061-T6, beam of hollow circular cross section (Fig.
P7.11) has an allowable bending strength σ all . Determine:
(a) The largest bending moment M about the z axis that may be applied based on a safety
factor n s with respect to yielding of the aluminum.
(b) The corresponding radius of curvature ρ of the beam.
Given: D = 2 in.,
d = 1.5 in.,
n s = 1.9
SOLUTION
σ y = 38 ksi
π
4
I=
=
64
π
64
E = 10 × 109 psi
(from Table B.4)
(D − d 4 )
(24 − 1.54 ) = 248.51(10−3 ) in.4
σy
38
= 20 ksi
ns 1.9
Mc
M (1)
σ all =
; 20(103 ) =
I
248.51(10−3 )
Solving,
M = 4.97 kip ⋅ in.
(a) c = 1 in.
(b)
1
ρ
or
=
σ all =
=
4.97(103 )
M
=
= 0.002
EI (10 × 106 )(248.51×10−3 )
ρ = 500 in = 41.67 ft
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________________________________________________________________________
PROBLEM (7.13) An S 150 x 26 steel I beam (see Table B.8) is simply supported on a
span L and carries a uniform load of intensity w.
(a) Determine the maximum value of w.
(b) Determine the width of a solid rectangular section of the same 152-mm depth to replace
the I shape.
(c) Compare the areas of the two sections.
Given: L = 10 in., σ all = 120 MPa
SOLUTION
S150x26
w
152 mm
10 m
wL
2
wL
2
As =3270 mm2
S =144x10-6
M max wL2 8
=
S
S
8Sσ all 8(144 ×10−6 )(120 × 106 )
= 1.38 kN m
w=
=
(10) 2
L2
(a) σ all =
(b)
S=
152 mm
bh 2
6
b(0.152) 2
,
144 × 10 =
6
−6
b
b = 37.4 mm
( c)
As = 3270 mm 2
Ar = 37.4(152) = 5685 mm 2
Thus,
Ar As = 1.74
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________________________________________________________________________
PROBLEM (7.14) A simply supported channel AB carries a concentrated load P at the
midpoint, as shown in Fig.P7.14. Compute the maximum tensile stress σ t and largest
compressive stress σ c in the member.
Given: L = 10 ft,
b = 14 in.,
h = 6 in.,
t = 0.5 in.,
P = 5 kips
SOLUTION
y=
∑ A y = 2A y + A y
2A + A
∑A
i
i
i
1
2
1
2
y
2
2(6 ×1)(3) + (12 ×1)(5.5)
=
2(6 × 1) + (12 × 1)
= 4.25 in.
1 in.
z
A2
A1
c2
c1
1 in.
C
y
6
1 in.
12
So c1 = 4.25 in. and c2 = 1.75 in.
1
I z = 2[ (1)(6)3 + (6 ×1)(1.25) 2 ] +
12
1
(12)(1)3 + (12 ×1)(1.25) 2 = 74.5 in.4
12
1
1
PL = (5)(10 × 12) = 300 kip ⋅ in.
4
2
Mc1 300(4.25)
σt =
=
= 17.11 ksi
74.5
Iz
Mc
300(1.75)
σc = − 2 = −
= −7.05 ksi
74.5
Iz
Mz = M =
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________________________________________________________________________
PROBLEM (7.15) A simple beam AB of a channel cross section is acted upon by a
concentrated load P at midpoint as shown in Fig. P7.14. Determine the maximum permissible
load P based upon an allowable normal stress σ all in the beam.
Given: L = 2 m,
b = 400 mm,
h = 150 mm,
t = 25 mm,
σ all = 50 MPa
SOLUTION
y=
∑ A y = 2A y + A y
2A + A
∑A
i
i
1
2
1
2
y
2
A2
A1
2(150 × 25)(75) + (350 × 25)(137.5)
=
c2
2(150 × 25) + (350 × 25)
z
C
y
25
c1
= 108.65 mm
25
So c1 = 108.65 mm and c2 = 41.35 mm
350 mm
1
3
2
I z = 2[ (25)(150) + (25 × 150)(33.65) ] +
12
1
(350)(25)3 + (350 × 25)(28.85) 2 = 30.29(106 ) mm 4
12
1
1
M z = M = PL = p (2) = P
2
2
i
25
Therefore
Mc
P(0.10192)
σ t = 1 ; 50(106 ) =
, P = 14.86 ksi
30.29(10−6 )
Iz
Mc2
P (0.0481)
σc =
50(106 ) =
, P = 31.49 kN
30.29(10−6 )
Iz
Hence Pall = 14.86 kN
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150 mm
________________________________________________________________________
PROBLEM (7.16) An unsymmetrical W beam is acted upon by a bending moment M z ,as
shown in Fig. P7.16. For the ratio of bending stresses of 5/8, at the top and bottom of the beam,
determine the width b of the bottom flange.
SOLUTION
150 mm
20 mm
y
100 mm
z’
cb
C
y
100 mm
15 mm
b
y=
5
(240) = 92.3 mm
13
8
ct = (240) = 147.7 mm
13
y = −120 + cb = −27.7 mm
cb =
ct
z
σt σb = 5 8
20 mm
Thus, y =
∑ A y = 150 × 20 ×110 − b(20)110 = −27.7
∑ A 150 × 20 + 200 ×15 + b(20)
i
∑Ay
∑A
i
i
i
i
i
Solving,
b = 301.5 mm
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________________________________________________________________________
PROBLEM (7.17) A rectangular beam of width b and depth h is to be cut from a circular
bar of diameter D, as shown in Fig. P7.17. Determine the ratio of h / b so that the beam will
carry a maximum moment M in pure bending.
SOLUTION
y
D
z
C
Mc 6M
= 2
I
bh
1
or
M = σ all bh 2
(1)
6
We have
bh 2 = b( D 2 − b 2 ) = bD 2 − b3
σ all =
h
D
b
In order bh 2 be maximum:
d (bh 2 ) d
=
(bD 2 − b3 ) = D 2 − 3b 2 = b 2 + h 2 − 3b 2 = h 2 − 2b 2 = 0
db
db
h b= 2
or
Hence, Eq. (1):
1
1
M max = σ all (b × 2b 2 ) = b3σ all
3
6
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________________________________________________________________________
PROBLEM (7.18) If a beam of the cross section shown in Fig. P7.18 is subjected to the
bending moment M z =20 kN ⋅ m, what is the total force acting on the lower flange of the beam?
SOLUTION
Dimensions shown are in millimeters.
A1
y
20
y
y
20
120
C
z
y B
180
z
A2
12.5
40
120
z
22.5
100
z’
A
120
A3
A1 y1 + A2 y2 120
+ A3 y3
y=
A1 + A2 + A3
120 × 20 × 170 + 120 × 20 × 100 + 120 × 40 × 20
=
= 77.5 mm
120 × 20 + 120 × 20 + 120 × 40
120(180)3
100(120)3
2
Iz =
+ 120 × 180(12.5) −
− 100 ×120(22.5) 2
12
12
6
4
= 41.22 × 10 mm
Mc
σA =
y
I
20 × 103 (77.5 ×10−3 )
=
41.22 × 10−6
x
= 37.6 MPa
B
77.5 mm
0.04
37.5
Fb
σ B = 37.6
= 18.19 MPa
A
77.5
1
σ avg = (σ A + σ B ) = 27.9 MPa
2
Total force on the bottom flange:
Fb = σ avg A = 27.9 ×106 (0.12 × 0.04) = 134 kN
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________________________________________________________________________
PROBLEM (7.19) A beam of the cross section depicted in Fig. P7.18 is subjected to a bending
moment Mz. Calculate the total force acting on the top flange of the beam.
Given: M z = 10 kN ⋅ m
SOLUTION
y
A
Ft
B
82.5mm
From solution of Prob. 7.18:
I z = 41.22 ×10−6 m 4
x
Mc
10 ×103 (0.1025)
=−
= −24.87 MPa
I
41.22 × 10−6
82.5
σ B = −24.87
= −20.2 MPa
102.5
1
σ avg = (σ A + σ B ) = −22.44 MPa
2
σA =
Total force on the top flange:
Ft = −22.44 × 106 (0.12 × 0.02) = −53.9 kN
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________________________________________________________________________
PROBLEM (7.20) A simple beam AB with a T-cross section carries a concentrated load P as
shown in Fig. P7.20. Determine the maximum tensile stress σ t and largest compressive stress σ c in the
member.
Given: P = 6 kN,
L = 3 m,
a = 1 m,
b = 75 mm,
b1 = 25 mm,
h = 100 mm,
h1 = 75 mm
SOLUTION
y=
∑Ay
∑A
i
i
i
(75 × 25)(12.5) + (75 × 25)(62.5)
= 37.5 mm
=
(75 × 25) + (75 × 25)
and c1 = 37.5 mm and c2 = 62.5 mm
1
I z = (75)(25)3 + (75 × 25)(25) 2
12
1
+ (25)(75)3 + (75 × 25)(25) 2
12
= 3.32(106 ) mm 4
M max =
y
25
c2
A2
C
z
75
100
y
c1
A1
75
6(1)
Pa
( L − a) =
(3 − 1) = 4 kN ⋅ m
L
3
M max c1 (4 × 103 )(0.0375)
σt =
=
= 45.18 MPa
Iz
3.32(10−6 )
M max c2
(4 × 103 )(0.0625)
σc =
=−
= −75.3 MPa
Iz
3.32(10−6 )
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________________________________________________________________________
PROBLEM (7.21) A beam with a T-cross section is supported and loaded as shown in Fig. P7.20.
Determine the maximum permissible load P based upon an allowable bending stress σ all in tension or
compression.
Given: L = 8 ft,
a = 2 ft,
b = 3 in.,
b = 1 in.,
h = 4 in.,
σ all = 15 ksi
h1 = 3 in. ,
SOLUTION
y=
∑Ay
∑A
i
i
i
(3 × 1)(0.5) + (3 ×1)(2.5)
=
= 1.5 in.
(3 × 1) + (3 ×1)
and c1 = 1.5 in. and c2 = 2.5 in.
1
I z = (3)(1)3 + (3 × 1)(1) 2
12
1
+ (1)(3)3 + (3 ×1)(1) 2
12
= 8.5 in.4
2
Pa
M max =
( L − a) = (8 − 2) P = 1.5 P
L
8
y
1 in.
c2
A2
C
z
3 in.
4 in.
y
c1
A1
3 in.
Hence
M c
1.5 P(1.5)
σ t = max 1 ; 15(103 ) =
, P = 56.67 kips
Iz
8.5
M c
1.5P(2.5)
σ c = max 2 ; 15(106 ) =
, P = 34 kips
Iz
8.5
So
Pall = 34 kips
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________________________________________________________________________
PROBLEM (7.22) A W 6 x 16 steel beam (see Table B-8) is loaded as shown in Fig. P7.22. On a
section located 3.6 ft from A, calculate:
(a) The bending stress at a point 1 in. above the bottom of the beam.
(b) The maximum bending stress on the section.
SOLUTION
y
4 kips
A
200 lb/ft
B
3.6 ft
C 4.5 ft
z
2.14 in.
1 in.
6 ft
W 6x16
RB
RA
6.28 in.
C
I=32.1 in.4
∑M = 0:
B
− RA (9) + 4(4.5) − 1.2(3) = 0,
∑ F = 0 : R = 3.6 kips
y
RA = 1.6 kips
B
M = 1.6 × 3.6 × 12 = 69.12 kip ⋅ in.
Mc 69.12 ×103 (2.14)
(a) σ =
=
= 4608 psi
I
32.1
(b) σ max = 4608(
3.14
) = 6761 psi
2.14
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________________________________________________________________________
PROBLEM (7.23) Redo Prob. 7.22 for an S 6 x 12.5 steel beam (see Table B.9).
SOLUTION
From solution of
Prob. 7.22:
M = 69.12 kip ⋅ in.
Mc 69.12 × 103 (2)
=
I
22.1
= 6255 psi
(a) σ =
3
(b) σ max = 6255( ) = 9383 psi
2
y
z
2 in.
1 in.
6 in.
S 6x12.5
I=22.1 in.4
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________________________________________________________________________
PROBLEM (7.24) A W 8 x 35 steel beam (see Table B.8) is loaded as shown in Fig. P7.22. On a
section 4.2 ft to the right of C, calculate:
(a) The flexural stress at a point located ¾ in. below the top of the beam.
(b) The maximum bending stress on the section.
SOLUTION
From solution of
Prob. 7.22, on a
Section 4.2 ft to
the right of C:
y
0.75 in.
z
M = 1.6(8.7) − 4(4.2)
= −2.88 kip ⋅ ft
= −34.56 kip ⋅ in.
(a) σ = −
3.31 in.
C
8.12 in.
4.06 in.
W 8x35
I=127 in.4
34.56 ×103 (3.31)
= −901 psi
127
(b) σ max = 901
4.06
= 1105 psi
3.31
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________________________________________________________________________
PROBLEM (7.25) A rectangular beam, b = 80 mm wide and h = 120 mm deep, is loaded as
depicted in Fig. P7.25. Calculate the magnitude and location of the maximum bending stress in the beam.
SOLUTION
∑ M = 0 : − 18(1.5) + R (5) = 0,
∑ F = 0 : R = 12.6 kN
A
B
y
RB = 5.4 kN
A
y
6 kN/m
A
z
B
C 2m
3m
RA=12.6 kN
120 mm
RB =5.4 kN
V, kN 12.6
x
xm
80(120)3
I=
= 11.52 × 10−6 m 4
12
-5.4
Mmax
M, kN ⋅ m
10.8
x
M max =
80 mm
From geometry:
x
12.6
= m ,
5.4 3 − xm
xm = 2.1 m
12.6 × 2.1
= 13.23 kN ⋅ m
2
Thus,
σ max =
Mc 13.23 ×103 (60 ×10−3 )
=
= 68.9 MPa
I
11.52 ×10−6
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PROBLEM (7.26) Redo Prob. 7.25 for the case in which an additional downward concentrated
force of 4 kN acts at C.
SOLUTION
∑ M = 0 : R (5) − 4(3) − 18(1.5) = 0,
∑ F = 0 : R = 14.2 kN
A
B
y
RB = 7.8 kN
A
4 kN
y
6 kN/m
A
B
C 2m
3m
RA=14.2 kN
z
M, kN ⋅ m
x
-3.8
Mmax
-7.8
15.6
x
M max =
120 mm
RB=7.8 kN
V, kN 14.2
xm
80 mm
I=
80(120)3
= 11.52 × 10−6 m 4
12
From geometry:
xm
3 − xm
=
, xm = 2.367 m
14.2
3.8
14.2 × 2.367
= 16.8 kN ⋅ m
2
Thus,
Mc 16.8 ×103 (60 × 10−3 )
σ max =
=
= 87.5 MPa
I
11.52 ×10−6
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PROBLEM (7.27) A beam of the cross section shown in Fig. P7.27 is loaded as shown in Fig.
P7.25. At the center span, calculate:
(a) The bending stress at point E.
(b) The bending stress at point D.
SOLUTION
y
From solution of
Prob. 7.18:
I = 41.22 × 10−6 m 4
20 mm
E
z
82.5 mm
C
77.5 mm
Refer to solution of Prob. 7.25. At midspan:
D
1
M = 12.6(2.5) − (6)(2.5) 2 = 12.75 kN ⋅ m
2
(a) σ E =
McE
12.75 × 103 (82.5 ×10−3 )
= −25.52 MPa
=−
41.22 × 10−6
I
(b) σ D = −25.52
77.5
= −23.97 MPa
82.5
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________________________________________________________________________
PROBLEM (7.28) As depicted in Fig. P7.28, two vertical concentrated forces are applied to an
S380 x 64 steel beam (see Table B.9) resting on the ground. Determine the largest bending stress in
the beam.
SOLUTION
y
120 kN
0.4 m C
120 kN
1.6 m
A
D
0.4 m
B
z
C
wr
S 380x64
Sz=977x10-6 m3
Moment at C and D:
1
M C = M D = (100 ×103 )(0.4) 2 = 8 kN ⋅ m
2
1
At midspan: M max = (100 × 103 )(1.2) 2 − 120 ×103 (0.8) = −24 kN ⋅ m
2
M
24 × 103
Thus, σ max = max =
= 24.6 MPa
Sz
977 × 10−6
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________________________________________________________________________
PROBLEM (7.29) A W 410 x 60 wide-flange beam (see Table B.8) is reinforced by two 240 x 10
mm steel cover plates of the cross section shown in Fig. P7.29. The beam carries a uniformly distributed
load w on a simply supported span of 10 m. For an allowable bending stress of 120 MPa, calculate the
largest permissible value of w.
SOLUTION
( I )WF=216x10-6 m4
y
Plate
w
A
5w
B
10 m
5w
z
C
407 mm
427 mm
WF410x60
240 mm
1
M max = w(10) 2 = 12.5w
8
The moment of inertia of cover plates about z:
240(4273 − 4073 )
Ip =
= 208.7 ×106 mm 4
12
Thus I = ( I )WF + I p = 424.7 ×10−6 m 4
and
M max = σ all I c gives:
120 ×106 (424.7 ×10−6 )
213.5 × 10−3
wall = 19.1 kN ⋅ m
12.5wall =
or
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PROBLEM (7.30) Repeat Prob. 7.29, with w = 0, for a beam carrying a concentrated load P
applied on a section 3 m from the left support.
SOLUTION
3P
∑ M = 0 : R = 10 ,
A
B
P
A
3m
7P
∑ F = 0 : R = 10
y
7m
C
RA=7P/10
V
A
B
RB=3P/10
7P/10
x
-3P/10
M
2.1P
From solution
of Prob. 7.29:
I = 424.7 × 10−6 m 4
c = 213.5 mm
x
Thus,
σ all = 120 × 106 =
Mc 2.1P(213.5 × 10−3 )
=
I
424.7 ×10−6
or
Pall = 113.3 kN
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________________________________________________________________________
PROBLEM (*7.31) An overhanging beam is loaded as shown in Fig. P7.31. The allowable
stresses in tension and compression are ( σ t )all and ( σ c )all , respectively. Determine the maximum
permissible value of the load w.
Given: ( σ t )all = 7 ksi,
( σ c )all = 18 ksi
*SOLUTION
y
1
w lb/ft
1
z
7.5 ft
A
RA=6.75w
3.75
3 ft
V
B
C
3 ft
RB=6.75w
1
3
2.53w
y =1.763
1.5
x
1
M C = −4.5w + (3.75) 2 w
2
= 2.53w
-3.75
M
2.237
Dimensions are in inches
C
-3w
3
wL2 16
x
-4.5w
y=
-4.5w
2(1× 4 × 2) + 1.5 × 1 + 0.5
= 1.763 in.
2(1× 4) + 1× 1.5
1 × 43
1.5 ×13
2
I = 2[
+ 1× 4(0.237) ] +
+ 1.5 × 1(1.263) 2
12
12
4
= 13.63 in.
At section A
σI
7 ×103 (13.63)
= 42.65 kip ⋅ in.
c
2.237
σ I 18 ×103 (13.63)
=
= 139.2 kip ⋅ in.
( M A )C =
c
1.763
( M A )t =
=
Tension governs. Hence,
4.5w = 42.65 × 103 12,
w = 790 lb ft
At section C
σI
7 × 103 (13.63)
= 54.12 kip ⋅ in.
c
1.763
σ I 18 ×103 (13.63)
( M C )c =
=
= 109.7 kip ⋅ in.
c
2.273
( M C )t =
=
Continued on next slide
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Tension governs:
2.53w = 54.12 × 103 12,
Allowable safe load is thus
wall = 790 lb ft
w = 1783 lb ft
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________________________________________________________________________
PROBLEM (7.32) For the beam loaded as shown in Fig. P7.31, using w = 800 kips/ ft, determine,
on a section 3 ft to the right of A:
(a) The bending stress at point E.
(b) The bending stress at point F.
SOLUTION
From solution of Prob. 7.31:
I = 13.63 in.4
y = 1.763 in.
y
F
800 lb/ft
2.237
z
3 ft A
3 ft O
5.4 kips
4.5 ft
C
B 3 ft
5.4 kips
E
1.763
0.763 in.
M O = −0.8 × 6 × 3 + 5.4 × 3 = 1.8 kip ⋅ ft = 21.6 kip ⋅ in.
Thus,
McE 21.6 × 103 (0.763)
(a) σ E =
=
= 1.21 ksi
13.63
I
2.237
(b) σ F = −1.21
= −3.55 ksi
0.763
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PROBLEM (7.33) A simply supported beam L = 15 ft long is constructed of two C 10 x 20 steel
channels (see Table B.10) bolted back to back. Using an allowable bending stress of 22 ksi, determine
the central concentrated load P that can be applied for two cases:
(a) The webs are horizontal (Fig. P7.33a).
(b) The webs are vertical (Fig. P7.33b).
SOLUTION
M max =
P
P
2
7.5 ft
C
7.5 ft
P
(7.5 × 12) = 45P
2
Refer to Table B.10
for properties of
a C10 × 20 section.
P
2
y
y
5 in.
z
2.739 in.
C
I z = 2(78.4) = 157.8 in.4
z
C
I z = 2[2.81 + 5.88(0.606) 2 ] = 9.94 in.4
(a) σ all =
Mc
45P(2.739)
,
; 22 ×103 =
Iz
9.94
(b) σ all =
Mc
45P (5)
,
; 22 × 103 =
Iz
157.8
P = 1.77 kips
P = 15.4 kips
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PROBLEM (*7.34) An S 200 x 34 steel beam (see Table B.9) is loaded as shown in Fig. P7.34.
Determine:
(a) The bending stress on section C and at a point 10 mm below the top of the beam.
(b) The maximum bending stress in the beam.
*SOLUTION
∑ M = 0 : R (5) − 16(3) − 8(1) = 0,
∑ F = 0 : R = 12.8 kN
A
RB = 11.2 kN
B
y
A
10 mm y
8 kN/m
4 kN/m
A
V, kN
1m
2m C
RA=12.8 kN
2m D
C1=120 mm
B
z
203 mm
RB=11.2 kN
12.8
S 200x34
4.8
x
M, kN ⋅ m
C
17.6
Mmax
xm
I=27x10-6 m4
-11.2
11.2
x
From geometry:
2 − xm
x
= m ,
4.8
11.2
Thus,
M max = 11.2 +
xm = 1.4 m
11.2 × 1.4
= 19.04 kN ⋅ m
2
M C c1
17.6 ×103 (91.5 × 10−3 )
(a) σ C =
=−
= −59.6 MPa
I
27 ×10−6
(b) σ max =
M max c
19.04 × 103 (101.5 × 10−3 )
=−
= 71.6 MPa
I
27 ×10−6
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________________________________________________________________________
PROBLEM (7.35) Rework Prob. 7.34 for the case in which end A is free and end B is fixed.
SOLUTION
8 kN/m
A
2m C
We have
M C = (4 × 2)1 = 8 kN ⋅ m
MB=64 kN ⋅ m
1m B
4 kN/m
2m D
M max = M B = 64 kN ⋅ m
RB=24 kN
From solution of Prob. 7.34:
I = 27 × 10−6 m 4
c1 = 91.5 mm
(a) σ C =
c = 101.5 mm
M C c1 8 ×103 (91.5 × 10−3 )
=
= 27 MPa
I
27 × 10−6
M max c 64 × 103 (101.5 × 10−3 )
(b) σ max =
=
= 240.6 MPa
I
27 × 10−6
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PROBLEM (7.36) A simple rectangular beam, with two vertical U-shaped 12- mm deep grooves
opposite each other on its edges at C (use Fig. 7.14), is loaded as shown in Fig. 7.36. Determine the
magnitude and location of the maximum bending stress in the beam for two cases:
(a) The grooves have a radius of r = 5 mm.
(b) The grooves have a radius of r = 10 mm.
SOLUTION
∑ M = 0 : R = 5.4 kN ,
A
B
∑ F = 0 : R = 12.6 kN
y
A
6 kN/m
80 mm
2m
3m
A
120 mm
96 mm
C
RA=12.6 kN
B
RB=5.4 kN
1
80(96)3
M C = 12.6(3) − (6)(3) 2 = 10.8 kN ⋅ m,
I=
= 5.9 × 10−6 m 4
2
12
M c 10.8 × 103 (48 × 10−3 )
σ nom = max =
= 87.9 MPa
I
5.9 ×10−6
(a)
(b)
5
r
D 120
=
= 0.052
=
= 1.25
d 96
d
96
K = 2.3
(Fig. 7.14)
σ max = 2.3(87.9) = 202.2 MPa
r 10
D 120
=
= 0.104
=
= 1.25
d 96
d
96
K = 1.82
(Fig. 7.14)
Thus
σ max = 1.82(87.9) = 160 MPa
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PROBLEM (7.37) A uniformly loaded simply supported rectangular beam has two 15-mm-deep
vertical U-shaped grooves opposite each other on its edges (use Fig. 7.13) at midspan, as shown in
Fig. 7.37. If the normal stress is limited to σ all = 127 MPa, calculate the smallest permissible radius
of the grooves.
SOLUTION
w=12 kN/m
80 mm
120 mm
96 mm
A
2m
C
2m
wL/2
B
wL/2
1
1
M max = wL2 = (10)(4) 2 = 20 kN ⋅ m
8
8
1
I = (80)(120)3 = 11.52 × 106 mm 4
12
Mc 20 × 103 (45 × 10−3 )
σ nom =
=
= 78.13 MPa
I
11.52 ×10−6
σ
127
= 1.62
K = all =
σ nom 78.13
Use Fig. 7.14.
For K = 1.62,
D 120
=
= 1.33 :
90
d
r
= 0.18
d
Thus,
rmin = 0.18(90) = 16.2 mm
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PROBLEM (7.38) For the cantilever spring shown in Fig. P7.38, Calculate the largest bending
stress for two cases:
(a) The fillet radius is r = 3/16 in.
(b) The fillet radius is r = ¾ in.
Given: b = ½ in.,
P = 80 lb
SOLUTION
At a section through B
M B = 80(12) = 960 lb ⋅ in.
M c
960(0.75)
σ nom = B =
= 5.12 ksi
1
I
3
(0.5)(1.5)
12
(a)
r 0.1875
=
= 0.125
d
1.5
D 2.25
=
= 1.5 :
d
1.5
K = 1.65
(Fig. 7.13)
σ max = 1.65(5.12) = 8.45 ksi
(b)
r 0.75
D 2.25
=
= 0.5
=
= 1.5 :
d 1.5
d
1.5
σ max = 1.22(5.12) = 6.25 ksi
K = 1.22
(Fig. 7.13)
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PROBLEM (7.39) Determine the largest permissible load P that the cantilever spring shown in
Fig. P7.38 can carry for an allowable stress of σ all = 20 ksi, fillet radius r = 3/8 in., and width b =
5/8 in.
SOLUTION
At a section through B
1
M B = 12 P ,
I = (0.625)(1.5)3 = 175.78 × 10−3 in.4
12
M Bc
12 P(0.75)
=
= 51.2 P
σ nom =
I
175.78 × 10−3
r 0.375
=
= 0.25
d
1.5
D 2.25
=
= 1.5 ,
d
1.5
K = 1.4
(Fig. 7.13)
Thus,
σ all = 1.4(51.2 P ) = 20 × 103
gives
Pall = 279 lb
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PROBLEM (*7.40) Based upon a factor of safety of ns = 1.5, determine the minimum
permissible width b of the beam shown in Fig. P7.38
Given:
P = 90 kips, σ all = 18 ksi,
r = 5/16 in.
*SOLUTION
At a section through B
M B = 12(90) = 1080 lb ⋅ in. ,
σ all = 18 1.5 = 12 ksi
I=
1
(b)(1.5)3 = 281.25 ×10−3 b in.4
12
r 0.9375
D 2.25
=
= 0.625
=
= 1.5 ,
d
1.5
d
1.5
1080(0.75)
2.881×103
σ nom =
=
281.25 × 10−3 b
b
K = 1.15
(Fig. 7.13)
Thus,
σ all = 12 × 103 = 1.15(2.881× 103 b)
yields
b = 0.276 in.
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PROBLEM (7.41) A hollow aluminum box beam with the rectangular cross section shown in Fig.
P7.41 is subjected to a shear force of V. Determine:
(a) The maximum shear stress τ max in the webs of the beam.
(b) The minimum shear stress
Given:
b = 100 mm,
τ min
in the webs of the beam.
h = 140 mm,
t = 10 mm,
t1 = 15 mm,
V = 200 kN
SOLUTION
1
1
(100)(140)3 − (80)(110)3
12
12
= 14(106 ) mm 4
I=
(a) Maximum shear stress (at N.A.):
Q = (100)(70)(35) − (80)(55)(27.5)
= 124(103 ) mm3
y
15
a
a
10
C
z
VQ 200(103 )(124 ×10−6 )
=
Ib
14(10−6 )(0.2)
= 8.86 MPa
τ max =
10
140
15
100
(b) Minimum shear stress (at section a-a):
Q = (100)(15)(70 − 7.5) = 93.75(103 ) mm3
VQ 200(103 )(93.75 × 10−6 )
=
= 6.7 MPa
τ min =
14(10−6 )(0.2)
Ib
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PROBLEM (7.42) A hollow square steel box beam with the cross section of uniform thickness t
shown in Fig. P7.42 is acted upon a shear force V. Calculate:
(a) The maximum shear stress τ max in the webs of the beam.
(b) The minimum shear stress τ min in the webs of the beam.
Given:
a = 8 in.,
t = ½ in.,
V = 20 kips
SOLUTION
y
t
a4 1
I z = − (a − 2t ) 4
12 12
84 1
= − (7) 4 = 141.25 in.4
12 12
A
A
z
C
a
t
a
(a) Maximum shear stress (at N.A.).
1
a a
a
a
Q = a( )( ) − (a − 2t )( − t )( − t )( )
2 4
2
2
2
1
= (8)(4)(2) − (7)(3.5)(3.5)( )
2
3
= 21.125 in.
VQ 20(103 )(21.125)
=
= 2.991 ksi
1
Ib
(141.25)(2 × )
2
(b) Minimum shear stress(at section A-A).
1
a t
Q = at ( − ) = (8)( )(3.75) = 15 in.3
2 2
2
3
20(10 )(15)
VQ
τ max =
=
= 2.12 ksi
Ib (141.25)(2 × 1 )
2
τ max =
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________________________________________________________________________
PROBLEM (7.22) A W 6 x 16 steel beam (see Table B-8) is loaded as shown in Fig. P7.22. On a
section located 3.6 ft from A, calculate:
(a) The bending stress at a point 1 in. above the bottom of the beam.
(b) The maximum bending stress on the section.
SOLUTION
y
4 kips
A
200 lb/ft
B
3.6 ft
C 4.5 ft
z
2.14 in.
1 in.
6 ft
W 6x16
RB
RA
6.28 in.
C
I=32.1 in.4
∑M = 0:
B
− RA (9) + 4(4.5) − 1.2(3) = 0,
∑ F = 0 : R = 3.6 kips
y
RA = 1.6 kips
B
M = 1.6 × 3.6 × 12 = 69.12 kip ⋅ in.
Mc 69.12 ×103 (2.14)
(a) σ =
=
= 4608 psi
I
32.1
(b) σ max = 4608(
3.14
) = 6761 psi
2.14
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PROBLEM (7.23) Redo Prob. 7.22 for an S 6 x 12.5 steel beam (see Table B.9).
SOLUTION
From solution of
Prob. 7.22:
M = 69.12 kip ⋅ in.
Mc 69.12 × 103 (2)
=
I
22.1
= 6255 psi
(a) σ =
3
(b) σ max = 6255( ) = 9383 psi
2
y
z
2 in.
1 in.
6 in.
S 6x12.5
I=22.1 in.4
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PROBLEM (7.24) A W 8 x 35 steel beam (see Table B.8) is loaded as shown in Fig. P7.22. On a
section 4.2 ft to the right of C, calculate:
(a) The flexural stress at a point located ¾ in. below the top of the beam.
(b) The maximum bending stress on the section.
SOLUTION
From solution of
Prob. 7.22, on a
Section 4.2 ft to
the right of C:
y
0.75 in.
z
M = 1.6(8.7) − 4(4.2)
= −2.88 kip ⋅ ft
= −34.56 kip ⋅ in.
(a) σ = −
3.31 in.
C
8.12 in.
4.06 in.
W 8x35
I=127 in.4
34.56 ×103 (3.31)
= −901 psi
127
(b) σ max = 901
4.06
= 1105 psi
3.31
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PROBLEM (7.25) A rectangular beam, b = 80 mm wide and h = 120 mm deep, is loaded as
depicted in Fig. P7.25. Calculate the magnitude and location of the maximum bending stress in the beam.
SOLUTION
∑ M = 0 : − 18(1.5) + R (5) = 0,
∑ F = 0 : R = 12.6 kN
A
B
y
RB = 5.4 kN
A
y
6 kN/m
A
z
B
C 2m
3m
RA=12.6 kN
120 mm
RB =5.4 kN
V, kN 12.6
x
xm
80(120)3
I=
= 11.52 × 10−6 m 4
12
-5.4
Mmax
M, kN ⋅ m
10.8
x
M max =
80 mm
From geometry:
x
12.6
= m ,
5.4 3 − xm
xm = 2.1 m
12.6 × 2.1
= 13.23 kN ⋅ m
2
Thus,
σ max =
Mc 13.23 ×103 (60 ×10−3 )
=
= 68.9 MPa
I
11.52 ×10−6
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PROBLEM (7.26) Redo Prob. 7.25 for the case in which an additional downward concentrated
force of 4 kN acts at C.
SOLUTION
∑ M = 0 : R (5) − 4(3) − 18(1.5) = 0,
∑ F = 0 : R = 14.2 kN
A
B
y
RB = 7.8 kN
A
4 kN
y
6 kN/m
A
B
C 2m
3m
RA=14.2 kN
z
M, kN ⋅ m
x
-3.8
Mmax
-7.8
15.6
x
M max =
120 mm
RB=7.8 kN
V, kN 14.2
xm
80 mm
I=
80(120)3
= 11.52 × 10−6 m 4
12
From geometry:
xm
3 − xm
=
, xm = 2.367 m
14.2
3.8
14.2 × 2.367
= 16.8 kN ⋅ m
2
Thus,
Mc 16.8 ×103 (60 × 10−3 )
σ max =
=
= 87.5 MPa
I
11.52 ×10−6
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PROBLEM (7.27) A beam of the cross section shown in Fig. P7.27 is loaded as shown in Fig.
P7.25. At the center span, calculate:
(a) The bending stress at point E.
(b) The bending stress at point D.
SOLUTION
y
From solution of
Prob. 7.18:
I = 41.22 × 10−6 m 4
20 mm
E
z
82.5 mm
C
77.5 mm
Refer to solution of Prob. 7.25. At midspan:
D
1
M = 12.6(2.5) − (6)(2.5) 2 = 12.75 kN ⋅ m
2
(a) σ E =
McE
12.75 × 103 (82.5 ×10−3 )
= −25.52 MPa
=−
41.22 × 10−6
I
(b) σ D = −25.52
77.5
= −23.97 MPa
82.5
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PROBLEM (7.28) As depicted in Fig. P7.28, two vertical concentrated forces are applied to an
S380 x 64 steel beam (see Table B.9) resting on the ground. Determine the largest bending stress in
the beam.
SOLUTION
y
120 kN
0.4 m C
120 kN
1.6 m
A
D
0.4 m
B
z
C
wr
S 380x64
Sz=977x10-6 m3
Moment at C and D:
1
M C = M D = (100 ×103 )(0.4) 2 = 8 kN ⋅ m
2
1
At midspan: M max = (100 × 103 )(1.2) 2 − 120 ×103 (0.8) = −24 kN ⋅ m
2
M
24 × 103
Thus, σ max = max =
= 24.6 MPa
Sz
977 × 10−6
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PROBLEM (7.29) A W 410 x 60 wide-flange beam (see Table B.8) is reinforced by two 240 x 10
mm steel cover plates of the cross section shown in Fig. P7.29. The beam carries a uniformly distributed
load w on a simply supported span of 10 m. For an allowable bending stress of 120 MPa, calculate the
largest permissible value of w.
SOLUTION
( I )WF=216x10-6 m4
y
Plate
w
A
5w
B
10 m
5w
z
C
407 mm
427 mm
WF410x60
240 mm
1
M max = w(10) 2 = 12.5w
8
The moment of inertia of cover plates about z:
240(4273 − 4073 )
Ip =
= 208.7 ×106 mm 4
12
Thus I = ( I )WF + I p = 424.7 ×10−6 m 4
and
M max = σ all I c gives:
120 ×106 (424.7 ×10−6 )
213.5 × 10−3
wall = 19.1 kN ⋅ m
12.5wall =
or
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PROBLEM (7.30) Repeat Prob. 7.29, with w = 0, for a beam carrying a concentrated load P
applied on a section 3 m from the left support.
SOLUTION
3P
∑ M = 0 : R = 10 ,
A
B
P
A
3m
7P
∑ F = 0 : R = 10
y
7m
C
RA=7P/10
V
A
B
RB=3P/10
7P/10
x
-3P/10
M
2.1P
From solution
of Prob. 7.29:
I = 424.7 × 10−6 m 4
c = 213.5 mm
x
Thus,
σ all = 120 × 106 =
Mc 2.1P(213.5 × 10−3 )
=
I
424.7 ×10−6
or
Pall = 113.3 kN
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PROBLEM (*7.31) An overhanging beam is loaded as shown in Fig. P7.31. The allowable
stresses in tension and compression are ( σ t )all and ( σ c )all , respectively. Determine the maximum
permissible value of the load w.
Given: ( σ t )all = 7 ksi,
( σ c )all = 18 ksi
*SOLUTION
y
1
w lb/ft
1
z
7.5 ft
A
RA=6.75w
3.75
3 ft
V
B
C
3 ft
RB=6.75w
1
3
2.53w
y =1.763
1.5
x
1
M C = −4.5w + (3.75) 2 w
2
= 2.53w
-3.75
M
2.237
Dimensions are in inches
C
-3w
3
wL2 16
x
-4.5w
y=
-4.5w
2(1× 4 × 2) + 1.5 × 1 + 0.5
= 1.763 in.
2(1× 4) + 1× 1.5
1 × 43
1.5 ×13
2
I = 2[
+ 1× 4(0.237) ] +
+ 1.5 × 1(1.263) 2
12
12
4
= 13.63 in.
At section A
σI
7 ×103 (13.63)
= 42.65 kip ⋅ in.
c
2.237
σ I 18 ×103 (13.63)
=
= 139.2 kip ⋅ in.
( M A )C =
c
1.763
( M A )t =
=
Tension governs. Hence,
4.5w = 42.65 × 103 12,
w = 790 lb ft
At section C
σI
7 × 103 (13.63)
= 54.12 kip ⋅ in.
c
1.763
σ I 18 ×103 (13.63)
( M C )c =
=
= 109.7 kip ⋅ in.
c
2.273
( M C )t =
=
Continued on next slide
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Tension governs:
2.53w = 54.12 × 103 12,
Allowable safe load is thus
wall = 790 lb ft
w = 1783 lb ft
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PROBLEM (7.32) For the beam loaded as shown in Fig. P7.31, using w = 800 kips/ ft, determine,
on a section 3 ft to the right of A:
(a) The bending stress at point E.
(b) The bending stress at point F.
SOLUTION
From solution of Prob. 7.31:
I = 13.63 in.4
y = 1.763 in.
y
F
800 lb/ft
2.237
z
3 ft A
3 ft O
5.4 kips
4.5 ft
C
B 3 ft
5.4 kips
E
1.763
0.763 in.
M O = −0.8 × 6 × 3 + 5.4 × 3 = 1.8 kip ⋅ ft = 21.6 kip ⋅ in.
Thus,
McE 21.6 × 103 (0.763)
(a) σ E =
=
= 1.21 ksi
13.63
I
2.237
(b) σ F = −1.21
= −3.55 ksi
0.763
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PROBLEM (7.33) A simply supported beam L = 15 ft long is constructed of two C 10 x 20 steel
channels (see Table B.10) bolted back to back. Using an allowable bending stress of 22 ksi, determine
the central concentrated load P that can be applied for two cases:
(a) The webs are horizontal (Fig. P7.33a).
(b) The webs are vertical (Fig. P7.33b).
SOLUTION
M max =
P
P
2
7.5 ft
C
7.5 ft
P
(7.5 × 12) = 45P
2
Refer to Table B.10
for properties of
a C10 × 20 section.
P
2
y
y
5 in.
z
2.739 in.
C
I z = 2(78.4) = 157.8 in.4
z
C
I z = 2[2.81 + 5.88(0.606) 2 ] = 9.94 in.4
(a) σ all =
Mc
45P(2.739)
,
; 22 ×103 =
Iz
9.94
(b) σ all =
Mc
45P (5)
,
; 22 × 103 =
Iz
157.8
P = 1.77 kips
P = 15.4 kips
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PROBLEM (*7.34) An S 200 x 34 steel beam (see Table B.9) is loaded as shown in Fig. P7.34.
Determine:
(a) The bending stress on section C and at a point 10 mm below the top of the beam.
(b) The maximum bending stress in the beam.
*SOLUTION
∑ M = 0 : R (5) − 16(3) − 8(1) = 0,
∑ F = 0 : R = 12.8 kN
A
RB = 11.2 kN
B
y
A
10 mm y
8 kN/m
4 kN/m
A
V, kN
1m
2m C
RA=12.8 kN
2m D
C1=120 mm
B
z
203 mm
RB=11.2 kN
12.8
S 200x34
4.8
x
M, kN ⋅ m
C
17.6
Mmax
xm
I=27x10-6 m4
-11.2
11.2
x
From geometry:
2 − xm
x
= m ,
4.8
11.2
Thus,
M max = 11.2 +
xm = 1.4 m
11.2 × 1.4
= 19.04 kN ⋅ m
2
M C c1
17.6 ×103 (91.5 × 10−3 )
(a) σ C =
=−
= −59.6 MPa
I
27 ×10−6
(b) σ max =
M max c
19.04 × 103 (101.5 × 10−3 )
=−
= 71.6 MPa
I
27 ×10−6
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PROBLEM (7.35) Rework Prob. 7.34 for the case in which end A is free and end B is fixed.
SOLUTION
8 kN/m
A
2m C
We have
M C = (4 × 2)1 = 8 kN ⋅ m
MB=64 kN ⋅ m
1m B
4 kN/m
2m D
M max = M B = 64 kN ⋅ m
RB=24 kN
From solution of Prob. 7.34:
I = 27 × 10−6 m 4
c1 = 91.5 mm
(a) σ C =
c = 101.5 mm
M C c1 8 ×103 (91.5 × 10−3 )
=
= 27 MPa
I
27 × 10−6
M max c 64 × 103 (101.5 × 10−3 )
(b) σ max =
=
= 240.6 MPa
I
27 × 10−6
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PROBLEM (7.36) A simple rectangular beam, with two vertical U-shaped 12- mm deep grooves
opposite each other on its edges at C (use Fig. 7.14), is loaded as shown in Fig. 7.36. Determine the
magnitude and location of the maximum bending stress in the beam for two cases:
(a) The grooves have a radius of r = 5 mm.
(b) The grooves have a radius of r = 10 mm.
SOLUTION
∑ M = 0 : R = 5.4 kN ,
A
B
∑ F = 0 : R = 12.6 kN
y
A
6 kN/m
80 mm
2m
3m
A
120 mm
96 mm
C
RA=12.6 kN
B
RB=5.4 kN
1
80(96)3
M C = 12.6(3) − (6)(3) 2 = 10.8 kN ⋅ m,
I=
= 5.9 × 10−6 m 4
2
12
M c 10.8 × 103 (48 × 10−3 )
σ nom = max =
= 87.9 MPa
I
5.9 ×10−6
(a)
(b)
5
r
D 120
=
= 0.052
=
= 1.25
d 96
d
96
K = 2.3
(Fig. 7.14)
σ max = 2.3(87.9) = 202.2 MPa
r 10
D 120
=
= 0.104
=
= 1.25
d 96
d
96
K = 1.82
(Fig. 7.14)
Thus
σ max = 1.82(87.9) = 160 MPa
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PROBLEM (7.37) A uniformly loaded simply supported rectangular beam has two 15-mm-deep
vertical U-shaped grooves opposite each other on its edges (use Fig. 7.13) at midspan, as shown in
Fig. 7.37. If the normal stress is limited to σ all = 127 MPa, calculate the smallest permissible radius
of the grooves.
SOLUTION
w=12 kN/m
80 mm
120 mm
96 mm
A
2m
C
2m
wL/2
B
wL/2
1
1
M max = wL2 = (10)(4) 2 = 20 kN ⋅ m
8
8
1
I = (80)(120)3 = 11.52 × 106 mm 4
12
Mc 20 × 103 (45 × 10−3 )
σ nom =
=
= 78.13 MPa
I
11.52 ×10−6
σ
127
= 1.62
K = all =
σ nom 78.13
Use Fig. 7.14.
For K = 1.62,
D 120
=
= 1.33 :
90
d
r
= 0.18
d
Thus,
rmin = 0.18(90) = 16.2 mm
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PROBLEM (7.38) For the cantilever spring shown in Fig. P7.38, Calculate the largest bending
stress for two cases:
(a) The fillet radius is r = 3/16 in.
(b) The fillet radius is r = ¾ in.
Given: b = ½ in.,
P = 80 lb
SOLUTION
At a section through B
M B = 80(12) = 960 lb ⋅ in.
M c
960(0.75)
σ nom = B =
= 5.12 ksi
1
I
3
(0.5)(1.5)
12
(a)
r 0.1875
=
= 0.125
d
1.5
D 2.25
=
= 1.5 :
d
1.5
K = 1.65
(Fig. 7.13)
σ max = 1.65(5.12) = 8.45 ksi
(b)
r 0.75
D 2.25
=
= 0.5
=
= 1.5 :
d 1.5
d
1.5
σ max = 1.22(5.12) = 6.25 ksi
K = 1.22
(Fig. 7.13)
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PROBLEM (7.39) Determine the largest permissible load P that the cantilever spring shown in
Fig. P7.38 can carry for an allowable stress of σ all = 20 ksi, fillet radius r = 3/8 in., and width b =
5/8 in.
SOLUTION
At a section through B
1
M B = 12 P ,
I = (0.625)(1.5)3 = 175.78 × 10−3 in.4
12
M Bc
12 P(0.75)
=
= 51.2 P
σ nom =
I
175.78 × 10−3
r 0.375
=
= 0.25
d
1.5
D 2.25
=
= 1.5 ,
d
1.5
K = 1.4
(Fig. 7.13)
Thus,
σ all = 1.4(51.2 P ) = 20 × 103
gives
Pall = 279 lb
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PROBLEM (*7.40) Based upon a factor of safety of ns = 1.5, determine the minimum
permissible width b of the beam shown in Fig. P7.38
Given:
P = 90 kips, σ all = 18 ksi,
r = 5/16 in.
*SOLUTION
At a section through B
M B = 12(90) = 1080 lb ⋅ in. ,
σ all = 18 1.5 = 12 ksi
I=
1
(b)(1.5)3 = 281.25 ×10−3 b in.4
12
r 0.9375
D 2.25
=
= 0.625
=
= 1.5 ,
d
1.5
d
1.5
1080(0.75)
2.881×103
σ nom =
=
281.25 × 10−3 b
b
K = 1.15
(Fig. 7.13)
Thus,
σ all = 12 × 103 = 1.15(2.881× 103 b)
yields
b = 0.276 in.
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PROBLEM (7.41) A hollow aluminum box beam with the rectangular cross section shown in Fig.
P7.41 is subjected to a shear force of V. Determine:
(a) The maximum shear stress τ max in the webs of the beam.
(b) The minimum shear stress
Given:
b = 100 mm,
τ min
in the webs of the beam.
h = 140 mm,
t = 10 mm,
t1 = 15 mm,
V = 200 kN
SOLUTION
1
1
(100)(140)3 − (80)(110)3
12
12
= 14(106 ) mm 4
I=
(a) Maximum shear stress (at N.A.):
Q = (100)(70)(35) − (80)(55)(27.5)
= 124(103 ) mm3
y
15
a
a
10
C
z
VQ 200(103 )(124 ×10−6 )
=
Ib
14(10−6 )(0.2)
= 8.86 MPa
τ max =
10
140
15
100
(b) Minimum shear stress (at section a-a):
Q = (100)(15)(70 − 7.5) = 93.75(103 ) mm3
VQ 200(103 )(93.75 × 10−6 )
=
= 6.7 MPa
τ min =
14(10−6 )(0.2)
Ib
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PROBLEM (7.42) A hollow square steel box beam with the cross section of uniform thickness t
shown in Fig. P7.42 is acted upon a shear force V. Calculate:
(a) The maximum shear stress τ max in the webs of the beam.
(b) The minimum shear stress τ min in the webs of the beam.
Given:
a = 8 in.,
t = ½ in.,
V = 20 kips
SOLUTION
y
t
a4 1
I z = − (a − 2t ) 4
12 12
84 1
= − (7) 4 = 141.25 in.4
12 12
A
A
z
C
a
t
a
(a) Maximum shear stress (at N.A.).
1
a a
a
a
Q = a( )( ) − (a − 2t )( − t )( − t )( )
2 4
2
2
2
1
= (8)(4)(2) − (7)(3.5)(3.5)( )
2
3
= 21.125 in.
VQ 20(103 )(21.125)
=
= 2.991 ksi
1
Ib
(141.25)(2 × )
2
(b) Minimum shear stress(at section A-A).
1
a t
Q = at ( − ) = (8)( )(3.75) = 15 in.3
2 2
2
3
20(10 )(15)
VQ
τ max =
=
= 2.12 ksi
Ib (141.25)(2 × 1 )
2
τ max =
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PROBLEM (7.43) A simple beam with two overhangs is loaded as shown in Fig. P7.43, with w = 1.2
kips/ft. Determine:
(a) The shearing stress at a point 1.5 ft to the left of B and ¾ in. below the top of the beam.
(b) The maximum shear stress in the beam.
SOLUTION
y
1 in. 1 in.
1.2 kips/ft
A
3 ft
7.5 ft
8.1 kips
V, kips
¾ in
1.5
ft
z
B
C
1.5 in.
1 in.
1.763 in.
x
-4.5
V=-2.7 kips
(from geometry)
From solution of Prob. 7.31: I = 13.63 in. 4
(a) τ =
2.237 in.
C
E
8.1 kips
3.6
4.5
-3.6
3 in.
y = 1.763 in.
VQ 2.7 × 103[0.75 ×1(2.237 − 0.375)]
=
= 277 psi
13.63(1)
Ib
VQ 4.5 × 103 [3 × 1(3 − 2.237)]
=
= 756 psi
(b) τ E =
13.63(1)
Ib
We have, at the neutral axis:
Q z = 1 × 2.237( 2.237 2) = 2.503 in.
τz =
Thus,
VQ 4.5 × 103 (2.503)
=
= 826 psi
13.63(1)
Ib
τ max = 826 psi
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PROBLEM (7.44) Calculate the ratio of the largest vertical shear force V1 to the maximum vertical
shear force V2 that can be carried by the I beam for the cases of web positioned vertical and horizontal
as shown in Figs. P7.44a and P7.44 b, respectively.
Assumption: The vertical shear force V2 acts through a point (S) so that only bending of the beam
occurs as is discussed in Sec. 7.16.
SOLUTION
Area properties, from solution of Prob. 7.18, are:
y = 77.5 mm
I z = 41.22 × 10 −6 m 4
Also
20 + 40
Iy =
(120)3
12
120
+
(20)3
12
= 8.72 × 10 6 mm 4
y
50 mm
20
40 mm
y =77.5 mm
z
C
102.5 mm
20 mm
120 mm
20 mm
120 mm
We have
τ max I z b
Vy =
Qz
, where for lower half of section:
Q z = 120 × 20 × 92.5 + 82.5 × 20 × 41.25 = 290.1 × 10 3 mm 3
So,
τ max (41.22 ×10−6 )(20 ×10−3 )
Vy =
Similarly,
Vz =
290.01×10−6
τ max I y b
Qy
= 0.00284τ max
, where for the upper shaded area:
Q y = 40 × 50 × 35 = 70 × 10 3 mm 3
Thus,
Vz =
τ max (8.72 ×10−6 )(40 ×10−3 )
70 ×10−6
= 0.00498τ max
Hence,
Vy
Vz
=
0.00284τ max
= 0.57
0.00498τ max
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PROBLEM (7.45) A beam of the cross section shown in Fig. P7.44a is loaded as shown in Fig.
P7.45. Determine:
(a) The shear stress at point E and 40 mm from the top of the beam.
(b) The maximum shearing stress in the beam
SOLUTION
∑ M = 0 : R (4) − 20(3) − 50(1) + 10(1) = 0,
∑ F = 0 : R = 55 kN
A
RB = 25 kN
B
y
A
10 kN
20 kN
25 kN/m
1m
1m
C
A
V, kN
2m
RA=55 kN
E 1m
D
B
RB =25 kN
45
x
-10
-5
-25
Refer to figure in solution of Prob. 7.44:
I z = 41.22 × 10 −6 m 4
Q z = Q max = 290.1 × 10 3 mm 3
For upper flange:
Q = 120 × 40 × 57.5 = 276 × 10 3 mm 3
25 × 103 (276 ×10−6 )
VQ
=
= 8.37 MPa
(a) τ E =
Ib 41.22 × 10−6 (20 × 10−3 )
(b) τ max =
Vmax Qmax
45 ×103 (290.1× 10−6 )
=
= 15.84 MPa
41.22 × 10−6 (20 × 10−3 )
Ib
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PROBLEM (7.48) A T-shaped beam with an overhang is supported and loaded as shown in Fig.
7.48. Calculate:
(a) The shear stress at a point 2 m from A and 25 mm from the top of the beam.
(b) The greatest shearing stress.
SOLUTION
Refer to Example 7.3:
V, kN
60
y
2m
5
3
E
B
1.25 m
VE
-7
20
x
25 30
z
102.5 mm
1.75 m
C
20
50
m
Dimensions are
in millimeters.
I = 1.36 × 10−6 m 4
From geometry:
0.75 1.75
=
,
VE
7
VE = 3 kN
(a) Q = 60 × 20 × 20 + 5 × 20 × 7.5 = 24.8 × 10 3 mm 3
3 × 103 (24.8 × 10−6 )
VQ
=
= 2.74 MPa
τE =
Ib 1.36 ×10−6 (20 × 10−3 )
(b) Q max = 20 × 50 × 25 = 25 × 10 3 mm 3
VQmax
7 ×103 (25 × 10−6 )
=
= 6.43 MPa
τ max =
1.36 ×10−6 (20 × 10−3 )
Ib
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PROBLEMS (7.49 through *7.51) A vertically symmetric beam of t =10 mm uniform wall
thickness as shown in Figs. P7.49 through P7.51, is acted upon by shear force V = 20 kN. Dimensions are
in millimeters. Determine:
(a) The shear stress at the sections indicated.
(b) The maximum shearing stress in the beam.
SOLUTION (7.49)
A1
c
38.48
6.52
c
y
c
z
41.52 A
3
z’
(c)
b1
80
C
a
15
b
A2
b1
A3
36.52
a
60
(b)
(a)
We have: V = 20 kN
t = 10 mm
A y + A2 y2 + A3 y3
y= 1 1
A1 + A2 + A3
60 ×10 × 75 + 2(60 ×10 × 40) + 2(25 × 10 × 5)
=
= 41.52 mm
60 ×10 + 2 × 60 × 10 + 2 × 25 ×10
60(10)3
10(60)3
I=
+ 60 × 10(33.48) 2 + 2[
+ 10 × 60(1.52) 2 ]
12
12
25(10)3
+2[
+ 25 × 10(36.52) 2 ] = 1.71× 106 mm 4
12
(a) Section a-a
Qa = 15 × 10 × 36.52 = 5.478 × 10 3 mm 3
τa =
VQ 20 × 103 (5.478 × 10−6 )
=
= 6.41 MPa
Ib 1.71×10−6 (10 × 10−3 )
Section b-b (Fig.b)
Qb = Qa + 80 × 10 × 1.52 = 6.694 × 10 3 mm 3
VQ 20 × 103 (6.694 × 10−6 )
=
= 7.83 MPa
τb =
Ib 1.71×10−6 (10 ×10−3 )
Continued on next slide
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Section c-c (Fig.c)
Qc = Qa + 70 × 10 × 6.52 = 10.042 × 10 3 mm 3
τc =
VQ 20 × 103 (10.042 × 10−6 )
=
= 11.75 MPa
1.71× 10−6 (10 × 10−3 )
Ib
(b) Refer to Fig. a:
Qmax = Qa + 41.52 × 10 × 20.76 = 14.098 × 10 3 mm 3
VQmax 20 × 103 (14.098 ×10−6 )
=
= 16.49 MPa
1.71× 10−6 (10 ×10−3 )
Ib
τ max =
SOLUTION(7.50)
y
30
a
40.16
a
A2
(a)
Dimensions are
in millimeters
A1
z
b
b
z’
25
A3
60
a
A2
55.16
c
24.84= y
7.34
b
c
(b)
19.84
c
(c)
We have: V = 20 kN
t = 10 mm
A y + A2 y2 + A3 y3
y= 1 1
A1 + A2 + A3
70 ×10 × 45 + 2(15 × 10 ×17.5) + 60 ×10 × 5
=
= 24.84 mm
70 ×10 + 2 × 15 × 10 + 60 ×10
I=
10(70)3
10(15)3
+ 70 × 10(20.16) 2 + 2[
+ 10 × 15(7.34) 2 ]
12
12
60(10)3
+
+ 60 × 10(19.84) 2 = 833.3 × 103 mm 4
12
(a) Section a-a (Fig. a)
Qa = 30 × 10 × 40.16 = 12.048 × 103 mm 3
τa =
VQ 20 × 103 (12.048 × 10−6 )
=
= 28.92 MPa
Ib 833.3 ×10−9 (10 × 10−3 )
Continued on next slide
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Section b-b (Fig.b)
Qb = 15 × 10 × 7.34 = 1.1 × 103 mm 3
τb =
20 ×103 (1.1× 10−6 )
VQ
=
= 2.64 MPa
Ib 833.3 ×10−9 (10 × 10−3 )
Section c-c (Fig.c)
Qc = Qb + 25 × 10 × 19.84 = 6.06 × 10 3 mm 3
τc =
20 × 103 (6.06 × 10−6 )
VQ
=
= 14.54 MPa
Ib 833.3 ×10−9 (10 × 10−3 )
55.16
) = 15.213 × 103 mm3
2
VQ
20 × 103 (15.213 ×10−6 )
= 36.51 MPa
τ max = max =
833.3 ×10−9 (10 × 10−3 )
Ib
(b) Qmax = 10 × 55.16(
SOLUTION (*7.51)
y
b
30
20
c
c
I=
40
c
d
40
c
d
d
(b)
We have: V = 20 kN
3
b
a
z
15
(a)
b
a
Dimensions are
in millimeters
d
50
(c)
(d)
t = 10 mm
3
60(80)
15(60)
] = 2.02 × 106 mm 4
− 2[
12
12
(a) Section a-a
At this section, we have τ a = 0
Section b-b (Fig.a)
Qb = 20 × 10 × 20 = 4 × 103 mm 3
τb =
20 × 103 (4 × 10−6 )
VQ
=
= 3.96 MPa
Ib 2.02 × 10−6 (10 × 10−3 )
Section c-c (Fig.b) – cut 2 surfaces:
Qc = 50 × 10 × 15 = 7.5 × 10 3 mm 3
Continued on next slide
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τc =
20 ×103 (7.5 × 10−6 )
VQ
=
= 3.71 MPa
Ib 2.02 ×10−6 (10 × 10−3 )2
Section d-d (Fig. c) – cut 2 surfaces:
Qd = 40 × 10 × 10 + 50 × 10 × 35 = 21.5 × 10 3 mm 3
20 ×103 (21.5 ×10−6 )
VQ
=
= 10.65 MPa
τd =
Ib 2.02 × 10−6 (10 ×10−3 )2
(b) Refer to Fig. d– cut 2 surfaces:
Qmax = 60 × 40 × 20 − 2(15 × 30 × 15) − 10 × 10 × 5 = 34 × 10 3 mm 3
τ max =
VQmax
20 × 103 (34 ×10−6 )
=
= 16.83 MPa
2.02 × 10−6 (10 × 10−3 )2
Ib
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PROBLEM (7.52) A beam of the cross section shown in Fig. P7.49 is loaded as seen in Fig. P7.52.
Determine the shear stress at section C, 10 mm from the bottom of the beam.
SOLUTION
From solution of Prob. 7.34:
Vc = 4.8 kN
From solution of Prob. 7.49:
I z = I = 1.71 × 10 −6 m 4
y = 41.52 mm (from bottom)
y
z
41.52 mm
10 mm
25 mm
Q = 10 × 25 × 36.52 = 9.13 × 10 3 mm 3
Thus,
VQ 4.8 × 103 (9.13 × 10−6 )
=
= 2.56 MPa
τc =
1.7 × 10−6 (10 × 10−3 )
Ib
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PROBLEM (7.53) As shown in Fig. P7.53, a beam is constructed of four wood boards nailed
together. Two of the boards, measure 20 x I00 mm and two 25 x 100 mm. Calculate:
(a) The shear force in each nail.
(b) The shearing stress 40 mm from the top of the beam.
Given: A longitudinal spacing between each nail of s = 100 mm and an allowable vertical shear force in
the beam of V = 3 kN.
SOLUTION
A1
y
20
40
70
A2
C
z
25
25
1
[100(140)3 − 50(100)3 ]
12
= 18.7 × 106 mm 4
I=
70
20
100
(a) Q1 = A1 y1 = 20 × 100 × 60 = 120 × 10 3 mm 3
VQ1 3 ×103 (120 ×10−6 )
=
= 19.25 kN m
18.7 ×10−6
I
We have q ⋅ s = 2 F or
1
F = (19.25 ×103 )(0.1) = 962.5 N
2
(b) Q = Q1 + 2 A2 y 2 = 120 × 10 3 + 2( 25 × 20 × 40)
= 160 × 10 3 mm 3
VQ
3 × 103 (160 ×10−6 )
=
= 513.4 kPa
τ=
Ib 18.7 × 10−6 (2 × 25 × 10−3 )
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PROBLEM (7.54) Redo Prob. 7.53, given that the cross section of the beam is turned 90°.
SOLUTION
A1
y
y
25
15
C
z
100
z
C
(b)
25
20
100
20
(a)
I=
Dimensions are
in millimeters
1
[140(100)3 − 100(50)3 ] = 10.67 × 106 mm 4
12
(a) Refer to Fig. (a): Q1 = A1 y1 = 100 × 25 × 37.5 = 93.75 × 10 3 mm 3
VQ 3 ×103 (93.75 × 10−6 )
=
= 26.36 kN m
q=
I
10.67 ×10−6
1
1
F = qs = (26.36 ×103 )(0.1) = 1.32 kN
2
2
(b) Refer to Fig. (b):
Q = 140 × 25 × 37.5 + 2(15 × 20 × 17.5) = 141.75 × 10 3 mm 3
τ=
VQ
3 × 103 (141.75 ×10−6 )
=
= 996 kPa
Ib 10.67 × 10−6 (2 × 20 × 10−3 )
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40
________________________________________________________________________
PROBLEM (7.55) For the beam of Prob. 7.53, if V = 2 kN and the allowable shear force in each
nail is 250 N, calculate the largest permissible longitudinal spacing between nails.
SOLUTION
From solution of Prob. 7.53:
I = 18.7 × 10 −6 m 4
Q = 120 × 10 3 mm 3
q=
VQ 2 × 103 (120 × 10−6 )
=
= 12.8 kN m
I
18.7 × 10−6
We have q ⋅ s = 2 F
or
2(250)
s=
= 39.06 × 10−3 m = 39.1 mm
3
12.8 × 10
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PROBLEM (7.56) Repeat Prob. 7.55 for the beam cross section turned 90°.
SOLUTION
From solution of Prob. 7.54:
I = 10.67 × 10 −6 m 4
Q = 93.75 × 10 3 mm 3
Thus,
VQ 2 × 103 (93.75 × 10−6 )
=
= 17.6 kN m
I
10.67 × 10−6
2F
2 × 250
s=
=
= 28.4 × 10−3 m = 28.4 mm
3
q 17.6 ×10
q=
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PROBLEM (7.57) A beam of the cross section shown in Fig. P7.53 is loaded as shown in Fig. P7.57.
Calculate the maximum shearing stress in the beam.
SOLUTION
∑ M = 0 : R (4) − 4(1)(3.5) − 20(3) + 15 = 0 ,
∑ F = 0 : R = 9.25 kN
A
R B = 14.75 kN
B
y
A
From solution of Prob. 7.53:
I = 18.7 × 10 3 m 4 .
y
20
20 kN 4 kN/m
15 kip ⋅ m
A
9.25 kN
1m
2m
C
D 1m
B
z
14.75 kN
9.25
100
C
25
25
20
100
V, kN
x
-10.75
-14.75
Dimensions are
in millimeters
Q max = 100 × 70 × 35 − 50 × 50 × 25 = 182.5 × 10 3 mm
τ max =
Vmax Qmax 14.75 ×103 (182.5 ×10−6 )
=
= 2.871 MPa
Ib
18.7 ×10−6 (2 × 25 ×10−3 )
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PROBLEM (7.58) A built-up beam, fabricated by gluing four planks as shown in Fig. P7.58, carries
a vertical shear force V = 45 kN. Determine:
(a) The maximum shearing stress in the beam.
(b) The shear stress at joint E.
(c) The shearing stress at joint F.
SOLUTION
y
y
F
E
E
z
z
C
40 mm
E
y
F
F
z
40 mm
240 mm
I = 121 ( 240 4 − 160 4 ) = 221.92 × 10 6 mm 4
(a) Q max = 240 × 120 × 60 − 160 × 80 × 40 = 1216 × 10 3 mm 3
τ max =
Vmax Qmax
45 × 103 (1216 × 10−6 )
=
= 3.1 MPa
Ib
221.92 × 10−6 (2 × 40 ×10−3 )
(b) Due to symmetry, τ E must be the same at both section shown (Fig. b).
QE = 240 × 40 × 100 = 960 × 103 mm 3
VQE
45 × 103 (960 × 10−6 )
τE =
=
= 2.4 MPa
Ib
221.92 × 10−6 (2 × 40 × 10−3 )
(c) Note that due to symmetry, τ F is the same at both sections shown
in Fig. c.
Q F = 160 × 40 × 100 = 640 × 10 3 mm 3
τF =
VQF
45 × 103 (640 × 10−6 )
=
= 1.6 MPa
Ib
221.92 × 10−6 (2 × 40 × 10−3 )
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PROBLEM (*7.59) A laminated beam is composed of three glued boards (see Fig. P7.59). Calculate:
(a) The highest permissible value for P.
(b) The corresponding maximum flexure stress in the beam.
Given: An allowable shearing stresses of 120 psi in the glue and 500 psi in the wood.
*SOLUTION
y
P
4P
3 ft
3
6 ft
6 ft
1.5 in
P
3 ft
4 in
z
3
1.5 in
3 in
2P
P
V
x
-
I=
3(7)3
= 85.75 in.4
12
-2P
M
(lb ⋅ ft)
9P
x
-3P
-3P
(a) Q = 3 × 1.5 × 2.75 = 12.375 in.3
VQ 2 P(12.375)
P = 1247 lb
τg =
=
= 120,
85.75(3)
Ib
Also,
3 V 3 2P
P = 3500 lb
τw =
=
= 500,
2 A 2 3× 7
Thus,
Pall = 1247 lb
(b) σ max =
Mc 9 ×12(1247)(3.5)
=
= 5.497 ksi
I
85.75
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________________________________________________________________________
PROBLEM (7.60) A beam is composed of two 20 x 100-mm steel plates welded together as shown
in Fig. P7.60. Determine:
(a) The largest permissible vertical shear force in the beam.
(b) The corresponding maximum shearing stress in the beam.
Given: An allowable shear force per unit length in each weld of 200 kN/m.
SOLUTION
y
25 mm
81.25 mm
z
100 mm
C
y =43.75 mm
25 mm
z’
100 mm
25 × 100 × 75 + 100 × 25 ×12.5
(a) y =
= 43.75 mm
100 × 25 + 100 × 25
25(100)3
100(25)3
I=
+ 100 × 25(31.25) 2 +
+ 100 × 25(31.25) 2
12
12
6
4
= 7.096 × 10 mm
At welds
Q = 100 × 25 × 31.25 = 78.13 × 10 3 mm 3
q=
VQ V (78.13 × 10−6 )
=
= 2(200 × 103 )
I
7.096 ×10−6
from which
V = 36.33 kN
(b) τ max =
VQ 2(200 ×103 )
=
= 16 MPa
Ib
25(10−3 )
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PROBLEM (7.61) Re-solve Prob.7.60 for a built-up beam of the cross section shown in Fig.
P7.61.
SOLUTION
y
10 mm
(a)
208.5 mm
C
407 mm
240 mm
10 mm
z
7.7 mm
From solution of
Prob. 7.29:
I = 424.7 × 10 −6 m 4
We have, at welds:
Q = 240 × 10( 208.5)
= 500.4 × 10 3 mm 3
VQ V (500.4 ×10−6 )
=
= 2(200 × 103 )
−6
I
424.7 × 10
Solving
V = 339.5 kN
q=
VQ 2(200 ×103 )
(b) τ max =
=
= 51.9 MPa
Ib
7.7(10−3 )
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PROBLEM (7.62) Redo Prob. 7.60 for a symmetrical built-up beam of the cross section shown in
Fig. P7.62.
SOLUTION
(a)
y
60 mm
120 mm
z
C
10 mm
60 mm
604 − 504
+ (602 − 502 )(902 )]
12
10(120)3
+
12
6
= 20.38 × 10 mm 4
I = 2[
60 mm
5 mm
At welds
Q = (60 2 − 50 2 )(90) = 99 × 10 3 mm 3
VQ V (99 × 10−6 )
=
= 2(200 ×103 )
q=
−6
I
20.38 ×10
from which:
V = 82.34 kN
(b) τ max =
VQ 2(200 ×103 )
=
= 40 MPa
Ib
10(10−3 )
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________________________________________________________________________
PROBLEM (*7.63) A beam with two overhang carries a uniform load of intensity w and
supported as shown in Fig. P7.63. Determine the maximum permissible value of the load w.
Given: Allowable tensile and compressive stresses of 6 and 10 ksi, respectively, and a shearing
stress not to exceed 1.2 ksi
*SOLUTION
y
2.237
C
z
1 in.
1.5 in.
σc =
1.763
1 in.
Let w be in lb in.
From solution of Prob. 7.31:
I = 13.63 in. 4
Vmax = 3.75 × 12 w = 45w lb ⋅ in.
We have
Mc 648w(2.237)
σt =
=
= 6 × 103
13.63
I
or
w = 56.4 lb in.
Mc 684w(1.763)
=
= 10 × 103
13.63
I
or
w = 119 lb in.
V Q
45w[1(2.237)(2.237 2)
τ max = max max =
= 1.2 × 103
Ib
13.63(1)
or
w = 145 lb in.
Thus,
wall = 56.4 lb in.
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________________________________________________________________________
PROBLEMS (7.64 through 7.66) The maximum shearing stress in a beam with circular, thinwalled circular, or triangular crosss section (Fig. P7.64 through P7.66) acted upon by a vertical force
V may be expressed as follows:
τ max = k
V
A
(P7.64)
Here A is the cross-sectional area of the beam and k represents a numerical factor, or shear
coefficient. Determine:
(a) The location of the point at which τ max occurs.
(b) The value of k.
(c) The maximum vertical shear force V the section may carry for c = 2 in., b = h = 4 in., r = 1
in., t = 0.1 in., and τ max = 9 ksi.
SOLUTION (7.64)
(a) Q is maximum at the center.
Thus τ max occurs at the center.
4c
3π
(b) I = πc 4 4
A = πc 2
1
1
4c
2
Qmax = Ay = π c 2 ( ) = c 3
2
2
3π
3
3
VQ
V (2c 3)
4 V
τ max = max =
=
4
(π c 4)(2c) 3 π c 2
Ib
4V
4
=
k=
3A
3
( c) V =
τ max A
k
=
2c
9 × 103 (π )(2) 2
= 84.82 kips
43
SOLUTION (7.65)
(a) Q is maximum at the center. So τ w occurs at the center.
(b) Refer to Table B.6.
For a thin tube:
I = π r 3t
A = 2π rt
For a semicircular thin tube.
2r
A = π rt
y=
b = 2t
π
Q = Ay = 2r 2t
y
r
C
y
z
t
Continued on next slide
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Therefore, for a thin tube:
VQ
V (2r 2t )
V
τ max =
=
=2
3
A
Ib (π r t )(2t )
where
k=2
τ max A
9(103 )
(2π ×1× 0.1)
K
2
= 2.83 kips
( c) V =
=
SOLUTION (7.66)
(a)
1 3
bh
36
y
t =b
h
V Q
τ=
I t
I=
b
y
h/3
h
C
z
t
y
2h 2 y
−
3
3
2y/3
y=
z’
1
( yt ) y
1
Q
= 2
= ( yh − y 2 )
3
t
t
In order τ be maximum:
dτ d VQ V d 1
[ ( yh − y 2 )] = 0
= ( )=
dy dy It
I dy 3
or
1
y= h
2
(b) For y = h 2 :
V1 h
h2
Vh 2 3V
( h− ) =
τ max =
=
4
12 I bh
I 3 2
3V
3
=
k=
2A
2
Continued on next slide
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( c) V =
τ max A
k
=
9 ×103 4 × 4
= 48 kips
32
2
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________________________________________________________________________
PROBLEM (7.67) A hollow box- beam supports the loads shown in Fig. P7. 67. Compute the
maximum value of P for σ all = 5 MPa and τ all = 1.2 MPa.
SOLUTION
∑ M = 0 : − 40(5) − P(2) + R (4) = 0
P
P
R = + 50
F = 0 : R = − 10
∑
2
2
A
B
B
y
A
y
40 kN
P
A
2m
C
C
200 mm
40
P/2-10
200 mm
50 mm
RB=P/2+50
RA=P/2-10
V, kN
z
B 1m
2m
50 mm
x
-P/2-10
M, kN ⋅ m
2P-20
x
40
1
[200(300)3 − 100(200)3 ] = 383.33 × 106 mm 4
12
Q max = 200 × 150 × 75 − 100 × 100 × 50 = 1750 × 10 3 mm 3
I=
We have
P
( − 10)103 (1750 ×10−6 )
VQ
; 1.2 ×106 = 2
τ all =
383.33 ×10−6 (2 × 50 ×10−3 )
Ib
or
1.2(38,333)
P
− 10 =
,
2
1750
P = 72.57 kN
Mc
(2 P − 20)103 (150 ×10−3 )
6
; 5 × 10 =
σ all =
I
383.33 ×10−6
or
2 P − 20 = 12.78,
P = 16.39 kN
Thus,
Pall = 16.39 kN
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________________________________________________________________________
PROBLEM (7.68) A cantilever beam, having a built-up cross section is loaded as shown in Fig.
P7.68. Determine:
(a) The maximum bending stress in the beam.
(b) The maximum shearing stress in the beam
Given: P = 800 N,
M = 200 N ⋅ m
SOLUTION
y
200 N ⋅ m
1000 N ⋅ m
0.5 m
800 N
800
160 mm
800
V, N
x
M, kN ⋅ m
-1000
(a) σ max =
240 mm
z
1m
x
-600
-800
2404 − 1604
I=
12
= 221.87 × 10 6 mm 4
Mc 1000(120 ×10−3 )
=
= 541 kPa
I
221.87 ×10−6
(b) Qmax = 120 × 240 × 60 − 80 × 160 × 40
= 1216 × 103 mm 4
τ max =
VQmax
800(1216 × 10−6 )
=
= 54.1 kPa
Ib
221.87 × 10−6 (80 ×10−3 )
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________________________________________________________________________
PROBLEM (7.69) As shown in Fig. P7.69, a rectangular beam 6 in. wide and 12 in. deep
supports a total distributed load of W pounds and a concentrated load of 2W pounds. If σ all = 1800
psi and τ all = 120 psi, determine the maximum value of W.
SOLUTION
∑ M = 0 : R (12) − 12W (9) − W (6) = 0,
∑F = 0: R =W
A
B
y
RB = 2W
A
2W
W (total)
A
W
9 ft
y
B
C 3 ft 2W
V
W
-0.25W
-1.75W
12 in.
z
6 in.
x
-2W
M
x
5.625W lb ⋅ ft
=67.5W lb ⋅ in.
1
(6)(12)3 = 864 in.4
12
Mc 67.5W (6)
W = 3840 lb
σ all =
=
= 1800,
864
I
3 V 3 2W
W = 1920 lb
τ all =
=
= 120,
2 A 2 6×8
Thus,
Wall = 1920 lb
I=
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________________________________________________________________________
PROBLEM (7.70) A beam of built-up bolted cross section is loaded as shown in Fig. P7.70.
Determine:
(a) The maximum bending stress at section B.
(b) The bolt size to be used with 15-mm spacing along the beam.
Assumptions: The part abcd is a plank and that the allowable shear stress in each bolt is 60 MPa.
SOLUTION
4
∑ M = 0 : − 3( 3 ) − 8(5) + R (5) = 0,
∑ F = 0 : R = 2.2 kN
A
B
y
A
3 kN
A
2.2 kN
RB = 8.8 kN
y
2 kN/m
49.17
1m
2m C
D 2 mB 2 m E
z
8.8 kN
V, kN
2.2
C
100
y = 90.83
100
20
4
x
-0.8
20
-4.8
y=
40
Dimensions are
in millimeters
[20 ×140(70)]2 + 100 × 40(120)
= 90.83 mm
(20 ×140)2 + 100 × 40
I = 2[
20(140)3
+ 20 ×140(20.83) 2 ]
12
100(40)3
+
+ 100 × 40(29.17) 2 = 15.51× 106 mm 4
12
1
(a) M B = (2)(2) 2 = 4 kN ⋅ m
2
Mc 4 × 103 (90.83 × 10−3 )
=
= 23.42 MPa
σ max =
15.15 ×10−6
I
(b) Q = 40 × 100 × 29.17 = 116.68 × 10 3 mm 3
VQ 4.8 ×103 (116.68 ×10−6 )
=
= 36.11 kN
q=
I
15.15 ×10−6
F = s ⋅ q = 0.015(36.11 × 10 3 ) = 542 N
Thus,
F 2 542 2
345.05
τ=
=
;
60 ×106 =
2
A
d2
πd 4
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or
d = 2.4 × 10 −3 m = 2.4 mm
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________________________________________________________________________
PROBLEM (7.71) Consider the cantilever beam of a triangular cross section subjected to a
variable load w = kx3 kN/m, as shown in Fig. P7.71. Determine:
(a) The maximum bending stress in the beam.
(b) The maximum shearing stress at the neutral axis of the beam.
SOLUTION
250 lb/ft
w=kx3
y
A
y
MB
2 in.
x
5 ft
z 1 in.
RB
250 = k (5)3
k =2
0.667 in.= y
2 in.
3 in.
34
I=
= 2.25 in.4
36
5
x4
VB = Vmax = ∫ (2 x )dx =
= 312.5 lb
0
2 0
5
3
5
x4
x5
M B = M max = ∫ Vdx = ∫
dx =
0
0 2
10 0
= 312.5 lb ⋅ ft = 3750 lb ⋅ in.
5
(a) σ max =
5
Mc 3750(2)
=
= 3.33 ksi
2.25
I
(b) Q = 12 ( 2 × 2)(0.667) = 1.334 in. 3
VQ 312.5(1.334)
τz =
=
= 92.6 psi
Ib
2.25(2)
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PROBLEM (7.72) Determine the ratio of the maximum shearing stress to the largest bending stress
for the simple rectangular beam in Fig.P7.72 subjected to a uniformly distributed load of intensity w.
SOLUTION
w
h
L
wL/2
V, kN
wL/2
b
wL/2
x
M
kN ⋅ m
Mmax= wL2/8
-wL/2
x
3 V 3 wL 2 3 wL
=
=
2 A 2 bh
4 bh
2
Mc wL 8(h 2) 3 wL2
σ max =
=
=
4 bh 2
I
bh3 12
Thus,
τ max σ max = h L
τ max =
(1)
(2)
(3)
For example, if L = 10h , the above ratio is 1 10 .
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________________________________________________________________________
PROBLEM (7.73) Determine the ratio of the largest shear stress to the maximum bending
stress for a simply supported beam of circular cross section carrying a concentrated load P at
its midspan (Fig. P7.73).
SOLUTION
P
L/2
L/2
L
P/2
V, kN
h=2c
P/2
I=
P/2
π c4
4
x
-P/2
M
kN ⋅ m
PL/4
x
From solution of Prob. 7.64:
4V 4 P 2 2 P
τ max =
=
=
3 A 3 π c2 3 π c2
Also
Mc 4 M PL
σ max =
= 3 = 3
I
πc πc
Thus,
τ max σ max = 2c 3L = h 3L
For example, if L = 10h , the above quotient is 1 30 .
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________________________________________________________________________
PROBLEM (7.74) A simple wooden beam is subjected to a uniform load of intensity w as
shown in Fig. P7.74. Calculate:
(a) The maximum permissible length L.
(b) The largest allowable distributed load w.
Given: σ all = 8.4 MPa, τ all = 1.2 MPa
SOLUTION
(a) Refer to Solution of Prob. 7.72.
Equation (3):
σ
8.4
L = h all = 0.15( ) = 1.05 m
τ all
1.2
(b) Equation (1):
wall =
4 bh
4 0.05 × 0.15
τ all =
(1.2 × 106 ) = 11.43 kN m
3 L
3 1.05
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________________________________________________________________________
PROBLEM (7.75) A cantilever beam supports a uniformly distributed load of intensity w
(Fig. P7.75). Determine:
(a) The largest allowable length L for the beam.
(b) The maximum permissible load w.
Given: σ all = 8.4 MPa, τ all = 1.2 MPa
SOLUTION
(a)
w
wL2
2
I=bh3/12
h
wL
L
b
Mc ( wL2 2)6
wL2
=
=
3
I
bh 2
bh 2
3 wL
τ all =
and σ all τ all = 2 L h
2 bh
or
σ
8.4
Lall = all h =
(150) = 525 mm
2τ all
2(1.2)
σ all =
(b) wall =
2 bh
2 0.05 × 0.15
τ all =
(1.2 × 106 ) = 11.43 kN m
3 L
3 0.525
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________________________________________________________________________
PROBLEM (7.76) Design the cross section of the wooden beam shown in Fig. P7.76.
Given:
σ all = 1800 psi, τ all = 150 psi
SOLUTION
w = 600 lb ft = 50 lb in.
Refer to solution of Prob. 7.72.
Equation (2):
σ all =
3 wL2
;
4 bh 2
h2 =
3(50)(12 ×12) 2
,
4(6)(1800)
h = 8.485 in.
Equation (1):
3 wL
;
4 bh
The required depth is
hall = 8.5 in.
τ all =
h=
3(50)(144)
= 6 in.
4(6)(150)
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________________________________________________________________________
PROBLEM (7.77) Determine the depth h needed in the rectangular beam loaded as seen in
Fig. P7.77.
σ all = 10 MPa, τ all = 600 kPa,
Given:
b = 150 mm
SOLUTION
From solution of Prob. 7.25:
Vmax = 12.6 kN
M max = 13.23 kN ⋅ m
Thus,
6(13.23 ×103 )
,
(150 ×10−3 )h 2
h = 230 mm
3(12.6 × 103 )
,
2(150 × 10−3 )h
h = 210 mm
σ all =
6M
;
bh 2
10 × 106 =
τ all =
3V
;
2bh
600 ×103 =
The required depth is
hall = 230 mm
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________________________________________________________________________
PROBLEM (7.78) If σ all = 140 MPa and τ all = 80 MPa, select the wide-flange section
(see Table B.8) that should be used to support the loading shown in Fig. P7.78. Use P = 16kN
and M = 2 kN ⋅ m.
SOLUTION
2kN m
22 kN m
0.5 m
16 kN
1m
16 kN
S=
M max
σ all
=
22 × 103
140 × 106
= 157 × 10 3 mm 3
From Table B.8, we select: W 150 × 24
and Aweb = 160 × 6.6 = 1056 mm 2 . Thus, using Eq.(7.25):
τ avg =
V
16 × 103
= 15.2 MPa < 80 MPa
=
Aweb 1056 ×10−6
O.K.
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________________________________________________________________________
PROBLEM (7.79) Redo Prob. 7.78 for the case in which an S-beam (see Table B.9) is to be
selected.
SOLUTION
From solution of Prob. 7.78:
M max = 22 kN ⋅ m
Vmax = 16 kN
S = 157 × 10 3 mm 3
From Table B.8, we select: S 150 × 24
and Aweb = 203 × 6.9 = 1401 mm 2 . Thus, using Eq.(7.25):
τ avg =
V
16 × 103
= 11.4 MPa < 80 MPa
=
Aweb 1401×10−6
O.K.
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________________________________________________________________________
PROBLEM (7.80) For σ all = 18 ksi, calculate the minimum width b of the steel beam
carrying the loads shown in Fig. P7.80.
SOLUTION
y
800 lb
400 lb
400 lb
0.375 in.
z
800 lb
V,
kips
3 ft
3 ft
3 ft
3 ft
800 lb
4 in.
0.375 in.
0.375 in.
800
400
x
-400
M,
kip ft
2.4
3.6
-800
2.4
x
M max = 3.6 × 12 = 43.2 kip ⋅ in.
0.375(4)3
b(0.375)3
I=
+ 2[
+ 0.375b(2.1875) 2 ] = 2 + 3.598b in. 4
12
12
Thus,
I M
2 + 3.598b 43.2 ×103
;
S= =
=
2.375
18 × 103
c σ all
or
2 + 3.598b = 5.7
Solving, the required width of the beam is:
b = 1.03 in.
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PROBLEM (*7.81) Given that σ all = 9 MPa and τ all = l.l MPa, design the cross section of
a rectangular wooden beam loaded as shown in Fig. P7.81.
Assumption: The beam is to be twice as deep as it is wide (h = 2b).
*SOLUTION
∑ M = 0 : 2(2.5) − 15 − 80(2) + R (4) = 0
∑ F = 0 : R = 39.5 kN
C
y
RD = 42.5 kN
D
C
2 kN
A
20 kN/m
15 kN m
1.5 m B
V, kN
C
4m
1m
39.5 kN
2b
D
42.5 kN
b
37.5
x
xm
-42.5
-4.8
M,
kN m
Mmax
12
8
x
-3
From geometry:
xm
4 − xm
=
xm = 1.875 m
,
37.5 42.5
1
M max = (42.5)(4 − 1.875) = 45.156 kN ⋅ m
2
Thus,
6M 3M max
σ all = 2 =
2b3
bh
3 45.156 × 103
6
9 ×10 =
,
b = 196 × 10−3 m
3
2
b
3V
τ all =
2A
3 42.5 ×103
6
1.1× 10 =
,
b = 170 ×10−3 m
2
2 2b
The required dimensions are
b = 196 mm
h = 392 mm
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PROBLEM (7.82) Design a rectangular cantilever beam of constant strength and width b, to
support a uniformly distributed load of intensity w (Fig. P7.82).
Assumption: Only the normal stresses owing to the bending need be considered; the allowable
stress is σ all .
SOLUTION
y
w
wL2/2
x
L
wL
Equation (7.31) becomes, at a distance x:
bh 2 wx 2 2
=
6
σ all
At the fixed end ( x = L and h = h0 ):
bh02 wL2 2
=
6
σ all
Divide Eq. (1) by Eq. (2),
h 2 h02 = x 2 L2
(1)
(2)
or
h = h0
x
L
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________________________________________________________________________
PROBLEM (*7.83) Design a simply supported rectangular beam of constant strength and
width b, carrying a uniformly distributed load of intensity w as shown in Fig. P7.83.
Assumption: Only the normal stresses due to the bending need be considered; the permissible
stress equals σ all .
*SOLUTION
We have
1
1
1
wLx − wx 2 = w( Lx − x)
2
2
2
Then, at a distance x, Eq. (7.31) becomes
1
w( Lx − x)
bh 2 2
=
σ all
6
At x = L 2 :
1
w L2
bh02 2 ( 4)
=
σ all
6
Divide Eq. (1) by Eq. (2),
h 2 Lx − x 2
x
x
− ( )2
or h = 2h0
= 2
2
h0
L 4
L L
Mx =
(1)
(2)
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________________________________________________________________________
PROBLEM (*7.84) Design a simple rectangular beam of constant strength and width b,
acted upon a downward triangular loading as depicted in Fig. P7.84.
Assumption: Only normal stresses due to bending need be considered; the allowable stress is
σ all .
*SOLUTION
Referring to Fig. P7.84, we write
1
1
RA = RB = w0 L , VC = 0
MC =
w0 L2
4
24
The magnitude of moment at x :
wx
w
1
1 x
M=
w0 L2 − ( 0 )( x)( )( ) = 0 ( L3 − 8 x 3 )
L2
24
2 3
24 L
S=
bh 2
6
M w0 ( L3 − 8 x3 )
=
4 Lbh 2
S
σ all =
or
12
⎡ w0 ( L3 − 8 x 3 ) ⎤
h=⎢
⎥
⎣ 4bLσ all ⎦
⎡ w0 L2
8 x3 ⎤
(1 − 3 ) ⎥
=⎢
L ⎦
⎣ 4bσ all
12
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________________________________________________________________________
PROBLEM (7.85) Design a cantilever beam of uniform strength having a rectangular cross
section with constant depth h and varrying width b, supporting a concentrated load P at its free
end as shown in Fig. P7.85.
Assumption: Only the normal stresses due to the bending need be considered; the permissible
stress is σ all .
SOLUTION
M = Px
bh 2
M 6 Px
= 2
S=
; σ all =
S
bh
6
or
6 Px
h 2σ all
Let width at fixed end (x = L):
6 PL
bo = 2
h σ all
Then
b x
x
=
or b = bo
bo L
L
b=
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________________________________________________________________________
PROBLEM (7.86) Design a cantilever beam of uniform strength and width b subjected to to
a linearly varying loading of intensity zero at the free end and w 0 at the fixed end (Fig. P7.86).
Assumption: Only the normal stresses owing to the bending need be considered; the allowable
stress is σ all .
SOLUTION
w0
y
L
w0 L
2
x1
1 w0 2
x
2 L
M=
x/3
Equation (7.31), at a distance x is
1 w0 3
x
bh 2 6 L
=
6
σ all
At x = L,
1 w0 3
L
bh02 6 L
=
6
σ all
Divide Eq. (1) by Eq. (2):
h2 x3
=
h02 L3
or
x 3
h = h0 ( ) 2
L
w0 L2
6
1 w0 3
x
6 L
(1)
(2)
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PROBLEMS (7.87 through 7.89) The composite beams of Figs. P7.87 through P7.89 are
fabricated by bonding a steel bar and a brass bar and are subjected to the bending moment M = 200
N ⋅ m. Dimensions are in millimeters. Determine the maximum bending stress in the brass and steel.
Given: Es = 200 GPa, Eb = 100 GPa
SOLUTION (7.87)
y
M = 200 N ⋅ m
20 mm
n=
40- y =23
30 mm
z
C
Es
=2
Eb
y =17 mm
10 mm
z’
20n=40 mm
Equivalent brass section
∑ A y = 20 × 30 × 25 + 40 ×10 × 5 = 17 mm
20 × 30 + 40 × 10
∑A
1
I = ∑ ( bh + Ad )
2
y=
i
i
i
3
2
t
1
1
(20)(30)3 + 20 × 30(8)2 + (40)(10)3 + 40 × 10(12)2
12
12
6
4
= 0.1443 ×10 mm
Thus,
Mc 200(23 ×10−3 )
(σ b ) max =
= 31.9 MPa
=
It
0.1443 × 10−6
=
(σ s ) max =
nMc 2(200)(17 × 10−3 )
= 47.1 MPa
=
It
0.1443 ×10−6
SOLUTION (7.88)
y
20 mm
40- y =22 mm
20 mm
z
20 mm
C
z’
M = 200 N ⋅ m
n=
Es
=2
Eb
y =18 mm
Equivalent brass section
5+n(10)+5=30 mm
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PROBLEMS (7.87 through 7.89) The composite beams of Figs. P7.87 through P7.89 are
fabricated by bonding a steel bar and a brass bar and are subjected to the bending moment M = 200
N ⋅ m. Dimensions are in millimeters. Determine the maximum bending stress in the brass and steel.
Given: Es = 200 GPa, Eb = 100 GPa
SOLUTION (7.87)
y
M = 200 N ⋅ m
20 mm
n=
40- y =23
30 mm
z
C
Es
=2
Eb
y =17 mm
10 mm
z’
20n=40 mm
Equivalent brass section
∑ A y = 20 × 30 × 25 + 40 ×10 × 5 = 17 mm
20 × 30 + 40 × 10
∑A
1
I = ∑ ( bh + Ad )
2
y=
i
i
i
3
2
t
1
1
(20)(30)3 + 20 × 30(8)2 + (40)(10)3 + 40 × 10(12)2
12
12
6
4
= 0.1443 ×10 mm
Thus,
Mc 200(23 ×10−3 )
(σ b ) max =
= 31.9 MPa
=
It
0.1443 × 10−6
=
(σ s ) max =
nMc 2(200)(17 × 10−3 )
= 47.1 MPa
=
It
0.1443 ×10−6
SOLUTION (7.88)
y
20 mm
40- y =22 mm
20 mm
z
20 mm
C
z’
M = 200 N ⋅ m
n=
Es
=2
Eb
y =18 mm
Equivalent brass section
5+n(10)+5=30 mm
Continued on next slide
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y=
∑ A y = 20 × 20 × 30 + 30 × 20 ×10 = 18 mm
20 × 20 + 30 × 20
∑A
i
i
i
1
1
(20) 4 + 20 × 20(12) 2 + (30)(20)3 + 30 × 20(8) 2
12
12
6
4
= 0.1293 ×10 mm
It =
Thus,
(σ b ) max =
Mc 200(22 × 10−3 )
=
= 34 MPa
It
0.1293 × 10−6
(σ s ) max =
nMc 2(200)(18 × 10−3 )
=
= 55.7 MPa
It
0.1293 ×10−6
SOLUTION (7.89)
M = 200 N ⋅ m
y
Brass
Steel
z
20 mm
20 mm
20 mm
Es
=2
Eb
I t = I b + nI s
20(40)
Continued on next =
slide
[
5 mm
n=
3
π (5) 4
π (5) 4
−
]
] + 2[
12
4
4
= 0.1072 × 106 mm 4
Transformed brass section
So,
(σ b ) max =
Mc 200(20 × 10−3 )
=
= 37.3 MPa
It
0.1072 × 10−6
(σ s ) max =
nMc 2(200)(5 ×10−3 )
=
= 18.66 MPa
0.1072 × 10−6
It
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________________________________________________________________________
PROBLEMS (7.90 through 7.92) For the composite beams with cross sections as shown in
Figs. P7.87 through P7.89, determine the maximum permissible bending moments.
Given: ( σ b )all = 120 MPa, ( σ s )all = 140 MPa , Eb = 100 GPa, Es = 200 GPa.
SOLUTION (7.90)
From solution of Prob. 7.87:
n=2
I t = 0.1443 × 106 mm 4
y
23 mm
C
z
17 mm
Assume 120 MPa is maximum
allowable stress:
I
0.1443 × 10−6
M = σ b t = 120 × 106
c
23 ×10−3
= 753 N ⋅ m
The corresponding maximum stress in steel is:
nMc 2(753)(17 × 10−3 )
σs =
=
= 177 MPa > 140 MPa
It
0.1443 × 10−6
Hence, steel stress controls:
2M (17 ×10−3 )
(σ s ) all = 140 × 106 =
0.1443 ×10−6
from which M = 594 N ⋅ m . Therefore,
M all = 594 N ⋅ m
SOLUTION (7.91)
From solution of Prob. 7.88:
n=2
I t = 0.1293 × 106 mm 4
y
20 mm
z
18 mm
M = σb
C
Assume 120 MPa is maximum
allowable stress:
It
0.1293 × 10−6
= 705 N ⋅ m
= 120 × 106
c
22 ×10−3
The corresponding maximum stress in steel is:
nMc 2(705)(18 × 10−3 )
σs =
=
= 196 MPa > 140 MPa
It
0.1293 ×10−6
Continued on next slide
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Hence, steel stress controls:
2 M (18 × 10−3 )
140 × 106 =
,
0.1293 × 10−6
M = 503 N ⋅ m
Thus, M all = 503 N ⋅ m
SOLUTION (7.92)
From solution of Prob. 7.89:
I t = 0.1072 ×106 mm 4
n = 2.
Assume 120 Mpa is maximum allowable stress:
0.1072 × 10−6
= 643 N ⋅ m
M = 120 × 106
20 × 10−3
The corresponding maximum stress in steel is
nMc 2(643)(5 ×10−3 )
σs =
= 60 MPa < 140 MPa
=
It
0.1072 ×10−6
Brass stress controls:
M all = 643 N ⋅ m
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PROBLEM (7.93) A 6-in.-wide by 10-in.-deep wood cantilever beam that is 5 ft long is
reinforced with 6-in.-wide and 1/8 -in.-deep steel plates on its top and bottom faces. The beam
supports to a uniform load of 150 Ib/ft over its entire length, as shown in Fig. P7.93. Determine the
maximum bending stresses in each material.
Given: Ew = 1.5 x 106 psi, Es = 30 x 106 psi
SOLUTION
y
z
C
y
1/8 in.
10 in.
C
z
Wood
6 in.
n = Es Ew = 20
1/8 in.
6x12=120 in.
Equivalent wood section
Maximum moment occurs at the fixed edge:
1
1 0.15
M max = wL2 = (
)(5 × 12) 2 = 22.5 kip ⋅ in.
2
2 12
6(10)3
120(0.125)3
It =
+ 2[
+ 120(0.125)(5.0625) 2 ] = 1268.9 in.4
12
12
Hence,
Mc 22.5 ×103 (5)
(σ w ) max =
=
= 88.66 psi
It
1268.9
(σ s ) max =
nMc 20(22.5 ×103 )(5.125)
=
= 1.82 ksi
It
1268.9
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________________________________________________________________________
PROBLEM (7.94) Redo Prob. 7.93, given that the beam is simply supported.
SOLUTION
Maximum moment occurs at midspan:
1
1 0.15
M max = wL2 = (
)(5 ×12) 2 = 5.625 kip ⋅ in.
8
8 12
From solution of Prob. 7.93:
n = 20
I t = 1268.9 in.4
Thus,
Mc 5.625 ×103 (5)
(σ w ) max =
=
= 22.17 psi
It
1268.9
(σ s ) max =
nMc 20(5.625 × 103 )(5.125)
=
= 454 psi
It
1268.9
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________________________________________________________________________
PROBLEM (7.95) Rework Prob. 7.93, given that the wood cantilever beam is reinforced
with 10-in.-deep and 1/8 in.-wide steel plates on both its side faces.
SOLUTION
From solution of Prob. 7.93: n = 20
Equivalent wood section
y
y
z
z
C
6 in.
1/8 in. 1/8 in.
It =
M max = 22.5 kip ⋅ in.
C
6 in.
1/8n=2.5 in. 1/8n=2.5 in.
1
(11)(10)3 = 916.67 in.4
12
Hence,
(σ w ) max =
Mc 22.5 ×103 (5)
=
= 122.7 psi
It
916.67
(σ s ) max =
nMc 20(22.5 ×103 )(5)
=
= 2.46 ksi
It
916.67
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________________________________________________________________________
PROBLEM (*7.96) An aluminum plate and a steel plate are bonded together to create the
composite beam shown in Fig. P7.96. Find the dimension b of the cross section of the beam for
( σ a )all = 80 MPa and ( σ s )all =240 MPa.
Given:
E a = 70 GPa,
E s = 210 GPa
*SOLUTION
3n=60 mm
y
180- y =90 mm
160 mm
z
20 mm
z’
C
y =90 mm
n = Es Ea = 3
Equivalent
aluminum
section
b
60 × 160 × 100 + 20b(10) 48 ×103 + 10b
=
60 × 160 + 20b
480 + b
3
48 × 10 + 10b 38.4 × 103 + 170b
180 − y = 180 −
=
480 + b
480 + b
y=
We have
3M (180 − y )
It
My
(σ a ) all = 80 ×106 =
It
(σ s ) all = 240 × 106 =
Divide the first by the second:
3(180 − y ) 3(38.4 × 103 + 170b)
=
3=
48 ×103 + 10b
y
or
48 × 10 3 + 10b = 38.4 × 10 3 + 170b , b = 60 mm
Then
48 × 103 + 10(60)
y=
= 90 mm
480 + 60
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________________________________________________________________________
PROBLEM (7.97) For the composite beam with the cross section shown in Fig. P7.97,
calculate the permissible bending moment M about the horizontal (z) axis.
Given: ( σ b )all = 17 ksi, ( σ s )all = 20 ksi, E b = 15 x 10 6 psi, E s = 30 x 10 6 psi
SOLUTION
Use Brass as reference material. Thus
E
30
n= s =
=2
Eb 15
For transformed section:
1
1
I t = (1.5)(1) 2 + (1.0)(1)3
12
12
−3
4
= 208(10 ) in.
y
Brass
z
1.0 in.
1 in.
1.5 in.
Trandformed brass section
Brass : y = 0.5 in.
(σ b ) all = 17 ksi
My
M (0.5)
σ all =
;
17(103 ) =
It
208(10−3 )
or
M = 7.07 kip ⋅ in.
Steel : y = 0.5 in.
(σ s ) all = 20 ksi
(2) M (0.5)
nMy
σ all =
;
20(103 ) =
I
208(10−3 )
or
M = 4.16 kip ⋅ in.
Hence
M all = 4.16 kip ⋅ in.
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PROBLEM (*7.98) A cantilever beam of length L = 4 ft is subjected to a uniformly
distributed load of intensity w = 3 kip/ft. The cross section of the beam is a hollow box with
wood flanges and aluminum webs, as shown in Fig. P7.98. Determine the required thickness t
of the aluminum plates
( σ a )all = 20 ksi
Given: E w = 1.6 x 10 6 psi,
( σ w )all = 1.2 ksi, E a = 10 x 10 6 psi,
*SOLUTION
t
4 in.
y
E
10
n= a =
= 6.25
Ew 1.6
1
1
I t = {[4 + (6.25 × 2t )](10)3} − [4(4)3 ]
12
12
4
= 312 + 1041.7t in.
3 in.
10 in.
Aluminum
z
3 in.
Wood
1 2 1
wL = (3)(4) 2
2
2
= 24 kip ⋅ in. = 288 kip ⋅ in.
M=
So
nMy
;
I
6.25(288)(5)
20 =
,
312 + 1041.7t
σa =
t = 0.132 in.
Likewise
288(5)
My
;
1.2 =
,
312 + 1041.7t
I
Stress in wood governs:
t = 0.85 in.
σw =
t = 0.85 in.
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________________________________________________________________________
PROBLEM (*7.99) A round copper tube of outside diameter d and an aluminum core of
diameter d/3 are bonded together to form a composite beam of the cross section shown in Fig.
P7.99. The allowable stress in the bronze is σ b . Determine the maximum bending moment M
that can be supported by the beam, in terms of E a , E b , σ b , and d, as needed. Calculate the value
of M for: E a = 70 GPa, E b = 110 GPa, σ b = 520 MPa, d = 75 mm.
*SOLUTION
n = Eb Ea = 1.571
Transform to bronze:
d
5π d 4
π
I b = [d 4 − ( ) 4 ] =
64
3
324
Aluminum Core: Each element of area has width reduced in ratio n.
1 π d 4
πd4
( ) =
Ia =
5,184n
n 64 3
and
1
1
πd4
πd4
It = Ib + I a =
(5 +
)=
(80 + )
n
324
16n 5184
We have
σ I
πd3
1
Mc
σb =
σ b (80 + )
,
M= b t =
It
d 2 2592
n
Introducing the given data:
π (0.075)3
1
M=
(520 ×106 )(80 +
)
2592
1.571
= 21.44 kN ⋅ m
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________________________________________________________________________
PROBLEM (7.100) The cross section of a bimetallic beam consists of strips of aluminum and
copper bonded together, as depicted in Fig. P7.100. Under the action of bending moment M
about the z axis, determine:
(a) The maximum stress in the aluminum strip.
(b) The maximum stress in the copper strip.
Given: E c = 105 GPa, E a = 75 GPa, M = 40kN ⋅ m
SOLUTION
Ec 105
=
= 1.4
Ea 75
Equivalent aluminum section:
We have n =
y
20 mm
3 mm
z
8 mm
5.162 mm= y
1.4x(20)=28 mm
(28)(8)(4) + (20)(3)(9.5)
= 5.162 mm
(28 × 8) + (20 × 3)
1
I t = (28)(8)3 + (28 × 8)(1.162) 2 +
12
1
(20)(3)3 + (20 × 3)(4.338)2 = 2.67(103 ) mm 4
12
y=
My
(40)(−5.162 ×10−3 )
(a) σ a = −
=−
= 77.33 MPa
It
2.67(10−9 )
(b) σ c = −
nMy
(1.4)(40)(5.838 × 10−3 )
=−
= −122.4 MPa
It
2.67(10−9 )
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________________________________________________________________________
PROBLEM (7.101) A reinforced concrete beam of the cross section shown in Fig. P7.101 is
acted upon a by bending moment M about the z axis. Calculate:
(a) The stress in the steel.
(b) The maximum stress in the concrete.
Given: E c = 3.75 x 10 6 psi, E s = 30 x 10 6 psi, M = 5 kip ⋅ in.,
A s = 4 in.
b = 10 in.,
d = 16 in.,
SOLUTION
n=
Es
30
=
= 8,
Ec 3.75
nAs = (8)(4) = 32 in.2
Locate the N.A. Use Eq.(7.43a):
2n
2n
(kd ) 2 + (kd )
As − dAs = 0
b
b
Substituting the data
(kd ) 2 + 6.4(kd ) − 102.4 = 0
Solving,
kd =
and
y
kd
z
d=16 in.
d(1-k)
−6.4 + (6.4) 2 − 4(1)(−102.4)
= 7.413 in.
2
2 in.
2
nAs=32 in.
b=10 in.
d − kd = 8.587 in.
Equation(7.43b):
1
7.413 2
I t = (10)(7.413)3 + (10 × 7.413)(
) + 0 + (32)(8.587) 2
12
2
= 339.469 + 1, 018.408 + 2359.6 = 3717.4 in.4
(a) σ s =
nMc 8(5 × 103 )(8.587)
=
= 92.4 ksi
It
3717.4
(b) (σ c ) max =
Mc (5 × 103 )(7.413)
=
= 9.97 psi
It
3717.4
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________________________________________________________________________
PROBLEM (*7.102) Determine the maximum permissible value of P for the simply
supported reinforced concrete beam, loaded as shown in Fig.P7.102.
Given: ( σ c )all = 8.4 MPa,
( σ s )all = 140 MPa, n = 8
*SOLUTION
Dimensions are
in millimeters
300
y
100
60
kd=180.3
d=310
z
50
d(1-k)=129.7
nAs=32x103 mm2
n=8
We have: M max = 2 P
kd
kd − 0.06
∑ Qz = 0 : kd (0.3) 2 − (kd − 0.06)(0.1) 2 = 32 ×10−3 (0.31 − kd )
Simplifying,
( kd ) 2 + 0.38( kd ) − 0.101 = 0
(1)
Solving Eq. (1):
(kd )1,2 =
Use
−0.38 ± (0.38) 2 + 0.404 −0.38 ± 0.741
=
2
2
kd = 0.1803 m = 180.3 mm,
d (1 − k ) = 310 − 180.3 = 129.7 mm
Referring to the figure:
300(180.3)3
It =
+ 300 ×180.3(90.15)
12
100(120.3)3
−[
+ 100(120.3)(60.15) 2 ] + 32 × 103 (129.7) 2
12
= 631.7 × 106 mm 4
Therefore,
Mc 2 P(180.3 ×10−3 )
(σ c ) max =
=
= 8.4 × 106
−6
631.7 ×10
It
Continued on next slide
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from which
P = 14.72 kN
Similarly,
(σ s ) max =
or
nMc 8(2 P)(129.7 ×10−3 )
=
= 140 × 106
631.7 ×10−6
It
P = 42.6 kN
Thus,
Pall = 14.72 kN
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________________________________________________________________________
PROBLEM (7.103) Consider the reinforced concrete beam of Fig. 7.103. For ( σ c )all = 10
MPa and ( σ s )all = 140 MPa., determine the largest bending moment M that may be applied to
about the z axis of the beam.
Given: b = 400 mm, d = 600 mm,
A s = 2500 mm2 ,
n=8
SOLUTION
Equation (7.43a) becomes
16
16
(kd ) 2 + (kd )
(2500) −
(600)(2500) = 0
400
400
or
( kd ) 2 + 100( kd ) − 60 × 10 3 = 0
Solving,
kd = 200 mm
600 − kd = 400 mm
From Eq. (d) of Example 7.15:
σ
kd 10 × 106
0.2
M c = c (bkd )(d − ) =
(0.4 × 0.2)(0.6 −
)
2
3
2
3
= 213 kN ⋅ m
kd
M s = σ s As (d − ) = 140 ×106 (2.5 × 10−3 )(0.533)
3
= 186.6 kN ⋅ m
Thus, M all = 186.6 kN ⋅ m
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________________________________________________________________________
PROBLEM (7.104) The beam of cross section shown in Fig. 7.103 is to support a moment
M z = 40 kN m about the z axis. Calculate:
(a) The minimum permissible value of width b.
(b) The minimum allowable value of area A s .
Given: ( σ s ) all = 150 MPa, (σ c ) all = 7.5 MPa,
d = 500 mm,
n = 10
SOLUTION
Referring to Fig. 7.38:
kd
d (1 − k )
=
;
(σ c ) all
σs n
(a) The first of Eqs. (7.44):
kd d (1 − k )
=
,
7.5 150 10
(σ c ) max = 7.5 × 106 =
k=
1
3
2(40 × 103 )
1
1
b(0.5) 2 ( )(1 − )
3
9
from which
b = 0.144 m
(b) The second of Eqs. (7.44):
σ s = 150 ×106 =
or
40 ×103
1
As (0.5)(1 − )
9
As = 600 mm 2
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________________________________________________________________________
PROBLEM (7.105) Determine the value of α for which the bending stress in the cantilever beam,
loaded as shown in Fig. P7.105, is a maximum.
Given: b = 6 in., h = 9 in.
SOLUTION
M z = − PL cos α
M y = − PL sin α
y
Psinα
z
Pcosα
D
A
C
9 in.
Maximum value of
B 6 in.
Bending stress occurs
at A(or B):
M z
M y
PL cos α (4.5) PL sin α (3)
+
σA = − z A + y A =
6(9)3 12
9(6)3 12
Iz
Iy
dσ A
PL(4.5) sin α PL(3) cos α
=−
+
=0
dα
6(9)3 12
9(6)3 12
or
α = 56.3o
tan α = 1.5,
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________________________________________________________________________
PROBLEM (7.106) A cantilever beam is loaded as shown in Fig. P7.105. Determine:
(a) The maximum normal stress.
(b) The orientation of the neutral axis.
Given: L = 2 m,
h = 2b = 100 mm,
α = 20°,
P = 500 N
SOLUTION
50(100)3
= 4.17 × 106 mm 4
12
100(50)3
Iy =
= 1.04 × 106 mm 4
12
M z = − PL cosα
M y = − PL sin α
Iz =
y
A
φ
z
C
B
100 mm
50 mm
(a) Maximum stress occurs at A (or B):
M z
M y
PL cos α (50) PL sin α (25)
+
σA = − z A + y A =
4.17(10−6 )
1.04(10−6 )
Iz
Iy
=
(b) tan φ =
N.A.
500 × 2 cos 20o (0.05) sin 20o (0.025)
[
+
] = 19.5 MPa
10−6
4.17
1.04
Iz
4.17
tan α =
tan 20o
1.04
Iy
or
φ = 55.6 o
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________________________________________________________________________
PROBLEMS (7.107 through 7.109) As depicted in the cross section in Figs. P7.107 through
P7.109, a beam is acted upon by a moment M with its vector forming an angle α with the horizontal
axis. Calculate the stress at points A, B, and D.
SOLUTION (7.107)
y
80(120)3
Iz =
= 11.52 × 106 mm 4
12
120(80)3
Iy =
= 5.12 × 106 mm 4
12
M z = 3 cos 25o = 2.72 kN ⋅ m
M y = − sin 25o = −1.27 kN ⋅ m
80 mm
D
C
z
120 mm
Mz
A
My B
Apply Eq.(7.45):
M z
2.72 × 103 (−0.06) 1.27 ×103 (0.04)
M y
σA = − z A + y A = −
−
11.52 ×10−6
5.12 × 10−6
Iz
Iy
= 14.2 − 9.92 = 4.28 MPa
σ B = 14.2 + 9.92 = 24.12 MPa
σ D = −14.2 − 9.92 = −24.12 MPa
SOLUTION (7.108)
y
My
5 mm
D
1
(100 × 1003 ) − 96 × 903
12
= 2.5 × 10 6 mm 4
Iz =
4 mm
z
Mz
100 mm
C
15 mm
A
B
100 mm
M z = 2 cos 60 o = 1 kN ⋅ m,
2
90(4)3
3
I y = (5 × 100 ) +
12
12
3
4
= 833.8 × 10 mm
5 mm
M y = 2 sin 60 o = 1.732 kN ⋅ m
1×103 (−0.05) 1.732 × 103 (−0.05)
M z yA M y zA
σA = −
+
=−
−
2.5 × 10−6
0.83 × 10−6
Iz
Iy
= 20 + 103.86 = 123.9 MPa
σ B = 20 − 103.86 = −83.86 MPa
σ D = −20 + 103.86 = 83.86 MPa
Continued on next slide
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SOLUTION (7.109)
y
D
80 mm
60 mm
z
C
100 mm
A
120 mm
1
(120 × 803 − 100 × 603 )
12
= 3.32 × 106 mm 4
Iz =
My
Mz
B
M z = −5 cos 30o = −4.33 kN ⋅ m,
1
(80 × 1203 − 60 × 1003 )
12
= 6.52 × 106 mm 4
Iy =
M y = 5 sin 30o = 2.5 kN ⋅ m
M z y A M y z A 4.33 × 103 (−0.04) 2.5 × 103 (0.06)
+
=
+
3.32 ×10−6
6.52 ×10−6
Iz
Iy
= −52.17 + 23.0 = −29.2 MPa
σ B = −52.17 − 23.0 = −75.2 MPa
σ D = 52.17 + 23.0 = 75.2 MPa
σA = −
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________________________________________________________________________
PROBLEM (*7.110 through 7.112) As shown in cross section in Figs. P7.110 through P7.112, a
beam is subjected to a moment M with its vector forming an angle α with the horizontal axis. Calculate:
(a) The orientation of the neutral axis.
(b) The maximum bending stress.
SOLUTION (*7.110)
N.A.
y
φ
1 in 2.25 in.
z
Mz
My
A
1 in.
(a) tan φ =
6(1)3
1(6)3
+ 6 ×1(1.75) 2 +
+ 1× 6(1.75) 2
12
12
4
= 55.25 in.
Iz =
6 in.
Iz
55.25
tan α =
tan15o ,
Iy
18.5
1(6)3 6(1)3
Iy =
+
= 18.5 in.4
12
12
M z = −100 cos15o = −96.6 kip ⋅ in.
M y = −100 sin 15o = −25.9 kip ⋅ in.
φ = 38.7 o
(b) Point A is farthest from N.A.
Thus,
M z
M y
σA = − z A + y A
Iz
Iy
−96.6 ×103 (−4.75) −25.9 × 103 (0.5)
=−
+
= −8.305 − 0.7
55.25
18.5
= −9.01 ksi
SOLUTION (7.111)
y
z =0.565 in.
z
φ
Mz
8 in.
C
My
N.A.
0.39 in.
Table B.10: C 8 × 18.75
I z = 44 in. 4 ,
I y = 1.98 in. 4
M z = 30 cos 30 o = 25.98 kip ⋅ in.
M y = −30 sin 30 o = −15 kip ⋅ in.
A
2.527 in.
(a) tan φ =
Iz
44
tan α =
tan(−30o ) ,
1.98
Iy
φ = −85.5o
Continued on next slide
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(b) Point A is farthest from N.A. Thus,
M z
M y
σ A = σ max = − z A + y A
Iz
Iy
25.98 × 103 (−4) −15 × 103 (−1.962)
+
44
1.98
= 2.36 + 14.86 = 17.22 ksi
=−
SOLUTION (7.112)
y
Table B.11: L8 × 8 × 3 4
A
y’
a =8 in.
z’A
A = 11.4 in. 2
rmin = 1.58 in.
I y = I z = 69.7 in. 4
y’A
My’
45o
z
y = z =2.28 in. φ '
Mz’
C
α
B
z’
N.A.
We have
2
I y ' = Armin
= 14.4(1.58) 2 = 28.46 in.2
I z ' = I 1 = I y + I z − I y ' = 110.94 in. 4
(a) tan φ ' =
Iz'
110.94
tan(45o + 20o ) =
tan 65o ,
I y'
28.46
φ ' = 83.2 o
(b) From geometry:
yA ' = 0
z B ' = 2 y = 3.224 in.
z A ' = −a sin 45o + 2 y cos 45o − 2.433 in.
y A ' = a cos 45o = 5.657 in.
M z ' = 200 sin(45o − 20 o ) = 84.52 kip ⋅ in.
M y ' = 200 cos(45o − 20o ) = 181.26 kip ⋅ in.
Apply Eq.(7.45):
M y ' M z '
σ A = − z' A + y' A
Iz'
I y'
84.52(5.657) 181.26(−2.433)
+
= −4.31 − 15.5
110.94
28.46
= −19.81 ksi
=−
Continued on next slide
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181.26(3.224)
= 20.53 ksi
28.46
Thus, σ max = 20.53 ksi
Similarly,
σB = 0+
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________________________________________________________________________
PROBLEM (7.113) Redo Prob. 7.111 for a C4 x 5.4 channel (refer to Table B . 1 0 ) with
M = 10 kip ⋅ in.
SOLUTION
N.A.
y
Table B.10: C 4 × 5.4
0.184 in.
4 in.
z
I z = 3.85 in. 4
I y = 0.319 in. 4
Mz
C
φ
z =0.458 in.
My
A
M z = 10 cos 30 o = 8.66 kip ⋅ in.
M y = −10 sin 30 o = −5 kip ⋅ in.
1.584 in.
3.85
(a) tan φ =
tan(−30o ),
0.319
φ = −81.8o
(b) Point A is farthest from N.A. Thus,
M z
8.66 × 103 (−2) −5 ×103 (−1.13)
M y
σ max = σ A = − z A + y A = −
+
3.85
0.319
Iz
Iy
= 4.5 + 17.71 = 22.21 ksi
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________________________________________________________________________
PROBLEM (7.114) Redo Prob. 7.112 for an L 5 x 3 x ¼ angle (refer to Table B 12) with
M = 15 kip-in.
SOLUTION
y A sin θ p
A
y’
z’A My’
y A cos θ p
φ'
z’
Table B.12: L5 × 3 × 1 4
z = 0.657 in.
y = 1.66 in.
I y = 1.44 in. 4
I z = 5.11 in. 4
tan θ p = 0.371
C
z
y
z A cos θ p
z A sin θ p
y
θ p = 20.4 o
Mz’
A = 1.94 in. 2
rmin = 0.663 in.
z
θp
N.A.
θ p + α = 20.4 + 20 = 40.4 o
M y ' = 15 sin 40.4 o = 9.72 kip ⋅ in. ,
2
I y ' = Armin
= 1.94(0.663) 2 = 0.853 in. 4 ,
M z ' = 15 cos 40.4 o = 11.42 kip ⋅ in.
I z ' = I z + I y − I y ' = 5.697 in. 4
Iz'
5.697
tan(40.4o ) =
tan(40.4o ) , φ = 80 o
0.853
I y'
(b) Point A is farthest from N.A. From geometry:
y A' = y A cosθ p + z A sin θ p
(a) tan φ ' =
z A' = − y A sin θ p + z A cosθ p
(1)
Substitute the numerical values into Eqs.(1):
y A' = (5 − 1.66) cos 20.4 o + 0.657 sin 20.4 o = 3.36 in.
z A' = −(5 − 1.66) sin 20.4 o + 0.657 cos 20.4 o = −0.548 in.
Thus,
M z
M y
σ A = − z ' A' + y ' A'
Iz'
I y'
11.42(3.36) 9.72(−0.548)
=−
+
= −6.74 − 6.25 = −13 ksi
5.697
0.853
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________________________________________________________________________
PROBLEMS (7.115 through 7.117) A beam of cross section shown (Figs. P7.115 through
P7.117) is acted upon by a moment M with its vector forming an angle α with the horizontal
axis. Determine:
(a) The orientation of the neutral axis.
(b) The maximum bending stress,
Given: M = 2 kN ⋅ m, α = 15°
For Z-section: I z = 3.11× 10 mm ,
6
4
I y = 12.43 × 106 mm 4 , I yz = 4.56 × 106 mm 4 .
SOLUTION (7.115)
y
My
φ
C
z
r
r = 110 mm
4r
y=
= 46.7 mm
3π
y
Mz
N.A.
A
4
I z = 0.11r = 0.11(110) 4 = 16.1 × 10 6 mm 4
Iy =
π r4
=
π
(110) 4 = 57.5 × 106 mm 4
8
8
M y = 2 × 10 3 sin 15o = 518 N ⋅ m
M z = 2 × 10 3 cos15o = 1932 N ⋅ m
(a) tan φ =
Iz
16.1
tan α =
tan15o ,
57.5
Iy
φ = 4.3o
(b) Point A is farthest from N.A. From geometry:
y A = − r cos 4.3o + 46.7 = −63 mm
z A = r sin 4.3o = 8.25 mm
Thus,
M z
M y
σA = − z A + y A
Iz
Iy
1932(−0.063) 518(0.00825)
+
= 7.56 + 0.074
16.1×10−6
57.5 × 10−6
= 7.63 MPa
=−
SOLUTION (7.116)
120 mm
y
A
My
N.A.
b=100 mm
z
φ
C
Mz
Continued on next slide
40 mm
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bh3 100(120)3
=
= 4.8 ×106 mm 4
36
36
3
hb 120(100)3
Iz =
=
= 2.5 × 106 mm 4
48
48
Iy =
M y = 2 × 103 sin 15o = 518 N ⋅ m ,
(a) tan φ =
Iz
2.5
tan α =
tan(−15o ) ,
4.8
Iy
M z = −2 × 103 cos15o = −1.932 N ⋅ m
φ = −7.9 o
(b) Point A is farthest from N.A. Thus,
M y M z 1932(0.05) 518(0.04)
σA = − z + y =
+
2.5 ×10−6
4.8 × 10−6
Iz
Iy
= 38.64 + 4.32 = 43 MPa
SOLUTION (7.117)
y
z’
N.A.
z A cos θ p
y’
y A sin θ p
60
Mz’
z
θp
C
15
My’
φ'
A
15
120
Dimensions are
In millimeters.
15
yA '
This part is already given (not required by student)
15(60)3
150(15)3
I z = 2[
+ 15 × 60(37.5) 2 ] +
= 3.11× 106 mm 4
12
12
60(15)3
15(150)3
I y = 2[
+ 15 × 60(67.5) 2 ] +
= 12.43 × 106 mm 4
12
12
I yz = 0 + 60(15)(37.5)(67.5) + 0 + 60(15)( −37.5)( −67.5)
= 4.56 × 10 6 mm 4
Continued on next slide
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r = (4.56) 2 + (4.66) 2
= 6.52
I yz
(106 )
(12.43, 4.56)
C
2θ p '
6
R
O
I ' = 7.77
I y , I z (10 )
tan 2θ p ' =
4.56
4.66
θ p ' = 22.2o
4.66
I y ' = I1 = 7.77 + 6.52 = 14.29 × 106 mm 4
I z ' = I 2 = 7.77 + −6.52 = 1.25 × 10 6 mm 4
M y ' = −2 sin(15o + 22.2 o ) = −1.21 kN ⋅ m
M z ' = 2 cos(15o + 22.2o ) = 1.59 kN ⋅ m
(a) tan φ ' =
Iz'
tan(−15o − 22.2o ) ,
I y'
φ = −3.8o
(b) Point A farthest from N.A.
z A ' = − z A cos θ p − y A sin θ p
= −60 cos 22.2o − 67.5sin 22.2o = −81.1 mm
y A ' = z A sin θ p − y A cos θ p
= 60sin 22.2o − 67.5cos 22.2o = −39.8 mm
Thus,
σA = −
M z ' yA ' M y ' z A '
+
Iz'
I y'
1.59 × 103 (−0.0398) −1.2 ×103 (−0.0811)
+
1.25 ×10−6
14.29 × 10−6
= 50.63 + 7.98 = 58.61 MPa
=−
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________________________________________________________________________
PROBLEMS (7.118 through 7.120) For the beam cross section shown in Figs. P7.115
through P7.117, determine the largest permissible value of the moment M if σ all = 100 MPa
and
α = 0o.
SOLUTION (7.118)
r = 110 mm
α =0
φ =0
y = 46.7 mm
y
y
z
C r
M
N.A.
A
From solution of Prob. 7.115:
I z = 16.1 × 10 6 mm 4
I y = 57.5 × 10 6 mm 4
We have
σ all = σ A =
M z yA
Iz
from which
σ all
100 ×106 (16.1×10−6 )
(0.11 − 0.0467)
yA
= 25.4 kN ⋅ m
Mz =
=
SOLUTION (7.119)
Refer to solution of Prob.7.116:
I y = 4.8 × 10 −6 m 4
I z = 2.5 × 10 −6 m 4
We now have M y = 0 and M z = − M . Thus,
σ all = σ A = −
Mz =
σ I
− My A
or M z = all z
I
yA
100 × 106 (2.5 ×10−6 )
= 5 kN ⋅ m
0.05
SOLUTION (7.120)
Refer to solution of Prob. 7.117:
I y ' = I 1 = 14.29 × 10 −6 m 4
I z ' = I 2 = 1.25 × 10 −6 m 4
θ P ' = 22.2o
y A ' = −39.8 mm
z A ' = −81.1 mm
We now have
Continued on next slide
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M y ' = − M sin 22.2 o = −0.378M
M z ' = − M cos 22.2o = 0.926 M
tan φ =
Iz'
1.25
tan(−22.2o ) =
tan(−22.2o ) ,
14.29
I y'
φ = −2 o
Point A is farthest from N.A. Thus,
M y M z 1932(0.05) 518(0.04)
+
σA = − z + y =
2.5 ×10−6
4.8 × 10−6
Iz
Iy
0.926M z (−0.0398) −0.378M (−0.0811)
100 × 106 = −
+
1.25 ×10−6
14.29 ×10−6
or
100 ×103 = 29.48M + 2.15M z
from which
M z = 3.16 kN ⋅ m
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________________________________________________________________________
PROBLEMS (7.121 through *7.123) For the beam cross section shown in Figs. P7.115
through P7.117, determine the largest permissible value of the moment M if σ all = 80 MPa
and
α = 30 o .
SOLUTION (7.121)
Refer to Solution of Prob.7.115:
r = 110 mm
y = 46.7 mm
I y = 57.5 × 10 −6 m 4
I z = 16.1 × 10 −6 m 4
We now have
M y = M sin 30 o = 0.5M
tan φ =
M z = M cos 30 o = 0.866 M
Iz
16.1
tan 30o =
tan 30o ,
57.5
Iy
φ = 9.2o
Point A is farthest from N.A.
y A = − r cos 9.2 o + y = −61.9 mm
z A = r sin 9.2 o = 17.6 mm
Therefore,
M z
M y
σA = − z A + y A
Iz
Iy
0.866 M (−0.0619) 0.5M (0.0176)
+
16.1×10−6
57.5 ×10−6
80 × 10 3 = 3.3M + 0.153M
800 ×106 = −
or
from which
M = 22.97 kN ⋅ m
SOLUTION (7.122)
See solution of Prob.7.116:
We now have
M y = M sin 30 o = 0.5M
tan φ =
I y = 4.8 × 10 −6 m 4
I z = 2.5 × 10 −6 m 4
M z = − M cos 30 o = −0.866 M
Iz
2.5
tan(−30o ) =
tan(−30o ) ,
4.8
Iy
Point A is farthest from N.A. Thus,
M z
M y
σA = − z A + y A
Iz
Iy
φ = −16.7 o
Continued on next slide
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80 ×106 =
0.866M (0.05) 0.5M (0.04)
+
2.5 ×10−6
4.8 × 10−6
or
80 × 103 = 17.32 M + 4.17 M
from which
M = 3.72 kN ⋅ m
SOLUTION (*7.123)
See solution of Prob.7.117:
I y ' = I1 = 14.29 ×10−6 m 4
θ p ' = 22.2o
I z ' = I 2 = 1.25 ×10−6 m 4
y A ' = −39.8 mm
We now have
I
1.25
tan φ ' = z ' tan(−30o ) =
tan(−30o ) ,
14.29
I y'
z A ' = −81.1 mm
φ ' = −2.9 o
M y ' = − M sin(22.2 o + 30 o ) = −0.79 M
M z = M cos(22.2 o + 30 o ) = 0.613M
Point A is farthest from N.A. Thus,
M y ' M z '
σ A = − z' A + y' A
Iz'
I y'
0.613M (−0.0398) −0.79 M (−0.0811)
80 ×106 = −
+
1.25 ×10−6
14.29 ×10−6
or
80 ×103 = 19.518M + 4.423M
Solving,
M = 3.34 kN ⋅ m
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________________________________________________________________________
PROBLEM (*7.124) As shown in Fig. P7.124, a simply supported beam, L meters long,
has a Z section of uniform thickness t for which
I y = 2tc 3 3
I y = 8tc 3 3
I yz = −tc 3
Determine the maximum bending stress in the beam if it is subjected to a load P at midspan
and c = 10t.
*SOLUTION
y
y’
P
θ '
z’ p
z
2θ p '
c
φ'
C
c
5
3
A
8
I z = tc3
3
R
O
c
t
2
I y = tc 3
3
I yz
N.A.
I yz = −tc 3
C
( 83 tc 3 , tc3 )
I1
I y , Iz
tc 3
c = 10t
R = tc3 12 + 12 = 2tc3
I z ' = I 1 = ( 53 + 2 )tc 3 = 3.081tc 3
I y ' = I 2 = ( 53 − 2 )tc 3 = 0.253tc 3
tan 2θ p ' =
1
1
θ p ' = 22.5o
M z ' = 0.25PL cos 22.5o = 0.231PL
M y ' = −0.25PL sin 22.5o = −0.096 PL
tan φ ' =
Iz'
tan(−22.5o ),
I y'
φ ' = −78.8o
Point A is farthest from N.A. From Example 7.17:
1
t
t
y A ' = − y A cos θ p '+ t sin θ p ' = −(c + ) cos 22.5o + sin 22.5o = −9.5t
2
2
2
t
t
t
z A ' = − y A sin θ p '− t cos θ p ' = −(c + ) sin 22.5o − cos 22.5o = −4.5t
2
2
2
Thus,
Continued on next slide
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σ A = σ max = −
M z ' yA ' M y ' zA '
+
Iz'
I y'
0.231PL(−9.5t ) −0.096 PL(−9.5t )
+
3.081tc3
0.253tc3
PL
PL
= 2.419 3 = 2.419 × 10−3 3
c
t
=−
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________________________________________________________________________
PROBLEM (*7.125) A beam that has singly symmetric section, carries a moment M in a plane
forming an angle α with the horizontal axis (Fig. P7.125). Calculate:
(a) The stress at point A.
(b) The stress at point B.
α = 25o
Given: M = 2 kN ⋅ m,
*SOLUTION
y
z My
A
h1
z
h2
C
Mz
b2
b1
z=
Iz =
B
1
(b)(h3 )
12
1
b
(h)(b3 ) + bh( ) 2
12
2
1
= (h)(b3 )
3
Iy =
∑ Az = (50 × 200)(25) + (100 × 50)(100) = 50 mm
(50 × 200) + (100 × 50)
∑A
1
1
(50)(200)3 + (100)(50)3 = 34.375(10−6 ) m 4
12
12
1
1
I y = (200)(50)3 + (50)(100)3 = 25(10−6 ) m 4
3
3
Iz =
y A = 25 mm yB = −100 mm
z A = −100 mm z B = 0
M z = M cos α = 2 cos 25o = 1.8126 kN ⋅ m
M y = M sin α = 2sin 25o = 0.8452 kN ⋅ m
(a) σ A = −
(1,826)(0.025) (845.2)(−0.1)
M z yA M y zA
+
=−
+
= −4.71 MPa
34.375(10−6 )
25(10−6 )
Iz
Iy
(b) σ B = −
(1,826)(−0.1) (845.2)(0)
M z yB M y zB
+
=−
+
= 5.31 MPa
34.375(10−6 )
25(10−6 )
Iz
Iy
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________________________________________________________________________
PROBLEM (7.128) A channel section of uniform thickness is loaded as shown in Fig. P7.128.
Calculate:
(a) The distance e to the shear center.
(b) The shearing stress at D.
(c) The maximum shearing stress.
Given: b = 120 mm,
h = 180 mm,
t = 5 mm,
Vy = 2 kN
SOLUTION
y
t
D
Vy
e
z
S
h/2
C
h/2
Equation (h) of Example 7.18:
1
1
I z = th3 + bth 2
12
2
1
1
= (5)(180)3 + (120 × 5)(180) 2
12
2
6
= 12.15 × 10 mm 4
b
(a) Equation (7.52):
b
120
e=
=
= 43.64 mm
2 + (h 3b) 2 + (180 240)
(b) τ D = Vy
bc 2 × 103 (0.12)(0.09)
=
= 1.78 MPa
I
12.15 ×10−6
(c) The first moment of the shaded section about the z-axis:
1
1
Qz = bt (h 2) + ht (h 4) = ht (4b + h)
2
8
Shear stress formula is thus:
V Q V (ht 8)(4b + h)
τ max = y z = y
1 2
I zt
th (6b + h)t
12
3V (4b + h)
= y
2th(6b + h)
(1)
(7.51)
Substitute the given data
3(2 ×103 )(4 × 0.12 + 0.18)
τ max =
= 244 kPa
2(0.05 × 0.18)(6 × 0.12 + 0.18)
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________________________________________________________________________
PROBLEM (7.129) For a C 6 x 10.5 channel, calculate the maximum shearing stress caused by a
Vy = 10 kips vertical shear force applied at the shear center S. (Refer to Table B.10.)
SOLUTION
Table B.10: C6x10
y
tavg=0.343 in.
Vy
6 in.
z
0.5 in.
C
0.314 in.
2.034 in.
We have
Qmax = 3 × 0.314 × 1.5 + 0.343 ×1.72(3 −
τ max =
Vy Qmax
I zb
=
0.343
) = 3.082 in. 3
2
10 ×103 (3.082)
= 6.46 ksi
15.2(0.314)
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________________________________________________________________________
PROBLEM (*7.130 through 7.132) Determine the distance e that locates the shear center S of
a thin-walled beam of constant thickness t for the cross section shown in Figs. P7.130 through P7.132.
SOLUTION (*7.130)
y
s
Vy e
z
From A to B:
s
ts 2
)=
QAB = st (
2 2
2 2
2
ta
QB =
2 2
B
s
a
A
C
S D
a
t
E
τB =
A2 = (τ D − τ B )a
2
3
τB
τB
τD
I zt
=
a 2Vy
2 2I z
A1 = 13 τ B a
FBD FAB
FBD
Vy QB
FAB = A1t =
a 3tVy
6 2I z
FA
(See: Table, B.6)
From B to D:
a
s
ta 2
ts
−
+
)=
(2a − s )
2 2 2
2 2 2 2
Vy QBD Vy a 2
s
τ BD =
= [
+
(2a − s )]
I zt
Iz 2 2 2 2
a 2Vy
τD =
2I z
QBD = QB + st (
5ta 3Vy
2
FBD = τ B at + (τ D − τ B )at =
3
6 2I z
∑M = 0:
s
2 FAB 2a
2F
2F
2a
(
+ e) = BD (
+ e) = BD e
2
2
2
2
2
or
e=
a
2 2
Continued on next slide
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SOLUTION (7.131)
y
FAB A
B
t
Vy
z
τ AB =
FDE
D
h/2
S
C
e
FBG
=
I zt
bh
=
Vy
2I z
h
FDE
FAB =
FA
G
Vy Qz
bt (h 2)Vy
I zt
bτ AB t b 2 ht
=
Vy
2
4I z
b
bh
Vy
4I z
τ DE =
τ DE =
b 2 ht
Vy
8I z
FBG = Vy
∑ M = 0 : V e = F h + F ( h 2)
s
y
AB
DE
or
e=
5 b 2 ht
tb 2 h 2
( + h2 ) =
4I z 4
16 I z
where,
th3
h
h
Iz ≈
+ 2[bt ( ) 2 + bt ( ) 2 ]
12
2
4
15
t
= (h3 − bh 2 )
12
2
SOLUTION (7.132)
y
FA
B
FBD
A
D
e Vy
z
S
h/2
h/2
Vy= FBE
t
FAB
E FBD
2b
b
bh
Vy
Iz
bh
τ DB =
Vy
2I z
1
FAB = τ AB (2b)t
2
b 2 ht
=
Vy
Iz
τ AB =
Continued on next slide
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b 2 ht
1
Vy
FBD = τ BD bt =
2
4I z
∑M = 0: V e = F h− F h
x
y
AB
BD
or
e=
3 b 2 h 2t
h 2t
(4b 2 − b 2 ) =
4I z
4 Iz
Where,
1 3
h
th + 2(2b + b)t ( ) 2
12
2
1
= h 2t (h + 18b)
12
Iz =
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________________________________________________________________________
PROBLEMS (*7.135 and 7.136) Determine the distance e that locates the shear center S of a
thin-walled beam of cross section shown in Figs. P7.135 and P7.136.
SOLUTION (*7.135)
y
Vy
e
z
FBD
2t
D
B
E
C
S
FDF F
τD
τD
FA
b
tA
s
G
FAB b
τB
D
B
E
A
F
G
τE
A1 = 13 τ B b
b/2 b/2
(See : Table B.6)
From A to B:
1
QAB = ts 2
2
τB =
τ AB =
b2
Vy
2I z
s2
Vy
2I z
τA = 0
1
b 3t
Vy
FAB = τ B bt =
6I z
3
From B to D:
5
b
b
QD = tb( ) + 2tb( ) = b 2t
2
2 2
2
5b t 2
5b 2
τD =
Vy =
Vy
4I z
I z (2t )
FBD =
V y 7b 3 t
1
1 b 2 5b 2
(τ B + τ D )b(2t ) = ( +
)2tb =
Vy
2
2 2
4
4I z
Iz
Then
∑ Fy = 0 : FDF − 2 FAB = Vy , FDF = (1 +
∑ M = 0 : F ⋅ e = F b + 2 F (b + e)
S
DF
BD
b3t
)Vy
3I z
AB
or
b( FBD + 2 FAB )
FDF − 2 FAB
Substituting the forces & simplifying:
25b 4t
e=
12 I z
e=
Continued on next slide
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1
10
where I z = 2[ (t )(2b)3 + b(2t )(b) 2 ] = b3t
12
3
5
e= b
Thus,
8
SOLUTION (7.136)
I =πr t
3
y
(by Table B.6)
θ
Q = ∫ (r sin φ )(t ⋅ rdφ ) = r 2t (1 − cos θ )
0
Hence
VQ V (1 − cos θ )
τ=
=
π rt
Ib
Thus
V
dA = t ⋅ rdθ
2π
Vr (1 − cos θ )
0
π
Tc = Ve = ∫ (τ dA)r ∫
θ
dφ
z
S
r
e
φ
C
= 2Vr
from which
e = 2r
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________________________________________________________________________
PROBLEM (7.137) Verify that the distance e that locates the shear center, point S, of a thin-walled
member having I an cross as section shown in Fig. P7.137 may be expressed in the form:
e=
h13t1b
h13t1 + h23t2
(P7.137)
Calculate the value of e for h1 = h2 = 120 mm, t1 = 20 mm, t2 = 40 mm, and b = 150 mm.
SOLUTION
Vy
t1
h1
t2
y
e
t3
z
S F2
F1
b
τ1
A1 = 23 τ 1h1
(See: Table B.6)
h2
τ2
A2 = 23 τ 2 h2
At the neutral axis(z):
Vy 1
Vy 2
1
( h1t1 ) h1 =
h1
I z t1 I z t1 2
4
8I z
In a similar manner, we have
Vy h22
τ2 =
8I z
τ1 =
Vy Q
=
We write ∑ Fy = 0 : Vy = F1 + F2
e=
∑M = 0: V e = Fb
B
F1b
F1 + F2
V
2
F1 = τ 1h1t2 = y h23t2
3
12 I z
Likewise
Vy 3
h2 t2
F2 =
12 I Z
Substitute Eqs. (2) & (3) into Eq. (1):
h 3t
e= 3 1 13 b
Q.E.D.
h1 t1 + h2 t2
Introducing the given data, we obtain
(120)3 (20)(150)
e=
= 50 mm
(120)3 (20) + (120)3 (40)
And
y
1
(1)
(2)
(3)
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________________________________________________________________________
PROBLEM (*7.138 ) A thin-walled half tube is loaded as shown in Pig. P7.138.
(a) Demonstrate that the location of shear center S is given by e ≈ 0.27r.
(b) Calculate the shear stress at D for r = 2 in., t = 1/8 in., and Vy = 100 lb.
*SOLUTION
θ
Qz = ∫ ydA = ∫ r cos θ ⋅ t ⋅ rdθ = r 2t sin θ
τ=
(a)
Vy Qz
I zt
0
2
=
Vy r t sin θ
(π r t 2)t
3
=
2Vy sin θ
π
2Vy sin θ
0
π rt
∑ T = 0 : V (e + r ) = τ rdA = ∫
o
y
(1)
π rt
r (rdθ ⋅ t ) =
4r
π
Vy
From which
4
e = r ( − 1) = 0.27 r
π
Q.E.D.
(b) At D, Eq. (1) becomes
2Vy sin 90o 2Vy
τ max =
=
π rt
π rt
Substituting the given data:
2(100)
τ max =
= 255 psi
π (2)(0.125)
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PROBLEM (7.139) A rectangular beam of the cross section shown in Fig. P7.139 is made of mild
steel with
σ y = 250 MPa. For bending about the z axis, calculate:
(a) The yield moment.
(b) The moment causing a 15-mm-thick plastic zone at the top and bottom of the beam.
Given:
b = 60 mm,
h = 80 mm
SOLUTION
Equation (7.53):
I 1 2
= bh σ y
c 6
0.06(0.08) 2
=
(250 ×106 ) = 16 kN ⋅ m
6
(a) M y = σ y
(b) Equation (7.54):
3
1 15
M = (16 × 103 )[1 − ( ) 2 ] = 22.875 kN ⋅ m
2
3 40
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PROBLEM (7.140) A steel rectangular beam (Fig. P7.139) is subjected to a moment 1.2 times
greater than My . Determine:
(a) The distance from the neutral axis to the point at which elastic core ends, y o .
(b) The residual stress pattern following release of loading.
(a) Equations (7.53) and (7.54):
M
3
1 y
= 1.2 = [1 − ( 0 ) 2 ]
My
2
3 h2
or
4 y2
0.8 = 1 − ( 02 ),
3 h
y0 = 0.387h
(b) The residual stress pattern will be as in Fig. 7.49c.
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PROBLEMS (7.141 and *7.142) A singly symmetric steel beam has the cross section shown in
Figs. 7.141 and 142. (Dimensions are in millimeters.) Calculate the plastic moment Mp .
Assumption: The steel is to be elastoplastic with a yield stress of σ y = 250 MPa.
SOLUTION (7.141)
A = 2(10)(40) + (10)(30) = 1100 mm 2
Neutral axis divides section into two equal areas:
A
y
(30)(10) + 2(10)(h1 ) = = 550 mm 2
2
10
Solving
h1
h2 = 27.5 mm
h1 = 12.5 mm
z
30
Therefore,
h2
A( y1 + y2 )
Z=
30
2
10
10
where
∑ Ai yi = 1 [2(h )(10)( h1 ) + 10(30)(h − 5)2 ] = 6.93 mm
y1 =
1
1
2
∑ Ai A 2
y2 =
h
1
[2(10)(h2 )( 2 )] = 13.75 mm
A2
2
So
1100
(6.93 + 13.75) = 11,374 mm3
2
M p = σ y Z = (250 × 106 )(11,374 ×10−6 )
Z=
= 2.84 kN ⋅ m
SOLUTION (*7.142)
y
A = (125)(10) + (200)(10) + (250)(10)
= 5.75 ×103 mm 2
250
10 mm
h1
Neutral axis divides section into two equal areas:
A
(250)(10) + (h1 − 10)(10) = = 2.875 × 103
2
Solving
h1 = 47.5 mm
h2 = 172.5 mm
Hence
z
200 mm
h2
10 mm
125 mm
Continued on next slide
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10 mm
Z=
A( y1 + y2 )
2
y1 =
∑ A y = 1 [(250)(10)(h − 5) + (10)(h − 10) ( 1 )] = 39.402 mm
2
∑A A 2
where
i
i
2
1
1
i
y2 =
1
1
[(125)(10)(h2 − 5) + 10(h2 − 10) 2 ( )] = 118.75 mm
A2
2
Thus
5750
(39.402 + 118.75) = 454.687(103 ) mm3
2
M p = σ y Z = (250 × 106 )(454.687 ×10−6 ) = 113.7 kN ⋅ m
Z=
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PROBLEM (7.143) Compute the shape factor f for a W 10 x 33 beam (see Table B.8).
SOLUTION
y
A1
A2
z
9.73 in.
8.86 in.
0.29 in.
0.435 in.
7.96 in.
W10x33
S = 35 in.3
Z = 2 A1 y1 + 2 A2 y2
9.73 − 0.435
2
8.86
8.86
+
(0.29)
]
2
4
= 37.88 in.3
= 2[(7.96 × 0.435)
Thus, Eq. (7.57):
Z 37.88
= 1.082
f = =
S
35
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PROBLEM (7.144) Determine the shape factor f for a W 150 x 30 beam (see Table B.8).
SOLUTION
y
A1
157 mm
z
138.4 mm
A2
6.6 mm
9.3 mm
153 mm
W150x33
S = 219 ×103 mm3
Z = 2 A1 y1 + 2 A2 y2
157 − 9.3
= 2[153 × 9.3 ×
2
138.4
138.4
+
(6.6)
]
2
4
= 241.8 × 103 mm3
Thus,
f =
241.8
= 1.104
219
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________________________________________________________________________
PROBLEM (7.145) Compute the shape factor f for an S 10 x 35 beam (see Table B.9).
SOLUTION
y
A1
W10x35
S = 29.4 in.3
Z = 2 A1 y1 + 2 A2 y2
10 − 0.491
2
9.018
9.018
+
(0.594)
]
2
4
= 2(11.542 + 6.038)
= 35.161 in.3
= 2[4.944 × 0.491
10 in.
z
9.018 in.
A2
0.594 in.
0.491 in. 4.944 in.
Thus,
f =
35.161
= 1.196
29.4
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PROBLEM (7.146 through 7.148) Determine the shape factor f for an elastoplastic beam of the
cross sections shown in Figs. P7.146 through P7.148.
SOLUTION (7.146)
y
y=
C
z
c
4c
3π
A=
π c2
2
S=
π c3
4
π c 4c
Z = 2 Ay = 2
2 3π
3
4c
=
3
2
Thus,
f =
Z 16
=
≈ 1.7
S 3π
SOLUTION (7.147)
See solution of Prob. 7.146:
4 3
c
3
Thus, for a hollow shaft:
4
Z = (c 3 − b 3 )
3
and
I π
S = = (c 4 − b 4 )
c 4c
Thus,
Z 16c(c3 − b3 )
f = =
S 3π (c 4 − b 4 )
Z=
Continued on next slide
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SOLUTION (7.148)
y
A1
h =
z
_
A2
h/2
b/2
b
bh h bh h
7bh 2
Z = 2 A1 y1 − 2 A2 y2 = 2[
]=
−
2 4 2(4) 8
32
2 1
I
b h
[bh3 − ( )3 ]
S=
=
2 2
h 2 h 12
=
bh 2
15
1
(1 − ) = bh 2
96
6
16
Thus,
f =
Z 7bh 2 96
=
= 1.4
S
32 15bh 2
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PROBLEMS (*7.149 through 7.151) Compute the shape factor f for an elastoplastic beam of
the cross sections shown in Figs. P7.149 through P7.151.
SOLUTION (*7.149)
For fully plastic case, N.A. divides the entire area into two equal parts:
b=2
h
5
= h
2.4 6
y
h
A1
H=2.4 in.
b
z’
A2
A3
0.703 in.
B=2 in.
At = A1 + A2 + A3
1
= (2 × 2.4) = 2.4 in.2
2
1
A1 = At = 1.2 in.2
2
Thus,
1
1 5
bh = ( h)h = 1.2
2
2 6
From which
h = 1.697 in.
b = 1.414 in.
Then
Z = A1 y1 + A2 y2 + A2 y3
1.414 × 1.697 1.697 1.414 × 0.703 0.703 2 × 0.703 (0.703)2
=
+
+
2
3
2
3
2
3
1
= (4.072 + 0.699 + 1.976) = 1.125 in.3
6
I BH 3 1
BH 2
S= =
=
36 2 H 3
24
c
=
2(2.4) 2
= 0.48 in.3
24
f =
Z 1.125
=
= 2.34
S 0.48
Thus,
Continued on next slide
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SOLUTION (7.150)
z = axis of elastic bending,
z ' = axis of plastic bending
y
20 mm
100 mm
c1 z’
h1
z
N.A.
c2
h2
A
2
160 mm
20 mm
Elastic bending
∑ Ai yi = 20(160)80 + 100(20)170 = 114.62 mm
c2 =
160(20) + 100(20)
∑ Ai
c1 = 180 − 114.62 = 65.38 mm
20
100
(20)3 + 20(100)(55.38) 2
I z = (160) 2 + 20(160)(34.62) 2 +
12
12
6
4
= 16.86 × 10 mm
I 16.86 × 106
= 14.71×104 mm3
S= =
c2
114.62
Plastic bending
A = 160 × 20 + 100 × 20 = 5.2 × 103 mm 2
A 5.2 × 103
=
, h2 = 130 mm
2
2
h1 = 180 − 130 = 50 mm
100 × 20(40) + 30 × (30 2)
= 34.23 mm
y1 =
100 × 20 + 30 × 20
130
y2 =
= 65 mm
2
h2 (20) =
A
5.2 ×103
Z = ( y1 + y2 ) =
(65 + 34.23) = 25.8 × 104 mm3
2
2
Thus,
Z 25.8
f = =
= 1.75
S 14.71
Continued on next slide
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SOLUTION (7.151)
Elastic bending
y
20 mm
80 mm
A1
A2
z
40 mm
20 mm
100 mm
1
(100 × 1203 − 60 × 803 )
12
= 11.84 × 106 mm 4
Iz =
I 11.84 × 106
=
c
60
= 19.73 × 104 mm3
S=
Plastic bending
Z = 2( A1 y1 + A2 y2 )
= 2(20 × 100 × 50 + 40 × 40 × 20) = 26.4 × 104 mm3
Thus,
Z 26.4
f = =
= 1.34
S 19.73
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________________________________________________________________________
PROBLEM (7.152) Redo Prob. 7.139 for the beam of the cross section shown in Fig. P7.152.
SOLUTION
y
15 mm
A2
A1
30 mm
120 mm
z
A3
A3
60 mm
F1
F2
F3
50 mm
100 mm
Iz
σy
c
13.5 ×106
=
250 × 10−6 = 56.25 kN ⋅ m
0.06
(a) M y =
(b) F1 = 250(100 × 15) = 375 kN
1
250
F2 = (250 +
)(100 ×15) = 312.5 kN
2
1.5
1
250
F3 = (0 +
)(50 × 30) = 125 kN
2
1.5
∑ M z = M = 2( F1 y1 + F2 y2 + F3 y3 )
or
M = 2(375 × 52.5 + 312.5 × 37.5 + 125 × 20) = 67.82 kN ⋅ m
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PROBLEM (7.153) Solve Prob. 7.139 for the beam of the cross section shown in Fig. P7.151.
SOLUTION
15 mm
60 mm
y
A1
F1
F2
A2
A3
z
40 mm
F3
20 mm
100 mm
From solution of Prob. 7.151:
I z = 11.84 × 106 mm 4
(a) M yp =
Iz
11.84 ×106
(250 × 106 ) = 49.333 kN ⋅ m
σ yp =
c
0.06
(b) F1 = 250(100 × 15) = 375 kN
1
8
F2 = [250 + (250)](100 × 5) = 118.1 kN
2
9
1
8
F3 = [0 + (250)](40 × 40) = 177.8 kN
2
9
∑ M = M = 2( F y + F y + F y )
z
or
1 1
2
2
3 3
M = 2(375 × 52.5 + 118.1× 42.5 + 177 × 26.67) = 58.9 kN ⋅ m
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PROBLEM (*7.154) A rectangular beam with b = 3 in. and h = 5 in. (Fig. P7.154) is subjected to
a plastic moment Mp. If σ y = 36 ksi for this beam of ductile material, calculate, for the case in which the
loading has been removed, the residual stresses at the upper and lower faces.
*SOLUTION
1
(3)(5)3
12
= 31.25 in.4
I
My = σy
c
31.25
=
(36 ×103 )
2.5
= 450 kip ⋅ in.
y
I=
5 in.
z
3 in.
3
M p = fM y = (450) = 675 kip ⋅ in.
2
Elastic rebound stress
Mc 675 × 103 (2.5)
=
σ 'max =
I
31.25
= 54 ksi
The results are sketched below.
y
-36 ksi
18 ksi
54 ksi
MP =0
MP
x
MP
=
+
36 ksi
-36 ksi
2.5 in.
36 ksi
Loading
-54 ksi
-18 ksi
Unloading
Residual
stresses
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________________________________________________________________________
PROBLEM (7.155) Rework Prob. 7.154 for the beam of the cross section shown in Fig. P7.152.
SOLUTION
From solution of Prob. 7.152:
I z = 13.5 ×10−6 m 4
M yp = 56.25 kN ⋅ m
From solution of Prob. 12.27: f = 1.4.
M p = fM yp = 1.4(56.25) = 78.25 kN ⋅ m
Elastic rebound stress
Mc 78.25 ×103 (0.06)
=
σ 'max =
I
13.5 ×10−6
The results are shown below.
y
-250 MPa
347.8 MPa
97.8 MPa
MP
x
=
+
MP
60 mm
250 MPa
Loading
347.8MPa
Unloading
-250
250
MP =0
-97.8 MPa
Residual
stresses
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________________________________________________________________________
PROBLEM (*7.156) A beam of the cross section shown in Fig. P7.150 is made of an elastoplastic
material for which E = 200 GPa and
bending strain of - 1250 μ
σ y = 250 MPa. Calculate the bending moment that produces a
on the lower face of the flange.
*SOLUTION
Since 1250 × 10−6 (200 × 109 ) = 250 MPa , lower face of the flange yields.
20 mm
1250 μ
100 mm
F1
F2
h
z
x
N.A.
160 mm
-250 MPa
F3
160-2h
20 mm
x
F4
250 MPa
Strains
Stresses
∑F = 0:
x
−250 × 106 (0.02 × 0.1) −
+
or
250 × 106
h × 0.02
2
250 ×106
h × 0.02 + 250 ×106 (0.16 − 2h)(0.02) = 0
2
h = 30 mm
F1 = 250(100 × 20) = 500 kN
250
F2 = − F3 =
(30 × 20) = 75 kN
2
F4 = 250(100 × 20) = 500 kN
∑M = M = ∑F y
z
i
i
or
2
2
M = 500 × 40 + 75( ⋅ 30) + 75( ⋅ 30) + 500 × 80
3
3
= 63 kN ⋅ m
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________________________________________________________________________
PROBLEM (7.157) A curved bar with a rectangular cross section is subjected to the bending
moment as shown in Fig. P7.157. Calculate:
(a) The stresses σ i and σ o of the inside and outside fibers, respectively, using the curved beam formula.
(b) Resolve item (a) using the flexure formula.
Given: M = 1.2 kN ⋅ m, b = 25 mm, h = 50 mm,
ri = 60 mm
SOLUTION
ro = ri + h = 60 + 50 = 110 mm
A = bh = (50)(25) = 1, 250 mm 2
We have
h
50
R=
=
= 82.4898
ro
110
ln
ln
60
ri
1
r = (ri + rθ ) = 85 mm
2
e = r − R = 2.5102 mm
(a) Using Eq.(7.67):
1200(82.4898 − 60)
σi = −
= −143.3 MPa
1250(10−6 )(0.06)(85 − 82.4898)
σo = −
1200(82.4898 − 110)
= 98.6 MPa
1250(10−6 )(0.11)(85 − 82.4898)
(b) σ o = −σ i =
Mc 1.2 × 103 (0.025)
=
= 66.7 MPa
1
I
3
(0.025)(0.06)
12
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________________________________________________________________________
PROBLEM (7.158) A rectangular bar is curved to a radius r along the centroidal axis and subjected
to end moments M (Fig. P7.157). Determine:
(a) The normal stress σ i of the inside fibers.
(b) The normal stress
Given:
σ o of the outside fibers.
M = 15 kip ⋅ in,
r = 10 in.,
b = 2 in.,
h = 3 in.
SOLUTION
h
= 10 − 1.5 = 8.5 in.
2
h
ro = r + = 10 + 1.5 = 11.5 in.
2
A = bh = (2)(3) = 6 in.2
ri = r −
Therefore
h
3
=
= 9.9245 in.
ro
11.5
ln
ln
8.5
ri
e = r − R = 10 − 9.9245 = 0.0755 in.
R=
(a) Applying Eq.(7.67):
M ( R − ri )
15 × 103 (9.9245 − 8.5)
σi = −
= −5.55 ksi
=−
6(0.0755)(8.5)
Aeri
(b) Use Eq. (7.67):
M ( R − ro )
15 ×103 (9.9245 − 11.5)
σo = −
= 4.48 ksi
=−
6(0.0755)(11.5)
Aero
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________________________________________________________________________
PROBLEM (7.159) A steel bar of square (b = h) cross section, curved to a radius r along the
centroidal axis, is acted upon by end moments M, as shown in Fig. P7.159. Calculate the maximum
permissible value of the bending moment M, if the allowable stress is σ all .
Given:
ri = 250 mm,
ro = 300 mm,
b = h = 50 mm,
σall = 140 MPa
SOLUTION
h
50
=
= 274.2407 mm
ro
300
ln
ln
250
ri
1
1
r = (ri + ro ) = (250 + 300) = 275 mm
2
2
A = (50)(50) = 2500 mm 2
e = r − R = 275 − 274.2407 = 0.7593 mm
R=
Using Eq.(7.67):
M ( R − ri )
M (274.20407 − 250)
σi = −
− 140(103 ) = −
M = 3.62 kN ⋅ m
;
Aeri
2500(10−6 )(250)
Similarly,
M ( R − ro )
M (274.20407 − 300)
σo = −
, M = 4.07 kN ⋅ m
; 140(103 ) = −
Aero
2500(10−6 )(300)
Therefore
M all = 3.62 kN ⋅ m
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PROBLEM (7.160) A machine frame has a T-shaped cross section and is subjected to end
moments M (Fig. P7.160). Determine the maximum stress in the member.
Given: M = 600 N ⋅ m,
ri = 80 mm
SOLUTION
Centroid of the cross section:
ro = 80 + 60 = 140 mm
20
A1
A = (40 × 20) + (50 × 20) = 1800 mm 2
40 mm
C
A2
A1r1 + A2 r2
20
mm
r=
A1 + A2
r
50 mm
80 mm
(40 × 20)(120) + (50 × 20)(90)
=
1800
= 103.3333 mm
1800
1800
A
= r
R=
=
= 100.6337 mm
r
100
140
dA
2 50 dr
3 20 dr
∑ ∫ r ∫r1 r + ∫r2 r 50 ln 80 + 20 ln 100
e = r − R = 2.6996 mm
Now apply Eq.(7.67):
M ( R − ri )
600(100.6337 − 80)
σi = −
=−
= −31.85 MPa
Aeri
1800(10−6 )(2.6996 ×10−3 )(80)
600(100.6337 − 140)
σo = −
= 34.96 MPa
1800(10−6 )(2.6996 ×10−3 )(140)
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PROBLEM (7.161) The curved frame with a T-shaped cross section is in pure bending as shown in
Fig. P7.160. If the member is made of a material having an allowable stress of σ all , determine the largest
moment M that can be applied.
Given: ri = 80 mm, σall = 150 MPa
SOLUTION
From the solution of Prob. 7.160, we have
ro = 140 mm
A = 1800 mm 2
r = 103.3333 mm
R = 100.6337 mm
e = 2.6996 mm
Assume tension failure:
M ( R − ro )
M (100.6337 − 140)
σo = −
;
150(106 ) = −
Aero
1800(10−6 )(2.6996 ×10−3 )(140)
or
M = 2.592 N ⋅ m
Assume compression failure:
M ( R − ri )
M (100.6337 − 80)
σi = −
− 150(106 ) = −
;
Aeri
1800(10−6 )(2.6996 ×10−3 )(80)
M = 2.826 N ⋅ m
or
Thus, tension controls,
M all = 2.592 N ⋅ m
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PROBLEM (*7.162) A curved beam of channel-shaped cross section is acted upon end moments M
(Fig. P7.162). Calculate the dimension b required so that the stress at the points A and B of the beam
are equal in magnitude.
*SOLUTION
A = 40b + 2(120 × 20) = 40b + 4800
We have rA = 120 mm and rB = 280 mm . Applying Eq. (7.67):
(− M )( R − rA ) (− M )( R − rB )
σ A = −σ B = −
=
AerA
AerB
from which
−280( R − 120) = 120( R − 280)
− rB ( R − rA ) = rA ( R − rB ) ,
R = 168 mm
or
A
Then R =
dA
∑∫ r
280 2(20) dr ⎤
8
14 ⎤
⎡
⎡ 160 bdr
A = 168 ⎢ ∫
or
= 168 ⎢b ln + 40 ln ⎥
+∫
⎥
160
r
6
8⎦
⎣
⎣ 120 r
⎦
= 48.3306b + 3760.6181
Hence 40b + 4800 = 48.3306b + 3760.6181
or
b = 124.8 mm
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PROBLEM (*7.163) A curved frame with a circular cross section of diameter d is fixed at one
end and carries a concentrated load P at the free end as shown in Fig. P7.163. Determine:
(a) The stress at point A.
(b) The stress at point B.
Given: a = 25 mm, b = 15 mm, d = 20 mm, P = 200 N
*SOLUTION
We have c =
P
1
d = 10 mm
2
r = b + c = 25 mm
1
R = [r + r 2 − c 2 ] (by Table 7.1)
2
1
= [25 + 252 − 102 ] = 23.9564 mm
2
e = r − R = 1.0436 mm
A = π c 2 = π (10) 2 = 314.16 mm 2
O
a
A
M=-P(a+ r )
B
b
P
(a) From Eq.(7.70):
(a + r )( R − rA )
P M ( R − rA ) P
−
= [1 +
]
A
AerA
A
erA
200
(50)(23.9564 − 15)
=
[1 +
] = 18.85 MPa
314.16
(1.0436)(15)
σA =
(b) Using Eq.(7.70):
(a + r )( R − rB )
P M ( R − rB ) P
σB = −
= [1 +
A
AerB
A
erB
200
50(29.1421 − 35)
=
[1 +
] = −8.55 MPa
314.16
1.0436(35)
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PROBLEM (*7.164) The curved frame shown in Fig. P7.163 with a circular cross section of
diameter d has the allowable tensile stress σall. Calculate the largest permissible distance a from the line
of action of the load P to the center of curvature O of the frame.
Given:
b = 5/8 in., d = 1 in.,
P = 40 lb, σall = 5 ksi
*SOLUTION
rA = b = 5 8 in.
1
c = d = 0.5 in.
r = 0.625 + 0.5 = 1.125 in.
2
1
R = [r + r 2 − c 2 ] (from Table 7.1)
2
1
= [1.125 + 1.1252 − 0.52 ] = 1.0664 in.
2
e = r − R = 0.0586 in.
A = π c 2 = π (0.5) 2 = 0.7854 in.2 ,
M = − P(a + r )
Largest tensile stress occurs at point A; σ all = σ A
Applying Eq.(7.70), we have
P M ( R − rA ) P ⎡ ( a + r )( R − rA ) ⎤ kP
= ⎢1 +
σA = −
⎥=
A
AerA
A⎣
erA
⎦ A
where
(a + r )( R − rA )
k = 1+
erA
This gives
σ A A (5 ×103 )(0.7856)
k=
=
= 98.2
P
40
From Eq.(1), we obtain
(k − 1)erA (97.2)(0.0586)(0.62)
a+r =
=
= 8.001
R − rA
1.0664 − 0.625
Hence a = 8.001 − 1.125 = 6.876 in.
(1)
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PROBLEM (*7.165) A split ring having an inner radius ri , outer radius ro , and a trapezoidal cross
section is loaded as shown in Fig. P7.165. Determine:
(a) The stress at point A.
(b) The stress at point B.
Given: ri = 4 in., ro = 10 in., b1 = 3 in., b2 = 2 in.,
P = 10 kips
*SOLUTION
Locate centroid :
A1r1 + A2 r2
A1 + A2
9(6) + (6)(8)
=
= 6.8 in.
9+6
6 in.
2 in.
C A1
A2
r=
3 in.
O
4 in.
r
1 2
h (b1 + b2 )
2
R=
r
(b1ro − b2 ri ) ln o − h(b1 − b2 )
ri
(0.5)(6) 2 (3 + 2)
= 6.3567 in.
10
[(3)(10) − (2)(4)]ln − (6)(3 − 2)
4
e = r − R = 0.4433 in.
=
(a) Use Eq.(7.71a),
r ( R − rA )
P M ( R − rA )
P
σA = − −
= − [1 +
]
A
AerA
A
erA
=−
P
10 ⎡ (6.8)(6.3567 − 4) ⎤
= −6.69 ksi
1+
15 ⎢⎣
(0.4433)(4) ⎥⎦
A
B
O
M=P r
(b) From Eq.(7.71b):
r ( R − rB )
P
σ B = − [1 +
]
A
erB
=−
P
10 ⎡ (6.8)(6.3567 − 10) ⎤
1+
⎥ = −3.06 ksi
15 ⎢⎣
0.4433(10)
⎦
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PROBLEM (7.166) A machine component with an elliptical cross section is fixed at one end and
supports a concentrated load P at the free end ( Fig. P7.166). Calculate the stress at points A and B.
Given: ri = 4 in., ro = 7 in. , a = 3 in., b = 1.5 in., P = 10 kips
SOLUTION
rA = 4 in.
rB = 7 in.
1
1
r = (ro + ri ) = (7 + 4) = 5.5 in.
2
2
From Table 7.1
A = π ab = π (3)(1.5) = 4.5π in.2
dA 2π b
2
2
∫ r = a [r − r − a ]
2π (1.5)
=
[5.5 − 5.52 − 32 ] = 2.7967 in.
3
Hence
A
4.5π
R=
=
= 5.0549 in.
dA 2.7967
∫A r
e = r − R = 0.4451 in.
P
A
P
M=-P r
B
Equation (7.70) with M = − PF :
r ( R − rA ) 10(103 ) ⎡ 5.5(5.0549 − 4) ⎤
P
σ A = [1 +
]=
1+
A
erA
4.5π ⎢⎣
(0.4451)(4) ⎥⎦
= 3.01 ksi
r ( R − rB ) 10(103 ) ⎡ 5.5(5.0549 − 7) ⎤
P
[1 +
]=
1+
A
erB
4.5π ⎢⎣
(0.4451)(7) ⎥⎦
= −1.721 ksi
σB =
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PROBLEM (7.167) A frame has an elliptical cross section as shown in Fig. P7.166. If the allowable
compressive and tensile stress is σall, determine the largest force P that can be applied to the member.
Given: r = 80 mm,
r = 200 mm, a = 60 mm,
b = 30 mm,
σall = 80 MPa
SOLUTION
rA = 80 mm
rB = 200 mm
1
r = (ro − ri ) = 140 mm
2
A = π ab = π (60)(30) = 1800π mm 2
2π b
dA
∫ r = a [r − r − a ]
=
2
2
P
2π (30)
[140 − 1402 − 602 ] = 42.4394 mm
60
A
1800π
=
= 133.2456 mm
dA 42.4394
∫A r
e = r − R = 6.7544 mm
R=
A
P
M=-P r
B
Using Eq.(7.70) with M = − PF :
r ( R − rB )
P
σ B = [1 +
]
A
erB
−80(106 ) =
⎡ (140)(133.2456 − 200) ⎤
Pall
1+
⎥
−6 ⎢
1800π (10 ) ⎣
(6.7544)(200)
⎦
Solving
Pall = 76.44 kN
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PROBLEM (*7.168) The allowable stress for the clamp shown in Fig. P7.168 is σall = 10 ksi.
Determine the maximum permissible load P that the member can resist.
*SOLUTION
A = 8(2) + 4(2) + 4(2) = 32 in.2
16(11) + 8(14) + 8(17)
= 13.25 in.
r=
32
P
B
Hence
A
32
= 12
dA
8dr 16 2dr 18 4dr
∑ ∫ r ∫10 r + ∫12 r + ∫16 r
32
=
= 12.7739 in.
12
16
18
8ln + 2 ln + 4 ln
10
12
16
e = r − R = 0.4765 in.
M = − P(13.25 + 2.5) = −15.75 P
R=
A
M
P
Apply Eq.(7.70);
σA =
P M ( R − rA )
P ⎡ 15.75(12.7739 − 10) ⎤
−
]; 10(103 ) = ⎢1 +
⎥ , P = 31.47 kips
32 ⎣
(0.4765)(10)
A
AerA
⎦
σB =
P M ( R − rB )
P ⎡ 15.75(12.7739 − 18) ⎤
−
]; −10(103 ) =
1+
⎥ , P = 37.22 kips
32 ⎢⎣
(0.4765)(18)
A
AerB
⎦
So
Pall = 31.47 kips
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PROBLEM (7.169) Redo Prob. 7.168, if the clamp has a box cross section as shown in Fig. P7.169.
SOLUTION
A = 8(6) − 4(4) = 32 in.2
12(11) + 2(4)(14) + 12(17)
= 14 in. (as intuitively expected)
r=
32
A
32
= 12
16 dr
18 dr
dA
dr
∑ ∫ r ∫10 6 r + ∫12 2 r + ∫16 6 r
32
=
= 13.7925 in.
12
16
18
6 ln + 2 ln + 6 ln
10
12
16
R=
e = r − R = 0.2075 in.
M = − P(13.25 + 2.5) = −15.75 P
P
B
A
M
P
Using Eq.(7.70);
σA =
P M ( R − rA )
P ⎡ 15.75(13.7925 − 10) ⎤
−
]; 10(103 ) = ⎢1 +
, P = 10.74 kips
A
AerA
32 ⎣
(0.2075)(10) ⎥⎦
σB =
P M ( R − rB )
P ⎡ 15.75(13.7925 − 18) ⎤
−
]; −10(103 ) =
1+
, P = 19.11 kips
A
AerB
32 ⎢⎣
(0.2075)(18) ⎥⎦
So
Pall = 10.74 kips
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PROBLEM (7.170) The triangular cross section of a curved bar is shown in Fig. P7.170. Derive the
expression for the radius R along the neutral axis. Compare the result with that given for Fig. C in Table
7.1.
SOLUTION
1
A = bh
2
The section width w varies linearly with r. Thus
w = c0 + c1r
(1)
Since
dr
w = b (at r = ri )
(2)
r
w = 0 (at r = ro )
Substituting Eq.(1) into Eq.(2);
w
b
bro
b
c1 = −
c0 =
h
h
ri
h
Then
rο
ro w
c0 + c1r
dA
∫A r = ∫ri r dr = r dr
Inserting c1 and c0 into this, after integrating and rearranging,
we have
r
r
dA
∫ r = b( ho ln roi − 1)
Therefore
1 h
A
2
=
R=
dA r
r
∫ r ho ln roi − 1
O
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PROBLEM (*7.171) The circular cross section of a curved beam is shown in Fig. P7.171. Derive
the expression for the radius R along the neutral axis and compare the result with that given for Fig. B in
Table 7.1.
*SOLUTION
r
A = πc
Through use of the polar coordinates we write:
w = 2c sin α
r = r − c cos α
dr = c sin α dα
dA = wdr = 2c 2 sin 2 α dα
2
r
C
π
2
π
r −c
2
2
tan −1
O
c
dA π 2c 2 sin 2 α
∫ r =∫0 r − c cos α dα
2
2
2
2
2
2
2
π c (1 − cos α )
π r − c cos α − ( r − c )
= 2∫
= 2∫
0
0
r − c cos α
r − c cos β
π
π
dα
= 2 ∫ (r + c cos α )dα − 2(r 2 − c 2 ) ∫
0
0 r − c cos α
= 2r α 0 + 2c sin α 0 − 2(r 2 − c 2 )
w
α
dr
ccosα
r 2 − c 2 tan
r +c
α
π
2
0
This gives
dA
2
2
∫ r = 2π (r − r − c )
Hence, it can be shown that
A
1
R=
= (r + r 2 + c 2 )
dA 2
∫r
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PROBLEM (*7.172) The trapezoidal cross section of a curved beam is shown in Fig. P7.172. Derive
the expression for the radius R along the neutral axis. Compare the result with that given for Fig. D in
table 7.1.
*SOLUTION
1
(b1 + b2 )h
2
The section width w varies
linearly with r as
A=
w = c0 + c1r
We have
w = b1 (at r = ri )
ri
h
w
b2
(1)
b1
O
r
dr
rο
(2)
w = b2 (at r = ro )
Introduce Eq.(1) into Eq.(2), then solve for c0 and c1
r b − rb
b −b
c0 = o 1 i 2
c1 = − 1 2
(3)
h
h
Then, we write
ro w
ro c + c r
r
dA
∫ r = ∫ri r dr = ∫ri 0 r 1 dr = c0 ln roi + c1 (ro − ri )
This gives, substituting Eqs.(3):
dA
∫r =
ro b1 − rb
r
i 2
ln o − (b1 − b2 )
h
ri
Hence
1 2
h (b1 + b2 )
A
2
R=
=
dA
r
∫ r (rob1 − rbi 2 ) ln roi − (b1 − b2 )
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CHAPTER 8
PROBLEM (8.1) A 60-mm by 40-mm plate of 5-mm thickness is subjected to uniformly distributed
biaxial tensile forces (Fig. P8.1). What normal and shear stresses exist along diagonal AC? As was done
in the derivations given in Sec. 8.2, use an approach based upon the equilibrium equations applied to
the wedge-shaped half ABC of the plate.
SOLUTION
∠ ACB = tan −1
50 ΔA cos56.3o
y
y’
B
100 ΔA sin56.3o
σx =
C
τ x ' y ' ΔA
56.3o
σ x ' ΔA
x
A
3
= 56.3o
2
30(103 )
= 100 MPa
0.06(0.005)
10(103 )
σy =
= 50 MPa
0.04(0.005)
x’
∑ F = 0 : σ ΔA − 50ΔA cos 56.3 − 100ΔA sin 56.3 = 0
2
x'
o
2
o
x'
σ x ' = 15.39 + 69.21 = 84.6 MPa
∑F = 0: τ
y'
x' y'
ΔA + 50ΔA sin 56.3o cos 56.3o − 100ΔA sin 56.3o cos 56.3o = 0
τ x ' y ' = −23.08 + 46.16 = 23.1 MPa
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PROBLEM (8.2) Redo Prob. 8.1 using Eqs. (8.4).
SOLUTION
From solution of Prob. 8.1:
σ x = 100 MPa
σ y = 50 MPa
50 MPa
y’
y
B
100 MPa
A
C
τ x' y'
56.3o
θ
θ = −(90 − 56.3) = −33.7o
σ x'
x
33.7o
x’
1
1
2
2
= 75 + 9.6 = 84.6 MPa
1
τ x ' y ' = − (100 − 50) sin(−67.4o ) = 23.1 MPa
2
σ x ' = (100 + 50) + (100 − 50) cos(−67.4o )
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PROBLEMS (8.3 through 8.5) The states of stress at three points in a loaded body are represented
in Figs. P8.3 through P8.5. Determine for each point the normal and shearing stresses acting on the
indicated inclined plane. As was done in the derivations given in Sec. 8.2, use an approach based upon
the equilibrium equations applied to the wedge shaped element shown in each figure.
SOLUTION (8.3)
y
τ x ' y ' ΔA
y’
σ x ' ΔA
50 ΔA cos15o
60 ΔA cos15
o
x’
15o x
60 ΔA sin15o
15o
80 ΔA sin15o
∑F = 0:
x'
σ x ' ΔA + 50 cos2 15o − 80ΔA sin 2 15o
− 2(60ΔA sin 15o cos15o ) = 0
σ x ' = −46.65 + 5.36 + 30 = −11.3 MPa
∑F = 0: τ
y'
x' y'
ΔA − 50ΔA sin 15o cos15o
− 80ΔA sin 15o cos15o − 60ΔA cos 2 15o
+ 60ΔA sin 2 15o = 0
τ x ' y ' = 12.5 + 20 + 55.98 − 4.02 = 84.5 MPa
SOLUTION (8.4)
x’
τ x ' y ' ΔA
y’
σ x ' ΔA
10 ΔA sin30o
30o
10 ΔA cos30o
25 ΔA cos30o
∑ F = 0 : σ ΔA − 25ΔA cos 30
2
x'
o
x'
+ 2(10ΔA sin 30 o cos 30 o ) = 0
σ x ' = 18.75 − 8.66 = 10.09 ksi
Continued on next slide
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∑F = 0: τ
y'
x' y'
ΔA + 25ΔA cos 30o sin 30o
+ 10ΔA cos 2 30o − 10ΔA sin 2 30 o = 0
τ x ' y ' = −10.83 − 7.5 + 2.5 = −15.83 ksi
SOLUTION (8.5)
40 ΔA cos22.5o
10 ΔA cos22.5o
10 ΔA sin22.5o
y’
22.5o
70 ΔA sin22.5
τ x ' y ' ΔA
σ x ' ΔA
o
x’
∑ F = 0 : σ ΔA + 40ΔA cos 22.5
2
x'
o
x'
− 70ΔA sin 2 22.5o − 2(10ΔA sin 22.5o cos 22.5o ) = 0
σ x ' = −34.14 + 10.25 + 7.07 = −16.8 MPa
∑ F = 0 : −τ
y'
x' y'
ΔA − 40ΔA sin 22.5o cos 22.5o
−70ΔA sin 22.5o cos 22.5o − 10ΔA cos 2 22.5o + 10ΔA sin 2 22.5o = 0
τ x ' y ' = 14.14 + 24.75 + 8.54 − 1.46 = 46 MPa
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PROBLEM (8.6 through 8.8) Redo Probs. 8.3 through 8.5 using Eqs. (8.4).
SOLUTION (8.6)
y
τ x' y'
y’
σ x'
50 MPa
x’
θ x
60 MPa
θ =15o
80 MPa
1
1
2
2
= 15 − 56.3 + 30 = −11.3 MPa
1
τ x ' y ' = − (−50 − 80) sin 30o + 60 cos 30o
2
= 32.5 + 51.96 = 84.5 MPa
σ x ' = (−50 + 80) + (−50 − 80) cos 30o + 60sin 30o
SOLUTION (8.7)
σ x'
x’
θ
θ = 30 + 90 = 120o
τ x' y'
10 ksi
30o
y’
25 ksi
1
1
2
2
= 12.5 + 6.25 − 8.66 = 10.09 ksi
1
τ x ' y ' = − (0 − 25) sin 240o + 10 cos 240o
2
= −10.83 − 5 = −15.83 ksi
σ x ' = (0 + 25) + (0 − 25) cos 240o + 10sin 240o
Thus, τ x ' y ' acts in opposite direction to that shown in the figure above.
Continued on next slide
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SOLUTION (8.8)
σ x'
x’
θ
τ x' y'
70 MPa
o
22.5
y’
10 MPa
x
θ =112.5o
40 MPa
1
1
2
2
= 15 − 38.89 + 7.07 = −16.8 MPa
1
τ x ' y ' = − (70 + 40) sin 225o − 10 cos 225o
2
= 38.89 + 7.07 = 46 MPa
σ x ' = (70 − 40) + (70 + 40) cos 225o − 10sin 225o
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PROBLEM (8.9) The state of stress at a point A of the airplane bracket under a service pull P is
represented by an element as illustrated shown in Fig. P8:9.
Given: σ x = σ y = 0, τ xy = 15 ksi
(pure shear)
Determine the stresses on all sides of an element rotated through an angle θ = 22.5o and show their sense
on the element.
SOLUTION
σx =σy = 0
τ xy = 15 ksi
θ = 22.5o
Equations (4.4):
σ x ' = 15 sin 45o = 10.61 ksi
τ x ' y ' = 15 cos 45o = 10.61 ksi
σ y ' = −15 sin 45o = −10.61 ksi
y’
10.61 ksi
10.61 ksi
x’
22.5o
x
10.61 ksi
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PROBLEM (8.10) At a point in a loaded member, the stresses are as shown in Fig. P8.10. The
normal stress at the point on the indicated plane is 7 ksi (tension). What is the magnitude of the shearing
stress τ ?
SOLUTION
y
τ
30o
θ
x’
6 ksi
x
y’
6+0 6−0
cos 300o − τ sin 300o
+
2
2
7 = 3 + 1.5 + 0.866τ
σ x' = 7 =
or
τ = 2.89 ksi
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PROBLEM (8.11) A triangular plate is subjected to stresses as shown in Fig. P8.11. Determine σ x ,
σ y , and τ xy and sketch the results on a properly oriented element.
SOLUTION
50 MPa
y
200 MPa
y’
x’
45o
x
τ x' y' = 0 = −
or
σ x −σ y
2
sin 90o + τ xy cos 90o
σx =σy
σ x ' = −200 = σ x + 0 + τ xy sin 90 o
(1)
σ y ' = −50 = σ x − 0 − τ xy sin 90o
(2)
Subtract (1) from (2):
τ xy = −75 MPa
Add (1) and (2):
σ x = σ y = −125 MPa
125 MPa
125 MPa
75 MPa
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PROBLEM (8.12) Calculate the normal and shear stresses acting on the plane indicated in Fig. P8.10
for τ = 4 ksi.
SOLUTION
y
4 ksi
4 ksi
30o
θ
x’
593 psi
6 ksi
x
7.96 ksi
y’
6+0 6−0
cos 300o − 4sin 300o
+
2
2
= 3 + 1.5 + 3.464 = 7.96 ksi
6−0
sin 300o − 4 cos 300o
τ x' y' = −
2
= 2.598 − 2 = 593 psi
σ x' =
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PROBLEM (8.13) At a critical point A in the loaded member in Fig. P8.13, the stresses on the
inclined plane are σ = 28 MPa and τ = 10 MPa, and the normal stress on the y plane is zero.
Calculate the normal and shear stresses on the x plane through the point. Show the results on a properly
oriented element.
SOLUTION
σ
x’
τ
θ = 120o
θ
30o
x
y’
σ = 28 =
σx +0 σx −0
2
+
2
cos 240o + τ xy sin 240o
σ x = 112 + 3.464τ xy
σx −0
o
(1)
τ = −10 = −
sin 240 + τ xy cos 240o
2
σ x = −23.094 + 1.155τ xy
Solving (1) and (2):
σ x = −90.6 MPa
(2)
τ xy = −58.5 MPa
y
58.5 MPa
A
90.6 MPa
x
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PROBLEM (8.14) At a particular point in a loaded machine part (use Fig. P8.9), the stresses are as
follows: σ x = 6 ksi, σ y = -4 ksi , and τ xy = 0. Determine the normal and shear stresses on the plane
whose normals are at angles of -30o and 120o with the x axis. Sketch the results on a properly oriented
elements.
SOLUTION
y’
x
At θ = −30 :
o
o
30
x’
4.33 ksi
3.5 ksi
6−4 6+4
cos(−60o ) + 0 = 1 + 2.5 = 3.5 ksi
+
2
2
6+4
sin(−60o ) + 0 = 4.33 ksi
τ x' y' = −
2
σ x' =
1.5 ksi
x’
4.33 ksi
At θ = 120o :
120o
y’
x
6−4 6+4
cos 240o = −1.5 ksi
+
2
2
6+4
sin 240o = 4.33 ksi
τ x' y' = −
2
σ x' =
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PROBLEM (8.15) At a point in a loaded structure, the normal and shear stresses have the magnitudes
and directions acting on the inclined element depicted in Fig. P8.15. Calculate the stresses σ x , σ y , and
τ xy on an element whose sides are parallel to the xy axes.
SOLUTION
Transform from θ = 30o to θ = 0 . For convenience in computations, Let
σ x = −200 MPa,
σ y = −60 MPa,
τ xy = 35 MPa and θ = −30o
Then
1
1
2
2
1
1
= (−200 − 60) + (−200 + 60) cos(−60o ) + 35sin(−60o )
2
2
= −195.3 MPa
σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ + τ xy sin 2θ
1
2
1
= − (−200 + 60) sin(−60o ) + 35cos(−60o )
2
= −43.1 MPa
τ x ' y ' = − (σ x − σ y ) sin 2θ + τ xy cos 2θ
So
σ y ' = σ x + σ y − σ x ' = −200 − 60 + 195.3 = −64.7 MPa
For θ = 0o :
y
64.7 MPa
43.1 MPa
195.3 MPa
x
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PROBLEM (8.16) At a point in a loaded steel bracket the stresses are σ x = 50 MPa, σ y = -15 MPa,
and τ xy = -30 MPa , as shown in Fig. P8.16. Determine the stresses acting on an element oriented at an
angle θ = 40o from the x axis. Sketch the result on a properly oriented element.
SOLUTION
1
1
2
2
1
1
= (50 − 15) + (50 + 15) cos80o − 30sin 80o = −6.45 MPa
2
2
σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ + τ xy sin 2θ
1
2
1
= − (50 + 15) sin 80o − 30 cos80o = −37.2 MPa
2
τ x ' y ' = − (σ x − σ y ) sin 2θ + τ xy cos 2θ
Hence,
σ y ' = σ x + σ y − σ x ' = 50 − 15 + 6.45 = 41.4 MPa
41.6 MPa
y’
29.8 MPa
6.4 MPa
x’
θ = 40o
x
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PROBLEM (8.17) Solve Prob. 8.16 for σ x = 20 ksi, σ y = 5 ksi, τ xy = 4 ksi, and θ = 25o ,
shown in Fig. P8.17.
SOLUTION
1
1
2
2
1
1
= (20 − 5) + (20 − 5) cos 50o + 4sin 50o = 15.38 ksi
2
2
σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ + τ xy sin 2θ
1
2
1
= − (20 − 5) sin 50o + 4 cos 50o = −3.17 ksi
2
τ x ' y ' = − (σ x − σ y ) sin 2θ + τ xy cos 2θ
Therefore,
σ y ' = σ x + σ y − σ x ' = 20 + 5 − 20.4 = 4.6 ksi
4.6 ksi
y’
3.17 ksi
15.38 ksi
x’
θ = 25o
x
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PROBLEM (8.18) Two triangular plates are weld together and subjected to biaxial stresses with σ x
= -5 MPa and σ y = 15 MPa, as depicted in Fig. P8.18. Determine:
(a) The normal stress σ w acting perpendicular to the weld.
(b) The shear stress τ w acting parallel to the weld.
SOLUTION
We have
x’
θ = 25 + 90 = 115
σ x = −5 MPa
σ y = 15 MPa
τ xy = 0
σ x'
τ x' y'
θ
o
5 MPa
o
25
y’
x
15 MPa
1
1
(a) σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ
2
2
1
1
= (−5 + 15) + (−5 − 15) cos 230o = 11.43 MPa
2
2
Thus,
σ w = σ x ' = 11.43 MPa
1
(b) τ x ' y ' = − (σ x − σ y ) sin 2θ
2
1
= − (−5 − 15) sin 230o = −7.66 MPa
2
τw
So
τ w = τ x ' y ' = −7.66 MPa
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PROBLEM (8.19) The stresses at a point in the enclosure plate of a boiler are as shown in the
element of Fig. P8.19. Calculate the normal and shear stresses at the point on the indicated inclined plane.
Show the results on a properly oriented element.
SOLUTION
4
= 53.1o
3
36 + 72 36 − 72
cos106.2o
+
σ x' =
2
2
= 54 + 5.02 = 59 MPa
36 − 72
sin106.2o = 17.3 MPa
τ x' y' = −
2
θ = tan −1
17.3 MPa
y’
59 MPa
x’
53.1o
x
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PROBLEM (8.20) Solve Prob. 8.19, with the state of the tensile stresses reversed so as to act in
compression.
SOLUTION
4
= 53.1o
3
σ x = −36 MPa
σ y = −72 MPa
θ = tan −1
τ xy = 0
36 + 72 −36 + 72
cos106.2o
+
2
2
= −54 − 5.02 = −59 MPa
36 + 72
sin106.2o = −17.3 MPa
τ x' y' = −
2
σ x' = −
17.3 MPa
y’
59 MPa
x’
53.1o
x
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PROBLEM (8.21) The stresses acting uniformly at the edges of a thick. rectangular plate are shown
in Fig. P8.21. Determine the stress components on planes parallel and perpendicular to a-a. Show the
results on a properly oriented element.
SOLUTION
θ = −30 o
σ x = σ y = −120 cos 45o = −84.85 MPa
τ xy = −120 sin 45o = −84.85 MPa
84.85 + 84.85 0
+ cos(−60o ) − 84.85sin(−60o )
2
2
= −84.85 + 73.48 = −11.4 MPa
σ y ' = −84.85 − 73.48 = −158.3 MPa
σ x' = −
τ x ' y ' = 0 − 84.85 cos(−60 o ) = −42.4 MPa
y’
x
o
30
42.4 MPa
x’
11.4 MPa
158.3 MPa
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PROBLEM (8.22) Figure P8.22 represents the state of stress at a point in an aircraft wing. Calculate
the normal and shear stresses at the point on the indicated inclined plane. Sketch the results on a properly
oriented element.
SOLUTION
21.9 MPa
3.5 MPa
x’
θ =75o
y’
75o
x
35 + 21 35 − 21
cos150o + 0
+
2
2
= 28 − 6.06 = 21.9 MPa
35 − 21
sin150o = −3.5 MPa
τ x' y' = −
2
σ x' =
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PROBLEM (8.23) At a point in a loaded structure, the stresses are as shown in Fig. P8.23. Determine
the allowable value of σ if the normal and shearing stresses acting simultaneously in the inclined plane
are limited to 5 and 3 ksi, respectively .
SOLUTION
We have θ = 15o and τ xy = 0
σ x' = 5 =
σ − 2.5 σ + 2.5
cos 30o
+
2
2
10 = σ − 2.5 + 0.866σ + 2.165 , σ = 5.54 ksi
σ + 2.5
sin 30o , σ = 9.5 ksi
τ x ' y ' = −3 = −
2
Thus, σ all = 5.54 ksi
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PROBLEM (8.24) Calculate the normal and shear stresses acting on the inclined plane in Fig. P8.23
for σ = 2ksi.
SOLUTION
σ x = 2 ksi
σ y = −2.5 ksi
τ xy = 0
θ = 15o
2 − 2.5 2 + 2.5
cos 30o = −0.25 + 1.949 = 1.699 ksi
+
2
2
2 + 2.5
sin 30o = −1.125 ksi
τ x' y' = −
2
σ x' =
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PROBLEM (8.25) Calculate and show the stresses on planes of maximum shearing stresses for an
element subjected to principal stresses :
(a) σ 1 = 6 ksi and σ 2 = 2 ksi (Fig. P8.25a)
(b)
σ 1 = 4 ksi and σ 2 = 1 ksi
SOLUTION
(Fig. 8.26b)
2 ksi
6−2
(a) τ max =
= 2 ksi
2
6+2
= 4 ksi
σ '=
2
σ'
6 ksi
τ max
σ'
4 +1
= 2.5 ksi
2
4 −1
= 1.5 ksi
σ '=
2
(b) τ max =
4 ksi
τ max
1 ksi
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PROBLEM (8.26) A welded plate is subjected to uniform biaxial tension as shown in Fig. P8.26.
Calculate the maximum stress σ for two cases:
(a) The weld has an allowable shear stress of 25 MPa.
(b) The weld has an allowable normal stress of 60 MPa.
SOLUTION
τ xy = 0
θ = 60 o
σ − 40
(a) τ x ' y ' = −25 = −
(b) σ x ' = 60 =
2
sin120o
σ + 40 σ − 40
2
+
2
σ = 97.7 MPa
cos120o
σ = 120 MPa
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PROBLEM (8.27) A thin plate in plane stress is subjected to stresses σ x = 70 MPa, σ y , and τ xy
= 25 MPa, as shown in Fig. P8.27. Calculate the largest value of σ y so that maximum in-plane shear
stress does not to exceed 100 MPa.
SOLUTION
τ max = [(
σ x −σ y 2
2
) + τ xy2 ] 2
1
this gives
σ x −σ y
2
2
= [τ max
− τ xy2 ] 2 ;
1
[
70 − σ y 1 2
] = 96.825 MPa
2
Solving,
σ y = 70 − 2(96.825) = −123.7 MPa
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PROBLEM (8.28) At a particular point in a machine, the material is in plane stress with σ x = 60
MPa, σ y = -0 MPa, and τ xy (Fig. P8.28). Calculate:
(a) The largest value of τ xy for which the maximum in-plane shear stress is limited to 80 MPa.
(b) The corresponding principal stresses.
SOLUTION
(a) τ max = [(
σ x −σ y 2
2
) + τ xy2 ] 2 ;
1
80 = [(
1
60 + 50 2
) + τ xy2 ] 2
2
or
τ xy = 802 − 552 = 58.1 MPa
1
1
(b) σ avg = (σ x + σ y ) = (60 − 50) = 5 MPa
2
2
σ 1 = σ avg + τ max = 5 + 80 = 85 MPa
σ 2 = 5 − 80 = −75 MPa
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PROBLEM (8.29) An element in plane stress is subjected to stresses with σ x = 65 MPa, σ y = 30
MPa, and τ xy =25 MPa (Fig. P8.29). Calculate:
(a) The principal stresses.
(b) The maximum shear stresses and associated normal stresses.
Sketch results in a properly oriented element.
SOLUTION
(a) tan 2θ p =
or
2θ p = 55o
2τ xy
σ x −σ y
=
2(25)
65 − 30
θ p = 27.5o
and
1
1
2
2
σ − σ y 2 2 12
) + τ xy ]
R = [( x
2
1
65 − 30 2
= [(
) + 252 ] 2 = 30.5 MPa
2
σ 1,2 = σ avg ± R = 47.5 ± 30.5 MPa
σ ' = σ avg = (σ x + σ y ) = (65 + 30) = 47.5 MPa
Thus,
σ 1 = 78 MPa
Note, for θ p = 27.5o :
1
2
σ 2 = 17 MPa
1
2
σ x ' = 78 MPa
σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ p + τ xy sin 2θ p , gives
(b) τ max = R = 30.5 MPa
σ ' = 47.5 MPa
17 MPa
47.5 MPa
30.52 MPa
78 MPa
x’
27.5o
x
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PROBLEM (8.30) Redo Prob. 8.29 for σ x = 6 ksi, σ y = -24 ksi, and τ xy = -7 ksi (see Fig.
P8.29).
SOLUTION
(a) tan 2θ p =
2τ xy
σ x −σ y
=
2(−7)
6 + 24
or
θ p = −12.51o
1
1
2
2
σ − σ y 2 2 12
) + τ xy ]
R = [( x
2
1
6 + 24 2
= [(
) + (−7) 2 ] 2 = 16.55 ksi
2
σ 1,2 = σ avg ± R = −9 ± 16.55 ksi
σ ' = σ avg = (σ x + σ y ) = (6 − 24) = −9 ksi
So
σ 1 = 7.55 ksi
σ 2 = −25.55 ksi
Note that for θ p = −12.51o :
1
2
1
2
σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ p + τ xy sin 2θ p ,
(b) τ max = R = 16.55 ksi
yields σ x ' = 7.55 ksi
σ ' = −9 ksi
25.55 ksi
9 ksi
7.55 ksi
16.55 ksi
12.51o
x
x’
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PROBLEMS (8.31 through 8.33) At three points in a structure subjected to plane stress, the
stresses have the magnitudes and directions shown in Figs. P8.31 through P8.33. For each point, obtain
the principal stresses and the related orientations. Show the results on properly oriented elements.
SOLUTION (8.31)
σ x = 3 MPa ,
σ y = 2 MPa,
τ xy = 4 MPa
3+ 2
3− 2 2
± (
) + 42 = 2.5 ± 4.03
2
2
σ 1 = 6.53 MPa
σ 2 = −1.53 MPa
2(4)
= 82.9o ; θ p = 41.4o
2θ p = tan −1
3− 2
3+ 2 3− 2
σ x' =
cos82.9o + 4sin 82.9o
+
2
2
= 6.53 MPa
θ p ' = 41.4o
Thus,
σ 1,2 =
41.4o
6.53 MPa
x
1.53 MPa
SOLUTION (8.32)
σ x = 15 ksi,
σ y = 6 ksi,
τ xy = 1 ksi
15 + 6
15 − 6 2 2
± (
) + 1 = 10.5 ± 4.61
2
2
σ 1 = 15.11 ksi
σ 2 = 5.89 ksi
2(1)
= 12.5o
2θ p = tan −1
15 − 6
15 + 6 15 − 6
σ x' =
+
cos12.5o + 1sin12.5o
2
2
= 15.11 ksi
θ p ' = 6.3o
Thus,
σ 1,2 =
5.98 ksi
15.11 ksi
6.3o
x
Continued on next slide
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SOLUTION (8.33)
σ x = 14 MPa,
σ y = 6 MPa,
τ xy = −2 MPa
14 + 6
14 − 6 2
) + (−2) 2 = 10 ± 4.5
± (
2
2
σ 1 = 14.5 MPa
σ 2 = 5.5 MPa
2(−2)
2θ p = tan −1
= −26.6o
14 − 6
14 + 6 14 − 6
σ x' =
+
cos(−26.6o ) + 2sin(−26.6o )
2
2
= 14.5 MPa
θ p ' = −13.3o
Thus,
σ 1,2 =
x
13.3o
14.5 MPa
5.5 MPa
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PROBLEM (8.34) For the state of stress given in Fig. P8.34, determine the magnitude and orientation of the principal stresses. Show the results on a properly oriented element.
SOLUTION
σ x = −50 MPa ,
σ y = 80 MPa ,
τ xy = 60 MPa
−50 + 80
−50 − 80 2
σ 1,2 =
) + 602
± (
73.5 MPa
2
2
= 15 ± 88.46
σ 1 = 103.5 MPa
σ 2 = −73.5 MPa
2(60)
= −42.7o
2θ p = tan −1
103.5 MPa
−50 − 80
−50 + 80 −50 − 80
σ x' =
+
cos(−42.7o ) + 60sin(−42.7o )
2
2
= 15 − 47.76 − 40.7 = −73.5 MPa
Thus, θ p " = −21.4o
x
21.4o
x’
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PROBLEM (8.35) Determine the maximum shearing stresses and the associated normal stresses for
the state of stress of Fig. P8.35. Sketch the results on a properly oriented element.
SOLUTION
σ x = 70 MPa ,
σ y = −40 MPa,
τ xy = −10 MPa
70 + 40 2
) + (−10) 2 = 55.9 MPa
2
70 + 40
2θ s = tan −1 (−
) = 79.7 o
2(−10)
70 + 40
τ x' y' = −
sin 79.7 o − 10 cos 79.7o
2
= −54.11 − 1.79 = −55.9 MPa
τ max = (
θ s " = 39.9o
Thus,
We also have
70 − 40
σ '=
= 15 MPa
2
y’
15 MPa
15 MPa
x’
39.9o
x
55.9 MPa
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PROBLEMS (8.36 through 8.38) The stresses at three points in a loaded member are represented
in Figs. P8.36 through P8.38. For each point, calculate and sketch:
(a) The principal stresses.
(b) The maximum shearing stresses with the associated normal stresses.
SOLUTION (8.36)
σ x = −40 MPa ,
σ y = 0,
τ xy = −60 MPa
−40
−40 2
) + (−60) 2
± (
2
2
= −20 ± 63.24
σ 1 = 43.2 MPa
σ 2 = −83.2 MPa
2(−60)
2θ p = tan −1
= 71.6o
−40
−40 −40
σ x' =
+
cos 71.6o − 60sin 71.6o
2
2
= −20 − 6.31 − 56.9 = −83.2 MPa
Thus, θ p " = 35.8o
(a) σ 1,2 =
43.2 + 83.2
= 63.2 MPa
2
−40 + 0
σ '=
= −20 MPa
2
(b) τ max =
43.2 MPa
20 MPa
63.2 MPa
83.2 MPa
x’
35.8o
x
SOLUTION (8.37)
σ x = −5 MPa,
σ y = −20 MPa,
τ xy = 10 MPa
−5 + 20 2
5 + 20
± (
) + 10 2
2
2
= −12.5 ± 12.5
σ1 = 0
σ 2 = 25 MPa
(a) σ 1,2 = −
Continued on next slide
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2(10)
= 53.1o
−5 + 20
5 + 20 −5 + 20
σ x' = −
+
cos 53.1o + 10sin 53.1o
2
2
= −12.5 + 4.5 + 8 = 0
Thus, θ p ' = 26.5o
25 MPa
2θ p = tan −1
0 + 25
= 12.5 MPa
2
−5 − 20
σ '=
= −12.5 MPa
2
(b) τ max =
12.5 MPa
x’
12.5 MPa
26.5o
x
SOLUTION (8.38)
σ x = 8 ksi,
σ y = 15 ksi,
τ xy = −7 ksi
15 + 8
15 − 8 2
) + (−7) 2 = 11.5 ± 7.826
± (
2
2
σ 1 = 19.33 ksi
σ 2 = 3.67 ksi
2(−7)
2θ p = tan −1
= −63.4o
15 − 8
15 + 8 15 − 8
σ x' =
+
cos(−63.4o ) − 7 sin(−63.4o )
2
2
= 11.5 + 1.567 + 6.259 = 19.33 ksi
11.5 ksi
Thus, θ p ' = −31.7 o
3.67 ksi
(a) σ 1,2 =
19.33 − 3.67
= 7.83 ksi
2
19.33 + 3.67
σ '=
= 11.5 ksi
2
(b) τ max =
x
31.7o
x’
7.83 ksi
19.33 ksi
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PROBLEM (8.39) Taking the τ xy of Prob. 8.36 to be 0, determine the maximum shear stresses
and the associated normal stresses.
SOLUTION
From Fig. 8.36 with τ xy = 0 :
σ1 = 0
σ 2 = −40 ksi
Therefore
σ −σ
0 − (−40)
= 20 ksi
τ max = 1 2 =
2
2
σ + σ 2 0 − 40
=
= −20 ksi
σ '= 1
2
2
20 ksi
40 ksi
20 ksi
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_______________________________________________________________________
PROBLEM (8.40) Given the stresses acting uniformly at the edges of a block (Fig. P8.40),
calculate:
(a) The stresses σ x , σ y , and τ xy .
(b) The maximum shearing stresses with the associated normal stresses.
Sketch the results on properly oriented elements.
SOLUTION
σ y = −5sin 30o = −2.5 MPa
τ xy = 5cos 30o = 4.33 MPa
(a)
2.5 ΔA (0.866)
15 ΔA
4.33 ΔA (0.866)
4.33 ΔA (0.5)
o
30
σ x ΔA (0.5)
ΔA
∑ F = 0 : −15ΔA + 0.5ΔAσ + 4.33ΔA(0.866) = 0 ,
x
x
22.5 + 2.5 2
) + 4.332 = 13.23 MPa
2
22.5 + 2.5
2θ s = tan −1[−
] = −70.9o
2(4.33)
σ ' = 12 ( 22.5 − 2.5) = 10 MPa
22.5 + 2.5
τ x' y' = −
sin(−70.9o ) + 4.33cos(−70.9)
2
= 11.81 + 1.42 = 13.23 MPa
Thus,
θ s ' = −35.5o
(b) τ max = (
σ x = 22.5 MPa
y’
10 MPa
x
o
35.5
x’
10 MPa
13.23 MPa
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PROBLEM (8.41) The shearing stress at a point in a loaded member is τ xy = 4 ksi (See Fig.
P8.41). The principal stresses at this point are σ 1 = 5 ksi and σ 2 = -8 ksi. Determine σ x and σ y and
indicate the principal and maximum shear stresses on an appropriate sketch.
SOLUTION
σ1 = 5 =
σx +σ y
σ 2 = −8 =
2
+ (
σx +σ y
σ x −σ y 2
− (
2
) + 42
(1)
σ x −σ y 2
2
2
Add Eqs. (1) and (2):
σ x + σ y = −3
Equation (1) becomes
σ −σ y 2 2
5 = −1.5 + ( x
) +4
2
or
σ x − σ y = 10.246
) + 42
(2)
(3)
(4)
Solving Eqs. (3) and (4):
σ x = 3.62 ksi
σ y = −6.62 ksi
We have
5+8
τ max =
= 6.5 ksi
2
2(4)
= 38o
2θ p = tan −1
3.62 + 6.42
3.62 − 6.62 3.62 + 6.62
σ x' =
+
cos 38o + 4sin 38o
2
2
= −1.5 + 4.04 + 2.46 = 5 ksi
Thus,
θ p ' = 19o
8 ksi
5 ksi
x’
6.5 ksi
19o
x
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PROBLEM (8.42) The state of stress at a point is shown in Fig. P8.42. For that point determine:
(a) The magnitude of the shear stress τ if the maximum principal stress is not to exceed 70 MPa.
(b) The corresponding maximum shearing stresses and the planes at which they act.
SOLUTION
0 + 60
0 − 60 2
+ (
) +τ 2
2
2
τ = 26.5 MPa
(a) σ 1 = 70 =
(b) τ max = (
−60 2
) + 26.52 = 40 MPa
2
60
= 30 MPa
2
0 − 60
2θ s = tan −1[−
] = 48.6o
2(26.5)
60
τ x ' y ' = sin 48.6o + 26.5cos 48.6o = 40 MPa
2
Thus,
θ s ' = 24.3o
σ '=
30 MPa
y’
30 MPa
x’
24.3o
40 MPa
x
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PROBLEM (*8.43) A structural member is subjected to two different loadings, each separately
producing stresses at point A, as indicated in Fig. P8.43. Calculate, and show on a sketch, the principal
planes and the principal stresses under the effect of the combined loading.
*SOLUTION
20 ksi
10 ksi
A
θ =-30o
x
10 + 20 10 − 20
+
cos(−60o ) = 15 − 2.5 = 12.5 ksi
2
2
σ y = 15 + 2.5 = 17.5 ksi
σx =
τ xy = −
10 − 20
sin(−60o ) = −4.33 ksi
2
17.5 ksi
17.5 ksi
A
12.5 ksi
+
A
A
=
5 ksi
4.33 ksi
12.5 ksi
0.67 ksi
12.5 + 17.5
12.5 − 17.5 2
) + 0.67 2 = 15 ± 2.59
± (
2
2
σ 1 = 17.59 ksi
σ 2 = 12.41 ksi
2(0.67)
2θ p = tan −1[
] = −15o
12.5 − 17.5
σ 1,2 =
12.5 − 17.5
cos(−15o ) + 0.67 sin(−15o )
2
= 15 − 2.41 − 0.173 = 12.41 ksi
σ x ' = 15 +
Thus,
θ p " = −7.5o
x
7.5o
A
12.41 ksi
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unlawful.
17.59is ksi
_______________________________________________________________________
PROBLEM (*8.44) Redo Prob. 8.43 for the case shown in Fig. P8.44.
*SOLUTION
1.4 ksi
2.1 ksi
A
θ = -45o
x
1.4 + 0 1.4 − 0
+
cos(−90o ) − 2.1sin(−90o ) = 0.7 + 0 + 2.1 = 2.8 ksi
2
2
σ y = 0.7 + 0 − 2.1 = −1.4 ksi
σx =
τ xy = −
1.4 − 0
sin(−90o ) − 2.1cos(−90o ) = 0.7 ksi
2
1.4 ksi
1.4 ksi
A
7 ksi
+
2.8 ksi
A
=
4.2 ksi
0.7 ksi
A
0.7 ksi
4.2 + 1.4
4.2 − 1.4 2
) + 0.7 2 = −2.8 ± 1.565
± (−
2
2
σ 1 = −1.235 ksi
σ 2 = −4.365 ksi
2(0.7)
2θ p = tan −1[
] = −26.6o
−4.2 + 1.4
4.2 + 1.4 4.2 − 1.4
σ x' = −
−
cos(−26.6o ) + 0.7 sin(−26.6o )
2
2
= −2.8 − 1.252 − 0.313 = −4.365 ksi
σ 1,2 = −
Thus,
θ p " = −13.3o
x
13.3o
A
4.365 ksi
1.235 ksi
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PROBLEM (*8.45) Redo Prob. 8.43 for the case shown in Fig. P8.45, where σ and θ are
known constants.
*SOLUTION
σ
A
Note:
cos(−θ ) = cos θ
sin(−θ ) = − sin θ
−θ
x
σx =
σy =
σ
+
2
σ
2
τ xy = −
−
σ
2
σ
2
σ
2
cos 2(−θ ) =
cos 2(−θ ) =
σ
2
σ
2
(1 + cos 2θ ) = σ cos 2 θ
(1 − cos 2θ ) = σ sin 2 θ
sin 2(−θ ) = σ sin θ cos θ
σ
σ (1 + sin 2 θ )
σ sin 2 θ
A
A
+
σ cos 2 θ
=
σ sin θ cos θ
A
σ cos 2 θ
σ sin θ cos θ
For the combined stress:
σ +σ y σ
σ '= x
= (cos 2 θ + 1 + sin 2 θ ) = σ
2
2
σ x −σ y σ
= (cos 2 θ − 1 − sin 2 θ ) = −σ sin 2 θ
2
2
σ sin θ cos θ
= − cotan θ
tan 2θ p =
−σ sin 2 θ
2θ p = 90o + θ ;
r2 = (
σ x −σ y 2
2
= σ sin 2 θ
θp ' =
θ
2
+ 45o
2
Thus,
σ1
) + τ xy2 = σ 2 sin 2 θ (sin 2 θ + cos 2 θ )
σ 1 = σ '+ r = σ + σ sin θ = σ (1 + sin θ )
σ 2 = σ '− r = σ − σ sin θ = σ (1 − sin θ )
A
θp '
σ2
x
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_______________________________________________________________________
PROBLEM (8.46) The state of stress on the horizontal and vertical planes at a point is only not
completely known, as shown in Fig. P8.46. However, at the point stresses σ 2 and τ max are prescribed
as -7 and 8 ksi, respectively. Determine stresses σ , τ , and σ 1 . Show the results on properly,
oriented elements.
SOLUTION
τ max = 8 =
σ1 + 7
2
;
We have
9 − 7 = 5+σ;
σ 1 = 9 ksi
σ = −3 ksi
5+3 2
τ = 6.93 ksi
) +τ 2 ;
2
2(6.93)
5−3
= 60o ,
σ '=
= 1 ksi
2θ p = tan −1
5+3
2
5+3
σ x' = 1+
cos 60o + 6.93sin 60o
2
= 1 + 2 + 6 = 9 ksi
Thus,
θ p ' = 30o
8= (
1 ksi
9 ksi
8 ksi
30o
7 ksi
x
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_______________________________________________________________________
PROBLEM (8.47) Consider a point in a loaded body subjected to the stresses shown in Fig.
P8.47. Determine:
(a) The principal stresses.
(b) The maximum shear stresses.
Sketch the results on properly oriented elements.
SOLUTION
−50 − 30
−50 + 30 2
) + 202 = −40 ± 22.4
± (
2
2
σ 1 = −17.6 MPa
σ 2 = −62.4 MPa
2(−20)
2θ p = tan −1
= 63.4o
−50 + 30
−17.6 + 62.4
(b) τ max =
= 22.4 MPa
2
−17.6 − 62.4
σ '=
= −40 MPa
2
50 + 30 50 − 30
σ x' = −
−
cos 63.4o − 20sin 63.4o
2
2
= −40 − 4.48 − 17.88 = −62.4 MPa
Thus,
θ p " = 31.7o
(a) σ 1,2 =
17.6 MPa
40 MPa
62.4 MPa
40 MPa
31.7o
22.4 MPa
x
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PROBLEM (*8.48) The state of stress at a point A in a structure is shown in Fig. P8.48.
Determine the normal stress
σ and the angle θ .
*SOLUTION
y’
16 MPa
x’
A
36 MPa
σ x + σ y = σ x' + σ y'
− 8 − 40 = −36 − σ ;
σ x ' = −36 =
θ
σ
x
σ = −12 MPa
−8 − 40 8 − 40
−
cos 2θ − 12sin 2θ
2
2
or
− 12 = 16 cos 2θ − 12 sin 2θ
−8 + 40
τ x ' y ' = −16 = −
sin 2θ − 12 cos 2θ
2
Multiply by − 3 4 :
12 = 12 sin 2θ + 9 cos 2θ
(1)
(2)
Add Eqs. (1) and (2),
0 = 25 cos 2θ
or
2θ = 90 o ,
θ = 45o
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_______________________________________________________________________
PROBLEM (*8.49) Redo Prob. 8.48 for the case shown in Fig. P8.49.
*SOLUTION
18 MPa
y’
σ
x’
14 MPa
θ
x
σ x + σ y = σ x' + σ y'
30 − 20 = −18 + σ ;
σ x ' = 28 =
σ = 28 MPa
30 − 20 30 + 20
cos 2θ + 10sin 2θ
+
2
2
or
23 = 25 cos 2θ + 10 sin 2θ
τ x ' y ' = −14 = −25 sin 2θ + 10 cos 2θ
Multiply this by − 2.5 :
35 = 62.5 sin 2θ − 25 cos 2θ
(1)
(2)
Add Eqs. (1) and (2),
58 = 72.5 sin 2θ , 2θ = 53.13o
or
θ = 26.6 o
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PROBLEM (*8.50) At a particular point in a loaded member (see Fig. P8.50) there exists on
the horizontal plane a normal stress σ y = 2 ksi and a negative shear stress. One of the principal
stresses at the point is 1 ksi (tensile), and the maximum shearing stress has a magnitude of 5 ksi.
Calculate:
(a) The unknown stresses on the horizontal and vertical planes.
(b) The unknown principal stress.
Show the principal stresses and maximum shear stresses on a sketch of a properly oriented
element.
*SOLUTION
(a) σ 1,2 =
σx +σ y
2
σ2 =1 =
±5
σx + 2
2
−5,
(1)
σ x = 10 ksi
10 − 2 2
) + (−τ xy ) 2
2
2
(5) = 16 + (−τ xy ) 2 , τ xy = −3 ksi
τ max = 5 = (
(b) From Eq. (1):
σ 1 = 11 ksi
We have
2(−3)
= −36.9o
10 − 2
10 + 2 10 − 2
σ x' =
cos(−36.9o ) − 3sin(−36.9o )
+
2
2
= 6 + 3.2 + 1.8 = 11 ksi
2θ p = tan −1
Therefore,
θ p ' = −18.5o
σ '=
10 + 2
= 6 ksi
2
1 ksi
x
5 ksi
18.5o
11 ksi
6 ksi
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PROBLEM (8.51) Using Mohr's circle, solve Prob. 8.3.
SOLUTION
τ
(MPa)
80 MPa
o
15
(80, 60)
15
60 MPa
O
α
50 MPa
R
C
σ (MPa)
o
(-50, -60) 30
x’
60
= 42.7 o
65
1
2
R = (65 + 602 ) 2 = 88.5
τ x ' y ' = sin 72.2 o (88.5) = 84.5 MPa
α = tan −1
σ x ' = cos 72.7 o (88.5) = −11.3 MPa
Sketch of results is as shown in solution of Prob. 8.3.
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PROBLEM (8.52) Using Mohr's circle, solve Prob. 8.4.
SOLUTION
τ (ksi)
x’
25 ksi
σ’=12.5
y
10 ksi
30o
o
81.3
O
(0, -10)
x
R
o
60
α
(25, 10)
σ (ksi)
C
y’
1
R = (12.52 + 102 ) 2 = 16
10
= 38.7o
α = tan −1
12.5
τ x ' y ' = 16 sin 81.3o = 15.83 ksi
σ x ' = 12.5 − 16 cos 81.3o = 10.09 ksi
Sketch of results is as shown in solution of Prob. 8.4.
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PROBLEM (8.53) Using Mohr's circle, solve Prob. 8.5.
SOLUTION
τ (MPa)
40 MPa
σ’=15
y’
R
10 MPa
o
22.5
70 MPa
x
O
(-40, -10)
55.3o
(σx’, σ y’)
(70, 10)
α
σ (MPa)
C
x’
10
= 10.3o
55
1
2
R = (10 + 552 ) 2 = 55.9
τ x ' y ' = 55.9 sin 55.3o = 46 MPa
α = tan −1
σ x ' = −55.9 cos 55.3o + 15
= −16.8 MPa
y’
16.8 MPa
46 MPa
70 MPa
o
22.5
x’
10 MPa
x
40 MPa
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PROBLEM (8.54) Using Mohr's circle, solve Prob. 8.10.
SOLUTION
τ (ksi)
σ’=3
τ
300o
6 ksi
O
o
60
x’
x
y
x
(6, τ)
60o α σ (ksi)
(7, −τ x’ y’ )
C
R
x’
y’
We have
4 = r cos(60 o − α )
where r = 3 cos α
3
4=
(0.5cos α + 0.866sin α )
or
cos α
This gives
2.5
α = tan −1
= 43.9o
2.596
Then R = 3 cos 43.9o = 4.16
Thus,
τ = 4.16 sin 43.9 o = 2.89 ksi
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PROBLEM (8.55) Using Mohr’s circle, solve Prob. 8.15.
SOLUTION
For θ = 30o :σ x ' = −200 MPa
τ x ' y ' = 35 MPa
R = [(200 − 130) 2 + 352 ]2 = 78.26 MPa, σ avg =
1
1
(−200 − 60) = −130 MPa
2
τ (MPa)
x
y’
R
60 β
α
o
-35
O
C
y
σ (MPa)
35
200 − 130
α = 26.56o
β = 33.44o
tan α =
x’
130
200
For θ = 0:
σ x = −130 − R cos β
σ x = −130 − (78.26) cos(33.44o ) = −195.3
τ xy = − R sin β = −(78.26) sin(33.44o ) = −43.1 MPa
σ y = −130 + R cos β
= −130 + (78.26) cos(33.44o ) = −64.7 MPa
Sketch of results is as shown in solution of Prob. 8.15.
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PROBLEM (8.56) Using Mohr’s circle, solve Prob. 8.16.
SOLUTION
σ x = 50 MPa σ y = −15 MPa τ xy = −30 MPa θ = 40o
τ (MPa)
1
2
σ avg = (50 − 15) = 17.5
50
x’ A’
β
O
x
A
80o
C α
σ (MPa)
R
B’
B
30
y’
17.5
R = (50 − 17.5) 2 + 302
= 44.23 MPa
30
tan α =
,
50 − 17.5
α = 42.71o
β = 180 − 80 − 42.71 = 57.29o
Point A '(θ = 40o ) :
σ x ' = 17.5 − R cos β
= 17.5 − (44.23) cos(57.21o ) = −6.45 MPa
τ x ' y ' = − R sin β = −(44.23) sin(57.29o ) = −37.2 MPa
σ y ' = 17.5 + R cos β
= 17.5 + (44.23) cos(57.29o ) = 41.4 MPa
Sketch of results is as shown in solution of Prob. 8.16.
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PROBLEM (8.57) Using Mohr’ circle, solve Prob. 8.17.
SOLUTION
σ x = 20 ksi σ y = 5 ksi τ xy = 4 ksi θ = 25o
1
2
σ avg = (20 + 5) = 12.5 ksi
τ (ksi)
β
β
O
α
y’
A’x’
50o
σ (ksi)
A
x
R = (20 − 12.5) 2 + 42
= 8.5 ksi
4
tan α =
,
20 − 12.5
α = 28.07o
12.5
20
β = 50 − α = 21.93o
σ x ' = 12.5 + R cos β
= 12.5 + (8.5) cos(21.93o ) = 20.4 ksi
τ x ' y ' = − R sin β = −(8.5) sin(21.93o ) = −3.17 ksi
σ y ' = 12.5 − R cos β
= 12.5 − (8.5) cos(21.93o ) = 4.61 MPa
Sketch of results is as shown in solution of Prob. 8.17.
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PROBLEM (8.58) Using Mohr’ circle, solve Prob. 8.18
SOLUTION
(a) σ x = −5 MPa σ y = 15 MPa τ xy = 0
τ (ksi)
x’
10
C
A
1
2
o
θ = 25 + 90o = 115o
1
R = (5 + 15) = 10 MPa
2
σ w = σ x ' = 5 + R cos 50o
σ avg = (−5 + 15) = 5 MPa
R
50o
B
O
230o
5
σ (ksi)
= 5 + 10 cos 50o
= 11.43 MPa
15
(b) τ w = τ x ' y ' = − R sin 50o
= −10sin 50o
= −7.66 MPa
τw
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PROBLEM (8.59) Using Mohr's circle, solve Prob. 8.19.
SOLUTION
τ (MPa)
σ' =5
x’
R=18
R
36 MPa
72
36
o
53.1
x
72 MPa
O
σ (MPa)
C
106.2o
x’
73.8o
(σx’, −τ x’ y’ )
τ x ' y ' = 18 sin 73.8o = 17.3 MPa
σ x ' = 54 + 18 cos 73.8o = 59 MPa
Sketch of results id as shown in solution if Prob. 8.19.
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PROBLEM (8.60) Using Mohr's circle, solve Prob. 8.22.
SOLUTION
τ (MPa)
x’
(σx’, τ x’ y’ )
35 MPa
60o
O
x’
21
o
15
C
35
σ (MPa)
R
y’
x
21 MPa
150o
σ’=28
1
R = (35 − 21) = 7
2
σ x ' = 28 − 7 cos 30o = 21.9 MPa
τ x ' y ' = −7 sin 30o = −3.5 MPa
Sketch of results is as shown in solution of Prob. 8.22.
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PROBLEM (8.61) Using Mohr's circle, solve Prob. 8.25.
SOLUTION
τ (ksi)
(a)
σ’=28
τmax
R
6
6−2
= 2 ksi
2
O
σ (ksi)
C
2
6+2
= 4 ksi
σ '=
2
Sketch of results is as shown in solution of Prob. 8.25a.
τ max =
(b)
τ (ksi)
1
2
τmax
τ max = (4 + 1) = 2.5 ksi
1
σ ' = (4 − 1) = 1.5 ksi
2
R
O
-1
C
4
σ’
σ (ksi)
Sketch of results is as shown in solution of Prob. 8.25b.
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PROBLEM (8.62) Using Mohr's circle, solve Prob. 8.26.
SOLUTION
(a)
τ (MPa)
x’
120o
1
R = (σ − 40)
2
σ
R
σ (MPa)
40 MPa
x’
σ
60o
O
40
C
x
τ x ' y ' = 25 =
σ − 40
2
(b) σ x ' = 60 = 40 +
σ = 120 MPa
sin 60o ;
σ − 40
2
σ = 97.7 MPa
(1 − cos 60o )
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PROBLEM (8.63) Using Mohr's circle, solve Prob. 8.27.
SOLUTION
σ x = 70 MPa
τ xy = 25 MPa
τ max = 100 MPa
R = 100 MPa
τ (MPa)
= (70 + σ avg ) 2 + 252
y B
R
α
σavg
Thus, R = 100
O
αA
C
25
x
σ (MPa)
70
or
2
σ avg
+ 140σ avg − 4475 = 0
−140 ± 1402 − 4(−4475)
2
= 26.83 MPa
σ avg =
25
,
α = 14.48o
70 + 26.83
Therefore,
σ y = σ avg + R cos α
tan α =
= −26.83 − (100) cos(14.48o ) = −123.7 MPa
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PROBLEM (8.64) Using Mohr's circle, solve Prob. 8.28.
SOLUTION
σ x = 60 MPa
σ y = −50 MPa
τ max = 80 MPa
(a)
R = 80 MPa
1
σ avg = (60 − 50) = 5 MPa
2
τ (MPa)
5
y B
C
B1
O
α
R
A1
A
σ (MPa)
τ xy = R 2 − (60 − 5) 2
τ xy = 802 − 552
= 58.1 MPa
60
(b) σ 1,2 = σ avg m R = 5 m 80
or
σ 1 = 85 MPa
σ 2 = −75 MPa
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PROBLEM (8.65) Using Mohr's circle, solve Prob. 8.29.
SOLUTION
σ x = 65 MPa
σ y = 30 MPa
τ xy = 25 MPa
1
2
= 47.5 MPa
25
tan α =
, α = 55o
17.5
σ avg = (65 + 30)
(a)
τ (MPa)
y
R
A1
C
α
O B1
25
σ (MPa)
A
47.5
x
17.5
65
R = 17.52 + 252
= 30.5 MPa
σ 1,2 = 47.5 ± 30.5
or
σ 1 = 78 MPa
σ 2 = 17 MPa
(b) τ max = R = 30.5 MPa
Sketch of results is shown in solution of Prob. 8.29.
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PROBLEM (8.66) Using Mohr's circle, solve Prob. 8.30.
SOLUTION
σ x = 6 ksi
σ y = −24 ksi
τ xy = −7 ksi
1
2
= −9 ksi
7
tan α = , α = 25.02o
15
σ avg = (6 − 24)
(a)
τ (ksi)
6
A x
B1
α
C
O
9
R
y B
A1
σ (ksi)
24
R = 152 + 7 2
= 16.55 ksi
σ 1,2 = −9 ± R = −9 ± 16.55
or σ 1 = 7.55 ksi
σ 2 = −25.55 ksi
(b) τ max = R = 16.55 ksi
Sketch of results is shown in solution of Prob. 8.30.
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PROBLEM (8.67) Using Mohr's circle, solve Prob. 8.36.
SOLUTION
τ (MPa)
20
(40, 60)
60 MPa
α
σ2
σ1
C
R
O
σ (MPa)
40 MPa
σ '=
α = tan −1 3 = 71.6o
−40 + 0
= −20 MPa
2
40 2
) + 602 = −20 ± 63.2
2
σ 1 = 43.2 MPa
σ 2 = −83.2 MPa
(a) σ 1,2 = −20 ± (−
(b) τ max = 63.2 MPa
Sketch of results is as shown in solution of Prob. 8.36.
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PROBLEM (8.68) Using Mohr's circle, solve Prob. 8.42.
SOLUTION
τ (MPa)
τmax
60 MPa
σ’
τ
R = 302 + τ 2
R
O α
C
70
σ (MPa)
(0, τ)
1
2
σ ' = (60 + 0) = 30 MPa
(a) 70 = 30 + 30 2 + τ 2
τ = 26.5 MPa
30
α = tan −1
= 48.6o
26.5
(b) τ max = 30 2 + 26.5 2 = 40 MPa
Sketch of results is as shown in solution of Prob. 8.42.
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PROBLEM (*8.69) Using Mohr's circle, solve Prob. 8.43.
*SOLUTION
1
2
σ avg = (10 + 20) = 15 ksi
τ (ksi)
1
(20 − 10) = 5 ksi
2
σ x = 15 − R cos 60o
R=
x A’
60o
O
R
C
A
R
10
= 15 − 5cos 60o = 12.5 ksi
τ xy = − R sin 60o
B σ (ksi)
= −5sin 60o = −4.33 ksi
σ y = 15 + R cos 60o
y
15
20
= 15 + 5cos 60o = 17.5 ksi
17.5 ksi
17.5 ksi
5 ksi
A
12.5 ksi
+
A
0.67 ksi
A
=
12.5 ksi
4.33 ksi
τ (ksi)
17.5
C
0.67
O
α
σ (ksi)
12.5
15
1
2
σ 1,2 = 15 ± R = 15 ± 2.59
σ avg = (17.5 + 12.5) = 15 ksi
σ 1 = 17.59 ksi
R = (15 − 12.5) 2 + (0.67) 2 = 2.59 ksi
σ 1 = 12.41 ksi
Sketch of results is shown in solution of Prob. 8.43.
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PROBLEM (*8.70) Using Mohr's circle, solve Prob. 8.44.
*SOLUTION
Rotate 2nd element 45o
τ(ksi)
α = tan −1 3 = 71.6 o
θ = 18.4o
(1.4, 2.1)
R
y
α
σ (ksi)
C
O
R = 0.7 2 + 2.12
= 2.214
θ
x
0.7
σ x = 0.7 + 2.214 cos18.4 o = 2.8 ksi
σ y = 1.4 ksi,
τ xy = 2.214 sin 18.4 o = 0.7 ksi
Thus,
1.4 ksi
1.4 ksi
τ
0.7 ksi
A
A
+
7 ksi
2.8 ksi
=
A
4.2 ksi
τ (ksi)
2.8
R
1.7
(0.7, 4.2 )
2θ
"
p
C
y
O
σ (ksi)
R = 0.7 2 + 1.42
= 1.565 ksi
1
0.7
θ p " = tan −1
= 13.3o
2
1.4
σ 1 = −2.8 + 1.566 = −1.235 ksi
σ 2 = −2.8 − 1.566 = −4.365 ksi
Sketch of results is as shown in solution of Prob. 8.44.
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PROBLEM (*8.71) Using Mohr's circle, solve Prob. 8.45.
*SOLUTION
Rotate 2nd element by θ
τ
σ x" =
R
O
τ x" y" =
σ
2
2θ
C
σ y" =
σ
(1 + cos 2θ ) = σ cos 2 θ
2
σ
(σ x " , − τ x " y " )
(1 − cos 2θ ) = σ sin 2 θ
2
sin 2θ = σ sin θ cos θ = σ (1 + sin 2 θ )
σ y"
σ
+ σ
A
σ y = σ (1 + sin 2 θ
τ x" y"
A
τ xy = σ sin θ cos
1
2
y
R
σ2
(σx, -τxy)
σ x = σ cos 2 θ
σ ' = (σ x + σ y )
σ’
x
A
=
x"
τ
O
σ
C
2θ p '
=
σ1
σ
2
(cos 2 θ + 1 + sin 2 θ ) = σ
σ
1
σ
(σ x − σ y ) = (cos 2 − 1 − sin 2 θ ) = −σ sin 2 θ
2
2
σ sin θ cos θ
= − cotan θ
tan 2θ p ' =
−σ sin 2 θ
Continued on next slide
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Continued on next slide
2θ p ' = θ + π ;
R2 = (
θp ' =
θ
2
+ 45o
σ x −σ y 2
) + τ xy2 = σ 2 sin 4 θ + σ 2 cos 2 θ
2
= σ sin 2 θ (sin 2 θ + cos 2 θ ) = σ 2 sin 2 θ
2
Thus,
σ 1 = σ '+ r = σ (1 + sin θ )
σ 2 = σ '− r = σ (1 − sin θ )
Sketch of results is as shown in solution of Prob. 8.45.
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PROBLEM (8.72) Using Mohr's circle, solve Prob. 8.46.
SOLUTION
τ (ksi)
σ
(σ, -τ)
τ
8
5 ksi
O
σ’
C
2θ p '
σ1 σ (ksi)
(5, -τ)
8 = (σ 1 + 7) 2
or
σ 1 = 9 ksi
9−7
4
σ '=
= 1 ksi
2θ p ' = cos −1 = 60o
2
8
o
σ = −8cos 60 + 1 = −3 ksi
τ = 8sin 60o = 6.93 ksi
Sketch of results is as shown in solution of Prob. 8.46.
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PROBLEM (8.73) Using Mohr's circle, solve Prob. 8.47.
SOLUTION
τ (MPa)
σ '=−
(−50, 20)
20
σ2
τ max = R = 102 + 202 = 22.4
R
C
10
2θ p = tan −1
σ1
2θ p
50 + 30
= −40 MPa
2
O
σ(MPa)
20
= 63.4 o
10
θ p = 31.7 o
40
σ 1 = −40 + 22.4 = −17.6 MPa
σ 2 = −40 − 22.4 = −62.4 MPa
Sketches of results are as shown in solution of Prob. 8.47.
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PROBLEM (*8.74) Using Mohr's circle, solve Prob. 8.48.
*SOLUTION
τ (MPa)
σ’=24
(-36, 16)
2θ
16
α
12 C
y
x
β
(-8, 12)
O
(σ, -16)
σ (MPa)
4
= 53.1o
3
σ − 36 = −40 − 8
σ = −12 MPa
α = tan −1
3
= 36.9o
4
2θ = 180 o − α − β = 90 o
θ = 45o
β = tan −1
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PROBLEM (*8.75) Using Mohr's circle, solve Prob. 8.50.
*SOLUTION
τ (ksi)
R=
σ’=6
2 ksi
1
(11 − 1) = 5 ksi
2
x
τxy
R
σx
4
O 1
C
2θ p '
11
σ (ksi)
y
(2, −τ xy)
1 + 11 = 2 + σ x ;
σ x = 10 ksi
τ xy = 52 − 4 2 = 3 ksi
2θ p ' = tan −1
3
= 36.9o
4
θ p ' = 18.5o
Sketch of results is as shown in solution of Prob. 8.50.
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PROBLEM (8.76) An element in three-dimensional stress is subjected to stresses σ x = 120 MPa
σ y = 60 MPa,
τ xy = -30 MPa,
σ z = -15 MPa
as shown in Fig. P8.76. Using Mohr’s circle, determine:
(a) The principal stresses.
(b) The absolute maximum shear stress.
SOLUTION
σ3
1
2
1
= (120 + 60) = 90 MPa
2
σ −σ y 2 2
R= ( x
) + τ xy
2
σ avg = (σ x + σ y )
τ (MPa)
O σ2
C
σ1 σ(MPa)
120 − 60 2
) + (−30) 2
= (
2
= 42.43 MPa
σavg =42.43
(a) σ 1,2 = σ avg ± R
σ 1 = 90 + 42.43 = 132.4 MPa
σ 2 = 47.6 MPa
σ 3 = −15 MPa
1
(b) (τ max ) a = (σ 1 − σ 3 )
2
1
= [132.4 − (−15)] = 73.7 MPa
2
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PROBLEM (8.77) Solve Prob. 76 for the case in which
σ x = 10 ksi
σy = 0
τ xy = 5 ksi
σ z = - 12 ksi
as depicted in Fig. P8.77.
SOLUTION
τ (ksi)
σavg
R
σ3
1
2
1
= (10 + 0) = 5 ksi
2
σ −σ y 2 2
R= ( x
) + τ xy
2
σ avg = (σ x + σ y )
O
C
σ1
σ2
σ (ksi)
10 − 0 2
) + (5) 2
= (
2
= 7.07 ksi
(a) σ 1,2 = σ avg ± R
σ 1 = 5 + 7.07 = 12.07 ksi
σ 2 = −2.07 ksi
σ 3 = −12 ksi
1
(b) (τ max ) a = (σ 1 − σ 3 )
2
1
= [12.07 − (−12)] = 12.04 ksi
2
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PROBLEM (8.78) Redo Prob. 8.76 for the given state of stress
σ x = 50 MPa
σ y = 10 MPa
τ xy = - 40 MPa
σ z = 25 MPa
as shown in Fig. P8.78.
SOLUTION
τ (MPa)
σavg =30
25
σ3
σ2 C
1
2
1
= (50 + 10) = 30 MPa
2
σ −σ y 2 2
R= ( x
) + τ xy
2
σ avg = (σ x + σ y )
σ1
σ (MPa)
50 − 10 2
) + (−40) 2
2
= 44.72 MPa
= (
(a) σ 1,3 = σ avg ± R
σ 1 = 30 + 44.72 = 74.7 MPa
σ 3 = 30 − 44.72 = −14.72 MPa
σ 2 = 25 MPa
1
(b) (τ max ) a = (σ 1 − σ 3 )
2
1
= [74.7 − (−14.72)] = 44.7 MPa
2
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PROBLEM (8.79) A point in a structure is subjected to state of stress with
σ x = 15 ksi
σy
τ xz = 10 ksi
acting as shown on a three-dimensional element in Fig. P8.79. Calculate two values of σ y for which the
maximum shear stress equals 13 ksi.
SOLUTION
τ (ksi)
τ=13
σavg =7.5
R
O
1
2
1
= (15 + 0) = 7.5 ksi
2
σ −σ z 2 2
R= ( x
) + τ xz
2
σ avg = (σ x + σ z )
C
σ1
σ (ksi)
15 − 0 2
) + 10 2
= (
2
= 12.5 ksi
τ=−13
We have
σ 1,3 = σ avg ± R
σ 1 = 7.5 + 12.5 = 20 ksi
σ 3 = 7.5 − 12.5 = −5 ksi
Assume: σ max = σ 1 = 20 ksi
Then σ y = σ min = σ 1 − 2τ max
= 20 − 2(13) = −6 ksi
Assume : σ min = σ 3 = −5 ksi
Then σ y = σ max = σ min + 2τ max
= −5 + 2(13) = 21 ksi
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PROBLEM (8.80) A point in a machine component is subjected to three-dimensional stress with
σ x = 150 MPa
σ y = 30 MPa
τ xy = 90 MPa
σz
as depicted in Fig. P8.105. Determine:
(a) The maximum shear stress for σ z = 40 MPa.
σ z = - 40 MPa.
(b) The maximum shear stress for
SOLUTION
1
2
1
= (150 + 30) = 90 MPa
2
σ −σ y 2 2
R= ( x
) + τ xy
2
σ avg = (σ x + σ y )
τ (MPa)
σavg =90
σ3
σ1
C
150 − 30 2
) + 90 2
= (
2
= 108.2 MPa
σ (MPa)
R
σ 1 = σ avg + R = 198.2 MPa
σ 3 = −18.2 MPa
(a) σ 2 = 40 MPa
σ 1 = 198.2 MPa
σ 3 = −18.2 MPa
1
(τ max ) a = (σ 1 − σ 3 )
2
1
= [198.2 − (−18.2)] = 108.2 MPa
2
σ 1 = 198.2 MPa
σ 2 = −18.2 MPa
(b) σ 3 = −40 MPa
1
(τ max ) a = (σ 1 − σ 3 )
2
1
= [198.2 − (−40)] = 119.1 MPa
2
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PROBLEM (8.81) A spherical vessel of radius r and wall thickness t is submerged in water having
density γ . Calculate the water depth h at which the circumferential stress in the sphere would be σ .
Assumption: A safety factor of ns is to be used.
Given: r = 2.5 ft, t = 1½ in., γ = 62.4 lb/ft3 ,
σ = 4200 psi ,
ns = 1.5
SOLUTION
p=
2tσ all 2(1.5)(4200) 1.5
=
= 280 psi
r
2.5 × 12
We have
62.4
h = 280
(12)3
or
h = 7754 in. = 646 ft
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PROBLEM (8.82) A compressed-air tank of uniform thickness t = 5-mm is subjected to an internal
pressure of p = 1.4 MPa (Fig. P8.77). Determine the maximum axial and circumferential stresses.
SOLUTION
∑F
axial
= 0 : σ a ⋅ Aw = p ⋅ A
where A = projected inside area
Aw =cross-sectional area of tank
or
σ a [250 + 150π ](5)2
1.5 mm
= 1.4(106 )[250 × 300 + π (150) 2 ]
L
from which
A
p
∑F
σt
p
0.15 m
0.25 m
vertical
σa
Aw
σt
σ a = 28.3 MPa
The maximum
circumferential
stress occurs on
the sides.
0.15 m
= 0 : 2(5) L ⋅ σ c = p (250 + 300) L
or
1.4 × 106 (550)
= 77 MPa
σt =
2(5)
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PROBLEM (8.83) The closed-ended cylindrical steel tank seen in Fig. P8.78 has a radius r and a
height h. It is completely filled with a liquid of density γ and is subjected to an additional internal gas
pressure of p. Calculate the wall thickness needed:
(a) At the top of the tank.
(b) At quarter-height.
(c) At mid-height.
Given: p = 400 kPa, h = 20 m, r = 5 m, γ = 15 kN/m3
Requirement: The allowable stress in cylinder walls is limited to 150 MPa.
SOLUTION
y
p
r
t
Total pressure at any depth:
p = 400(103 ) + γ (h − y )
(a) At y = h :
p = 400(103 ) Pa
h
x
Liquid
pressure
Thus,
pr 400(103 )5000
t=
=
= 13.33 mm
σ all
150(106 )
(b) At y = h 4 :
p = 400(103 ) + 15(103 )(15) = 625(103 ) Pa
t=
625(103 )5000
= 20.8 mm
150(106 )
(c) At y = h 2 :
p = 400(103 ) + 15(103 )(10) = 550(103 ) Pa
t=
550(103 )5000
= 18.3 mm
150(106 )
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PROBLEM (8.84) Solve Prob. 8.83, assuming that the gas pressure is p = 100 kPa.
SOLUTION
Total pressure at any depth:
p = 100(103 ) + γ (h − y )
(a) At y = h :
p = 100(103 ) Pa
t=
pr
σ all
=
100(103 )5000
= 3.3 mm
150 × 106
(b) At y = h 4 :
p = 100(103 ) + 15(103 )15 = 325(103 ) Pa
325(103 )5000
= 10.8 mm
150 × 106
(c) At y = h 2 : :
p = 100(103 ) + 15(103 )10 = 250(103 ) Pa
t=
t=
250(103 )5000
= 8.33 mm
150 ×106
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PROBLEM (8.85) Determine the required thickness t of a cylindrical vessel of 3.6 ft in diameter
under a pressure of p = 150 psi.
Given: A material strength of 25 ksi.
Assumption: A safety factor of ns = 1.5 will be used.
SOLUTION
The thickness for circumferential stress:
pr 150(1.8 × 12)
=
= 0.1944 in.
t=
σ all 25(103 ) 1.5
The thickness for axial stress:
0.1944
pr
=
= 0.0972 in.
t=
2σ all
2
13
Thus, treq = 0.1944 ≈
in.
64
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PROBLEM (8.86) An underwater vehicle having a living space that can be approximated as a sphere
of radius r and thickness t is to operate at depths h (Fig. P8.81). It is filled with air at pressure p . The
material of the sphere is steel with yield strength σ y . The factor of safety is to be ns . Determine the
required thickness t of the sphere.
Given:
r = 2.5 m,
p = 4 MPa,
Assumptions:
σ y = 600 MPa, h = 1000 m, ns = 2.5
Sea water weighs about γ = 10 kN/m3. Buckling of the vessel will not occur.
SOLUTION
σy
600
= 240 MPa
ns
2.5
The maximum pressure acting on the sphere is
p = pa − pw = pa − γ h
σ all =
=
= 4(106 ) − (10 ×103 )(1000) = −6 MPa
where the minus sign means outer pressure. Hence
−6(2.5)
pr
σ all = ;
− 240 =
t
t
This gives
treq = 0.0625 m = 62.5 mm
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PROBLEM (8.87) A spherical vessel of 1.6-m inner diameter is constructed by joining two
hemispheres with 40 equally spaced bolts (Fig. P8.82). The vessel will operate at an internal pressure of
600 kPa. Calculate the bolt diameter d and the vessel thickness t.
Given: The allowable stresses for the bolts and sphere wall equal 100 and 50 MPa, respectively.
SOLUTION
Free Body: Hemisphere
y
Initial tightening
of bolts are omitted.
pA
0.8 m
Fb
Fb
t
∑ F = 0 : 40 F = pA ;
y
Solving
We have
b
d = 19.6 mm
t=
pr
2σ all
=
π
40[100 × 106 ( d 2 )] = 600 ×103 [π (0.8) 2 ]
4
600 ×103 (800)
= 4.8 mm
2(50 × 106 )
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PROBLEM (8.88) A penstock, a pipe for conveying water (specific weight γ = 9.81 kN/m3 ) to a
turbine, operates at a head of 120 m. The penstock has a 0.9-m diameter and a wall thickness of t.
Determine the minimum required value of t for a material strength of 100 MPa.
Assumption: A safety factor of ns = 1.6 will be used.
SOLUTION
σt =
treq =
pr
t
pr
σ all
p =γh
=
γ hr 9.81(103 )(120)(450)
=
= 8.48 mm
100(106 ) 1.6
σ all
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PROBLEM (8.89) Redo Prob. 8.88 for the case in which the allowable stress is 80 MPa.
SOLUTION
treq =
pr
σ all
=
γ hr 9.81(103 )(120)(450)
=
= 6.62 mm
80(106 )
σ all
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PROBLEM (8.90) A closed cylindrical tank fabricated of 10-mm-thick plate is subjected to an
internal pressure of p = 6 MPa. Determine:
(a) The maximum diameter if the maximum shear stress is limited to 30 MPa.
(b) The limiting value of tensile stress, for the diameter found in part (a).
SOLUTION
σ1 − σ 2
pr pd
1 pr pr
= ( − )=
=
2
2 t
2t
4t
8t
6
8tτ
8 × 10 × 30 ×10
= 400 mm
d = max =
P
6 × 106
(a) τ max =
(b) σ 1 =
pr 6 × 106 × 200
=
= 120 MPa
t
10
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PROBLEM (8.91) A closed cylindrical vessel constructed of a thin plate is l.2 m in diameter and
subjected to an external pressure of 1 MPa. Calculate:
(a) The wall thickness t if the maximum allowable shear stress is set at 20 MPa.
(b) Tthe corresponding maximum principal stress.
σ 1 = pr t = 2σ 2
σ −σ
pr
τ max = 1 2 =
2
(a) t =
pr
4τ max
=
4t
1(106 )600
= 7.5 mm
4(20 × 106 )
pr 1(106 )600
(b) σ 1 =
=
= 80 MPa
t
7.5
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PROBLEM (8.92) The side wall of the cylindrical steel pressure vessel has butt-welded seams (Fig.
P8.92). The allowable tensile strength of the joint is 90 % of that of steel. Determine the maximum value
of the seam angle φ if the tension in the steel is to be limiting.
SOLUTION
σ 1 = 2σ
y
θ
x’
φ
σ2
θ = φ + 90o
x
y’
σ x ' = 0.9σ 1 =
σ 2 + σ1 σ 2 − σ1
2
+
2
3
1
cos 2θ = −0.6, θ = 63.4 o
0.9σ 1 = σ 1 − σ 1 cos 2θ
4
4
Thus,
φ = 63.4 − 90 = −26.6 o = 26.6 o
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PROBLEM (8.93) A cylindrical vessel of internal diameter 200 mm and wall thickness 2 mm has a
welded helical seam angle of φ = 60o (see: Fig. P8.92). The allowable tensile stress in the weld is 100
MPa. Determine, the maximum value of internal pressure p and the corresponding shear stress in the
weld, using:
(a) Equations (8.4).
(b) An approach based on the equilibrium equations applied to a wedge shaped stress element.
(c) Mohr’s circle.
SOLUTION
(a)
σ 1 = 50P
θ
x’
60o
σ2 =
pr 100 P
=
= 25 P
2t
2(2)
θ = 150o
x
y’
σ x ' = 100 =
τ x' y' = −
(b)
p
p
(25 + 50) + (25 − 50) cos 300o ,
2
2
p = 3.2 MPa
3.2
(25 − 50) sin 300o = −34.6 MPa = τ w
2
∑F = 0:
x'
100ΔA − 25ΔA(0.866) 2 p
− 50ΔA(0.5) 2 p = 0
or
p = 3.2 MPa
100 MPa
x’
τ
ΔA
60o
y’
25P
50P
∑F = 0:
y'
τΔA − 3.2[25ΔA(0.5 × 0.866) − 50ΔA(0.5 × 0.866)] = 0
or
τ = 34.6 MPa = τ w
Continued on next slide
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( c)
σ1 =
pr p (100)
=
= 50 p
t
2
σ2 =
σ1
2
= 25 p
τ
x’
o
60
120o
25p
O
25p
C
50p
σ (MPa)
R
50p
R =12.5p
σ x ' = 100 = 37.5 p − 12.5 p cos 60 o , p = 3.2 MPa
τ x ' y ' = −12.5 p sin 60o = −34.6 MPa = τ w
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PROBLEM (8.94) A closed cylindrical pressure vessel of inner diameter d and wall thickness t is
constructed with a helical welding that make an angle φ with the longitudinal axis as depicted in Fig.
P8.94. The vessel is subjected to an internal pressure of p. Determine:
(a) Tangential and axial stresses.
(b) The maximum in-plane shearing stress.
(c) The absolute maximum shear stress.
(d) The normal stress σ w and shear stress τ w acting on planes perpendicular and parallel to the weld.
Sketch the results on a properly oriented element.
Given: d = 2r = 4 ft,
t = ½ in.,
φ = 55o ,
p = 500 psi
SOLUTION
(a) Principal stresses
pr (500)(2 × 12)
=
= 24 ksi
σt =
12
t
σa =
σt
2
= 12 ksi
(b) In-plane maximum shear stress
1
τ max = (σ t − σ a ) = 6 ksi
2
(c) Absolute maximum shear stress
1
(τ max ) a = (σ t − 0) = 12 ksi
2
(d) σ x = σ a = 12 ksi
σ y = σ t = 24 ksi
τ xy = 0
For θ = −90 + 55 = −35 :
1
1
σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ = 18 − 4.10 = 13.9 ksi = σ w
2
2
1
τ x ' y ' = − (σ x − σ y ) sin 2θ = −5.2 ksi = τ w
2
o
y’
24 ksi
12 ksi
5.2 ksi
x
35o
Weld
13.9 ksi
x’
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PROBLEM (8.95) Redo Prob. 8.94 for a vessel (see Fig. P8.94) with the following given numerical
values:
d = 2r = 1.2 m,
φ = 40o ,
t = 12 mm,
p = 2 MPa
SOLUTION
(a) Principal stresses
pr (2 ×106 )(0.6)
=
= 100 MPa
σt =
t
0.012
σa =
σt
2
= 50 MPa
(b) In-plane maximum shear stress
1
τ max = (σ t − σ a ) = 25 MPa
2
(c) Absolute maximum shear stress
1
τ max = (σ t − 0) = 50 MPa
2
(d) σ x = σ a = 50 MPa
σ y = σ t = 100 MPa
τ xy = 0
For θ = −90o + 40o = −50o :
1
1
σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ = 75 + 4.34 = 79.34 MPa = σ w
2
2
1
τ x ' y ' = − (σ x − σ y ) sin 2θ = −24.6 MPa = τ w
2
σ y ' = σ x + σ y − σ x ' = 91.1 MPa
y’
100 MPa
24.6 MPa
50 MPa
50o
Weld
58.9 MPa
x
x’
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PROBLEM (8.96) A steel boiler of inner diameter d and thickness t is welded using a spiral seam
that makes an angle of φ with respect to axial axis (see Fig. P8.94). The boiler is subjected to an intrnal
pressure of p. Determine:
(a) The normal stress σ w perpendicular to the weld and shear stress τ w parallel to the weld.
Sketch the results on a properly oriented element.
(b) The absolute maximum shear stress in the boiler.
Given: d = 2r = 1.5 m,
t = 15 mm, φ = 60o ,
p = 800 kPa
SOLUTION
σx = σa
σ y = σt
τ xy = 0
θ = −90 + 60 = −30o
Principal stresses:
pr (800 × 103 )(750)
=
= 40 MPa
σt =
t
15
σa =
σt
= 20 MPa
2
1
1
(a) σ w = (σ x + σ y ) + (σ x − σ y ) cos 2θ
2
2
y’
40 MPa
1
1
= (20 + 40) + (20 − 40) cos(−60) = 25 MPa
2
2
20 MPa
1
τ w = − (σ x − σ y ) sin 2θ
x
2
30o
25 MPa
1
= − (20 − 40) sin(−60) = −8.66 MPa
Weld 8.66 MPa
2
x’
1
40
(b) (τ max ) a = (σ t − 0) =
= 20 MPa
2
2
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PROBLEM (8.97) A cylindrical tank of inner diameter d and wall thickness t has butt welded
helical seam forming an angle φ with the longitudinal axis, as shown in Fig. P8.97. The allowable
normal stress in the weld equals σ w . Determine the largest internal pressure p that can be applied in the
tank, using Mohr’s circle.
Given: d = 2r = 1.6 ft,
φ = 55o ,
t = 8 mm,
σ w = 10 ksi
SOLUTION
pr
t
σt =
σa =
σt
θ = 55o
2
σ t = 2σ a =
τ
y’
σavg
O
C
σa
1
2
R
σt
σ
1
1
R = (σ t − σ a ) = σ t
2
4
σ w = σ x ' = σ avg − R cos 70o
3
1
= σ t − σ t cos 70
4
4
pr
= 0.664σ t = 0.664
t
x’
10(103 ) = 0.664
3
4
σ avg = (σ t + σ a ) = σ t
110o
Introducing the data:
pr
t
p (1.6 ×12)
,
0.008
p = 6.28 ksi
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PROBLEM (*8.98) A cylindrical tankof inner diameter d is constructed with a steel plate having
thickness t welded along a helix forming an angle φ with the longitudinal axis (Fig. P8.97). Calculate
the internal pressure p that will cause a shear stress τ w parallel to the weld, using Mohr’s circle.
Given:
d = 2r = 600 mm,
*SOLUTION
pr
σt =
t
τ
σa =
t = 10 mm,
σt
φ = 70o ,
θ = −70o
2
σavg
σa
O
τ w = 40 MPa
σ t = 2σ a =
C
σt
R
140o
1
2
3
4
σ avg = (σ t + σ a ) = σ t
1
1
R = (σ t − σ a ) = σ t
2
4
x’
τ w = R sin 40o = 0.161σ t = 0.161
σ
pr
t
pr
t
Therefore,
40(106 ) = 0.161
p(300)
,
10
p = 8.28 MPa
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PROBLEM (8.99) A cylindrical pressure tank of inner diameter d is fabricated by shaping steel
plates of thickness tc and welding the plates along helical arcs of angle φ with a transverse plane
(Fig.P8.99). The end caps are spherical and have uniform wall thickness t s . The maximum internal
pressure in the tank is p. Calculate:
(a) The normal and absolute maximum shear stresses in the caps.
(b) The normal and absolute maximum shear stresses in the cylinder.
(c) The normal stress σ w and shear stress τ w acting on planes perpendicular and parallel to the weld.
Sketch the results on a properly oriented element.
Given: d = 2r = 2.4 m, tc = 15 mm,
ts = 10 mm,
φ = 30o ,
p = 1.4 MPa
SOLUTION
(a) Stress in the sphere
pr 1.4(1.2)
σ=
=
= 84 MPa
2t 2(0.01)
1
σ
(τ max ) a = (σ − 0) = = 48 MPa
2
2
(b) Principal Stresses
pr 1.4(1.2)
σt =
=
= 112 MPa
t
0.015
σa =
σt
2
= 56 MPa
Absolute maximum shear stress
1
(τ max ) a = (σ t − 0) = 56 MPa
2
(c) σ x = σ a = 56 MPa
σ y = σ t = 112 MPa
τ xy = 0
For θ = 30 :
1
1
σ w = σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ = 84 − 14 = 70 MPa
2
2
1
τ w = τ x ' y ' = (σ x − σ y ) sin 2θ = 24.2 MPa
2
o
y’
x’
70 MPa
Weld
30o
x
24.2 MPa
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PROBLEM (8.100) A pressure vessel of inner radius r and wall thickness t is fabricated from a
welded pipe with helix angle φ and equipped with two end plates, as shown in Fig. P8.100. If pressure
inside the vessel is p, determine:
(a) The normal stress σ w perpendicular to the weld.
(b) The shear stress τ w parallel to the weld.
Given:
r = 4 ft,
φ = 38o ,
t = ½ in.,
p = 200 psi
SOLUTION
σt =
pr
t
σa =
pr
2t
θ = −90 + 38 = −52o
Principal stresses
pr (200)(4 ×12)
=
= 19.2 ksi
σt = σ y =
12
t
σa = σx =
σt
2
= 9.6 ksi
τ (ksi)
σavg
104o
σa
O
C
x’
α = 180 − 2θ = 76o
R
α
σ avg = (σ t + σ a ) = 14.4 ksi
1
2
σt
R
σ(ksi)
1
R = (σ t − σ a ) = 4.8 ksi
2
(a) σ w = σ x ' = σ avg + R cos α
= 14.4 + 4.8cos 76 = 15.56 ksi
(b) τ w = − R sin α = −4.8(sin 76o ) = −4.66 ksi
19.2 ksi
y’
9.6 ksi
Weld
4.66 ksi
76o
15.56 ksi
x
x’
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PROBLEM (8.101) A thick-walled, closed-ended cylinder of inner radius a and outer radius b is
subjected to an internal pressure p only (Fig. P8.101). The cylinder is made of a material with
permissible tensile strength σ all and shear strength τ all . Calculate the allowable value of pi.
Given: a = 0.8 m,
b = 1.2 m,
σ all = 100 MPa,
τ all = 60 MPa
SOLUTION
Axial Stress, Eq. (8.32):
0.82 pi
pi a 2
σa = 2 2 = 2
= 0.8 pi = 100 ,
pi = 125 MPa
1.2 − 0.82
b −a
Tangential stress, Eq. (8.30a):
b2 + a 2
1.22 + 0.82
(σ t ) max = 2
pi = 38.5 MPa
=
p
pi = 2.6 pi = 100 ,
i
1.22 − 0.82
b − a2
Shear Stress, Eq. (8.31b):
pi b 2
1.22
τ max = 2 2 = 2
pi = 60
b − a 1.2 − 0.82
Solving
pi = 33.3 MPa = pall
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PROBLEM (8.102) A thick-walled steel cylindrical tank of internal radius a is subject to an internal
pressure pi (Fig. P8.101). The ultimate strength of the material in tension and compression is equal to
σ all . factor of safety is to be ns . Calculate the wall thickness.
Given: a = 3 ft,
p = 2 ksi ,
σ all = 40 ksi,
ns = 1.5
SOLUTION
The maximum normal stress is given by Eq. (8.30a).
Thus, with a = 3 ×12 = 36 in.:
σu
a2 + b2
ns
b2 − a 2
Substituting the given numerical values
40(103 )
362 + b 2
= (2 × 103 ) 2
,
b = 38.81 in.
1.5
b − 362
Hence
t = b − a = 38.81 − 36 = 2.81 in.
= (σ t ) max = pi
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PROBLEM (8.103) A thick-walled cylindrical tank of inner radius a and outer radius b is made of
ASTM A-48 cast iron (see Table B.4) having an ultimate strength in tension
σ u , modulus of elastcity
E, and Poisson’s ratio of ν . Determine the maximum radial displacement of the tank, if it is subjected to
an internal pressure of pi , as shown in Fig. P8.101.
σ u = 170 MPa, E = 70 GPa,
ν = 0.3
Given: a = 0.5 m,
b = 1 m,
pi = 10 MPa,
SOLUTION
The maximum radial displacement umax occurs at the inner edge of the tank.
So, Eq. (8.29) for r=a, results in
ap
a 2 + b2
+ν )
umax = i = ( 2
E
b − a2
Introducing the given data
(0.5) 0.52 + 1.02
(
umax =
+ 0.3) = 0.14(10−3 ) m = 0.14 mm
70(109 ) 1.02 − 0.52
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PROBLEM (*8.104) A cylindrical thick-walled pipe having an inner radius a and outer radius b
is subjected to an internal pressure pi (Fig. P8.101). Determine:
(a) The ratio of the wall thickness to the inner radius, if the internal pressure is one-half of the maximum
tangential stress: pi = ½ ( σ t )max
.
(b) The increase in inner radius of the pipe, if a = 2 ft, pi =1.2 ksi, E = 30 x 106 psi, and ν = 0.3.
*SOLUTION
(a) From Eq. (8.30a),
(σ t ) max a 2 + b 2
= 2
=2,
b = 1.732a
pi
b − a2
Then
b a+t
t = 0.732a
=
= 1.732;
a
a
So
t 0.732a
=
= 0.732
a
a
(b) Using Eq. (8.29), with r = a = 2 ×12 = 24 in. and b = 1.732a = 41.57 in. :
api a 2 + b 2
+ν )
(
u=
E b2 − a 2
24(1.2 ×103 ) 242 + 41.57 2
(
=
+ 0.3) = 2.21(10−3 ) in.
6
2
2
30 × 10
41.57 − 24
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PROBLEM (8.105) A thick-walled cylinder of inner radius a and outer radius b is subjected to an
internal pressure pi , as shown in Fig. P8.101. Determine:
(a) The maximum tangential and radial stresses in the cylinder.
(b) The minimum tangential and radial stresses in the cylinder.
(c) Tha axial stress and the maximum shear stress in the cylinder.
Given: a = 25 mm,
b = 75 mm,
pi = 35 MPa
SOLUTION
(a) Through the use of Eq. (8.30a) and Eq. (8.27) at r=a (see Fig. 8.27):
b2 + a 2
752 + 252
(σ t ) max = pi 2
35
=
= 43.75 MPa
752 − 252
b − a2
(σ r ) max = − pi = −35 MPa
(b) By Eq Eq. (8.30a) and Eq. (8.27) at r=b (see Fig. 8.27):
2a 2
2(252 )
(σ t ) min = pi 2
35
=
= 8.75 MPa
752 − 252
b − a2
(σ r ) min = 0
(c) From Eqs. (8.32) and 8.31b):
pi a 2
(35)(252 )
σa = 2 2 = 2
= 4.375 MPa
75 − 252
b −a
p b2
(35)(752 )
τ max = 2 i 2 = 2
= 39.375 MPa
75 − 252
b −a
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PROBLEMS (8.106 through 110) The state of strain at particular points in a structure is given in
the table below. Determine the state of strain associated with axes x’ and y’ rotated through the specified
angle θ (Figs. P8.106 through P8.110). Use Eqs. (8.35).
------------------------------------------------------------------------------------Problem
εx
εy
εz
θ
------------------------------------------------------------------------------------800 μ
1500 μ
30o
8.106
-400 μ
8.107
500 μ
1200 μ
-1000 μ
-45o
8.108
200 μ
410 μ
100 μ
40o
8.109
-300 μ
420 μ
200 μ
10o
-200 μ
240 μ
-30o
8.110
-720 μ
--------------------------------------------------------------------------------------
SOLUTION (8.106)
−400 + 800 −400 − 800
cos 60o + 750sin 60o
+
2
2
= 200 − 600(0.5) + 750(0.866) = 550 μ
ε y ' = 200 + 300 − 650 = −150 μ
ε x' =
γ x ' y ' = −( −400 − 800) sin 60o + 1500 cos 60 o = 1789 μ
SOLUTION (8.107)
500 + 1200
+ 0 − 500sin(−90o )
2
= 850 + 500 = 1350 μ
ε y ' = 850 − 500 = 350 μ
ε x' =
γ x ' y ' = −(500 − 1200) sin( −90 o ) + 0 = −700 μ
Continued on next slide
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SOLUTION (8.108)
200 + 410 200 − 410
cos80o + 50sin 80o
+
2
2
= 305 − 18 + 49 = 336 μ
ε y ' = 305 + 18 − 49 = 274 μ
ε x' =
γ x ' y ' = −( 200 − 410) sin 80 o + 100 cos 80 o = 224 μ
SOLUTION ( 8.109)
−300 + 420 −300 − 420
cos 20o + 100sin 20o
+
2
2
= 60 − 338 + 34 = −244 μ
ε y ' = 60 + 338 − 34 = 364 μ
ε x' =
γ x ' y ' = −( −300 − 420) sin 20 o + 200 cos 20 o
= 246 + 188 = 434 μ
SOLUTION ( 8.110)
−720 − 200 −720 + 210
cos(−60o ) + 120sin(−60o )
+
2
2
= −460 − 130 − 104 = −694 μ
ε y ' = −460 + 130 + 104 = −226 μ
ε x' =
γ x ' y ' = −( −720 + 200) sin( −60 o ) + 240 cos( −60 o )
= −330 μ
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PROBLEM (8.111) Using Mohr's circle, redo Prob. 8.106.
SOLUTION
−400 + 800
= 200 μ
2
7.5
= 51.34o
α = tan −1
6
β = 180 − 60 − 51.34
ε'=
γ
2 (μ)
200
y’
y
R
O
α
x
(-400, -750)
ε (μ)
C
60o
x’
= 68.66 o
R = 6002 + 7502 = 960.5 μ
ε x ' = 200 + 960.5 cos 68.66o
= 200 + 350 = 550 μ
ε y ' = 200 − 350 = −150 μ
γ x ' y ' = 2(960.5 sin 68.22 o )
= 1789 μ
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PROBLEM (8.112) Using Mohr's circle, redo Prob. 8.107.
SOLUTION
γ
2 (μ)
x
(500, 500)
α
O
500 + 1200
= 850 μ
2
5
α = tan −1
= 55o
3.5
β = 180 − 90 − 55
= 35o
ε'=
β
C
y’
ε'
x’
ε (μ)
R
y
R = 5002 + 3502 = 610 μ
ε x ' = 850 + 610 cos 35o
= 850 + 500 = 1350 μ
ε y ' = 850 − 500 = 350 μ
γ x ' y ' = −2(610 sin 35o ) = −700 μ
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PROBLEM (8.113) Using Mohr's circle, redo Prob. 8.108.
SOLUTION
200 + 410
= 305 μ
2
50
α = tan −1
= 25.5o
105
β = 180 − 25.5 − 80
= 74.5o
ε'=
γ
2 (μ)
ε'
y’
y
R
α
O
x
(200, -50)
C
80o
(410, 50)
β
x’
ε (μ)
R = 1052 + 502 = 116 μ
γ x ' y ' = 2(116 sin 74.5o ) = 224 μ
ε x ' = 305 + 116 cos 74.5o = 336 μ
ε y ' = 305 − 116 cos 74.5o = 274 μ
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PROBLEM (8.114) Using Mohr's circle, redo Prob. 8.109.
SOLUTION
ε'=
−300 + 420
= 60 μ
2
α = tan −1
10
= 15.5o
36
γ
2 (μ)
ε'
α
o
x 20
(-300, -100)
x’
y’
y
O
C
R
ε (μ)
R = 3602 + 1002 = 373.6 μ
γ x ' y ' = 2[373.6 sin( 20o + 15.5o )]
= 434 μ
ε x ' = −373.6 cos 35.5o + 60
= −304 + 60 = −244 μ
ε y ' = 304 + 60 = 364 μ
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PROBLEM (8.115) Using Mohr's circle, redo Prob. 8.110.
SOLUTION
γ
2 (μ)
σ'
x’
β
α
R
y
60o
O
C
x
(-720, -120)
y’
ε (μ)
720 + 200
= −460 μ
2
12
α = tan −1 = 24.8o
26
β = 60 − 24.8 = 35.5o
ε'=
R = 2602 + 1202 = 286.4 μ
γ x ' y ' = −2( 286.4 sin 35.2 o ) = −330 μ
ε x ' = −286.4 cos 35.2 o − 460 = −234 − 460 = −694 μ
ε y ' = 234 − 460 = −226 μ
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PROBLEMS (8.116 through 120) The state of strain at a point in a thin steel plate is given in the
following table. Calculate:
(a) The in-plane principal strains and the maximum in-plane shear strain.
(b) The absolute or true maximum shearing strain (ν = 0.3).
Sketch the results found in part (a) on properly oriented deformed elements.
--------------------------------------------------------------------------------------Problem
εx
εy
γ xy
--------------------------------------------------------------------------------------100 μ
200 μ
8.116
400 μ
8.117
-900 μ
400 μ
-200 μ
8.118
-720 μ
0 μ
300 μ
8.119
200 μ
600 μ
600 μ
-100 μ
150 μ
8.120
500 μ
----------------------------------------------------------------------------------------
SOLUTION (116)
400 + 100
= 250 μ
2
1
10
θ p ' = tan −1 = 16.8o
2
15
θ s " = 16.8 + 45 = 61.8o
ε'=
(a)
γ
2 (μ)
ε'
y
R
O ε2
ε1
C
2θ p '
ε (μ)
x
(400, -100)
y’
R = 1002 + 1502 = 180.3 μ
γ max = 2 R = 361 μ
ε 1 = 250 + 180.3 = 430 μ
ε 2 = 250 − 180.3 = 70 μ
x’
430 μ
y’
70 μ
250 μ
x’
61.8o
16.8o
x
250 μ
360 μ
x
Continued on next slide
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(b) Using Eq. (8.2)
0.3
(400 + 100) = −214 μ
ε3 = ε z = −
0.7
(γ max ) a = 430 + 214 = 644 μ
SOLUTION (117)
−900 + 400
= −250 μ
2
1
10
θ p " = tan −1 = 4.4o
2
65
θ s " = 4.4 + 45 = 49.4o
ε'=
(a)
γ
2 (μ)
ε'
(-900, -100)
x
R
ε 2 2θ " C
p
y’
O
ε1
y
ε (μ)
R = 1002 + 6502 = 657.6 μ
γ max = 2 R = 1315 μ
ε 1 = 657.6 − 250 = 408 μ
ε 3 = −657.6 − 250 = −908 μ
908 μ
x’
408 μ
4.4o
x’
x
250 μ
250 μ
49.4o
1316 μ
x
(b) Using Eq. (8.2):
0.3
(−900 + 400) = 214 μ
ε2 = εz = −
0.7
Thus,
(γ max ) a = 1315 μ
Continued on next slide
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SOLUTION (8.118)
−720
= −360 μ
2
1
15
θ p " = tan −1 = 11.3o
2
36
θ s ' = 45 − 11.3 = 33.7o
ε'=
(a)
γ
2 (μ)
ε'
y
ε3
R
C
(-720, -150)
x
y’
2θ p "
O ε2
ε (μ)
750 μ
y
30 μ
11.3o
(b) ε1 = ε z = −
Thus,
R = 1502 + 3602 = 390 μ
γ max = 2r = 780 μ
ε 2 = 390 − 360 = 30 μ
ε 3 = −390 − 360 = −750 μ
x
x’
x’
360 μ
360 μ
780 μ 33.7o
x
0.3
(−720) = 308 μ
0.7
(γ max ) a = 308 + 750 = 1058 μ
Continued on next slide
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SOLUTION (8.119)
(a)
ε'=
γ
2 (μ)
ε'
y
R
ε2
O
2θ p "
C
ε1
ε (μ)
200 + 600
= 400 μ
2
R = 2002 + 3002 = 361 μ
1
3
θ p " = tan −1 = 28.2o
2
2
θ s " = 45 + 28.2 = 73.2o
γ max = 2 R = 722 μ
ε1 = 400 + 361 = 761 μ
ε 3 = 400 − 361 = 39 μ
(200, -300)
x
400 μ
y’
761 μ
722 μ
x
28.2o
39 μ
(b) ε 3 = ε z = −
Thus,
x’
y’
73.2o
x
400 μ
x’
0.3
(500 − 100) = −343 μ
0.7
(γ max ) a = 761 + 343 = 1104 μ
Continued on next slide
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SOLUTION (8.120)
500 − 100
= 200 μ
2
1
75
θ p ' = tan −1
= 7o
2
300
θ s " = 45 + 7 = 52o
ε'=
(a)
γ
2 (μ)
ε'
y
ε2 O
y’
ε1
R
C
2θ p '
x ε (μ)
(500, -75)
R = 752 + 3002 = 309 μ
γ max = 2 R = 618 μ
ε1 = 200 + 309 = 509 μ
ε 2 = −390 + 200 = −109 μ
509 μ
x’
109 μ
220 μ
x’
7o
(b) ε 3 = ε z = −
Thus,
x
200 μ
618 μ
52o
x
0.3
(500 − 100) = −172 μ
0.7
(γ max ) a = 509 + 172 = 681 μ
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_______________________________________________________________________
PROBLEM (8.121) A 60 x 80-mm rectangular aluminum plate of modulus elasticity E and
Poisson’s ratio ν is subjected to the uniform stresses shown on Fig. P8.121. Determine:
(a) The deformation of diagonal AC.
(b) The deformation of diagonal BD.
ν = 1/3
Given: E = 210 GPa,
SOLUTION
∠DAC = θ = tan −1
6
= 36.9o
8
LAC = LBD = 602 + 802 = 100 mm,
G=
3
E
= E
2(1 + 1 3) 8
1
150 350(106 )
(300 +
)=
3
E
E
1
300
250(106 )
)=−
ε y = (−150 −
3
E
E
6
100 800(10 )
γ xy =
=
3E
G
(a) ε x =
Taking x ' along AC:
γ
1
1
ε x ' = (ε x + ε y ) + (ε x − ε y ) cos 73.8o + sin 73.6o
2
2
2
6
10
= (50 + 83.7 + 128)
= 261.71(106 ) E
E
ΔLAC = 100
261.71(106 )
= 124.6 ×10−3 mm
9
210(10 )
(b) θ 1 = 180 − 36.9 = 143.1o
Taking x ' along BD:
ε x ' = (50 + 300 cos 286.2o + 133.2sin 286.2o )
= (50 + 83.7 − 128)
ΔLBD = 100
106
E
106
= 5.7(106 ) E
E
5.7(106 )
= 2.7 × 10−3 mm
210(109 )
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PROBLEM (8.122) An element at a point on a loaded frame has strains as follows:
ε x = 600 μ
ε y = 1000 μ
γ xy = -1400 μ
Calculate:
(a) The principal strains.
(b) The maximum shear strain.
(c) The absolute maximum shear strain.
SOLUTION
γ
2 (μ)
1
2
1
= (600 + 1000) = 800 μ
2
ε avg = (ε x + ε y )
600
R
O
ε2
C
ε1
ε (μ)
800
R=
(800 − 600) 2
+ 7002
2
= 728 μ
(a) ε1,2 = ε avg ± R
ε1 = 800 + 728 = 1528 μ
ε 2 = 800 − 728 = 72 μ
(b) γ max = 2 R = 2(728) = 1456 μ
(c) (γ max ) a = ε1 − ε 3 = ε1 − 0 = 1528 μ
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PROBLEM (*8.123) At a point A on the surface of a vessel made of structural steel, the strains
ε x ' and ε y ' are measured in the x’ and y’ directions oriented at an angle θ to the x and y axes,
respectively (Fig. P8.123). Calculate:
(a) The components of strain ε x , ε y , and γ x ' y ' .
(b) Poisson’s ratio ν for the steel.
Given:
ε x ' = 250 μ ,
ε y ' = 400 μ , γ xy = 0, θ = 35o
Assumption: The stresses remain below yield strength of the material.
*SOLUTION
(a) γ xy = −(ε x ' − ε y ' ) sin(−2θ o ) + γ x ' y ' cos(−2θ o )
0 = −(250 − 400) sin(−70o ) + γ x ' y ' cos(−70o ) ,
γ x ' y ' = 412.1 μ
Then
1
1
1
2
2
2
1
1
1
= (250 + 400) + (250 − 400) cos(−70o ) + (412.1) sin(−70o )
2
2
2
= 105.8 μ
1
1
1
ε y = (ε x ' + ε y ' ) − (ε x ' − ε y ' ) cos(−2θ o ) − γ x ' y ' sin(−2θ o )
2
2
2
1
1
1
= (250 + 400) − (250 − 400) cos(−70o ) − (412.1) sin(−70o )
2
2
2
= 544.3 μ
ε x = (ε x ' + ε y ' ) + (ε x ' − ε y ' ) cos(−2θ o ) + γ x ' y ' sin(−2θ o )
(b) Hooke’s Law, with ε t = ε y , ε a = ε x , and σ t = 2σ a
εx =
So
σx
(1 − 2ν )
εy =
σx
(2 −ν )
E
E
ε y 544.3
2 −ν
=
= 5.145 =
,
ν = 0.34
1 − 2ν
ε x 105.8
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_______________________________________________________________________
PROBLEM (*8.124) A steel shaft of radius r is loaded by a torque T, as shown in Fig. P8.124. A
strain gage is placed at point A measures the strain ε φ at an angle of φ to the axis of the shaft. What
is the value of the torque T ?
Given: r = 2 in.,
G = 11.5 x 106 psi ,
ε φ = 500 μ ,
φ = 25o
*SOLUTION y
τ xy =
Tc
J
J=
π
2
c3 γ xy =
x
τ
Mohr’s circle for strain:
1
ε avg = (ε x + ε y ) = 0
2
γ
2
1
R = γ xy
2
1
2
ε φ = ε avg + R sin 2θ = γ xy sin 2θ
This gives
T=
τ xy
2G
sin 2φ =
r=c
G
σ x = σ y = 0 ε x = ε y = 0 θ = φ = 25o
A
=
τ xy
Tc
sin 2φ
2GJ
x
A’
x’
2φ
R
O
ε
y
2GJ ε φ
c sin 2φ
Substitute the given data:
π (11.5 ×106 )(2)3 (500 × 10−6 )
= 94.3 kip ⋅ in.
T=
(2) sin 50o
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_______________________________________________________________________
PROBLEM (8.125) A metallic plate of width a and thickness t is subjected to a uniform axial
force P (Fig. P8.125). Two strain gages placed at point A measures the strains ε φ and
at 30o and 60o, respectively, to the axis of the plate. Determine:
(a) The normal strains ε x and ε y
(b) The normal strains
(c) The shear strain
ε x ' and ε y ' .
γ x' y' .
For axial loading, take γ xy = 0.
Assumption:
Given: a = 50 mm,
t = 5 mm,
E = 210 GPa,
ν = 0.3,
P = 20 kN
SOLUTION
(a) σ x =
20(103 )
P
=
= 80 MPa
A (0.05 × 0.005)
σx
80(106 )
= 381 μ
E 210(109 )
ε y = −νε x = −(0.3)(381× 10−6 ) = −114.3 μ
εx =
=
1
1
(b) ε x ' = (ε x + ε y ) + (ε x − ε y ) cos 2θ
2
2
o
For θ = −30 :
1
1
ε x ' = (381 − 114.3) + (381 + 114.3) cos(−60o ) = 257 μ
2
2
1
1
ε y ' = (381 − 114.3) − (381 + 114.3) cos(−60o ) = 9.5 μ
2
2
(c) For θ = −30o :
γ x ' y ' = −(ε x − ε y ) sin 2θ
= −(381 + 114.3) sin(−60o ) = 429 μ
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_______________________________________________________________________
PROBLEM (8.126) Show that for plane strain, ε x + ε y = ε x ' + ε y ' = ε1 + ε 2 .
SOLUTION
Adding Eqs. (8.35a) and (8.35b):
ε x' + ε y' = ε x + ε y
Similarly, addition of Eqs. (8.36) yields
ε x + ε y = ε1 + ε 2
Thus,
ε x' + ε y' = ε x + ε y = ε1 + ε 2
Q.E.D.
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_______________________________________________________________________
PROBLEM (8.127) The 60o strain rosette (Fig. P8.127) is mounted on the surface of an
automobile frame. The following readings are obtained during a static test:
ε a = 1,110 μ , ε b = 420 μ , ε c = -240 μ
Determine:
(a) The principal strains.
(b) The maximum shear strain.
Sketch the results obtained in (a) on a properly oriented deformed element.
SOLUTION
ε a = 1,110 μ
ε b = 420 μ
ε c = −240 μ
o
θa = 0
θb = −60
θ c = −120o
Thus
ε a = ε x cos 2 θ a + ε y sin 2 θ a + γ xy sin θ a cos θ a
1,110 μ = ε x cos 2 0o + ε y sin 2 0o + γ xy sin 0o cos 0o ,
ε x = 1,100 μ
Similarly,
420 μ = ε x cos 2 (−60o ) + ε y sin 2 (−60o ) + γ xy sin(−60o ) cos(−60o )
or
420 μ = 0.25ε x + 0.75ε y − 0.443γ xy
(1)
Likewise,
−240 μ = ε x cos 2 (−120o ) + ε y sin 2 (−120o ) + γ xy sin(−120o ) cos(−120o )
or
−240 μ = 0.25ε x + 0.75ε y + 0.443γ xy
Subtract Eq. (2) from Eq. (1):
660 μ = −0.866 γ xy ,
(2)
γ xy = −762 μ
Then Eq. (2) gives ε y = −250 μ
Continued on next slide
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(a)
γ
2 (μ)
430
ε2
O
R
1
2
1
= (1110 − 250) = 430 μ
2
ε avg = (ε x + ε y )
1110
2θ p'
C
381
ε1
ε (μ)
R = (1110 − 430) 2 + (381) 2
= 779.5 μ
ε1,2 = ε avg ± R
ε1 = 430 + 779.5 = 1210 μ
ε 2 = 430 − 779.5 = −350 μ
(b) γ max = ε1 − ε 2 = 1210 + 350 = 1560 μ
Sketch:
381
tan 2θ p ' =
,
θ p ' = 14.6o
1110 − 430
y
y’
350 μ
x
14.6o
1210 μ
x’
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_______________________________________________________________________
PROBLEM (*8.128) The strain gage readings from a 60o rosette mounted at a point A on a loaded
clamp (Fig. P8.128) are as follows:
ε a = 920 μ , ε b = 360 μ ,
ε c = -80 μ
Determine the principal strains. Show results on a properly oriented deformed element.
*SOLUTION
ε a = 920 μ (θ a = 0o ),
ε b = 360 μ (θ b = −60o ),
ε c = −80 μ (θ c = −120o )
Therefore
ε a = ε x cos 2 θ a + ε y sin 2 θ a + γ xy sin θ a cos θ a
920(10−6 ) = ε x cos 2 0o + ε y sin 2 0o + γ xy sin 0o cos 0o ,
ε x = 920 μ
Similarly,
ε b = ε x cos 2 θb + ε y sin 2 θ b + γ xy sin θb cos θ b
360(10−6 ) = ε x cos 2 (−60o ) + ε y sin 2 (−60o ) + γ xy sin(−60o ) cos(−60o )
360(10−6 ) = 0.25ε x + 0.75ε y − 0.443γ xy
(1)
and,
ε c = ε x cos 2 θ c + ε y sin 2 θ c + γ xy sin θ c cos θ c
−80(10−6 ) = ε x cos 2 (−120o ) + ε y sin 2 (−120o ) + γ xy sin(−120o ) cos(−120o )
−80(10−6 ) = 0.25ε x + 0.75ε y + 0.443γ xy
(2)
Subtract Eq. (2) from Eq. (1):
440 μ = −0.833 γ xy
So
γ xy = −528.2 μ
ε y = −105 μ
γ
2 (μ)
920
1
2
2θ p ' 264.1
1
O
C
ε (μ)
= (920 − 105) = 407.5 μ
R
2
y
528.2 2 12
R = [(920 − 407.5) 2 + (−
) ]
2
= 576.6 μ
407.5
A
x
ε avg = (ε x + ε y )
ε1,2 = ε avg ± R
ε1 = 407.5 + 576.6 = 984 μ
ε 2 = −169 μ
Continued on next slide
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Continued on next slide
tan 2θ p ' =
264.1
,
920 − 407.5
x’
θ p ' = 13.63o
x
13.63o
984 μ
y
y’
A
169 μ
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_______________________________________________________________________
PROBLEM (8.129) During a static test of an aircraft panel, a 45° rosette measures the following
normal strains on the free surface (Fig. P8.129):
ε a = -400 μ , ε b = -500 μ ,
ε c = 200 μ
Calculate the principal strains. Show the results on a properly oriented deformed element.
SOLUTION
Using Eqs. (8.44), we have
ε x = −400 μ ,
ε y = 200 μ ,
γ xy = 2( −500) − ( −400 + 200) = −800 μ
Equations (8.36) are thus,
−400 + 200
600 2
800 2
) = −100 ± 500
ε1,2 =
± (−
) + (−
2
2
2
ε 1 = 400 μ ,
ε 2 = −600 μ
−800
] = 26.6o
θ p = 12 tan −1[
−400 − 200
ε x ' = −100 − 300 cos 53.1o − 400 sin 53.1o = −600 μ = ε 2
Thus,
θ p " = 26.6o
y’
600 μ
400 μ
x’
26.6o
x
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_______________________________________________________________________
PROBLEM (8.130) Using a 60° rosette, we find the following strains at a critical point on the
surface of a loaded beam shown in Fig. P8.130:
ε a = -200 μ ,
ε b = -350 μ ,
ε c = -500 μ
Determine:
(a) The maximum in-plane shear strains and the accompanying normal strains.
(b) The true maximum shear strain. Use ν = 0.3.
Sketch the results found in part (a) on a properly oriented distorted element.
SOLUTION
Using Eq. (8.45):
1
3
ε x = −200 μ
γ xy =
ε y = [2(−850) + 200] = −500 μ
2
(−350 + 500) = 173 μ
3
(a) γ max = 2 (
300 2 173 2
) +(
) = 346 μ
2
2
1
2
ε ' = − (200 + 500) = −350 μ
1
2
θ s = tan −1[−
300
] = −30o
173
ε 2 = −350 + 173 = −177 μ ,
ε 3 = −350 − 173 = −523 μ
Check:
γ x ' y ' = −( −200 + 500) sin( −60 o ) + 173 cos( −60 o ) = 346 μ = γ max
Thus,
θ s ' = 30o
350 μ
x
30o
346 μ
x’
350 μ
0.3
(−700) = 300 μ
0.7
(γ max ) a = 300 + 523 = 823 μ
(b) ε z = ε1 = −
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_______________________________________________________________________
PROBLEM (8.131) Verify that, for a 45° rosette (Fig. P8.129), the principal strains are expressed
as follows:
ε1,2 =
εa + εc
2
1
1
± [2ε a (ε a − 2ε b ) + 2ε c (ε c − 2ε b ) + 4ε b2 ] 2
2
(8.131)
SOLUTION
Equation (8.44):
ε x = εa
ε y = εc
γ xy = 2ε b − (ε a + ε c )
Introducing the above into Eqs. (8.36), we have
ε +ε
ε −ε
2ε − ε − ε
ε1,2 = a b ± ( a b ) 2 + ( b a c ) 2
2
2
2
ε +ε 1 2
= a b±
ε a − 2ε aε c + ε c2 + 4ε b2 − 4ε bε a − 4ε bε c + ε a2 + ε c2
2
2
ε +ε 1
2ε a (ε a − 2ε b ) + 2ε c (ε c − 2ε b ) + 4ε b2 Q.E.D.
= a b±
2
2
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_______________________________________________________________________
PROBLEM (8.132) At a point on the free surface of a steel member of modulus of elasticity E and
Poisson’s ratio ν , subjected to plane stress, a 60° rosette (Fig. P8.132) measures the strains
ε a = 1200 μ ,
ε b = -650 μ ,
ε c = 500 μ
Determine:
(a) The principal strains and their directions.
(b) The corresponding principal stresses and the maximum shear stresses.
Sketch the results found in part (b) on a properly oriented deformed element.
SOLUTION
Equation (8.45):
1
3
ε x = 1200 μ
γ xy =
ε y = [2(−650 + 500) − 1200] = −500 μ
2
(−650 − 500) = −1328 μ
3
(a)
1
2
ε ' = (1200 − 500) = 350μ
γ
2 (μ)
ε'
x
R
ε2
O
(1200, 664)
C 2θ p ' ε1
ε (μ)
R = 6642 + 8502 = 1079 μ
γ max = 2r = 2158 μ
1
2
664
] = −19o = 19o
850
y
θ p ' = tan −1[−
ε 1 = 350 + 1079 = 1429 μ
ε 2 = 350 − 1079 = −729 μ
(b) Equation (3.16):
200(103 )
[1429 + 0.3(−729)] = 266 MPa
σ1 =
1 − 0.09
200(103 )
[−729 + 0.3(1429)] = −66 MPa
σ2 =
1 − 0.09
200(103 )
τ max =
(2158) = 166 MPa
2(1 + 0.3)
Check:
100 MPa
266 − (−66)
= 166 MPa
τ max =
2
We also have
266 − 66
= 100 MPa
σ '=
2
66 MPa
x
19o
x’
266 MPa
166 MPa
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_______________________________________________________________________
PROBLEM (8.133) Redo Prob. 8.132 for the case in which ε a = 400 μ , ε b = 500 μ , and
ε c = -700 μ .
SOLUTION
Equation (8.45):
1
3
ε x = 400 μ
γ xy =
ε y = [2(−200) − 400] = −267 μ
2
(500 + 700) = 1386 μ
3
1
2
ε ' = (400 − 267) = 67 μ
γ
(a)
2 (μ)
y
R
ε2
ε'
C
O
2θ p '
x
ε1
ε (μ)
R = 3332 + 6932 = 769 μ
1
693
= 32.2 o
θ p ' = tan −1
2
333
(400, 693)
ε 1 = 67 + 769 = 836 μ
ε 2 = 67 − 769 = −702 μ
γ max = 2r = 1538 μ
(b) Equation (3.16):
200(103 )
[836 + 0.3(−702)] = 137.5 MPa
0.91
200(103 )
[−702 + 0.3(836)] = −99.2 MPa
σ2 =
0.91
200(103 )
(1538) = 118.3 MPa
τ max =
2.6
σ1 =
137.5 MPa
x’
19.2 MPa
Check:
τ max =
137.5 + 99.2
= 118.3 MPa
2
We also have
137.5 + 99.2
= 19.2 MPa
σ '=
2
32.2o
99.2 MPa
x
118.3 MPa
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PROBLEM (8.134) At a point A on the surface of a loaded bracket, the strain readings are
ε a = -200 μ , ε b = -500 μ , ε c = -900 μ
θb = 120°, and θ c = 240° (see Fig. P8.134). Calculate:
for θ a = 0o,
(a) The in-plane principal strains.
(b) The in-plane maximum shear strains.
Show the results on properly oriented deformed elements.
SOLUTION
ε x = −200 μ
Equations (8.43) become
− 500 = ε x (0.25) + ε y (0.75) + γ xy ( −0.5 × 0.866)
(1)
− 900 = ε x (0.25) + ε y (0.75) + γ xy (0.866 × 0.5)
(2)
Subtract Eq. (2) from Eq. (1): γ xy = −462 μ
− 500 = −50 + 0.75ε y + 200 ,
Equation (1) yields then:
(a)
ε'=−
γ
(1200, 664)
2 (μ)
2θ p '
ε2
R
ε1 O
C
ε (μ)
ε
ε y = −866 μ
866 + 200
= −533 μ
2
R = 3332 + 2312 = 405 μ
1
231
= 17.4o
θ p ' = tan −1
2
333
ε 1 = −533 + 405 = −128 μ
ε 2 = −533 − 405 = −938 μ
(b) γ max = 2r = 810 μ
Check:
γ max = −128 + 938 = 810 μ
533 μ
128 μ
838 μ
x’
533 μ
x
o
17.4
810 μ
27.6o
x
x’
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PROBLEM (*8.135) The strain readings at a point on the free surface of a steel bracket subjected
to plane stress (see Fig. 8.134) are
ε a = 400 μ
ε b = 350 μ
εc
= 800 μ
for θ a = 0o, θb = 45°, and θ c = 135°. Calculate:
(a) The maximum in-plane shearing strains.
(b) The true maximum shear strain (ν = 1/3).
Show the results found in part (a) on properly oriented distorted element.
*SOLUTION
Equations (8.43): ε x = 400 μ
350 = ε x (0.5) + ε y (0.5) + γ xy (0.5)
(1)
800 = ε x (0.5) + ε y (0.5) + γ xy ( −0.5)
(2)
γ xy = −450 μ
Subtract Eq. (2) from Eq. (1):
Equation (1) yields
350 = 200 + 0.5ε y − 225 ,
ε y = 750 μ
(a)
γ
2 (μ)
ε'
(400, 225)
ε'=
x
ε3
O
ε2
2θ p "
10
R = 1752 + 2252 = 285 μ
ε1
C
R
400 + 750
= 575 μ
2
ε(μ)
y
1
225
= 26.1o
2
175
γ max = 2r = 570 μ ,
θ s " = 18.9o
θ p " = tan −1
ε 1 = 575 + 285 = 860 μ
ε 2 = 575 − 285 = 290 μ
Continued on next slide
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13
(400 + 750) = −575 μ
1−1 3
(γ max ) a = 860 + 575 = 1435 μ
(b) ε 3 = ε z = −
575 μ
570 μ
18.9o
575 μ
x
x’
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PROBLEM (9.1) A cylinder with external diameter D and internal diameter d is subjected to an
axial compressive load P and a torque T (Fig. P9.1). Calculate the maximum principal stress and the
maximum shearing stress. Show the results on a properly oriented element.
Given: D = 6 in., d = 4 in., P = 20 π kips, T = 15 π kip ⋅ in.
SOLUTION
State of stress on an element at the cylinder’s surface is
σx =
σx =
Tc
τ xy =
J
P
A
τ =−
−20π ×103
= −4 ksi
π (32 − 22 )
15π × 103 (3)
π
2
= −1.385 ksi
(3 − 2 )
4
4
Equation (9.1):
4
4
2
2
σ 1 = 0.433 ksi
σ 2 = −4.433 ksi
τ max = 2.433 ksi
Equation (9.3):
2(−1.385)
tan 2θ p =
,
2θ p = 34.7o
−4
Equation (4.4a) gives
1
1
σ x ' = (−4) + (−4) cos 34.7o − 1.385sin 34.7o
2
2
= −4.433 ksi
Thus
θ p " = 17.35o
σ 1,2 = − ± (− ) 2 + (−1.385) 2 = −2.433
2.433 ksi
4.433 ksi
17.35o
x
0.433 ksi
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_______________________________________________________________________
PROBLEM (9.2) A hollow outboard motor propeller shaft with an outer diameter D and inner
diameter d is simultaneously acted upon by a torque T and an axial thrust P (Fig. P9.1) . What is the
maximum shearing stress in the shaft?
Given: D = 8 in.
d = 4 in.
T = 400 kip ⋅ in. P = 250 kips
SOLUTION
A=
π
4
(82 − 42 ) = 37.7 in.2
J=
π
32
(84 − 44 ) = 377 in.4
250
P
=−
= −6.63 ksi
37.7
A
Tr 400(4)
τ xy = =
= 4.24 ksi
377
J
Thus
σx = −
τ max = (
τ xy
σx
σx 2
2
) + τ xy2
= (−3.315) 2 + (4.24) 2 = 5.38 ksi
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_______________________________________________________________________
PROBLEM (9.3) A shaft of diameter d is subjected to an axial compressive load (P) and two
torques (T1 , T2 ) as shown in Fig. P9.3. What is the maximum shear stress at point A on the surface of the
shaft?
Given: d = 120 mm,
P = 500 kN, T1 = 4 kN ⋅ m,
T2 = 6 kN ⋅ m
SOLUTION
At point A, T = 10 kN and P = 500 kN
4(500)
P
σx = − = −
= −44.21 MPa
A
π (0.12)2
16T
16(10)
τ xy = 3 =
= 29.47 MPa
πd
π (0.12)3
τ xy
A
σx
Hence
τ max = (
σx 2
2
) + τ xy2
= (−22.105) 2 + (29.47) 2
= 36.84 MPa = R
σ avg =
σx
2
τ
y
R
= −22.105 MPa
tan 2θ p " =
σ 2 2θp"
29.47
,
22.105
θ p = 26.56o
C
O
σavg
σ1
σ
x
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_______________________________________________________________________
PROBLEM (9.4) Determine the largest value of the axial load P that can be carried by the stepped
steel shaft shown in Fig. P9.4 for allowable bending and shear shear stresses σ all and τ all , respectively.
Given:
σ all = 100 MPa ,
τ all = 60 MPa,
T = 0.01P N ⋅ m.
SOLUTION
ACD =
J CD =
J AB =
π
4
(582 − 502 ) = 216π mm 2
π
(584 − 504 ) = 0.158π × 106 mm 4
32
π
(504 ) = 0.195π × 106 mm 4
32
Since J AB > J CD , the critical section is in CD. Thus,
σx =
σx =
τ=
TcCD
J CD
P
ACD
τ =−
=−
P
4630 P
=
−6
π
216π × 10
0.01P(29 ×10−3 )
0.158π × 10−6
1835P
π
Equation (9.1):
4630 P
4630 P 2 −1835 P 2
)
σ1 =
) +(
+ (
π
2π
2π
2315P 2954 P 5269 P
=
+
=
π
τ max =
π
π
2954 P
π
We have
100 × 106 =
60 × 106 =
5269 P
π
2954 P
π
,
,
P = 59.6 kN
P = 63.8 kN
Thus,
Pall = 59.6 kN
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_______________________________________________________________________
PROBLEM (9.5) Repeat Prob. 9.4 for the case in which the 50-mm-diameter hole in the shaft is
eliminated.
SOLUTION
Critical section is in AB.
σx =
τ=
4P
2
π d AB
σx =
4P
1600 P
=
2
π (0.05)
π
τ =−
16(0.01P)
1280 P
=−
3
π (0.05)
π
16T
3
π d AB
Equation (9.1):
800 P
800 P 2 −1280 P 2
)
σ1 =
) +(
+ (
π
=
π
π
800 P 1509 P 2309 P
+
=
π
π
π
100 × 106 =
2309 P
We have
60 × 106 =
π
1509 P
π
,
,
P = 136 kN
P = 125 kN
Thus,
Pall = 125 kN
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_______________________________________________________________________
PROBLEM (9.6) Determine the maximum shearing stress and its orientation in the stepped shown in
Fig. P9.4.
Given: P = 30π kN,
T = 0.01P kN ⋅ m
SOLUTION
See solution of Prob. 9.4:
−1835 P
2954 P
4630 P
τ max =
, τ=
, σx =
π
π
π
Thus,
τ max =
2954(30π × 103 )
= 88.6 MPa
π
1835(30π ×103 )
= −55.1 MPa
τ =−
π
4630(30π × 103 )
= 138.9 MPa
σx =
π
τ
(MPa)
(50.56, 23.4)
σ ' = 69.45 MPa
2θs"
σ (MPa)
C
R
88.6
tan 2θ s " =
69.45
55.1
θ s " = 25.8o
σ'
69.45 MPa
88.6 MPa
25.8o
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_______________________________________________________________________
PROBLEM (9.7) A rectangular shaft of depth a and width b is subjected to an axial tensile load P
and a torque T, as shown in Fig. P9.7. What is the largest permissible value of T if the allowable tensile
stress in the shaft is σ all ?
Given:
a = 3 in.,
b = 2 in.,
P = 30 kips,
σ all = 20 ksi
SOLUTION
P
30
=
= 5 ksi
A 3× 2
From Table 5.1, with a b = 1.5 :
T
T
τ=
=
= 0.3608T
2
α ab (0.23)(3)(2) 2
σ=
σ1 =
σ
(T in kip ⋅ in.)
σ
+ ( ) 2 + τ 2 = σ all
2
2
= 2.5 + (2.5) 2 + (0.3608Tall ) 2 = 20
or
Tall = 48 kip ⋅ in.
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_______________________________________________________________________
PROBLEM (9.8) A shaft with an a x b rectangular cross section is acted upon by an axial tensile load
P and a torque T (Fig. P9.7). Determine the largest permissible value of P if the allowable tensile
stress in the member is limited to σ all .
Given:
a = 100 mm,
b = 50 mm,
T = 5 kN ⋅ m,
σ all = 120 MPa
SOLUTION
By Table 5.1, with a b = 2 :
5(103 )
T
=
= 81.3 MPa
α ab 2 (0.246)(0.1)(0.05) 2
P
σ=
= 200 P MPa
( P in MN )
(0.1)(0.05)
Hence
τ=
σ1 =
σ
σ
+ ( ) 2 + τ 2 = σ all
2
2
= 100 P + (100 P) 2 + (81.3) 2 = 120
(120 − 100 P) 2 = 10, 000 P 2 + 6, 609.7
14, 400 − 24, 000 P + 10, 000 P 2 = 10, 000 P 2 + 6, 609.7
or
Pall = 0.3246 MN = 325 kN
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_______________________________________________________________________
PROBLEM (9.9) A turbine shaft of diameter d is under a thrust load P and a torque T (Fig. P9.9).
Calculate the largest permissible value of the load P if the allowable normal stress is not to exceed σ all .
Given:
σ all = 220 MPa
T = 2 kN ⋅ m,
d = 60 mm,
SOLUTION
y
z
C
A
Maximum normal stress
occurs at a point A.
d
We have
16T 16(2 ×103 )
τ= 3=
= 47.16 MPa
πd
π (0.06)3
4P
4P
σ= 2=
= 353.7 P (P in MN)
πd
π (0.06)2
Hence
σ1 =
σ
σ
+ ( ) 2 + τ 2 = σ all
2
2
220 = 176.8 P + (176.8P) 2 + (47.16) 2
(220 − 176.8P) 2 = 31, 258 P 2 + 2, 224
or
48, 400 − 77, 792 P + 31, 258 P 2 = 31, 258 P 2 + 2, 224
Solving,
P = 0.594 MN = 594 kN
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_______________________________________________________________________
PROBLEM (9.10) A thin-walled cylindrical vessel of radius r and thickness t is acted upon by an
internal pressure p as well as an axial compressive load P applied to the vessel through the rigid end
plates as shown in Fig. P9.10. What is the magnitude of P needed to produce pure shear in the cylinder
wall ?
SOLUTION
We have d=2r. For pure shear,
σ 1 = −σ 2
pr
pr
P
=−
+
2t 2π rt
t
Solving
σ 1 = pr t
σ2 =
pr
P
−
2t 2π rt
P = 3π pr 2
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_______________________________________________________________________
PROBLEM (9.11) A thin-walled cylindrical tank of radius r and thickness t is subjected to an
internal pressure p, axial compression P , and a torque T applied to the tank through the rigid end plates
(Fig. P9.11). Calculate the maximum principal stress in the cylinder wall.
Given: d = 10 in,
t = 0.2 in.,
p = 400 ksi, P = 5 kips, T = 250 kip ⋅ in.
SOLUTION
d = 2r
A = 2π rt
J = 2π r 3t
y
σy
Stresses are:
P pr
σx = − +
x
A 2t
5
0.4(5)
=−
+
= 4.2 ksi
2π (5)(0.2) 2(0.2)
pr 0.4(5)
σy =
=
= 10 ksi
t
0.2
250
Tr
T
τ xy = − = −
=−
= −7.96 ksi
2
J
2π r t
2π (5) 2 (0.2)
Maximum principal stress is then
4.2 + 10
4.2 − 10 2
) + (−7.96) 2 = 15.57 ksi
σ1 =
+ (
2
2
σx
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_______________________________________________________________________
PROBLEM (9.12) A closed-ended cylinder of radius r and wall thickness t is subjected to an
internal pressure p and an axial tension P. Determine for the shell wall:
(a) The maximum tensile stress.
(b) The maximum shearing stresses and the orientation of their planes.
Given: r = 10 in.,
t = ½ in.,
p = 500 psi,
P = 100 kips
SOLUTION
A = 2π rt = 2π (10)(0.5)
= 31.42 in.2
500(10)
σt =
= 10 ksi
0.5
(a)
pr
σt =
t
σa =
P pr
+
A 2t
100 ×103
+ 5 = 8.18 ksi
σa =
31.42
1
(b) τ max = (10 − 8.18) = 0.91 ksi
2
1
2
σ ' = (8.15 + 10) = 9.08 ksi
9.08 ksi
0.91 ksi
45o
9.08 ksi
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_______________________________________________________________________
PROBLEM (9.13) A thin-walled cylinder of radius r and wall thickness t is acted upon by an internal
pressure p and a torque T. Calculate the largest shearing stress and its orientation in the wall of this
closed-ended vessel.
Given: r = 250 mm,
t = 10 mm,
p = 3 MPa, T = 40 kN ⋅ m
SOLUTION
J = 2π r 3t = 2π (250)3 (10) = 981.7 × 106 mm 4
σt =
pr
t
σx =
τ xy =
τ
(MPa)
3 × 106 (250)
= 37.5 MPa
2(10)
σ t = 2σ x = 75 MPa
σx =
pr
2t
Tc
J
τ=
σ ' = 56.25 MPa
(37.5, 10.19)
2θs"
σ (MPa)
C
O
−40 ×103 (250 ×10−3 )
= −10.19 MPa
981.7 ×10−6
R
(75, -10.19)
σ'
R = (18.75) 2 + (10.19) 2
= 21.34
tan 2θ s " =
18.75
,
10.19
θ s " = 30.7o
τ max = 21.34 MPa
30.7o
21.34 MPa
56.25 ΜPa
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_______________________________________________________________________
PROBLEM (9.14) Calculate the normal and shearing stresses on the spiral weld of the steel shaft of
diameter d acted upon by an axial load P and a torque T as shown in Fig. P9.14.
Given: P = 200 kN, T = 2 kN ⋅ m,
d = 50 mm,
φ = 60°
SOLUTION
θ = α + 90 = 60 + 90 = 150 o
x'
P
σx =
A
x
θ
τ xy =
y'
σx =
200 × 103
π
= 102 MPa
2
(0.05)
4
16(2 × 10−3 )
τ=
= 81.5 MPa
π (0.05)3
16T
πd3
Equations(8.4):
σ w = σ x' =
102 102
cos 300o + 81.5cos 300o = 147.1 MPa
+
2
2
and
τ w = τ x' y' = −
102
sin 300o − 81.5cos 300o = 44.2 − 40.75 = 3.45 MPa
2
x’
150 o
147.1 MPa
x
3.45 MPa
y’
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_______________________________________________________________________
PROBLEM (9.15) A cantilever steel shaft of diameter d is subjected to an axial load P and a torque
T as shown in Fig. P9.14. Determine the permissible value of T. The allowable normal and shear stresses
on the spiral weld are σ all and τ all , respectively.
Given:
σ all = 20 ksi,
τ all = 12 ksi,
d = 4 in.,
φ = 50o ,
P = 40π kips
SOLUTION
θ = α + 90 = 50 + 90 = 140 o
x'
P
σx =
A
θ
x
y'
τ xy =
16T
πd3
σx =
40π ×103
π
= 10 ksi
2
(4)
4
−16T
τ=
= −0.0796T ksi
π (4)3
Equations(8.4):
10 10
+ cos 280o − 0.0796T sin 280o
2 2
T = 180 kip ⋅ in.
= 5 + 0.8682 + 0.0784T ,
10
τ x ' y ' = −12 = − sin 280o − 0.0796T cos 280o
2
T = 1225 kip ⋅ in.
= 4.924 − 0.01382T ,
σ x ' = 20 =
Thus,
Tall = 180 kip ⋅ in.
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_______________________________________________________________________
PROBLEM (9.16) A thin-walled cylindrical pressure vessel of radius r = 0.4 m and wall thickness
t = 8 mm has a welded spiral seam at an angle of φ =60o with the axial direction. The shell is subjected
to an internal pressure of p pascals and an axial tensile load of P = 10π kN applied through rigid end
plates. Calculate the permissible value p if the allowable normal and shear stresses acting simultaneously
in the plane of welding are not to exceed σ all = 25 and τ all = -10 MPa, respectively.
SOLUTION
60o
P
P
σt =
x'
θ
θ = 60 + 90 = 150 o
A = 2π rt
pr
t
σx =
x
P pr
+
A 2t
σ y = σt =
p (400)
= 50 p
8
y'
10π × 103
p (400)
+
= 1.5625 ×106 + 25 p
2π (0.4 × 0.008)
2(8)
Equations(8.4):
p
p
σ x ' = 25 ×106 = (25 + 50) + (25 − 50) cos 300o
2
2
1.5625 6 1.5625 6
(10 ) +
(10 ) cos 300o
+
2
2
p = 0.763 MPa
1.5625 6
p
(10 ) sin 300o
τ x ' y ' = −10 × 106 = − (25 − 50) sin 300o −
2
2
p = 0.986 MPa
Thus, pall = 763 kPa
σx =
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PROBLEM (9.17) Rework Prob. 9.16 for the case in which the vessel is subjected to an axial
compression load P = -10 π kN.
SOLUTION
Refer to solution of Prob. 9.16. We now have
σ x = −1.5625(106 ) + 25 p
σ y = 50 p
Equations (8.4):
σ x ' = 25(106 ) =
or
p = 838 kPa
τ x ' y ' = −10 ×106 = −
p
p
(25 + 50) + (25 − 50) cos 300o
2
2
6
1.5625(10 ) 1.5625(106 )
cos 300o
−
−
2
2
1.5625(106 )
p
(25 − 50) sin 300o +
sin 300o
2
2
Solving,
p = 866 kPa
So
pall = 836 kPa
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PROBLEM (9.18) A helical spring of R = 4-in. mean radius is fabricated of d = 1-in.-diameter bar
having an allowable shear stress of τ all = 40 ksi (see Fig.9.2a). Determine the largest value of applied
load P using:
(a) Equation (9.4).
(b) Equation (9.5).
SOLUTION
d
16 PR
(1 +
)
3
4R
πd
16(4) Pall
1
40 ×103 =
(1 +
),
3
4× 4
π (1)
(a) τ max =
2(4)
=8
1
16 Pall R 4m − 1 0.615
+
(
)
τ max =
π d 3 4m − 4
m
16(4) Pall 31 0.615
40 ×103 =
( +
),
8
π (1)3 28
Pall = 1.85 kips
(b) m =
Pall = 1.66 kips
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PROBLEM (9.19) Redo Prob. 9.18 for the case in which d = ¼ in. and R = 2 ½ in.
SOLUTION
d
16 PR
(1 +
)
3
4R
πd
16(2.5) Pall
0.25
40 ×103 =
(1 +
),
3
4 × 2.5
π (0.25)
(a) τ max =
Pall = 47.9 lb
16 PR 4m − 1 0.615
5(2.5)
m=
+
(
),
= 20
3
m
0.25
π d 4m − 4
16(2.5) Pall 39 0.615
Pall = 45.9 lb
40 ×103 =
( +
),
10
π (0.25)3 36
(b) τ max =
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PROBLEM (9.20) A helical brass spring (G = 35 GPa) having a number of coils N is subjected to
a load P = 800 N (see Fig. 9.6a). Calculate:
(a) The maximum shearing stress in the spring using Eqs. (9.4) and (9.5).
(b) The deflection of the spring.
(c) The spring constant.
Given: N = 15, d = 10 mm,
R = 40 mm
SOLUTION
Using Eq.(9.4):
16(800)(0.04)
10
(1 +
) = 173.2 MPa
(a) τ max =
3
4 × 40
π (0.01)
Equation (9.5):
m = 80 10 = 8
16(800)(0.04) 31 0.615
( +
) = 192.9 MPa
τ max =
28
8
π (0.01)3
(b) δ =
64 NPR 3 64(15)(800)(0.04)3
=
= 141 mm
d 4G
(0.01) 4 (35 × 109 )
( c) k =
P
δ
=
800
= 5.7 kN m
0.141
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PROBLEM (9.21) Rework Prob. 9.20 for a helical spring made of copper (G = 41 GPa) with an R
of 30 mm.
SOLUTION
(a) Equation (9.4):
16(800)(0.03)
10
(1 +
) = 132.4MPa
τ max =
3
120
π (0.01)
Equation (9.5):
m = 60 10 = 6
16(800)(0.03) 23 0.615
( +
) = 153.1 MPa
τ max =
20
6
π (0.01)3
64 NPR 3 64(15)(800)(0.03)3
(b) δ =
=
= 50.1 mm
d 4G
(0.01) 4 (41× 109 )
( c) k =
P
δ
=
800
= 16 kN m
0.0501
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PROBLEM (9.22) A steel helical spring fits within a brass helical spring (see Fig. 9.7). Each spring
has ends constrained to deflect an identical amount. Calculate:
(a) The total permissible load P the two springs can sustain jointly.
(b) The ratio of the spring rates.
Given: The properties of each spring are as follows.
---------------------------------------------------------------------------------------------------------Outer (brass) spring
Inner (steel) spring
---------------------------------------------Mean diameter, 2R
200 mm
140 mm
Wire diameter, d
20 mm
20 mm
Number of coils, N
10
10
Modulus of rigidity, G
40 GPa
80 GPa
Allowable shear stress, τ all
150 MPa
250 MPa
-----------------------------------------------------------------------------------------------------------
SOLUTION
(a) δ b = δ s
64 NPb Rb3 64 NPs Rs3
;
=
d 4Gb
d 4Gs
Pb Rb3 Ps Rs3
=
Gb
Gs
Substitute the given data:
Pb (0.1)3 Ps (0.07)3
Ps = 5.83Pb
,
=
40 × 109
80 × 109
Assume steel controls:
16 Pb (0.07)
20
(1 +
)
τ s = 250 ×106 =
3
4 × 70
π (0.02)
Ps = 5.24 kN
Pb = 0.9 kN
Check steel spring's stress:
16(0.9 ×103 )(0.1)
20
τb =
(1 +
)
3
π (0.02)
4 × 100
= 60.2 MPa < 150 MPa O.K.
Thus, Ptotal = 5.24 + 0.9 = 6.14 kN
Pb d 4Gb
Ps d 4Gs
k
=
,
s
64 NPs Rs3
δ b 64 NPb Rb3
Therefore,
k s Gs Rb3 80 100 3
) = 5.83
=
= (
kb Gb Rs3 40 70
(b) kb =
Pb
=
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PROBLEM (9.23) Determine the maximum normal stress and the maximum shear stress at a point A
of the aluminum bar of diameter d loaded as shown in Fig. P9.23.
Given: a = 5 ft, d = 2 in.,
R = 40 lb, P = 20 kips, T = 6 kip ⋅ in.
SOLUTION
4(20 ×103 ) 32(5 × 12 × 40)
+
π (2) 2
π (2)3
= 6.37 + 3.06 = 9.43 ksi
σx =
A
σx =
4 P 32M
+
πd2 πd3
−16(6 ×10 )
τ = 16T π d 3
= −3.82 ksi
3
π (2)
Equation (9.1):
9.43
9.43 2
σ1 =
+ (
) + (−3.82) 2 = 4.715 + 6.068 = 10.78 ksi
2
2
τ max = 6.068 ksi
τ=
3
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PROBLEM (9.24) Resolve Prob. 9.23 for the case in which the axial load is compressive, P = - 20
kips.
SOLUTION See solution of Prob.9.23:
A
σ x = −6.37 + 3.06 = −3.31 ksi
τ = 3.82 ksi
Equation (9.1):
3.31 2
−3.31
m (−
σ 1,2 =
) + (−3.82) 2 = −1.655 m 4.163
2
2
σ 1 = 2.51 ksi
σ 2 = −5.89 ksi
τ max = 4.16 ksi
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PROBLEM (9.25) A circular bar is fixed at one and loaded as illustrated in Fig. P9.23. Find the
required diameter d of the shaft.
Given: P = 0, R = 50 π N,
T = 120 π N ⋅ m,
L = 2 m,
σ all = 120 MPa,
τ all = 70 MPa
SOLUTION
D
T = 120π N ⋅ m
E
2m
σx =
D
32M
πd3
τ = 16T π d 3
50π N
32(2 × 50π ) 3200
= 3
πd3
d
1920
−16(120π )
τt =
=− 3
πd3
d
σx =
Equation (9.1):
120 × 106 =
1600
1600
−1920
+ ( 3 )2 + ( 3 )2
3
d
d
d
120 × 106 =
1
(1600 + 2499.2798),
d3
or
Equation (9.2):
70 × 10 6 = 2499.2798 d 3 ,
Similarly,
σx = 0
E
τ=
1920
+τ d
d3
d = 0.0325 m
d = 0.0329 m
From Example 9.4:
4V
VQ
τd =
=−
3A
Ib
4 50π
−267
=−
= 2
2
3 πd 4
d
Equation (9.1):
1
1920 276 2
− 2 ) = 3 (1920 + 267 d )
3
d
d
d
Solve by trial & error :
d = 0.0252 m.
Equation (9.2):
1
70 ×106 = 3 (0) 2 + (−1920 − 267d ) 2
d
Solve by trial & error :
d = 0.0302 m.
Thus,
d all = 25.2 mm
120 × 106 = 0 + (−
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PROBLEM (9.26) A hollow shaft with outer diameter D and inner diameter d is acted upon by an
axial tensile load P, a torque T, and a bending moment M, as portrayed in Fig. P9.26. Use Mohr’s
circle to determine the principal stresses and the plane on which they act.
Given: D = 50 mm,
d = 30 mm,
P = 40 kN, T = 500 N ⋅ m,
M = 200 N ⋅ m
SOLUTION
M
M
T
P
P
T
A
A = π ( 25 2 − 152 ) = 1256.6 mm 2
σx =
A
τ=
P Mc
+
A I
J=
π
(254 − 154 ) = 534 × 103 mm 4
2
I = J 2 = 267 × 10 3 mm 4
Tc
J
500(25 × 10−3 )
= −23.4 MPa
τ =−
534 × 10−9
40 × 103
200(25 × 10−3 )
+
= 31.83 + 18.73 = 50.56 MPa
σx =
1256.6 × 10−6
267 ×10−9
τ
(MPa)
(50.56, 23.4)
2θp’
σ2
σ1
C
R
σ (MPa)
R = (25.28) 2 + (23.4) 2
= 34.45 MPa
σ ' = 25.28 MPa
σ'
tan 2θ p ' =
23.4
, θ p ' = 21.4o
25.28
σ 1 = 25.28 + 34.45 = 59.73 MPa
σ 2 = 25.28 − 34.45 = −9.17 MPa
21.4 ο
59.73 MPa
A
9.17 MPa
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PROBLEM (9.27) Redo Prob. 9.26 for a D = 40 mm shaft made of solid steel and subjected to a zero
axial load.
SOLUTION
M
M
T
T
A
(31.83, 39.8)
σ ' = 15.92 MPa
2θp’
R = (15.92) 2 + (39.8) 2 = 42.87
σx =
A
16T
πd3
τ=
τ
(MPa)
σ'
σ2
32M
πd3
32(200)
= 31.83 MPa
π (0.04)3
−16(500)
τ=
= −39.8 MPa
π (0.04)3
Ο
σx =
σ (MPa)
σ1
C
R
σ 1 = 15.92 + 42.87 = 58.8 MPa
σ 2 = 15.92 − 42.87 = −27 MPa
tan 2θ p ' =
39.8
, θ p ' = 34.1o
15.92
27 MPa
34.1ο
58.8 MPa
A
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PROBLEM (*9.28) Determine the principal stresses and the maximum shearing stress at point D in
the L-shaped bracket of the uniform thin-walled cross section shown in Fig. P9.28.
*SOLUTION
x
Equation (5.41):
T
τ=
2 Amt
Am = 55 × 115
120 mm
T=1.5P
5 mm
M=0.5P
= 6.325 × 10 3 mm 2
P
D
60 mm
z
y
120(60)3 − 110(50)3
12
= 1.01 × 10 6 mm 4
Thus
1.5(600)
τ =−
2(6.325 ×10−3 )(0.005)
= −14.2 MPa
Iz =
σ
Mc −0.5(600)(0.03)
=
I
1.01× 10−6
= −8.9 MPa
σ =−
D
τ
Equation (9.1):
8.9
8.9 2
) + (−14.2) 2 = −4.45 ± 14.88
σ 1,2 = −
± (−
2
2
σ 1 = 10.4 MPa
σ 2 = −19.3 MPa
τ max = 14.9 MPa
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PROBLEM (9.29) A steel shaft. d = 2 in. in diameter and rotating at n = 2400 rpm is acted upon by
a bending moment of M = 5 kip-in. Calculate the torque and the horsepower that can also be applied
simultaneously to the shaft.
Given: τ all = 9 ksi,
σ all = 14 ksi
SOLUTION
Mc 32 M 32(5 × 103 )
σx =
=
=
= 6.37 ksi
πd3
π (2)3
I
τ
(ksi)
(637, τ )
R
C
6.37 ksi
O
τ=
3.185
Tc
J
σ1
σ
(ksi)
9
For Mohr's circle:
τ = 9 2 − 3.1852 = 8.42 ksi
σ 1 = 3.185 + 9 = 12.185 ksi < 14 ksi
O.K.
Thus,
T=
τJ
=
τπ c3
=
8.42 ×103 (π )(1)3
= 13.23 kip ⋅ in.
2
2
c
Formula (5.26)
TN
13.23 × 103 (2400)
=
= 504 hp
P=
63, 000
63, 000
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PROBLEM (9.30) Determine the maximum principal stress and the maximum shearing stress for a d
= 50-mm-diameter shaft in equilibrium shown in Fig. P9.30.
Assumption: The bearings at A and B are taken to be frictionless; they act as simple support.
SOLUTION
4 kN
A
0.4 m
0.6 m
RA = 2.4 kN
∑T = 0 :
B
RB = 1.6 kN
T = 2 × 10 3 (0.2) = 400 N ⋅ m
M c = M max = 2.4 × 10 3 (0.4) = 960
σx =
C
τ=
Equation (9.1):
32 M
16 M 2
16T
16
) + (− 3 ) 2 =
(M ± M 2 + T 2 )
σ 1,2 =
± (
3
3
2π d
πd
πd
πd3
16
=
(960 ± 9602 + 4002 ) = (39.114 ± 42.373)10 6
π (0.05)3
Thus,
σ 1 = 81.5 MPa
σ 2 = −3.26 MPa
τ max = 42.4 MPa
32M
πd3
16T
πd3
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PROBLEM (9.31) A solid shaft of diameter d = 40 mm is subjected to a bending moment of M =
200π N ⋅ m. Determine the torque T that can also act if σ all = 120 MPa and τ all = 80 MPa.
SOLUTION
σx =
τ=
32M 32(200π )
=
= 100 MPa
π d 3 π (0.04)3
16T
= 79.577 × 10−3 T MPa
πd3
Equation (9.1):
100
100 2
120 =
) + (79.577 × 10−3 T ) 2
+ (
2
2
4900 = 2500 + 6333 × 10 −6 T 2 ,
T = 616 N ⋅ m
Equation (9.3):
80 = 2500 + 6333 × 10 −6 T 2 ,
Thus,
Tall = 616 N ⋅ m
T = 785 N ⋅ m
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PROBLEM (9.32) Determine the maximum shearing stress at point E on the surface at midspan of
the D = 3 in.-diameter shaft of the assembly shown in Fig. P9.32. Bearings A and B are taken to be
frictionless simple supports.
SOLUTION
1.6 kips
1.6 kips
3 ft
A
1.5 ft C
D 1.5 ft
E
1.6 kips
B
1.6 kips
T = 800(6) = 4.8 kip ⋅ in.
σ=
M E = 2.4 kip ⋅ ft = 28.8 kip ⋅ in.
32M
πd3
τ = 16T π d 3
τ max = (σ 2) 2 + τ 2 =
=
16
M 2 +T2
3
πd
16
(28.8) 2 + (4.8) 2 = 5.51 ksi
π (3)3
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PROBLEM (9.33) Redo Prob. 9.32 for the case in which the shaft is hollow, with a D = 3-in.
outer diameter and a d = 2-in. inner diameter.
SOLUTION
See solution of Prob. 9.32:
T = 4.8 kip ⋅ in.
M E = 28.8 kip ⋅ in.
We now have
16
τ max =
M E2 + T 2
3
3
π (d o − di )
16
=
(28.8) 2 + (4.8) 2 = 7.83 ksi
3
3
π (3 − 2 )
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PROBLEM (9.34) A circular shaft is supported on frictionless bearings and loaded as
illustrated in Fig. P9.34. What is the smallest permissible shaft diameter d if the
allowable shear stress is limited to τ all ?
Given:
a = 6 in.,
T = 5 kip ⋅ in.,
r = 4 in.,
τ all = 10 ksi
SOLUTION
I=
πd4
64
J = 2I =
πd4
32
∑ T = 0 : T = Fr ,
F=
x
T 5
= = 1.25 kips
r 4
F
C
z
A
B
a
T
(kip ⋅ in.)
∑F = 0:
z
F
= 0.625 kips
2
M C = 0.625(6) = 3.75 kip ⋅ in.
RB = RD =
3.75
x
5
Critical section is at C.
x
T (d 2) 16T
=
J
πd3
σ
16
Hence τ max = τ all = ( ) 2 + τ 2 =
M 2 +T2
3
πd
2
or
16
16
(3.75) 2 + (5) 2 = 1.471 in.
d all = 3
M 2 +T2 = 3
πτ all
π (10)
σ=
M (d 2) 32 M
=
I
πd3
x
RD
RB
M
(kip ⋅ in.)
D
a
τ=
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PROBLEM (9.35) Resolve Prob. 9.34 for the case in which the allowable normal stress is not to
exceed σ all .
Given:
a = 180 mm,
r = 100 mm,
T = 800 N ⋅ m,
σ all = 120 MPa
SOLUTION
T
800
∑ T = 0 : T = Fr , F = r = 0.1 = 8 kN
x
F
C
z
B
A
D
a
a
720
M
(N ⋅ m)
F
= 4 kN
2
M C = 4(0.18) = 720 N ⋅ m
RB = RD =
RD
RB
∑F = 0:
z
x
x
800
T
(N ⋅ m)
x
32M
πd3
Therefore
σ=
τ=
σ 1 = σ all =
Critical section is at C.
16T
πd3
σ
σ
+ ( )2 + τ 2
2
2
from which
d all = 3
16
πσ all
(M + M 2 + T 2 ) = 3
16
(720 + 7202 + 8002 )
6
π (120 ×10 )
= 0.0424 m = 42.4 mm
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PROBLEM (*9.36) The shaft ABC shown in Fig. P9.36 is driven by a motor, which delivers
through the gears 30 kW at a speed of n = 500 rpm to a machine connected to the shaft DE. Calculate
the required diameter d of shaft ABC if the allowable shear stress is not to exceed τ all .
Given:
a = 120 mm,
r = 100 mm.
τ all = 50 MPa,
n = 600 rpm
*SOLUTION
9550 P 9550(30)
=
= 477.5 N ⋅ m
12
600
Tangential force F at gear A:
T 477.5
F= =
= 4775 N
r
0.1
Bending moment at B:
M B = Fa = 4775(0.12) = 573 N ⋅ m
Mc M (d 2) 32 M
16T
σ=
=
=
τ= 3
4
3
I
π d 64 π d
πd
Hence
16
τ max = τ all = 3 M 2 + T 2
πd
or
16
d all = 3
πτ all M 2 + T 2
T=
16
5732 + 477.52
6
π (50 ×10 )
= 0.042.4 m = 42.4 mm
=3
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PROBLEM (9.37) A circular shaft AB, simply supported on bearings at its ends, is subjected to
torque (T) and two loads (F1 , F2 ) acting as shown in Fig. P9.37. The shaft has diameter d = 2 in.
Determine:
(a) The maximum tensile stress.
(b) The maximum shear stress.
SOLUTION
∑ F = 0 : 0 and ∑ M = 0 yield
y
A
RAy = 400 lb RBy = 200 lb
12
Mz
(kip ⋅ in.)
RAz = 150 lb RBz = 300 lb
6
A
C
My
(kip ⋅ in.)
2.25
B
x
4.5
x
A
D
B
M C = 122 + 2.252 = 12.2 kip ⋅ in.
M D = 4.52 + 62 = 7.5 kip ⋅ in.
T = 10 kip ⋅ in.
Critical section is at C.
We have
τ=
(a) σ =
16T 16(10)
=
= 6.366 ksi
π d 3 π (2)3
32M 32(12.2)
=
= 15.54 ksi
πd3
π (2)3
σ
(b) τ max = ( ) 2 + τ 2 = (7.77) 2 + (6.366) 2 = 10 ksi
2
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PROBLEM (9.38) If the allowable normal stress at a point D in the simple beam of Fig. P9.38 is
10 MPa, determine the maximum permissible value of P.
SOLUTION
From Example 9.5:
I = 3.6 × 10 −6 m 4
Q D = 33.75 × 10 −6 m 3
P
1m
D
1m
120 m
30 mm
P/2 0.75 m
P/2
25 mm
M D = 0.375P
VD = 0.5 P
Mc 0.375P(0.03)
σD =
=
= 3125 P
I
3.6 ×10−6
VQ 0.5 P(33.75 ×10−6 )
=
= 187.5 P
τD =
3.6 ×10−6 (0.025)
Ib
Equation (9.1):
3125 P
3125 2
) + (187.5) 2
+P (
2
2
= 1562.5 P + 1573.7 P = 3136.2 P
(σ 1 ) D = 10 ×106 =
or
Pall = 3.19 kN
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PROBLEM (9.39) An overhanging beam of rectangular cross section, 6 in. deep by 4 in. wide,
carries a uniform load of intensity w = 10 kips/in. as shown in Fig. P9.39. Determine the principal
stresses at point D. Sketch the results on a properly oriented element.
SOLUTION
∑ M = 0 : − 10(7.5)(3.75) + R (4.5) = 0 ,
A
R B = 62.5 kips
B
∑F = 0:
R A = 12.5 kips
y
y
10 kip/ft
3 in.
A
3 ft
D 1.5 ft B
3 ft
6 in.
z
C
D
62.5 kips
12.5 kips
1 in.
4 in.
V D = 10(3) − 12.5 = 17.5 kips
1
M D = 12.5(3) − (10)(3)3 = −7.5 kip ⋅ ft = −90 kip ⋅ in.
2
1
I = (4)(6)3 = 72 in.4
12
Q = 4(2)2 = 16 in.3
Therefore,
VQ 17.5 × 103 (16)
=
= 972 psi
τD =
72(4)
Ib
σD =
Mc −90 × 103 (1)
=
= −1250 psi
I
72
τ (psi)
625
D
σD
τD
σ2
2θp"
C
O
σ (psi)
σ1
R
(-1250, -972)
R = 6252 + 9722 = 1160
σ 1 = 1160 − 625 = 535 psi
σ 2 = −1160 − 625 = −1785 psi
972
θ p " = 28.6o
tan 2θ p " =
,
625
1785 psi
28.6o
D
535 psi
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PROBLEM (9.40) Repeat Prob. 9.39 for a beam of S 8 x 23 cross section. (Refer to Table B.9.)
SOLUTION
From solution of Prob.9.39:
V D = 17.5 kips
M D = −90 kip ⋅ in.
y
I = 64.9 in. 4
Q = 4.17 × 0.425 × 3.7875
1.575
= 1.575 × 0.441(
+ 2)
2
= 8.65 in. 3
0.441 in.
8 in.
C
D
z
2 in.
0.425 in.
4.171 in.
S 8x23
Hence,
VQ 17.5 ×103 (8.65)
=
= 5.29 ksi
τD =
64.9(0.441)
Ib
σD =
Mc −90 × 103 (2)
=
= −2.77 ksi
I
64.9
τ
(psi)
τD
D
σD
tan 2θ p " =
R
C
σ2
2θp"
(-2.77, -5.29)
R 2 = (1.385) 2 + (5.29) 2 ,
R = 5.468
σ 1 = 5.468 − 1.385 = 4.08 ksi
σ 2 = −5.468 − 1.385 = −6.85 ksi.
O
σ1
σ
(psi)
5.29
1.385
θ p " = 37.7o
1.385
6.85 ksi
D
37.7o
4.08 ksi
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PROBLEM (9.41) For a beam loaded as shown in Fig. P9.41, by taking into account the beam's
own weight of 500 N/m, calculate the maximum normal and the maximum shearing stresses at point D
of the cross section at midspan. Show the results on a properly oriented element.
SOLUTION
∑ M = 0 : R (2) − 0.5(2)1 − 20(0.5) = 0 ,
∑ F = 0 : R = 15.5 kN
A
R B = 5.5 kN
B
y
A
y
20kN
0.5 kN/m
80
A
15.5 kN
C
0.5 m
z
B
1.5 m
5.5 kN
160
D
50
80
V, kN
15.5
15.25
1m
-4.75
Vm=5
-5.5
x
Dimensions
are in
millimeters
Mm = 5.52+5 (1)=5.25
M, kN ⋅ m
x
1
(80)(160)3 = 27.31×106 mm 4
12
Q = 80 × 50 × 55 = 220 × 10 3 mm
I=
τD =
VQ 5 ×103 (220 ×10−6 )
=
= 0.504 MPa
Ib 27.31× 10−6 (0.08)
Mc 5.25 ×103 (0.03)
σD =
=
= 5.77 MPa
I
27.31×10−6
τ
(MPa)
τD
D
σD
2.885
R = (2.885) 2 + (0.504) 2 = 2.93 MPa
σ2
σ1
R
O
C 2θ ’
p
σ (psi)
(5.77, -0.504)
Continued on next slide
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τ max = 2.93 MPa
σ 1 = 2.93 + 2.885 = 5.82 MPa
σ 2 = −2.93 + 2.885 = −0.05 MPa
0.05 MPa
and
tan 2θ p ' =
θ p ' = 4.95
0.504
2.885
2.93 MPa
D
5.82 MPa
4.95o
x
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PROBLEM (9.42) If the allowable shearing stress at a point B in the cantilever beam shown in Fig. P9.42 is 80 MPa, determine the largest permissible value of P.
SOLUTION
See Example 9.6:
Q B = 78.6 × 10 −6 m 3
c B = 69.7 mm
M = 2P
V = −P
I = 13.4 × 10 −6 m 4
t b = 6.6 mm
Mc 2 P(69.7 ×10−3 )
σB =
=
= 10, 403P
I
13.4 × 10−6
B
τB =
− P(78.6 ×10−6 )
VQ
=
= −888.7 P
Ib 13.4 × 10−6 (6.6 × 10−3 )
Equation (9.3):
or
10, 403 2
(τ B ) all = 80 ×106 = P (
) + (−888.7) 2 = 5276.9 P
2
Pall = 15.2 kN
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PROBLEM (9.43) A simply supported rolled-steel beam of S 250x52 section (see table B.9) is under
a uniform load of intensity w as depicted in Fig. P9.43. What is the maximum tensile stress at point D?
Given: a = 0.4 m,
L = 1.8 m,
w = 150 kN/m
SOLUTION
y
∑ F = 0 : R = R = 135 kN
y
A
B
At a section C:
V = 81 kN
1
M = 135(0.4) − (150)(0.4) 2
2
= 42 kN ⋅ m
12.5 mm
Table B.9:
I z = 61.2(106 ) mm 4
254
yD =
− 12.5 = 114.5 mm
2
Stress Components:
MyD 42 ×103 (0.1145)
σ=
=
= 78.6 MPa
Iz
61.2(10−6 )
z
C
d=254 mm
15.1 mm
D
126 mm
VQ 81×103 [0.126 × 0.0125(0.127 − 0.00625)]
=
61.2(10−6 )(0.0151)
I zb
= 16.67 MPa
τ=
σ1 =
σ
σ
+ ( )2 + τ 2
2
2
= 40.41 + (39.3) 2 + (16.67) 2 = 83.1 MPa
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PROBLEM (9.44) A simple metallic beam having a box cross section of uniform thickness t = 20
mm supports a concentrated load P as shown in Fig. P9.44. Determine the largest permissible value of P,
if the allowable compression stress at point D is not to exceed σ all = 250 MPa.
SOLUTION
Reaction at B, as found from equilibrium conditions (see Fig. P9.44):
RB = 0.375P
y
At a cross section through the point D:
V = 0.375 P
M = 0.375 P(0.6) = 0.225 P
D
Also
100 mm
20 mm
1
1
3
3
N.A.
I = (150)(200) − (110)(160)
z
12
12
C
100 mm
= 62.453(106 ) mm 4
Q = Ay = (150 × 20)(100 − 10)
150 mm
= 270(103 ) mm3
So
0.225 P(0.08)
My
σx = −
=−
= −288.217 P
62.453(10−6 )
I
τ xy =
0.375 P (270 × 10−6 )
VQ
=
= 40.53P
Ib 62.453(10−6 )(2 × 0.02)
Hence
σ1 =
σx
2
+ (
σx 2
2
) + τ xy2 = σ all
250(10 ) = −144.108 P + (−144.108 P) 2 + (40.53P ) 2
6
Solving
Pall = 44.7 kN
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PROBLEM (*9.45) A simply supported beam AB constructed of a W24 x 84 wide-flange section
(refer to Table B.8) is acted upon by three concentrated loads P = 40 kips (Fig. P9.45). Calculate the
maximum principal stress at the junction of a flange and web of the beam.
Assumption: The maximum shear stress is taken as uniformly distributed over the web area and
estimated as τ avg = Vmax / d tw by Eq. (7.25).
*SOLUTION
From Table B.8:
I 2 = 2370 in.4 d = 24.10 in. tw = 0.47 in. t f = 0.77
40 kips
A
40 kips
40 kips
2 ft
D 2 ft
RA =60 kips
60
V
kips
20
B
C
2 ft
E
2 ft
RB =60 kips
x
-20
160
M
(kip ⋅ ft)
120
-60
120
x
The reactions, as obtained from equations of statics, are indicated on the load
diagram.
At the junction of center C:
M (d 2 − t f ) (160 × 12)(11.28)
σ x = max
=
= 9.138 ksi
I2
2370
V
20
τ xy = τ avg = max =
= 1.766 ksi
(24.10)(0.47)
dt
σy = 0
R= (
σ1 =
σx 2
σx
2
2
) + τ xy2 = (4.569) 2 + (1.766) 2 = 4.898 ksi
+ R = 4.569 + 4.898 = 9.47 ksi
Continued on next slide
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At junction of point D (or E):
120
σx =
(9.138) = 6.854 ksi
160
τ xy =
40
(1.766) = 3.532 ksi
20
R = (3.427) 2 + (3.532) 2 = 4.921 ksi
σ1 =
σx
2
+ R = 3.427 + 4.921 = 8.35 ksi
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_______________________________________________________________________
PROBLEM (9.46) A structural member is loaded as illustrated in Fig. P9.46. Determine the
maximum tensile and compressive stresses for two cases:
(a) The eccentricity is a = 2h.
(b) The eccentricity is a = h/4.
SOLUTION
P
P
A
Pa/L
2L/3
C
h
B
L/3
Pa/L
P
b
I = bh3 12
A = bh
P
x
M
2Pa/3
x
-Pa/3
Just to the left of C (i.e., M = 2Pa 3 ), at the bottom & top edges, respectively:
P Mc P 4 Pa
P 4 Pa
σ bot = +
=
+ 2 ,
σ top = − 2
A I
bh bh
bh bh
Just to the right of C (i.e., M = 2Pa 3 ):
Mc
Mc
2 Pa
2 Pa
σ bot = −
=− 2 ,
σ top =
=− 2
I
bh
I
bh
(a) To the left of C, for a = 2h :
(σ bot )tens. = 9 P A
(σ top )comp. = − 7 p A
(b) To the left of C, for a = h 4 :
(σ bot )tems. = 2 P A
To the right of C, for a = h 4 :
(σ bot )comp. = − 0.5 P A
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PROBLEM (9.47) A closed-ended cylinder of r = 6 in. radius and t = ½ in. wall thickness
is subjected to an internal pressure of p = 500 psi and an eccentricity e = 4-in. off-center axial
load P (Fig. P9.47). Determine the largest permissible value of P if ( σ t ) all = 6 ksi and ( σ c ) all
= 4.2 ksi.
SOLUTION
1m
A
1m
P 4 in.
t
P
B
2r
Since r t ≥ 10 , cylinder is thin-walled:
J = 2π r 3t = 2 I
A
σ1 =
pr P Mc
+ +
2t A I
σ2 =
pr P Mc
+ −
2t A I
A = 2π rt
pr
t
B
We have
pr 500(6)
=
= 3 ksi
2t 2(0.5)
P
P
=
= 0.053P
A 2π (6)(0.5)
Mc
P(4)(6.5)
=
= 0.077 P
π (6.0)3 (0.5)
I
Tension at A:
6 ×103 = 3 × 103 + 0.053P + 0.077,
P = 23.08 kips
Compression at B:
−4.2 × 103 = 3 ×103 + 0.053P − 0.0779 P,
Thus,
Pall = 23.08 kips
P = 300 kips
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PROBLEM (9.48) A link with a T cross section is loaded as shown in Fig. P9.48.
Determine at section A-B
(a) The maximum stress.
(b) The location of the neutral axis.
SOLUTION
y
B
y
20
B
M
47
P
yo
P=48
70
A
20 mm
C
z
10
y
A
60
70 × 20(45) + 60 ×10(5)
= 33 mm
70 × 20 + 60 × 10
y 0 = 20 + y = 53 mm
y=
A = 20 × 70 + 60 × 10 = 2 × 10 3 mm 3
20(70)3
60(10)3
+ 20 × 70(12) 2 +
+ 60 × 10(28) 2
12
12
6
= 10 (0.572 + 0.202 + 0.005 + 0.47)
= 1.249 × 10 6 mm 4
M = 48 × 0.053 = 2.544 kN ⋅ m
I=
P 48 ×103
(a) σ a = =
= 24 MPa
A 2 × 10−3
Mc −2544(0.047)
(σ b ) B =
=
= −95.7 MPa
I
1.249 ×10−6
33
= 67.2 MPa
(σ b ) A = 95.7
47
σ A = 24 + 67.2 MPa
σ A = 24 + 67.2 = 91.2 MPa
σ B = 24 − 95.7 = −71.7 MPa
Continued on next slide
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(b) Apply Eq.(9.13):
2544 yn
P Myn
= 24 ×106 −
0= −
A
I
1.249 × 10−6
y n = 0.0118 m = 11.8 mm
σb
B
N.A
x
yn
A
σ
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PROBLEM (9.49) A cold-rolled beam of W 410 x 60 section (see Table B.8) supports an
axial tensile load P =400 kN, which does not pass through the centroid. What is the largest
eccentricity e y ?
Given:
ez = 0,
( σ t ) all = 120 MPa,
( σ c ) all = 80 MPa
SOLUTION
y
B
P
ey
P = 400 kN ,
x
ey
P
z
D
C
407 mm
A
M = 400ey
W410x60
At point B:
P Mc
σB = + B
A
I
400 ×103 400ey (0.2035)
+
80 ×106 =
7.61× 10−3
216 × 10−6
ey = 72.8 mm
80 = 52.6 + 376.9ey ,
At point A:
σ A = −120 = 52.6 − 376.9ey ,
A = 7.61× 10−3 m 2
I = 216 × 10−6 m 4
ey = 179 mm
Thus,
(ey ) max = 72.8 mm
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PROBLEM (9.50) For the bracket of b by h rectangular cross section, loaded as shown in Fig. P9.50,
determine:
(a) The maximum normal stress.
(b) The location of the neutral axis.
Given: P = 8 kips,
b = 1in., h = 4 in.
SOLUTION
1(4)3
= 5.33 in.4
12
P = 8 kips
M = 8 × 10 3 ( 2) = 16 kip ⋅ in.
(a) At the top fibers:
P Mc 8 × 103 16 × 103 (2)
σt = −
=
−
4
5.33
A I
= 2 − 6 = −4 ksi
A = bh = 1(4) = 4 in.2
I=
At the bottom fibers:
σ b = 2 + 6 = 8 ksi
(b) Equation (9.13):
0 = 2 × 103 −
y n = 0.667 in.
16 ×103 yn
5.33
σt
N.A
yn
x
σb
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PROBLEM (9.51) Determine the largest load P that the bracket of Fig. P9.50 can carry.
Given:
h = 6b = 6 in.,
σ all = 18 ksi
SOLUTION
A = bh = 1(6) = 6 in. 2
I = bh 3 12 = 1(6) 3 12 = 18 in. 4
At the bottom fibers:
P Mc P 3P (3)
σb = +
= +
A I
6
18
Solving,
18 × 10 3 = 0.1667 P + 0.5P
Pall = 27 kips
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PROBLEM (*9.52) A typical cantilever steel beam with modulus of elasticity E = 200 GPa is
loaded as shown in Fig. P9.52. Compute the magnitude of load P and distance d.
Given: The normal strains ε A = 600 μ and ε B = -200 μ occur at the extreme fibers at points A
and B.
*SOLUTION
1
(120 × 1603 − 100 × 1203 ) = 26.56 × 106 mm 4
12
A = 120(160) − 100(120) = 7.2 × 10 3 mm 2
I=
y
1.5 m
20
z
0.8 P
20
M
C
120
20
d
120
M = (0.8d − 0.9) P
0.8 P M (0.08)
σ A = Eε A =
+
A
I
0.8 P M (0.08)
σ B = Eε B =
−
A
I
0.8P
(1)
0.6P
(2)
(3)
Add Eqs. (2) and (3):
1.6 P
1.6 P
; 200 ×106 (400 × 10−6 ) =
E (ε A + ε B ) =
7.2 ×10−3
A
P = 360 kN
Subtract Eq.(3) from Eq.(2):
0.16M
0.16M
; 200 × 106 (400 ×10−6 ) =
E (ε A + ε B ) =
26.56 ×10−6
I
M = 26.56 kN ⋅ m
Thus, Eq.(1) yields
26.56 × 10 3 = (0.8d − 0.9)(360 × 10 3 )
or
d = 1.217 m
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PROBLEM (9.53) For the beam loaded as illustrated in Fig. P9.53, determine:
(a) The maximum bending stress.
(b) The location of the neutral axis.
Given:
P = 20 kN, d = 0.2 m
SOLUTION
See solution Prob.9.52:
M = (0.8d − 0.9) P = (0.8 × 0.2 − 0.9)20 × 10 3 = −14.8 kN ⋅ m
(a) Maximum stress occurs at B:
0.8 P M (0.08)
σB =
+
A
I
0.8(20 ×103 ) 14.8 × 103 (0.08)
=
+
7.2 × 10−3
26.56 ×10−6
σ B = 2.22 + 44.59 = 46.8 MPa
σ A = 2.22 − 44.59 = −42.37 MPa
(b) Use Eq.(9.13):
0.8P Myn
−
0=
;
A
I
y n = −3.99 mm
0.8(20 × 103 ) −14.8 × 103 yn
−
0=
7.2 ×10−3
26.56 ×10−6
B
σB
yn
x
N.A
σA
A
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PROBLEM (9.54) Figure P9.54 shows a cast-iron C clamp. What is the largest permissible
load P that may be applied?
Given: ( σ t ) all = 35 MPa, ( σ c ) all = 50 MPa,
h = 2b = 100 mm
SOLUTION
x
P
A
P
M
B
h
y
b
z
b=50 mm
h=100 mm
A = 50 × 100 = 5 × 103 mm 2
0.25 mm
σ A = 35 × 106 =
or
Iz =
50(100)3
= 4.17 × 106 mm 4
12
P Mc
P
0.25P(0.05)
+
=
+
−3
A Iz
5 ×10
4.17 × 10−6
35 = 0.0002 P + 0.003P,
P = 10.94 kN
σ B = −50 = 0.0002 P − 0.003P,
P = 17.86 kN
So,
Pall = 10.94 kN
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PROBLEM (9.55) The C clamp in Fig. P9.54 is tightened until P = 4 kN. If h = 1.6b = 80 mm.
calculate:
(a) The maximum normal stress.
(b) The location of the neutral axis.
SOLUTION
We now have: b = 50 mm, h = 80 mm
P = 4 kN
See the figure in solution of Prob.9.54:
1
I z = (50)(80)3 = 2.13 ×106 mm 4
A = 80 × 50 = 4 × 10 3 mm 2
12
M = 0.25P = 0.25( 4) = 1 kN ⋅ m
(a) σ A =
P Mc 4 × 103 1× 103 (0.04)
+
=
+
4 ×10−3 2.13 × 10−6
A Iz
= 1 × 10 6 + 18.76 × 10 6 = 19.78 MPa
σ B = 1 − 18.78 = −17.78 MPa
(b) Equation (9.13):
P Myn
1 −0.25 yn
0= −
;
0 = P( −
)
A Iz
A
Iz
I
2.13 × 10−6
= −2.13 mm
yn = − z = −
0.25 A
0.25(4 ×10−3 )
σA
y
x
N.A
B
A
σB
yn
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PROBLEM (9.56) Figure P9.54 shows a steel C clamp. Determine the required dimension b
of the cross section.
Given: P = 20 kN,
h = 60 mm,
σ all = 140 MPa
SOLUTION
See the figure in solution of Prob. 9.54:
b(0.06)3
= 18 × 10−6 b m 4
M = 0.25( 4 × 10 3 ) = 1 kN ⋅ m
A = 0.06b
Iz =
12
Maximum stress occurs at A. Thus,
P Mc 2 × 103 1×103 (0.03)
6
σ all = 140 ×10 = +
=
+
0.06b
18 × 10−6 b
A I
or
1
50
140 =
+
,
b = 12.14 mm
30b 30b
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PROBLEM (9.57) A 50-mm wide by 100-rnm deep bracket is acted upon by a load of P =
20 kN, as shown in Fig. P9.57. Determine the principal stresses and the maximum shearing stress
at point A. Show the results on a properly oriented element.
SOLUTION
0.32 m
y
100 mm
z
A
12 kN
M
50 mm
16 kN
0.03 m
B
0.09 m
12 kN
3
16 kN
4
Figure S9.57
A = 50(100) = 5 × 10 3 mm 2
I=
M = 12(0.09) + 16(0.32) = 6.2kN ⋅ m
1
(50)(100)3 = 4.17 × 106
12
P Mc
+
A I
−12 × 103 6.2 ×103 (0.03)
=
+
5 ×10−3
4.17 × 10−6
= (−2.4 + 44.6)106 = 42.2 MPa
σx = −
σx
A
τ
3V
3 16 ×103
τ =−
=−
= −4.8 MPa
2A
2 5 ×10−3
So,
42.2
42.2 2
) + (−4.8) 2
σ1 =
+ (
21.6 MPa
2
2
= 21.1 + 21.64 = 42.7 MPa
σ 2 = 21.1 − 21.64 = −0.54 MPa
τ max = 21.6 MPa
2τ 2(−4.8)
θ p ' = −6.4o
tan 2θ p ' =
=
,
σx
42.2
6.4o
A
42.7 MPa
0.54 MPa
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PROBLEM (9.58) Calculate the largest load P that can be applied to the bracket in Fig.
P9.57 if the allowable normal stress at point B is σ all = 90 MPa.
SOLUTION
See: Figure, S9.57 of Prob. 9.57:
M = Px (0.09) + Py (0.32) = 0.6 P(0.09) + 0.8P(0.32) = 0.31P
We have
or
σ all = σ x
−90 × 106 = −
P
0.31P(0.05)
−
−3
5 ×10
4.17 × 10−6
B
from which
Pall = 23 kN
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PROBLEM (9.59) Member ABC of the frame shown in Fig. P9.59 is made of a C 6 x 8.2 steel
channel (see Table B.10) having σ y =36 ksi. Determine the largest load P that can be
supported by the member, based upon a factor of safety of ns = 1.2. Neglect the effect of direct shear.
SOLUTION
∑ M = 0 : F = 2.357 P
∑ F = 0 : R = 1.667 P
∑ F = 0 : R = 0.667 P
A
BD
x
Ax
y
Ay
A
RAx
RAy
6 ft
45
y
P
B
z
4 ft
o
C
0.512 in.
FBD
N
1.667P
x
V
1.92 in.
C6x8.2
I = 0.692 in.4
A = 2.4 in.2
P
x
-0.667P
M
x
4P
Maximum stress at B occurs at the bottom of channel:
−36 ×103 N Mc 1.667 P 4(1.408) P
= −
=
−
σB =
1.2
2.4
0.692
A
I
or
− 30 × 10 3 = 0.659 P − 8.139 P
from which
Pmax = 4.03 kips
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PROBLEM (9.60) Determine the maximum normal stress in the member ABC of the
structure shown in Fig. P9.59 for P = 2 kips.
SOLUTION
See solution of Prob. 9.59:
1.667 P 4 P(1.408)
σ max = σ B =
−
2.4
0.692
For
P = 2 kips :
1.667(2) 4(2)(1.408)
σ max =
−
= 1.389 − 16.219 = −14.89 ksi
2.4
0.692
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PROBLEM (9.61) The cantilever beam with a rectangular cross section is subjected to loads
P, Q, and R as illustrated in Fig. P9.61. What is the maximum shearing stress τ max at point A?
Given:
P = 100 kN,
Q = 15 kN,
R = 10 kN
SOLUTION
At the section that includes points A and B, internal forces and moments (Fig. a):
P = 100 kN
Vy = 15 kN
Vz = 10 kN
M z = −Q(0.3 − 0.08) = −15(0.22) = −3.3 kN ⋅ m
80 mm
y
M y = R(0.3) = 10(0.3) = 3 kN ⋅ m
A
Cross-sectional properties :
A = 80 ×120 = 9600 mm 2
1
I z = (80)(120)3 = 11.52(106 ) mm 4
12
1
I y = (120)(80)3 = 5.12(106 ) mm 4
12
Mz
R
z
P Q
3 Vz 3 10 ×10
=
= 1.563 MPa
2 A 2 9600(10−6 )
τ xy
A
σx
3
Hence
τ max = (
B
My
Figure (a)
At point A (Fig. b):
100(103 ) 3.3(103 )(0.06)
P M y
σx = − z A =
+
9600(10−6 )
11.52(10−6 )
A
Iz
= 27.60 kip
τ xy =
120 mm
C
Figure (b)
σx 2
2
) + τ xy2 = (13.8) 2 + (1.563) 2 = 13.89 MPa
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PROBLEM (9.62) For the cantilever beam described in Prob. 9.61 and shown in Fig. P9.61,
determine the principal stresses at point A. Show the result on a properly oriented element.
SOLUTION
τ xy
Refer to the solution of Prob. 9.61.
The state of stress at point A is found as
σ x = 27.60 MPa
σx
27.6
σ avg =
=
= 13.8 MPa
2
2
R= (
2
) +τ
σx
τ xy = 1.563 MPa
From the corresponding
Mohr’s circle, we have:
σx 2
A
τ (MPa) σavg
y
σ2
R
2
xy
= (13.8) 2 + (1.563) 2 = 13.89 MPa
Thus
σ 1,2 = σ avg ± R
or
σ 1 = 13.8 + 13.89 = 27.7 MPa
σ 2 = 13.8 − 13.89 = −90 MPa
and
1.563
tan 2θ p ' =
,
θ p ' = 3.23o
13.8
σ1
C
x
σ (MPa)
2θ p'
σx
90 Pa
A
27.7MPa
3.23o
x’
x
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PROBLEM (9.63) Reconsider the cantilever beam illustrated in Fig. P9.61and described in
Prob. 9.61 to calculate he maximum shear stress τ max at point B.
SOLUTION
The internal forces and moments and area properties at the section containing
points A and B are as found in the solution of Prob. 9.61.
τ xy
At point B (Fig. a):
yB = −(60 − 25) = −35 mm
z B = −40 mm
M z
P M y
σx = − z B + y B
A
Iz
Iy
B
σx
Figure (a)
100(103 ) 3.3(103 )(−0.035) 3(103 )(−0.04)
=
+
+
9600(10−6 )
11.52(10−6 )
5.12(10−6 )
= 10.417 − 10.026 − 23.438 = −23.05 MPa
See Figure (b):
VQ
τ xy = y
I zb
Where
Vy = 15 kN
I 2 = 11.52 kN ⋅ m
b = 80 mm
Q = A * y = (80 × 25)(35 +
25
)
2
y
= 95(103 ) mm3
15 kN
C
z
and
15(103 )(95 ×10−6 )
τ xy =
= 1.546 MPa
11.52(10−6 )(0.08)
Therefore
τ max = (
σx 2
) + τ xy2 = (11.525) 2 + (1.546) 2
y
60 mm
B 35 mm
80 mm
A*
Figure (b)
2
= 11.63 MPa
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PROBLEM (9.62) For the cantilever beam described in Prob. 9.61 and shown in Fig. P9.61,
determine the principal stresses at point A. Show the result on a properly oriented element.
SOLUTION
τ xy
Refer to the solution of Prob. 9.61.
The state of stress at point A is found as
σ x = 27.60 MPa
σx
27.6
σ avg =
=
= 13.8 MPa
2
2
R= (
2
) +τ
σx
τ xy = 1.563 MPa
From the corresponding
Mohr’s circle, we have:
σx 2
A
τ (MPa) σavg
y
σ2
R
2
xy
= (13.8) 2 + (1.563) 2 = 13.89 MPa
Thus
σ 1,2 = σ avg ± R
or
σ 1 = 13.8 + 13.89 = 27.7 MPa
σ 2 = 13.8 − 13.89 = −90 MPa
and
1.563
tan 2θ p ' =
,
θ p ' = 3.23o
13.8
σ1
C
x
σ (MPa)
2θ p'
σx
90 Pa
A
27.7MPa
3.23o
x’
x
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PROBLEM (9.63) Reconsider the cantilever beam illustrated in Fig. P9.61and described in
Prob. 9.61 to calculate he maximum shear stress τ max at point B.
SOLUTION
The internal forces and moments and area properties at the section containing
points A and B are as found in the solution of Prob. 9.61.
τ xy
At point B (Fig. a):
yB = −(60 − 25) = −35 mm
z B = −40 mm
M z
P M y
σx = − z B + y B
A
Iz
Iy
B
σx
Figure (a)
100(103 ) 3.3(103 )(−0.035) 3(103 )(−0.04)
=
+
+
9600(10−6 )
11.52(10−6 )
5.12(10−6 )
= 10.417 − 10.026 − 23.438 = −23.05 MPa
See Figure (b):
VQ
τ xy = y
I zb
Where
Vy = 15 kN
I 2 = 11.52 kN ⋅ m
b = 80 mm
Q = A * y = (80 × 25)(35 +
25
)
2
y
= 95(103 ) mm3
15 kN
C
z
and
15(103 )(95 ×10−6 )
τ xy =
= 1.546 MPa
11.52(10−6 )(0.08)
Therefore
τ max = (
σx 2
) + τ xy2 = (11.525) 2 + (1.546) 2
y
60 mm
B 35 mm
80 mm
A*
Figure (b)
2
= 11.63 MPa
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PROBLEM (9.64) A rolled steel cantilever beam of S12 x 50 section (see Table B.9) is
subjected to an eccentric axial load P at its free end (Fig. P9.64). What is the largest permissible
value of P if the allowable stress is limited to σ all = 15 ksi.
SOLUTION
y
From Table B.9:
A = 14.7 in.2
I 2 = 305 in.4
I y = 15.7 in.4
z
Mz
t f = 0.659 in.
C
My
We have (see Fig. P9.64):
M z = 5.7 P
M y = −2 P
y A = yB = −6 in.
A
P
B
1
z A = − z B = t f = 0.3295 in.
2
Then
5.7 P(−6) 2 P(0.3295)
P M z yA M y zA
P
−
+
=
−
−
14.7
305
14.7
A
Iz
Iy
= 0.0680 P + 0.1121P − 0.0448 P = 0.1353P
σ B = 0.0680 P + 0.1121P + 0.0448 P = 0.225P
σA =
Hence
σ all = σ B ;
15 = 0.8369 Pall ,
P = 66.7 kips
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PROBLEM (9.65) A metal beam ABC with a wide flange cross section is simply supported at
A and B and has an overhang AB as shown in Fig. P9.65. The beam supports a concentrated load
P at its free end . Determine the value of the maximum permissible load P based on an
allowable shear stress of τ all .
I z = 5.464 in. 4,
Given:
A = 1.911 in. 2 ,
τ all = 14 ksi
SOLUTION
Reactions, as obtained from equations of equilibrium, are indicated in Fig. (a).
y
P
C
B
A
RCx
2 kips
RB =3P/2
RCy
Figure (a)
We observe that the critical point is at D of a section just to the left of support B:
V = P and M = (4 × 12) P = 48 P
Stress at point D:
2(103 ) 48P(2 − 0.23)
σx =
+
1.911
5.464
= 1046.6 + 15.549 P
0.23
)]
2
5.464(0.15)
y
P[(3 × 0.23)(2 −
VQ
=−
Ib
= −1.5869P
τ xy = −
σx
D
τ xy
x
Therefore, with τ max = τ all ; we have
τ all = (
σx 2
2
) + τ xy2
or
1
(14 ×103 ) 2 = (1, 095,376.6 + 32,547.2 P + 241.77 P 2 ) + (−1.5869 P) 2
4
from which
P 2 + 129, 2349 P − 3,108, 674 = 0
Solving this quadratic equation:
Pall = 3.399 lb = 3.4 kips.
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PROBLEM (*9.66) Loads P and R are applied at the free end of the post having a diameter d shown
in Fig. P9.66. Calculate the principal stresses and the maximum shearing stress:
(a) At point A.
(b) At point B.
Indicate the results on properly oriented elements.
R = 800π ⋅ N, d = 50 mm
Given: P = 400π ⋅ N,
*SOLUTION
x
T=0.4P
R
C
My=0.4R
d = 50 mm
P = 400π N
R = 800π N
B
A
Mz=0.6P
P
y
z
16T 16(0.4 × 400π )
= 20.48 MPa
=
πd3
π (0.05)3
800π
32(0.6 × 400π )
R M y
σx = + z = −
−
2
A
I
π (0.025)
π (0.05)3
= −1.28 − 61.44 = −62.7 MPa
(a) τ t =
62.7
62.7 2
) + (20.48) 2
± (−
2
2
= −31.35 ± 37.45
σ 1 = 6.1 MPa
σ 2 = −68.8 MPa
τ max = 37.5 MPa
2τ 2(20.48)
=
θ p ' = −16.6o
tan 2θ p ' =
,
σx
−62.7
σ 1,2 = −
x
A
τt
σx
Continued on next slide
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τ (MPa)
68.8 MPa
x
σ2
2θ
'
p
16.6o
y
σ1
C
(−62.7, −20.48 )
σ
(MPa)
A
6.1 MPa
37.5 MPa
(b)
x
σx =
R Myz
+
A
I
B
τ = τt +τ d
4P
VQ 4 V
=
=
(Example 9.4)
Ib 3 A 3π r 2
4 400π
=
= 0.85 MPa
3 π (0.025) 2
and
τ = τ t + τ d = 20.48 + 0.85 = 21.33 MPa
τd =
Similarly,
32(0.4 × 800π )
π (0.05)3
= −1.28 − 81.92 = −83.2 MPa
σ x = −1.28 −
Continued on next slide
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48.4 MPa
Thus,
83.2
83.2 2
) + (21.33) 2
m (−
2
2
= −41.6 m 46.75
σ 1 = 5.15 MPa
σ 2 = −88.4 MPa
σ 1,2 = −
13.6o
B
5.15 MPa
46.8 MPa
τ max = 46.8 MPa
2τ 2(21.33)
=
,
tan 2θ p ' =
σx
−83.2
θ p ' = −13.6o
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PROBLEM (9.67) Redo Prob. 9.66 for the case in which the vertically applied load R is zero.
SOLUTION
Refer to solution of Prob. 9.66
(a)
x
τ=
Tc
= 20.48 MPa
J
A
σ x = 61.44 MPa
Therefore,
61.44
61.44 2
) + (20.48) 2
σ 1,2 = −
± (−
2
2
= −30.72 ± 36.92
σ 1 = 6.2 MPa
σ 2 = −67.6 MPa
τ max = 36.9 MPa
tan 2θ p ' =
2τ
σx
=
2(20.48)
,
−61.44
θ p ' = −16.9o
67.6 MPa
x
16.9o
A
36.9 MPa
6.2 MPa
Continued on next slide
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(b) See solution of Prob. 9.66:
τ
σx = 0
σ2
B
τ = 21.33 MPa
Thus,
τ max = 21.33 MPa
O
τ max
σ1
σ
σ 1 = −σ 2 = 21.3 MPa
21.3 ksi
B
21.3 ksi
21.3 ksi
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PROBLEM (*9.68) A sign of weight W is supported by a pipe having an outer-diameter D and
inner-diameter d as shown in Fig. P9.68. For a wind load of P on the sign, determine the factor of safety
ns against failure by permanent deformation.
Given:
W = 1.5 kN,
D = 125 mm,
σ y = 250 MPa,
d = 100 mm,
P = 2 kN
*SOLUTION
T = 1.2 P
x
T
P
W
J=
My
π
M y = 1.2W
(1254 − 1004 )
32
= 14.15 × 10 6 mm 4
I = J 2 = 7.08 × 10 6 mm 4
C
3m
A
A=
(1252 − 1002 )
4
= 4.416 × 10 3 mm 2
B
z
y
π
Figure S9.68
Point A:
σx = −
W Mzy
−
A
I
M z = 3P
A
τ=
Tc
J
1.2(2 × 103 )(0.0625)
= 10.6 MPa
14.15 × 10−6
1.5 ×103
3(2 ×103 )(0.0625)
σx = −
−
4.416 × 10−3
7.08 ×10−6
= (0.34 − 52.97)106 = −53.31 MPa
τt =
53.31
53.31 2
) + (10.67) 2
± (−
2
2
= −26.7 ± 28.7
σ 1 = 2 MPa
σ 2 = −55.4 MPa
τ max = 28.7 MPa
σ 1,2 = −
Thus,
250 ×106
= 55.4 ×106 ,
ns
ns = 4.5
Continued on next slide
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Point B:
W M z
σx = − − y
A
I
B
τ = τt +τ d
For solid semi-circle:
π r 2 4r 2 3
= r
Q = Ay =
2 3π 3
For half thin-tube:
2
Q = (62.53 − 503 )
3
= 79.427 × 10 3 mm 3
PQ 2 × 103 (79.427 ×10−6 )
=
= 0.897 MPa
7.08 ×10−6 (0.025)
Ib
τ = 10.6 + 0.897 = 11.5 MPa
τd =
1.2(1.5 × 103 )(0.0625)
σ x = −0.34 −
= −16.23 MPa
7.08 ×10−6
16.23
16.23 2
) + (11.5) 2 = −8.12 ± 14.08
σ 1,2 = −
± (−
2
2
σ 1 = 5.96 MPa
σ 2 = −22.2 MPa
So,
250 ×106
= 22.2 × 106 ,
ns
We have
(ns ) all = 4.5
ns = 11.3
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PROBLEM (9.69) The sign of Fig. P9.68 is supported by a steel pipe of an outer-diameter D and
inner-diameter d. Determine the maximum shearing stress in the pipe.
Given: D = 100 mm, d = 75 mm,
P = 800 N, W = 500 N
SOLUTION
Refer to solution of Prob.9.68:
x
σx =
W Mzy
+
A
I
x
σx =
A
W M yz
+
A
I
B
Tc
J
τ = τt +τ d
(1004 − 754 ) = 6.71×106 mm 4
I = J 2 = 3.355 × 10 6
τ=
We now have
J=
A=
π
32
π
(1002 − 752 ) = 3.43 ×103
4
Point A:
1.2(800)(0.05)
τt = −
= −7.15 MPa
6.71×10−6
500
3(800)(0.05)
σx = −
−
= −35.92 MPa
−3
3.43 ×10
3.355 ×10−6
−35.92 2
) + (−7.15) 2 = 19.33 MPa
τ max = (
2
Point B:
For semi-circle: Q = Ay =
π r 2 4r 2 3
⋅
= r
2 3π 3
Thus, for half thin-tube:
2
Q = (503 − 37.53 ) = 48.18 × 103 mm3
3
800(48.18 × 10−6 )
VQ
τd = −
=−
= −0.46 MPa
3.355 ×10−6 (0.025)
Ib
τ = −7.15 − 0.46 = −7.61 MPa
1.2(500)(0.05)
σ x = −0.146 −
= 9.09 MPa
3.355 × 10−6
9.09 2
) + (−7.61) 2 = 8.86 MPa
τ max = (−
2
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PROBLEM (9.70) For the member loaded as shown in Fig. P9.70, calculate the maximum normal
stress:
(a) At point A.
(b) At point B.
(c) At point D.
x
SOLUTION
1.2 kips
M y = 1.2(10) + 0.8( 24)
= 31.2 kip ⋅ in.
A = 4(3 4) = 3 in.
Iy =
My
A
1.25 in.
D
0.8 kips
B
Figure S9.70
3
(3 4)(4)
= 4 in.4
12
y
0.75 in.
4 in.
z
P M y z −1.2 × 103 31.2 ×103 (2)
+
=
+
3
4
A
I
= −0.4 + 15.6 = 15.2 ksi
(a) σ A = −
(b) σ B = −0.4 − 15.6 = −16 ksi
31.2(0.75)
= 5.45 ksi
4
VQ 0.8(1.25 × 0.75 ×1.375)
τD =
= 0.344 ksi
=
4(0.75)
Ib
(c) σ D = −0.4 +
Thus,
5.45
5.45 2
) + (0.344) 2 = 2.725 + 2.747
+ (
2
2
= 5.47 ksi
(σ D ) max =
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PROBLEM (9.71) The tube shown in Fig. P9.71 is of constant t = 20-mm thickness. Determine the
maximum permissible load P.
Given:
d = 30 mm,
( σ t )all = 80 MPa,
( σ c )all = 120 MPa,
SOLUTION
t = 20 mm
d = 30 mm
M y = 0.04 P
M z = 0.05P
x
100(140)3 − 60(100)3
12
= 17.87 × 10 6 mm 4
140 mm
Iy =
100 mm
d
P
t
D
A = 140(100) − 100(60)
= 8 × 10 3 mm 2
P
E
Mz
A
My
z
y
B
140(100)3 − 100(60)3
12
= 9.87 × 10 6 mm 4
Iz =
Figure S9.71
Maximum tensile stress occurs at D:
P M z M y
σD = − + y + z
A
Iy
Iz
0.04 P(0.07) 0.05P(0.05)
P
+
+
80 ×106 = −
0.008 17.87 ×10−6
9.87 × 10−6
P = 283.7 kN
80 × 10 3 = −0.125P − 0.157 P − 0.25P ;
Maximum compressive stress is at B:
− 120 × 10 3 = −0.125P − 0.157 P − 0.25P ;
Thus,
Pall = 255.6 kN
P = 255.6 kN
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PROBLEM (9.72) The tube shown in Fig. P9.71 is of t = 20-mm uniform thickness. What is the
minimum permissible distance d?
Given: P = 250 kN,
( σ t )all = 90 MPa,
( σ c )all = 130 MPa
SOLUTION
See solution of Prob.9.71:
M z = 0.05P = 0.05( 250) = 12.5 kN ⋅ m
M y = P (0.07 − d ) = 250 × 10 3 (0.07 − d )
= 17.5 × 10 3 − 250 × 10 3 d
Maximum tensile stress is at D:
P M z M y
σD = − + y + z
A
Iy
Iz
90 ×106 =
−250 × 103 (17.5 − 250d )103 (0.07) 12.5 × 103 (0.05)
+
+
8 × 10−3
17.87 ×10−6
9.87 ×10−6
or
90 = −31.25 + 68.55 − 979.29d + 63.32 ,
Maximum compressive stress is at B:
− 130 = −31.25 − 68.55 + 979.29d − 63.32 ,
Thus,
d min = 33.8 mm
d = 10.8 mm
d = 33.8 mm
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PROBLEM (9.73) Determine the normal stress at points A, B, D, and E of the tube seen in Fig.
P9.71,which has a constant t = 10-mm thickness.
Given: d = 50 mm,
P = 150 kN
SOLUTION
See Figure S9.71 (Solution of Prob. 9.71).
We now have,
t = 10 mm
d = 50 mm
P = 150 kN
M z = 0.05P = 7.5 kN ⋅ m
M y = 0.02 P = 3 kN ⋅ m
Therefore,
A = 140(100) − 120(80) = 4.4 × 10 3 mm 2
1
I y = [100(140)3 − 80(120)3 ] = 11.35 ×106 mm 4
12
1
I z = [140(100)3 − 120(80)3 ] = 6.55 × 106 mm 4
12
Then
P M z M y
σx = − ± y ± z
A
Iy
Iz
yields
150 × 103 3 × 103 (0.07) 7.5 ×103 (0.05)
σA = −
+
−
4.4 ×10−3 11.35 ×10−6
6.55 × 10−6
= −34.1 + 18.5 − 57.25 = −72.85 MPa
σ B = −34.1 − 18.5 − 57.25 = −109.9 MPa
σ D = −34.1 + 18.5 + 57.25 = −41.65 MPa
σ E = −34.1 − 18.5 + 57.25 = 4.65 MPa
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PROBLEM (9.74) Referring to Sec. 9.7, show that for a solid circular section of radius r, the core is
contained in a circle of radius r/4.
SOLUTION
x
σA = 0 = −
P
A
ro
y
z
P Mc
+
A I
I = πr 4 4
A = πr 2
M = Pr0
c=r
Thus,
0=−
P 4 Pr0
+
π r2 π r3
from which
r
Q.E.D.
4
Due to the symmetry, kern is within the circle of radius r0
r0 =
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PROBLEM (9.75) A short column constructed of a C 100 x 11 channel section (see Table B.10)
supports an eccentric axial load of P (Fig. P9.75). Determine the largest permissible value of the distance
e, if the maximum compressive stress is not to exceed σ all .
Given:
σ all = 40 MPa
P = 20 kN,
SOLUTION
For C 6 × 25 ×12 rolled-steel section (Table B.10):
I z = 1.91(106 ) mm 4
I y = 0.18(106 ) mm 4
w f = 43 mm
d = 102 mm
We have
d
y A = − = −51 mm
2
z A = − z = −11.7 mm
A = 1.37(103 ) mm 2
z = 11.7 mm
M z = − Pe = −20 ×103 e
t
M y = Pz p = P ( z − w ) = 152 N ⋅ m
2
tw = 8.2 mm
y
tw
z
d
e
P
So
σA = −
P M z yA M y z A
−
+
A
Iz
Iy
Mz = −
I z M y zA P
[
− −σ A]
yA I y
A
A
z
or
1.91× 10−6 (152)(−11.2 ×10−3 ) 20 ×103
[
−
− (−40 × 106 )]106
−3
−6
−3
0.18(10 )
1.37(10 )
−51× 10
= −607.834 N ⋅ m
=
Hence
e=−
Mz
−607.834
=−
= 30.4 × 10−3 m = 30.4 mm
3
P
20 ×10
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PROBLEM (*9.76) A short column made of a W6 x 25 x 12 wide-flange section(see Table B.8) is
subjected to an eccentric axial load of P as shown in Fig. P9.76. If the maximum compressive stress is
limited to σ all , what is the largest permissible values of the distance e?
Given:
σ all = 15 ksi
P = 10 kips,
*SOLUTION
For W 6 × 25 × 12 steel section (Table B.8):
I z = 22.1 in.4
I y = 2.99 in.4
A = 3.55 in.2
w f = 4 in.
d = 6.03 in.
y
We have
d
= 3.015 in.
2
wf
zA =
= 2 in.
2
M z = 2.9 P = 2.9(10)
= 29 kip ⋅ in.
M y = − Pe = −10e
yA = −
A P
e
z
C
d
wf
Hence
σA = −
P M z yA M y z A
−
+
A
Iz
Iy
or
I y M z yA P
[
+ +σ A]
zA I z
A
2.99 29(3.015) 10
=
[
+
+ (−15)] = −12.299 kip ⋅ in.
2
22.1
3.55
My =
Then
e=−
My
P
=−
−12.299
= 1.23 in.
10
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PROBLEM (9.77) Figure P9.77 shows a d = 50-mm-diameter steel shaft of yield strength in
tension σ y and yield strength in shear τ y subjected to a load R and P = Q = 0. If the factor of
safety is ns = 2, calculate the largest permissible value of R in accordance with:
(a) The maximum shear stress theory.
(b) The von Mises theory.
σ y = 260 MPa, τ y = 140 MPa
Given:
SOLUTION
At the fixed end A(see Fig. P9.77):
T = 1.5R
M z = 2R
Vy = − R
The effect of Vy may be neglected. Thus, at a point on D the top of the bar at the
fixed end:
σx =
A
τ =−
32 M
32(2 R)
=
= 162,975 R
3
πd
π (0.05)3
16T
16(1.5 R)
=−
= −61,116 R
3
πd
π (0.05)3
Then
162,975 R
162,975 R 2
) + (−61,116 R) 2
± (
2
2
σ 1 = 18.34 ×104 R
σ 2 = −2.04 ×104 R
τ max = 10.19 × 104 R
σ 1,2 =
130 ×106
(a)
= 10.19 × 104 R,
2
R = 638 N
(b) σ 12 − σ 1σ 2 + σ 22 = (σ yp ns ) 2
1
[18.342 − 18.34(−2.04) + (−2.04) 2 ] 2 (104 ) R = 130 × 106
or
R = 669 N
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PROBLEM (9.78) Redo Prob. 9.77 for P = 40R and Q = 0.
SOLUTION
y
2m
σx
A
A
B
z
50 mm
40 R
τ
x
At the fixed end:
T = 1.5 R
Px = 40 R
C
R
M z = 2R
Vy = − R
The effect of Vy may be neglected. Therefore, at point A:
Px
40 R
=
= 2.037 × 104 R
A π (0.05) 2
4
32
32(2 R)
σx ' = 3 =
= 16.298 × 104 R
3
πd
π (0.05)
16T
16(1.5R)
τ =− 3 =−
= −6.116 × 104 R
πd
π (0.05)3
σx ' =
18.34
18.34 2
) + (−6.116) 2 ]
± (
2
2
σ 1 = 20.196 × 104 R
σ 2 = −1.852 ×104 R
σ 1,2 = 104 R[
130 ×106
(a)
= 11.024 × 104 R,
2
1
(b) [σ 12 − σ 1σ 2 + σ 22 ] 2 =
τ max = 11.024 × 104 R
R = 590 N
σy
2
1
104 R[(20.196) 2 − (20.196)(−1.852) + (−1.852) 2 ] 2 = 130 × 106
Solving,
R = 614 N
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PROBLEM (9.79) A thin-walled, closed-ended metal tube having ultimate strengths in
tension σ u and compression σ u ' , outer and inner diameters D and d, respectively, is subjected
to an internal pressure of p and a torque of T. Determine the factor of safety ns according
to the maximum principal stress theory.
Given: σ u = 36 ksi, σ u ' = 74 ksi, D = 8 in.,
d = 7 ½ in ,
p = 800 psi,
T = 400 kip ⋅ in.
SOLUTION
σt
τ
J ≈ 2π r 3t
pr 800(3.875)
σt =
=
t
0.25
= 12 ksi
pr
σa =
= 6 ksi
2t
σa
Tc
400 × 103 (4)
= −19.32 ksi
τ =− =−
2π (3.75)3 (0.25)
J
6 + 12
6 − 12 2
) + (19.32) 2
± (
2
2
= 9 ± 19.55
σ 1,2 =
σ 1 = 28.6 ksi
σ 2 = −10.6 ksi
σ
36
Thus, σ 1 = u ;
28.6 = ,
ns
and
σ2 =
σu '
ns
ns
;
− 10.6 =
−74
,
ns
ns = 1.3
ns = 6.98
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PROBLEM (9.80) A steel circular bar ( σ y = 10 MPa) of d = 50-mm diameter is acted upon
by combined moments M and axial compressive loads P at its ends. If M = 1.2 kN m, calculate,
on the basis of the maximum energy of distortion theory of failure, the largest permissible
value of P.
SOLUTION
σ1 = 0
50 mm
M
A
P
P
P
A
σ2
M=1.2 kN ⋅ m
P Mc
4P
32(1.2 × 103 )
−
=−
−
π (0.05) 2
π (0.05)3
A I
= −509.3P − 97.79 × 106
σ2 = −
We have
σ 12 − σ 1σ 2 + σ 22 = σ y2
or
σ2 = σ y
So,
509.3P + 97.79 × 106 = 210 ×106 ,
Pall = 220 kN
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PROBLEM (9.81) A cantilever beam of width b, depth h, and length L is subjected to a
concentrated load of P at its free end. For σ y = 240 MPa. what is the factor of safety ns ,
assuming failure occurs in accordance with the maximum energy of distortion theory?
Given: b = 60 mm, h = 80 mm, L = 1.2 m,
P = 2 kN
SOLUTION
P=2 kN
VA A
B
MA
80 mm
1.2 m
60 mm
I = 60 (80)3 12 = 2.56 × 106 mm 4
Vmax = VA = 2 kN
M max = M A = 2.4 kN ⋅ m
Point A:
2.4 ×103 (0.04)
σ 1 = Mc I =
= 37.5 MPa
2.56 × 10−6
σ
240
σ 12 − σ 1 σ 2 + σ 22 = ( y ) 2 ;
37.5 =
,
ns
ns
Point B:
3V
3 2 × 103
τ max = max =
= 0.625 MPa
2 A
2 4.8 ×10−3
τ max = σ 1 = −σ 2
Thus, σ 12 − σ 1σ 2 + σ 22 = (
1.73(0.625) =
σy 2
ns
240
,
ns
) ;
2
3τ max
=(
ns = 6.4
σy 2
ns
)
ns = 222
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PROBLEM (9.82) Referring to Fig. P9.77, design a solid steel shaft ( τ y = 60 MPa) subjected
to loads of R = 600 N, Q = 200 N, and P = 0. Apply the maximum shear stres theory using a
factor of safety ns = 1.2.
SOLUTION
At the fixed end:
M y = 200 × 2 = 400 N ⋅ m
M z = −600 × 2 = −1.2 kN ⋅ m
T = 600 ×1.5 = 900 N ⋅ m
A load-diagram is shown in Fig. S9.82a.
y
My
T
z
R
Mz
A
My
Q
d
2 m
B
x
T
400 N ⋅ m
x
Mz
x
-1.2 kN ⋅ m
900 N ⋅ m
T
x
Figure S9.82
From the moment and torque diagrams we observe that the critical point
is just to the right of A.
Thus,
M y2 + M z2 + T 2 = 4002 + 12002 + 9002
= 1552 N ⋅ m
Since
60
= 50 MPa
1.2
Equation (9.30) is thus
16(1552)
= 54.1× 10−3 = 54.1 mm
d=3
6
π (50 ×10 )
Use a
55-mm diameter shaft.
σ all =
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PROBLEM (9.83) Design the cross section of a rectangular beam b meters wide by 2b meter ,
deep, loaded as shown in Fig. P9.83, if
σ all = 18 ksi. Use the maximum principal stress theory of
failure.
SOLUTION
∑ M = 0 : 10(7.5)(3.75) − R (4.5) = 0 ,
∑ F = 0 : R = 12.5 kips
A
y
RB = 62.5 kips
B
A
y
10 kips/ft
z
A
12.5 kips
4.5 ft
3 ft
B
2b
C
b
62.5 kips
V, kips
30
12.5
x
-32.5
M, kip ⋅ in.
x
-540
At B at the upper outermost fiber:
M c 540 × 103 810 × 103
σ max = max =
=
1
I
b3
3
b(2b)
12
Thus,
810 × 103
σ max = σ all ;
= 18 × 103 ,
b = 3.56 in.
3
b
At B at the neutral axis axis:
3V 3 32.5 × 103 24.375 ×103
τ max =
=
=
2 A 2 2b 2
b2
And
τ max = σ 1 = −σ 2
Continued on next slide
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Thus,
24.375 × 103
b = 1.35 in.
= 18 ×103 ,
b
Use a 3 9 16 in. by 7 1 8 in. rectangular beam.
τ max = σ all ;
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PROBLEM (9.84) A 6-m-long steel shaft of allowable strength σ all = 120 MPa carries a torque of
T = 400 N ⋅ m and its own weight. Calculate the required shaft diameter d in accordance with the von
Mises theory of failure.
Assumptions: Use ρ = 7.86 Mg/m3 as the mass per unit volume for steel (see Table B.4).
The shaft is supported by frictionless bearings that act as simple supports at its ends.
SOLUTION
w = 7.86(9.81)(π d 2 4) = 60.6d 2 kN m
w=60.6d2
T=400 N ⋅ m
A
C
3m
B
T
3m
C
σx =
32M
πd3
τ = 16T π d 3
32( wL2 8) 32(60.6d 2 )(36)103 2.78 ×106
=
=
d
8π d 3
πd3
3
16(400)
2.04 × 10
τ =−
=−
3
d3
πd
2
Equation (9.19), σ x2 + 3τ 2 = σ all
σx =
2.78 × 106 2
−2.04 ×106 2
) + 3(
) = (120 × 106 ) 2
3
d
d
or
7.7284 ×106 12.4848
+
= 144 ×108
2
6
d
d
Solving by trial & error:
d = 34.2 mm.
Use a 35 − mm diameter shaft.
(
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PROBLEM (*9.85) As shown in Fig. P9.85. a solid shaft AB is to transmit 20 kW at 180 rpm
from the motor and gear to pulley D, where 8 kW is taken off, and to pulley C where the remaining 12
kW is taken off. If τ all = 45 MPa, determine the required diameter d of the shaft AB according to the
maximum shear stress theory of failure.
Assumption: The ratios of the pulley tensions are to be 'T1 / T2
= 3 and T3 / T 4 = 3.
*SOLUTION
159 P
f
GEAR E
T=
f = 180 60 = 3 Hz
159(20)
= 1060 N ⋅ m
3
The tooth force:
T
1060
( FE ) z = − E = −
= −10.6 kN
RE
0.1
Torque: TE =
PULLY D
159(8)
= 424 N ⋅ m
3
= (3T2 − T2 )(0.15) = 0.3T2
Torque: TD =
or
T2 = 1.41 kN
T1 = 4.23 kN ; T1 + T2 = 5.64 kN
Thus,
(T1 + T2 ) z = 5.64sin 30o = 2.82 kN ;
(T1 + T2 ) y = 5.64 cos 30o = 4.88 kN
PULLY C
159(12)
= 636 N ⋅ m
3
= (3T4 − T4 )(0.2) = 0.4T4
T4 = 1.59 kN
T3 = 4.77 kN ; T3 + T4 = 6.36 kN
Torque: TC =
or
The load, moment, and torque diagrams are shown below.
From statics:
Ay = 7.74 kN
By = 3.5 kN
Az = 0.36 kN
Bz = 7.42 kN
Continued on next slide
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y
A
z
Mz , kN ⋅ m
4.88 kN
Az 6.36 kN
0.2 m
Ay
C 0.4 m
D
636 N ⋅ m
424 N ⋅ m
10.6 kN
Bz
x
B
E
B
1060 N ⋅ m y
(a)
2.1
1.548
0.7
My, kN ⋅ m
-0.072
-0.216
x
(b)
x
(c)
x
(d)
-1.488
T, kN ⋅ m
0.636
1.06
Figure S9.85
Point C:
M y2 + M z2 + T 2 = 1.5482 + 0.0722 + 0.6362 = 1.675 kN ⋅ m
Point D:
2.12 + 0.2162 + 1.062 = 2.362 kN ⋅ m
Point E:
0.7 2 + 1.4882 + 1.062 = 1.956 kN ⋅ m
The critical point is thus just to the right of D. Apply Eq. (9.30):
16(2362) 13
d =[
] = 64.4 ×10−3 m = 64.4 mm
6
π (45 ×10 )
Use a 65 − mm diameter shaft.
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PROBLEM (*9.86) Repeat Prob. 9.85 for the case in which tensions T1 and T2 on pulley D are
in the horizontal (z) direction.
*SOLUTION
Now Fig. S9.85a (see: Solution on of Prob. 9.85) becomes as shown below. From
statics:
Ay = 5.3 kN
By = 1.06 kN
Az = −1.05 kN
Az 6.36 kN
y
A
z
Bz = 6.01 kN
5.64 kN
C 0.4 m
D
636 N ⋅ m
424 N ⋅ m
0.2 m
Ay
Mz , kN ⋅ m
1.06
My, kN ⋅ m
10.6 kN
Bz
x
B
E
B
1060 N ⋅ m y
0.636
0.212
(a)
x
(b)
x
(c)
x
(d)
0.63
0.21
-1.20
T, kN ⋅ m
0.636
1.06
Figure S9.86
The moment and torque diagrams are shown in Figs. S9.86b, c, and d.
Point C:
M y2 + M z2 + T 2 = 1.062 + 0.212 + 0.6362 = 1.254 kN ⋅ m
Point D:
0.6362 + 0.632 + 1.062 = 1.387 kN ⋅ m
Point E:
0.2122 + 1.22 + 1.062 = 1.615 kN ⋅ m
The critical point is therefore just to the left of E. Applying Eq. (9.30):
16(1.615)103
d=
= 57 ×10−3 m = 57 mm
6
π (45 ×10 )
Use a 60 − mm diameter shaft.
3
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PROBLEM (9.87) Design the steel shaft ( τ y = 60 MPa) of the system shown in Fig. P9.87. Use the
maximum shear stress theory of failure and a factor of safety ns = 1.2.
SOLUTION
From statics:
Ay = 2.4 kN
y
4 kN
A
z
0.4 m
By = 1.6 kN
400 N ⋅ m
0.6 m
C
400 N ⋅ m
B x
(a)
By
Ay
960
Mz , kN ⋅ m
T, kN ⋅ m
x
(b)
x
(d)
400
The critical point is thus just to the right of C:
M z2 + T 2 = 9602 + 4002 = 1040 N ⋅ m
Equation (9.30):
16 ×1040
d=3
= 47.3 × 10−3 m = 47.3 mm
π (50 ×106 )
Use a 50 − mm diameter shaft.
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PROBLEM (9.88) For the shaft of the system shown in Fig. P9.88, determine the diameter d,
applying the maximum principal stress theory of failure.
σ all = 20 ksi.
Given:
SOLUTION
Ay = By = 1.6 kips
From statics:
1.6 kips
y
A
z
1.5 ft
Ay
1.6 kips
C
4.8 kips ⋅ in.
3 ft
D
1.5 ft
4.8 kips ⋅ in.
B x
(a)
By
2.4x12=28.8 kip ⋅ in.
Mz
T
4.8 kipin.
x
(b)
x
(d)
Figure S9.88
The critical section is between C and D. Thus
M z2 + T 2 = 28.82 + 4.82 = 29.197 kip ⋅ in.
Apply Eq. (9.32):
d=3
16 × 103
(28.8 + 29.197) = 2.45 in.
π (20 ×103 )
Use a 2 1 2 − in. diameter shaft.
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PROBLEM (*9.89) Redo Prob. 9.88, assuming that the tensions on the pulley C are in the horizontal
(z) direction.
*SOLUTION
Now Fig. S9.88a (see: Solution of Prob. 9.88) becomes as shown below.
From statics:
Ay = 0.4 kips
By = 1.2 kips
Az = 1.2 kips
Bz = 0.4 kips
y
B
A
1.5 ft
z
1.6 kips
1.6 kips
Az Ay
C
4.8 kip ⋅ in.
3 ft
D
4.8 , kip ⋅ in. Bz By
x
(a)
x
(b)
x
(c)
x
(d)
Mz
0.6x12=7.2 kip ⋅ in.
1.8x12=21.6 kip ⋅ in.
21.6 kip ⋅ in.
My
T
4.8 kip ⋅ in.
The critical section is just to the right of C (or just to the left of D). Thus
M z2 + M y2 + T 2 = 21.62 + 7.22 + 4.82 = 23.27 kip ⋅ in.
Applying Eq. (9.32):
16 × 103
d=
(21.6 + 23.21) = 2.25 in.
π (20 × 103 )
Use a 2 1 4 − in. diameter shaft.
3
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PROBLEM (9.90) At a critical point in a loaded ASTM-A36 structural steel bracket, the plane
stresses have the magnitudes and directions shown on element A in Fig. P9.90. Determine whether the
loading will cause bracket to fail, on the basis of a safety factor of ns = 1.2, according to:
(a) The maximum shear stress theory.
(b) The maximum energy of distortion theory.
SOLUTION
σ y = 120 MPa
τ xy = −100 MPa
σ y = 250 MPa (Table B.4)
σx = 0
Hence
σ 1,2 =
σy
2
± (
−σ y 2
) + τ xy2
2
= 60 ± (−60) 2 + (−100) 2 = 60 ± 116.6
or
σ 1 = 176.6 MPa
σ 2 = −56.6 MPa
(a) Maximum shear stress memory:
σ1 − σ 2 ≤
σy
ns
176.6 − (−56.6) ≤
233.2 > 208.3
Failure will occur.
250
1.2
(b) Maximum energy of distortion theory:
σ 12 − σ 1σ 2 + σ 22 ≤ (
σy
ns
)
(176.6) 2 − (176.6)(−56.6) − (−56.6) 2
12
≤ 208.3
194.9 < 208.3
Failure will not occur.
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PROBLEM (9.91) A cast iron ASTM grade A-48 shaft of circular cross section is subjected
simultaneously to a torque T and a vertical load P (Fig. P9.91). Calculate the diameter d of the shaft
based on the of a safety factor of ns, employing:
(a) The maximum normal stress theory of failure.
(b) The Coulomb-Mohr theory of failure.
σ u = 25 ksi, σ u ' = 95 ksi (from Table B.4), L = 10 in., T = 5 kip ⋅ in., ns = 2
Given:
Assumption: The effect of transverse shear may be disregarded.
SOLUTION
32M
16T
τ xy = 3
3
πd
πd
M = 200(10) = 2 kip ⋅ in.
25
= 12.5 ksi
(σ u ) all =
2
95
= 42.5 ksi
(σ u ') all =
2
Also
σx =
σ 1,2 =
σx
2
± (
σx 2
2
) + τ xy2 =
16
(M ± M 2 + T 2 )
3
πd
Substituting the given data
16
σ 1,2 = 3 (2 ± 22 + 52 )
πd
or
37.61
17.24
σ1 = 3
σ2 = − 3
d
d
(a) Maximum Principal Stress Theory
37.61
= 12.5
or
d = 1.44 in.
d3
(b) Coulomb-Mohr Theory:
σ1
σ2
−
= 1;
(σ u ) all (σ u ') all
or
37.61 −17.24
−
=1
12.5d 3 42.5d 3
d = 1.5 in.
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PROBLEM (9.92) The C clamp shown in Fig.P9.92 is made of a cast-aluminum alloy having an
ultimate strength in tension σ u = 8 ksi and in compression σ u ' = 16 ksi. Using Coulomb-Mohr theory of
failure, determine the maximum load P for which rupture should be expected.
SOLUTION
A = 1(0.5) = 0.5 in.2
I=
M = 3P
1
(0.5)(1)3 = 41.667(10−3 ) in.4
12
P
P Mc
+
A I
P
3P (0.5)
=
+
0.5 41.667(10−3 )
= 2 P + 36 P = 38P
σ B = 2 P − 36 P = −34 P
σA =
We thus have
σ 1 = 38P
σ 2 = −34 P
Then
σ2 = 0
σ1 = 0
A
M
B
P
at point A
at point B
σ1
38P
=
= 1, P = 210.5 lb
σ u 8(103 )
σ2
34 P
=
= 1, P = 470.6 lb
σ u ' 16(103 )
Hence
Pmax = 210.5 lb
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PROBLEM (*9.93) The gear A attached to the steel shaft AB with an allowable shear strength τ all
= 10 ksi is acted upon by a vertical load Fy = 300 lb (Fig. P9.93). Determine:
(a) The value of the torque T required to support the loading.
(b) The required shaft diameter d using the maximum shear stress theory of failure.
Assumption: Bearings at C and B act as simple supports.
*SOLUTION
(a) Reactions as obtained
from equilibrium
conditions, are marked
in Figure (a).
T= 800 N ⋅ in.
2 in.
E
RBz=285.7 lb
x
B
400 lb
y
RCz=114.3 lb
300 lb
C
A
RBy=128.6 lb
3 in.
D
RCy=428.6 lb
4 in.
z
(b) From Fig. (a):
T = 800 N ⋅ m
Critical section is at C. Equation (9.30):
16
d=3
M 2 +T2
πτ all
Substituting the numerical
values (see Fig.b):
1
16
d =[
(2.186) 2 + (1.2) 2 ] 3
π (10)
= 1.083 in.
Figure (a)
Mz
(kip ⋅ in.)
A
2.186
0.625
C
D
My
(kip ⋅ in.)
A
T
(kip ⋅ in.)
B
x
0.237
B
x
1.2
x
Figure (b)
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PROBLEM (9.94) Redo Prob. 9.93 if the shaft AB is made of an ASTM- A47 gray cast iron (see
Table B.4), applying the maximum normal stress theory of failure.
SOLUTION
Refer to solution of Prob. 9.93.
From Table B.4:
σ all = σ u = 25 ksi
Equation (9.32):
1
16
d =[
( M + M 2 + T 2 )] 3
πτ all
Substituting the numerical values:
1
16
d =[
(2.186 + (2.186) 2 + (1.2) 2 )] 3
π (25)
= 0.984 in.
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PROBLEM (9.95) A solid circular shaft-pulley assembly is supported by frictionless supports at C
and D and acted upon by tensile forces through the belts as depicted in Fig. P9.95. If the allowable shear
stress is τ all = 20 ksi, what is the required minimum shaft diameter d according to the maximum shear
stress criterion of failure?
SOLUTION
The reactions and the torques, as found by the equations of statics, are marked in
Fig. a. Moment and torque diagrams are shown in Fig.(b). Critical sections is at C.
y
A
C
z
D 60 lb ⋅ in.
95 lb 60 lb ⋅ in.
190 lb
B
Figure (a)
Mz
(kip ⋅ in.)
100 lb
x
95 lb
950
A
My
(kip ⋅ in.)
C
D
300
600
x
B
x
T
(kip ⋅ in.)
C
-60
B
x
Figure (b)
Eq. (9.32) is thus
16
d=3
πτ all
=[
or
M z2 + M y2 + T 2
1
16
( 9502 + 3002 + (−60) 2 )] 3
3
π (20 ×10 )
d = 0.633 in.
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PROBLEM (9.96) Resolve Prob. 9.95, employing the maximum energy of distortion theory of failure
with an allowable normal stress of σ all = 30 ksi.
SOLUTION
Refer to solution of Prob. 9.95. Now apply Eq. (9.31):
d=3
=[
or
32
πσ all
3
M z2 + M y2 + T 2
4
1
32
3
2
2
2
3
+
+
−
(
950
300
(
60)
)
]
π (30 ×103 )
4
d = 0.697 in.
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PROBLEM (*9.97) A bracket arm is acted upon by a vertical load P = 6 kN at its free end, as
depicted in Fig.P9.97. Calculate the factor of safety ns for the rod of diameter d = 50 mm, according to
the maximum shear stress theory of failure.
Given: The material is an aluminum alloy with a yield strength of τ y = 140 MPa (see Table B.4).
*SOLUTION
M = 6 × 0.2 = 1.2 kN ⋅ m
T = 6 × 0.25 = 1.5 kN ⋅ m
An element at point A:
32M
32(1.2)
σx =
=
= 97.78 MPa
3
πd
π (0.05)3
16T
16(1.5)
τ xz = 3 =
= 61.12 MPa
πd
π (0.05)3
z B
So
τ max = (
σx 2
τy
2
ns
) + τ xz2 =
= (48.89) 2 + (61.12) 2 = 78.28 MPa
140
=
ns
A
T
C
P
y
x
Solving,
ns = 1.79
An element at point B:
y
σx
A
τ xy
4 P 16T
Figure (a)
z
+
3A π d 3
4(4 × 6)
16(1.5)
140
=
+
= 65.19 MPa =
2
3
ns
3π (0.05) π (0.05)
B
τ xy
τ max = τ d + τ t =
or
ns = 2.15
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x
_______________________________________________________________________
PROBLEM (9.98) A steel shaft of yield strength σ y and diameter d is subjected to an axial load
P, torque T, and bending moment M as portrayed in Fig. P9.98. Calculate the safety factor ns with
respect to yield based on the maximum energy of distortion criterion of failure.
Given: d = 40 mm,
P = 20 N, T = 500 N ⋅ m,
M = 350 N ⋅ m,
σ y = 250 MPa
SOLUTION
The critical stress occurs at a point A.
P Mc
σx = +
A I
4 P 32 M
=
+
πd2 πd3
4(20)
32(350)
=
+
= 71.62 MPa
2
π (0.04) π (0.04)3
16T 16(500)
τ xy = 3 =
= 39.79 MPa
πd
π (0.04)3
Principal stresses are then
σ 1,2 =
σx
M
d
P T
A
y
τ xy
A
σx
x
σx 2
) + τ xy2 = 35.81 ± (35.81) 2 + (39.79) 2
2
2
= 35.81 ± 53.53
σ 1 = 89.34 MPa
σ 2 = −17.72 MPa
Therefore
± (
σ 12 − σ 1σ 2 + σ 22 = (
σy 2
ns
) ;
(89.34) 2 − (89.34)(−17.72) + (−17.72) 2 =
(250) 2
ns2
or
ns = 2.51
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x
_______________________________________________________________________
PROBLEM (9.99) Resolve Prob. 9.98 for the case in which the shaft is made of cast iron ( σ u = 160
MPa, σ u ' = 320 MPa) using the Coulomb-Mohr theory of failure.
SOLUTION
From the solution of Prob. 9.98, we have
σ 1 = 89.34 MPa
σ 2 = −17.72 MPa
Thus
σ1
σ
89.34 −17.72 1
− ' 2 = 1;
−
=
σ u ns σ u ns
ns
160
320
or
ns = 1.63
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PROBLEM (9.100) A shaft of diameter d is mounted between bearings and loaded by two pulleys
of pitch diameters D as shown in Fig.P9.100. Note that the pitch diameter refers to the diameter where belt
forces T1 and T2 act. Determine the factor of safety ns corresponding to yield, applying the maximum
energy of distortion theory of failure.
Assumptions: Bearings act as simple frictionless supports. The shaft is made of cold-rolled steel with an
allowable strength of σ all .
Given:
d = 45 mm,
D = 150 mm,
T1 = 2T2 ,
σ y = 600 MPa
SOLUTION
Reactions, as found from the conditions of equilibrium, are worked in Fig. a.
The critical region is between A and B.
30 kN
y
75 mm
30 kN
100 mm
B
30 kN A T=1.5 kN ⋅ m
Mz
(N ⋅ m)
x
30 kN
2250
x
Figure (a)
Hence, Eq. (9.31):
σy
32
3
=
M2 + T2
3
4
ns π d
600(106 )
32
3
=
(2250) 2 + (1500) 2
3
ns
π (0.045)
4
ns = 2.07
Solving,
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PROBLEM (9.101) Redo Prob.9.100 if the shaft is made of ASTM class 50 gray cast iron ( σ u =
365 MPa, σ u ' = 1135 MPa) by employing:
(a) The maximum normal stress criterion.
(b) Coulomb-Mohr criterion.
SOLUTION
The state of stress at point A(or B)
is shown in Fig. a. Refer to solution
of Prob. 9.100.
16
(M ± M 2 + T 2 )
πd3
16
=
(2.25 ± (2.25) 2 + (1.5) 2 )(103 )
3
π (0.045)
σ 1,2 =
or
(a)
(b)
σ 1 = 276.9 MPa
σ1
= 1;
σ1
−
σ u ns
σ u ns
σ u ' ns
= 1;
τ xy
Figure (a)
σ 2 = −25.38 MPa
276.9
= 1,
365 ns
σ2
σx
A
ns = 1.32
1 276.9 −25.38
=
−
,
ns
365
1135
ns = 1.28
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PROBLEM (9.102) At a particular point in a cast aluminum alloy ( σ u = 10 ksi, σ u ' = 26 ksi)
structure subjected to plane stress, the stresses have the magnitudes and directions shown in Fig. P9.102.
Determine whether failure occurs at the point according to:
(a) The maximum principal stress theory.
(b) Coulomb-Mohr theory.
SOLUTION
We have σ y = −25 ksi
τ xy = 30 ksi
σ x = 0.
Thus
σ 1,2 = −
σy
± (
σy 2
2
σ 1 = 5.54 ksi
2
) + τ xy2 = −3 ± (3) 2 + (8) 2
σ 2 = −11.54 ksi
(a) Maximum principal stress theory:
But
σ1
= 1;
σu
5.54 < 10
σ2
= 1;
σu
11.54 > 10
(b) Coulomb-Mohr Theory:
σ1 σ 2
−
= 1;
σ u σ u'
gives
(failure occurs)
5.54 −11.54
−
=1
10
26
0.554 + 0.443 = 0.998 < 1
(no fracture)
Comment: Coulomb-Mohr criteria is more reliable when σ u >> σ u ' , as in this
problem.
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_______________________________________________________________________
PROBLEM (9.103) A plate of thickness t = 5 mm is made of a cast-aluminum alloy for which σ u
= 70 MPa and σ u ' = 175 MPa. Calculate the force P required to punch a d = 15 mm-diameter hole
through the plate, as depicted in Fig. P9.103 and discussed in Sec. 2.5. Apply:
(a) The maximum normal stress theory of failure.
(b) Colulomb-Mohr theory of failure.
Assumption: The shear force is taken to be uniformly distributed through the plate thickness.
SOLUTION
Uniform shear stress τ acts on a typical element shown in Fig. (a).
(a) σ 1 = σ u or
σ3 = σu
P
= σu
π td
P = π tdσ u = π (0.005)(0.015)(70 ×106 )
or
P = 16.49 kN
(b)
σ1 σ 3
−
= 1;
σu σu '
τ (1 +
τ = P π dt
σ 1 = −σ 3 = τ
σu
) = σu
σu '
From which
P=
tdσ u
1+
or
σu '
σu
=
(0.005)(0.015)(70 ×106 )
175
1+
70
P = 1.5 kN
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PROBLEMS (10.1 through 10.4) A cantilever beam is loaded as shown in Figs. P10.1 through
10.4. Using the double-integration method, determine:
(a) The equation of the elastic curve.
(b) The slope at the free end.
(c) The deflection at the free end.
SOLUTION (10.1)
1
(a) EIv " = − wx 2
2
1
EIv ' = − wx 3 + C1
6
1
EIv = − wx 4 + C1 x + C2
24
Boundary conditions:
1
v '( L) = 0 : C1 = wL3
6
1
1
v( L) = 0 : − wL4 + wL4 + C2 ,
24
6
Thus,
w
( x3 − L3 )
v' = −
6 EI
w
( x 4 − 4 L3 x + 3L4 )
v=−
24 EI
(b) x=0 in Eq. (1): θ A =
y
w
A
V
x
1
M = − wx 2
2
1
C2 = − wL4
8
(1)
(2)
wL3
6 EI
(c) x=0 in Eq. (2): v A = −
wL4 wL4
=
↓
8EI 8 EI
SOLUTION (10.2)
M = − P( L − x) = P( x − L)
(a)
y
C
A
2EI
PL
L/2
P
EI
L/2
x
B
P
Segment AC
2 EIv1 '' = P( x − L)
Continued on next slide
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1
P( x − L) 2 + C1
2
1
2 EIv1 = P( x − L)3 + C1 x + C2
6
2 EIv1 ' =
Segment CB
EIv2 '' = P( x − L)
1
EIv2 ' = P( x − L) 2 + C3
2
1
EIv2 = P( x − L)3 + C3 x + C4
6
Boundary conditions:
1
1
v1 '(0) = 0 : C1 = − PL2 ,
v1 (0) = 0 : C2 = PL3
2
6
Thus
P
v1 ' =
( x 2 − 2 xL)
4 EI
P
v1 =
( x 3 − 3 x 2 L)
12 EI
(1)
(2)
Boundary conditions:
L
L
3
1
5
v1 '( ) = v2 '( ) : − PL2 = PL2 + C3 , C3 = − PL2
2
2
16
8
16
L
L
5
1
5
1
v1 ( ) = v2 ( ) : − PL3 = − PL3 − PL3 + C4 , C4 = − PL3
2
2
96
48
32
8
So,
P
5
[( x − L) 2 − L2 ]
2 EI
8
P 1
L3
5
v2 =
[ ( x − L)3 − L2 x + ]
2 EI 3
8
4
v2 ' =
(3)
(4)
(b) x=L in Eq. (3):
θB = −
5 PL2 5 PL2
=
16 EI 16 EI
(c) x=L in Eq. (4):
3 PL3 3 PL3
θB = −
=
↓
16 EI 16 EI
Continued on next slide
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SOLUTION (10.3)
(a)
x2
w0
2L
1
EIv " = −
w0 x 3
6L
1
EIv ' = −
w0 x 4 + C1
24 L
1
EIv = −
w0 x 5 + C1 x + C2
120 L
y
x
w0
L
A
M =−
x/3
Boundary conditions:
1
v '( L) = 0 : C1 =
w0 L3
24
1
1
v( L) = 0 : −
w0 L5 + w0 L4 + C2 = 0
120
24
Therefore
w0
v' =
( x 4 − L4 )
24 EIL
w0
( x 5 − 5L4 x + 4 L5 )
v=−
120 EIL
x
V
C2 =
1
w0 L3
24
x3
w0
6L
(1)
(2)
(b) x=0 in Eq. (1):
w L2
θA = 0
24 EI
(c) x=0 in Eq. (2):
vA = −
w0 L4 w0 L4
=
↓
30 EI 30 EI
SOLUTION (10.4)
y
(a)
MA =
3 2
wa
2
w
A
a
C
a
B
x
RA= wa
Segment AC
Continued on next slide
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3
wa
(2 x − 3a )
M 1 = − wa 2 + wax =
2
2
wa
EIv1 '' =
(2 x − 3a)
2
wa 2
EIv1 ' =
( x − 3ax) + C1
2
wa x 3 3 2
EIv1 =
( − ax ) + C1 x + C2
2 3 2
Boundary conditions:
v1 '(0) = 0 : C1 = 0,
Thus
wax
v1 ' =
( x − 3a)
2 EI
wax 2
v1 =
(2 x − 9a)
12 EI
(0 ≤ x ≤ a )
v1 (0) = 0 : C2 = 0
(1)
(2)
Segment BC
w
w
M 2 = − (2a − x) 2 = (− x 2 + 4ax − 4a 2 )
2
2
w
EIv2 '' = (− x 2 + 4ax − 4a 2 )
2
w x3
EIv2 ' = (− + 2ax 2 − 4a 2 x) + C3
2
3
w x4 2
EIv2 = (− + ax3 − 2a 2 x 2 ) + C3 x + C4
2 12 3
Continuity conditions:
7
1
v1 '(a) = v2 '(a) : − wa 3 = − wa 3 + C3 ,
C3 = wa 3
6
6
7
13
1
v1 (a) = v2 (a) : − wa 4 = − wa 4 + C4 ,
C4 = − wa 4
12
24
24
Thus
w
(−2 x3 + 12ax 2 − 24a 2 x − 2a 3 )
12 EI
w
v2 =
(− x 4 + 8ax3 − 24a 2 x 2 + 4a 3 x − a 4 )
24 EI
v2 ' =
(3)
(4)
Continued on next slide
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(b) x=2a in Eq. (3):
7 wa 3 7 wa 3
θB = −
=
6 EI
6 EI
(c) x=2a in Eq. (4):
41 wa 4 41 wa 4
vB = −
=
↓
21 EI 21 EI
________________________________________________________________________
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PROBLEMS (10.5 and 10.6) Solve Probs. 10.1 and 10.2, using the multiple-integration method.
SOLUTION (10.5)
y
w
A
L
B
x
EIv '''' = − w
EIv ''' = − wx + C1
1
EIv '' = − wx 2 + C1 x + C2
2
1
1
EIv ' = − wx3 + C1 x 2 + C3
6
2
1
1
1
EIv = − wx 4 + C1 x 3 + C2 x 2 + C3 x + C4
24
6
2
(1)
(2)
Boundary conditions:
EIv '''(0) = −VA = 0,
C1 = 0
EIv ''(0) = M A = 0,
EIv '( L) = 0 : C3 = wL3 6
EIv( L) = 0,: C3 = − wL4 8
Then, Eqs.(1) & (2) give the results of the solution of Prob. 10.1.
SOLUTION (10.6)
y
P
EI
MA=PL
2EI
L/2
A
RA=P
C
Segment AC
2 EIv1 ''' = −V = P
2 EIv1 '' = Px + C1
1
2 EIv1 ' = Px 2 + C1 x + C2
2
1
1
2 EIv1 = Px 3 + C1 x 2 + C2 x + C3
6
2
Segment BC
EIv2 ''' = P
EIv2 '' = Px + C4
L/2
x
B
(1)
(2)
Continued on next slide
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1 2
Px + C4 x + C5
2
1
1
EIv2 = Px 3 + C4 x 2 + C5 x + C6
6
2
Boundary conditions:
2 EIv1 ''(0) = − PL : C1 = − PL,
2 EIv1 '(0) = 0 : C2 = 0
EIv1 (0) = 0 : C3 = 0
EIv2 ' =
(3)
(4)
Conditions of continuity:
L
L
PL PL
=
+ C4 ,
C4 = − PL
2 EIv1 ''( ) = EIv2 ''( ) : −
2
2
2
2
L
L
3PL2
3PL2
3
v1 '( ) = v2 '( ) : −
=−
+ C5 ,
C5 = PL2
2
2
16
8
16
3
3
L
L
PL
PL3
5PL
v1 ( ) = v2 ( ) : −
=−
+ C6 ,
C6 = −
2
2
96
96
24
Then, Eqs. (1) through (4) yield the results given in solution of Prob. 10.2.
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PROBLEM (10.7 and 10.8) Redo Probs. 10.3 and 10.4, using the multiple-integration method.
SOLUTION (10.7)
(x/L)w0
y
w0
A
B
x
L
x
x2
EIv '''' = − w0
EIv ''' = −
w0 + C1
L
2L
x3
EIv '' = −
w0 + C1 x + C2
6L
x4
1
EIv ' = −
w0 + C1 x 2 + C2 x + C3
24 L
2
5
x
1
1
EIv = −
w0 + C1 x 3 + C2 x 2 + C3 x + C4
120 L
6
2
(1)
(2)
Boundary conditions:
EIv '''(0) = 0 : C1 = 0
EIv ''(0) = 0 : C2 = 0,
EIv '( L) = 0 : C3 =
1
w0 L3
24
1
1
1
w0 L4 + w0 L4 + C4 = 0,
C4 = − w0 L4
120 L
24
30
Then, Eqs.(1) and (2) yield the results given in solution of Prob. 10.3.
EIv( L) = 0 : −
SOLUTION (10.8)
y
w
3
MA= wa2
2
A
a
C
a
B
x
RA=wa
Continued on next slide
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Segment AC
EIv1 '''' = 0
EIv1 ''' = C1
EIv1 '' = C1 x + C2
1
EIv1 ' = C1 x 2 + C2 x + C3
2
1
1
EIv1 = C1 x 3 + C2 x 2 + C3 x + C4
6
2
(1)
(2)
Boundary conditions:
EIv1 '''(0) = −VA = wa : C1 = wa
3
3
EIv1 ''(0) = − wa 2 : C2 = − wa 2
2
2
EIv1 '(0) = 0 : C3 = 0,
EIv1 (0) = 0, C4 = 0
Segment BC
EIv2 '''' = − w
EIv2 ''' = − wx + C5
1
EIv2 '' = − wx 2 + C5 x + C6
2
1
1
EIv2 ' = − wx 3 + C5 x 2 + C6 x + C7
6
2
1
1
1
EIv2 = − wx 4 + C5 x 3 + C6 x 2 + C7 x + C8
24
6
2
Continuity conditions:
v1 '''(a) = v2 '''(a ) : C1 = − wa + C5 ,
(3)
(4)
C5 = 2 wa
wa 2
wa 2
=−
+ 2 wa 2 + C6 ,
C6 = −2 wa 2
2
2
wa 3
1
v2 '(a) = v2 '(a) : − wa 3 = −
+ wa 3 − 2wa 3 + C7 ,
C7 = wa 3
6
6
v1 (a) = v2 (a) :
v1 ''(a) = v2 ''(a) : −
−
wa 4 wa 4 wa 4
7
wa 4 = −
+
+
+ C8 ,
12
24
3
6
C8 = −
1
wa 4
24
Then, Eqs.(1) through (4) yield the results given in solution of Prob.10.4.
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________________________________________________________________________
PROBLEM (10.9) Calculate the maximum slope and maximum deflection of the beam in Figure
P10.1. The beam is an S 100 x 14 steel (see Table B.9):
Given: w = 1.2 kN/m, L = 2 m,
E = 200 GPa
SOLUTION
See solution of Prob. 10.1:
wL3
wL4
θA =
vA =
6 EI
8EI
From Table B.9: I = 2.83 × 106 mm 4
Thus,
1.2 ×103 (2)3
= 2.827 × 10−3 rad
θA =
9
−6
6(200 × 10 )(2.83 × 10 )
1.2 × 103 (2) 4
= 4.24 mm ↓
vA =
8(200 × 109 )(2.83 × 10−6 )
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________________________________________________________________________
PROBLEMS (10.10 and 10.11) A cantilever beam is loaded as depicted in Fig. P10.10 and
P10.11. Using the double-integration approach, determine:
(a) The equation of the deflection curve.
(b) The slope at the free end.
(c) The deflection at the free end.
SOLUTION (10.10)
(a)
y
PL
P
A
L/2
C
L/2
B
x
M 1 = PL
(0 ≤ x ≤ L 2)
3
L
M 2 = PL − Px
( ≤ x ≤ L)
2
2
Segment AC
EIv1 '' = PL
EIv1 ' = PLx + C1
1
EIv1 = PLx 2 + C1 x + C2
2
Segment BC
3
EIv2 '' = PL − Px
2
3
1
EIv2 ' = PLx − Px 2 + C3
2
2
3
1
EIv2 = PLx 2 − Px3 + C3 x + C4
4
6
Boundary conditions:
3 2 1 2
v2 '( L) = 0 :
PL − PL + C3 = 0, C3 = − PL2
2
2
3 3 1 3
5
v2 ( L) = 0 :
PL − PL − PL3 + C4 = 0, C4 = PL3
4
6
12
Thus,
P
(3Lx − x 2 − 2 L2 )
v2 ' =
2 EI
P
(9 Lx 2 − 2 x 3 − 12 L2 x + 5 L3 )
v2 =
12 EI
(1)
(2)
Continued on next slide
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Continuity conditions:
1 2
3
L
L
v1 '( ) = v2 '( ) :
PL + C1 = − PL2 ,
2
2
2
8
5
L
L
PL3
,
v1 ( ) = v2 ( ) : − PL3 + C2 = −
2
2
16
12
7
C1 = − PL2
8
19
C2 = − PL2
48
Therefore,
P
(−8 Lx − 7 L2 )
8 EI
P
(24 Lx 2 − 42 L2 x + 19 L3 )
v1 =
48 EI
v1 ' =
(3)
(4)
(b) x=0 in Eq. (3):
7 PL2 7 PL2
θA = −
=
8 EI 8 EI
(c) x=0 in Eq. (4):
19 PL3
vA = −
↑
48 EI
SOLUTION (10.11)
y
(a)
w
MA=2wa2
A
2a
C
a
B
x
RA=2wa
M1 =
1
w(−4a 2 + 4ax − x 2 )
2
M2 = 0
(0 ≤ x ≤ 2a )
L
( ≤ x ≤ L)
2
Continued on next slide
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Segment AC
w
EIv1 '' = (−4a 2 + 4ax − x 2 )
2
w
x3
EIv1 ' = (−4a 2 x + 2ax 2 − ) + C1
2
3
2 3 x4
w
2 2
EIv1 = (−2a x + ax − ) + C1 x + C2
2
3
12
Boundary conditions:
v1 '(0) = 0 : C1 = 0,
v1 (0) = 0 : C2 = 0
Thus,
w
(−12a 2 x + 6ax 2 − x 3 )
v1 ' =
6 EI
w
(−24a 2 x 2 + 8ax 3 − x 4 )
v1 =
24 EI
Segment BC
EIv2 '' = 0,
EIv2 ' = C3 ,
(1)
(2)
EIv2 = C3 x + C4
Continuity conditions:
4
v1 '(2a) = v2 '(2a) : − wa 3 = C3
3
8
v1 (2a ) = v2 (2a ) : − 2wa 4 = − wa 4 + C4 ,
3
2
C4 = − wa 4
3
Therefore,
4 wa 3
3 EI
wa 3
(−4 x + 2a)
v2 =
3EI
v2 ' = −
(3)
(4)
(b) x=3a in Eq. (3):
4 wa 3 4 wa 3
θB = −
=
3 EI
3 EI
(c) x=3a in Eq. (4):
10 wa 4 10 wa 4
vB = −
=
↓
3 EI
3 EI
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________________________________________________________________________
PROBLEM (10.12) Calculate the deflection at the center of the beam shown in Fig. P10.10 if the
beam is a solid rod of diameter d.
Given:
d = 2 in., P = 80 lb,
L = 3 ft,
E = 10 x 106 psi.
SOLUTION
See solution of Prob.10.10:
P
(24 Lx 2 − 42 L2 x + 19 L3 )
v1 =
48 EI
For x=L/2:
1 PL3 1
80(3 ×12)3
= 39.6 ×10−3 in.
=
vC =
12 EI 12 10 ×106 (π × 24 64)
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_______________________________________________________________________
PROBLEMS (10.13 and 10.14) Solve Probs. 10.10 and 10.11, using the multiple-integration
approach.
SOLUTION (10.13)
y
PL
P
A
L/2
C
L/2
B
x
Segment AC
EIv1 '''' = 0
EIv1 ''' = C1
EIv1 '' = C1 x + C2
1
EIv1 ' = C1 x 2 + C2 x + C3
2
1
1
EIv1 = C1 x 3 + C2 x 2 + C3 x + C4
6
2
(1)
(2)
Segment BC
EIv2 '''' = 0
EIv2 ''' = C5
EIv2 '' = C5 x + C6
1
EIv2 ' = C5 x 2 + C6 x + C7
2
1
1
EIv2 = C5 x 3 + C6 x 2 + C7 x + C8
6
2
(3)
(4)
Boundary conditions:
v1 '''(0) = 0 : C1 = 0
v1 ''(0) = PL : C2 = PL
L
v2 '''( ) = − P : C5 = − P
2
3
L
v2 ''( ) = PL : C6 = PL
2
2
1 2 3 2
v2 '( L) = 0 : − PL + PL + C7 = 0,
C7 = − PL2
2
2
3
5
PL 3 3
v2 ( L) = 0 : −
+ PL − PL3 + C8 = 0,
C8 = PL3
6
4
12
Continued on next slide
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Continuity conditions:
3
7
L
L
PL2
v1 '( ) = v2 '( ) :
+ C3 = − PL2 ,
C3 = − PL2
2
2
2
8
8
3
3
3
19
L
L
PL 7 PL
PL
,
v1 ( ) = v2 ( ) :
−
+ C4 =
C4 = − PL3
2
2
8
16
12
48
Then, Eqs.(1) through (4) yield the results given in solution of Prob. 10.10.
SOLUTION (10.14)
y
w
MA=2wa2
A
2a
C
a
B
x
RA=2wa
Segment AC
EIv1 '''' = − w
EIv1 ''' = − wx + C1
1
EIv1 '' = − wx 2 + C1 x + C2
2
1 3 1
EIv1 ' = − wx + C1 x 2 + C2 x + C3
6
2
1
1
1
EIv1 = − wx 4 + C1 x 3 + C2 x 2 + C3 x + C4
24
6
2
Segment BC
EIv2 '''' = 0
EIv2 ''' = C5
EIv2 '' = C5 x + C6
1
EIv2 ' = C5 x 2 + C6 x + C7
2
1
1
EIv2 = C5 x 3 + C6 x 2 + C7 x + C8
6
2
(1)
(2)
(3)
(4)
Continued on next slide
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Boundary conditions:
EIv1 '''(0) = −V = 2aw : C1 = 2aw
EIv1 ''(0) = −2aw2 : C2 = −2aw2
EIv1 '(0) = 0 : C3 = 0
EIv1 (0) = 0 : C4 = 0
Continuity conditions:
v1 '''(2a) = v2 '''(2a ) : C5 = 0
v1 ''(2a) = v2 ''(2a) : C6 = 0
4
v1 '(2a) = v2 '(2a) : − wa 3 = C7
3
8
2
v1 (2a ) = v2 (2a ) : − 2wa 4 = − wa 4 + C8 ,
C8 = wa 4
3
3
Then, Eqs.(1) through (4) yield the results given in solution of Prob. 10.11
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________________________________________________________________________
PROBLEM (*10.15) A cantilever beam is loaded as shown in Fig. P10.15. Using the multiple-integration method, determine:
(a) The equation of the elastic curve.
(b) The deflection at the free end.
(c) The reactions at the fixed support.
*SOLUTION
(a) EIv '''' = − w0 cos
EIv ''' = −(
2 w0 L
πx
2L
πx
+ C1
2L
4 w L2
πx
+ C1 x + C2
EIv '' = ( 02 ) cos
π
2L
8w L3
πx 1
+ C1 x 2 + C2 x + C3
EIv ' = ( 03 ) sin
π
2L 2
4
16w0 L
1
πx 1
+ C1 x 3 + C2 x 2 + C3 x + C4
) cos
EIv = −(
4
π
2L 6
2
π
) sin
Boundary conditions:
16w0 L4
v(0) = 0 : C4 =
4
π
v '(0) = 0 : C3 = 0
2w L
v(0) ''' = 0 : C1 = 0
π
v(0) '' = 0 : C2 = −
2 w0 L2
π
Therefore,
w
πx
− π 3 Lx3 + 3π 3 L2 x 2 − 48 L4 ]
v = − 40 [48L4 cos
3π EI
2L
(b) Let x=L in the above expression:
w
w L4
vB = − 20 (−π 3 L4 + 3π 3 L4 − 48L4 ) = 0.04795 0 ↓
3π EI
EI
(c) Expressions for shear & moment are
2w L
πx
V = 0 (1 − sin )
π
2L
2w0 L
πx
M = 2 (cos
+ π x − π L)
π
2L
Continued on next slide
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These result in, for x=0:
2w L
RA = 0 ↑
π
2(π − 2) w0 L2
MA =
π2
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________________________________________________________________________
PROBLEMS (10.16 through *10.19) A simple beam is loaded as shown in Figs. P10.16
through P10.19. Using the double-integration method, determine:
(a) The equation of the elastic curve.
(b) The slope at the end A.
(c) The deflection at midspan.
SOLUTION (10.16)
(a)
w0
y
L−x
( L − x)2
w0
L
2L
wo
A
L
2
w0L/3
x
C
L
2
M
B
O
V
w0L/6
w0L/6
w0
1
L-x
∑ M = 0 : − M + 6 w L( L − x) − 6 L ( L − x) = 0
o
3
0
or
M=
w0 x 3 w0 x 2 w0 Lx
−
+
6L
2
3
Thus,
w0 x3 w0 x 2 w0 Lx
−
+
6L
2
3
4
3
w x w x w Lx 2
EIv ' = 0 − 0 + 0
+ C1
24 L
6
6
w x 5 w x 4 w Lx 3
EIv = 0 − 0 + 0
+ C1 x + C2
120 L
24
18
EIv '' =
(1)
(2)
Boundary conditions:
v(0) = 0 : C2 = 0
v( L) = 0 :
w0 L4 w0 L4 w0 L4
−
+
+ C1 L,
120
24
18
C1 = −
Equations (1) and (2):
w0
(15 x 4 − 60 Lx 3 + 60 L2 x 2 − 8 L4 )
v' =
360 EIL
w0
(3x 5 − 15Lx 4 + 20 L2 x 3 − 8L4 x)
v=
360 EIL
w0 L3
45
(3)
(4)
Continued on next slide
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(b) Make x=0 in Eq. (3):
w0 L3
w0 L3
θA = −
=
45 EI 45 EI
(c) Make x=L/2 in Eq. (4):
5w L4 5w0 L4
vC = − 0 =
↓
768EI 768EI
SOLUTION (10.17)
y
A
w
L/2
C
L/2
RA=wL/8
1
L
(0 ≤ x ≤ )
M 1 = wLx
8
2
1
1
L 2
M 2 = wLx − w( x − )
8
2
2
5
1 2 1 2
= wLx − wx − wL
8
2
8
B
x
RB=3wL/8
L
( ≤ x ≤ L)
2
(a) Segment AC
1
EIv1 '' = wLx
8
1
EIv1 ' = wLx 2 + C1
16
1
EIv1 =
wLx 3 + C1 x + C2
48
Segment BC
5
1
1
EIv2 '' = wLx − wx 2 − wL2
8
2
8
5
1
1
EIv2 ' = wLx 2 − wx3 − wL2 x + C3
16
6
8
5
1
1
EIv2 =
wLx3 − wx 4 − wL2 x 2 + C3 x + C4
48
24
16
(1)
(2)
(3)
Continued on next slide
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Boundary and continuity conditions:
v1 (0) = 0 : C2 = 0
wL3
L
L
wL3
wL3
C3 = C1 +
v1 '( ) = v2 '( ) :
+ C1 = −
+ C3 ,
48
2
2
64
192
4
4
4
4
L
L
wL C1 L 5wL wL wL C1 L wL4
v1 ( ) = v2 ( ) :
+
=
−
−
+
+
+ C4 ,
2
2
384
2
384 384 64
2
96
wL4
C4 = −
384
4
4
5wL wL wL4
wL3
wL4
C3 =
v2 ( L) = 0 :
−
−
+ C3 L +
= 0,
384
48
24
16
384
3
Equation (a) gives C1 = − 7 wL 384 .
Equations (1) through (3):
w
(24 Lx 2 − 7 L3 )
v1 ' =
384 EI
w
(8 Lx 3 − 7 L3 x)
v1 =
384 EI
w
(40 Lx3 − 16 x 4 − 24 L2 x 2 + L3 x − L4 )
v2 =
384 EI
(a)
(4)
(5)
(b) Make x=L/2 in Eq. (4):
7 wL3
7 wL3
θA = −
=
384 EI 384 EI
(c) Make x=0 in Eq. (5):
5wL4
5wL4
vC = −
=
↓
768EI 768EI
Continued on next slide
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SOLUTION (10.18)
(a)
y
M0
A
a
b
C
B
x
L
RA=M0/L
RB= M0/L
M0
(0 ≤ x ≤ a)
x
L
M
M2 = − 0 x + M0
(0 ≤ x ≤ a )
L
Segment AC
M
EIv1 '' = − 0 x
L
M
EIv1 ' = − 0 x 2 + C1
2L
M
EIv1 = − 0 x 3 + C1 x + C2
6L
M1 = −
(1)
(2)
Segment BC
M
M
EIv2 '' = − 0 x + M 0
EIv2 ' = − 0 x 2 + M 0 x + C3
2L
L
M0 3 M0
EIv2 = −
x +
x + C3 x + C4
6L
2
Boundary conditions:
v1 (0) = 0 : C2 = 0
v2 ( L) = 0 :
M 0 L2
+ C3 L + C4 = 0
3
Continuity conditions:
v1 '(a) = v2 '(a ) : C1 = M 0 a + C3
1
v1 (a ) = v2 (a) : C1a = M 0 a 2 + C3 a + C4
2
Solving Eqs.(a) through (c):
M L
M a2
C1 = − 0 + M 0 a − 0
3
2L
M L M a
M a2
C3 = − 0 − 0
C4 = − 0
3
2L
2
(3)
(a)
(b)
(c)
Continued on next slide
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Then, Eqs.(1) through (3):
M0
(3x 2 + 2 L2 − 6aL + 3a 2 )
v1 ' = −
6 EIL
M0
( x 3 + 2 L2 x − 6aLx + 3a 2 x)
v1 = −
6 EIL
M0
( x3 − 3Lx 2 + 2 L2 x + 3a 2 x − 3L2 a )
v2 = −
6 EIL
(4)
(5)
(6)
(b) Make x=0 in Eq. (4):
M
θ A = − 0 (2 L2 − 6 La + 3a 2 )
6 EIL
(c) Make x=L/2 in Eq. (5):
M 9 L2
v=− 0 (
− 6aL + 3a 2 )
6 EI 4
*SOLUTION (10.19)
(a)
y
k
P
A
a
C
P/2
a
B
x
P/2
1
Px
(0 ≤ x ≤ a)
2
1
1
M 2 = Px − P( x − a) = − Px + Pa
2
2
M1 =
Segment AC
1
EIv1 '' = Px
2
1
EIv1 ' = Px 2 + C1
4
1
EIv1 = Px 3 + C1 x + C2
12
( a ≤ x ≤ 2a )
(1)
(2)
Continued on next slide
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Segment BC
1
EIv2 '' = − Px + Pa
2
1
EIv2 ' = − Px 2 + Pax + C3
4
1
1
EIv2 = − Px3 + Pax 2 + C3 x + C4
12
2
(3)
Boundary and continuity conditions:
P
PEI
v1 (0) = − : C2 = −
2k
2k
3
C4 2 Pa 2
2 Pa
3
C3 = − −
v2 (2a ) = 0 : −
+ 2 Pa + 2aC3 + C4 = 0 ,
2a
3
3
2
2
2
Pa
Pa
Pa
+ C3
C1 =
v1 '(a) = v2 '(a) :
+ C1 = −
+ Pa 2 + C3 ,
2
4
4
Pa 3
PEI
Pa 3 Pa 3
Pa 3
v1 (a) = v2 (a) :
+ C1a −
=−
+
+ C1a −
+ C4 ,
12
2k
12
2
2
Pa 3 PEI
C4 =
−
6
2k
Solve Eqs.(a) and (b):
3
PEI
C3 = − Pa 2 +
4
4ak
2
Pa
PEI
C1 = −
+
4
4ak
Then, Eqs.(1) through (3):
P
EI
( x2 − a2 + )
v1 ' =
4 EI
ak
3EI
6 EI
P
( x3 − 3a 2 x +
)
v1 =
x−
12 EI
ak
k
3EI
6 EI
P
(− x 3 + 6ax 2 − 9a 2 x +
)
v2 =
x + 2a 3 −
12 EI
ak
k
(a)
(b)
(4)
(5)
Continued on next slide
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(b)
Make x=0 in Eq. (4):
P
EI
(−a 2 + )
θA = −
4 EI
ak
(c) Make x=a in Eq. (5):
3EI
P
(2a 3 +
)↓
vC =
12 EI
k
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________________________________________________________________________
PROBLEMS (10.20 and 10.21) Redo Probs. 10.16 and 10.17, using the multiple-integration
method.
SOLUTION (10.20)
y
w0
x
( L − x) = w0 − w0
L
L
wo
A
w0 L
3
x
x
B
L-x
w0 L
6
x
x2
w0 ,
EIv ''' = − w0 x +
w0 + C1
2L
L
1
x3
EIv '' = − w0 x 2 +
w0 + C1 x + C2
2
6L
1
1
x4
3
EIv ' = − w0 x +
w0 + C1 x 2 + C2 x + C3
6
24 L
2
5
1
x
1
1
EIv = − w0 x 4 +
w0 + C1 x3 + C2 x 2 + C3 x + C4
24
120 L
6
2
EIv '''' = − w0 +
(1)
(2)
Boundary conditions:
wL
wL
v '''(0) = −V = 0 : C1 = 0
3
3
v ''(0) = 0 : C2 = 0
v(0) = 0 : C4 = 0
w0 L4 w0 L4 w0 L4
w L3
+
+
+ C3 L = 0,
C3 = − 0
24
120
18
45
Then, Eqs.(1) and (2) yield the results given in solution of Prob.10.16.
v( L) = 0 : −
Continued on next slide
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SOLUTION (10.21)
y
A
wL/8
w
L/2
C
L/2
B
x
3wL/8
Segment AC
EIv1 '''' = 0
EIv1 ''' = C1
EIv1 '' = C1 x + C2
1
EIv1 ' = C1 x 2 + C2 x + C3
2
1
1
EIv1 = C1 x 3 + C2 x 2 + C3 x + C4
6
2
(1)
(2)
Segment BC
EIv2 '''' = − w0
EIv2 ''' = − w0 x + C5
1
EIv2 '' = − w0 x 2 + C5 x + C6
2
1
1
EIv2 ' = − w0 x 3 + C5 x 2 + C6 x + C7
6
2
1
1
1
EIv2 = − w0 x 4 + C5 x 3 + C6 x 2 + C7 x + C8
24
6
2
Boundary and continuity conditions:
EIv1 (0) = 0 : C4 = 0
1
1
EIv1 '''(0) = −V = wL : C1 = wL
8
8
EIv1 ''(0) = 0 : C2 = 0
3
3
EIv2 '''( L) = − wL : − w0 L + C5 = wL,
8
8
1 2
EIv2 ''( L) = 0 : C6 = − wL
8
(3)
5
C5 = wL
8
Continued on next slide
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and
1
L
L
wL3
wL3
v1 '( ) = v2 '( ) :
+ C3 = −
wL3 + C7 ,
C7 = C3 +
(4)
2
2
64
192
48
L
L
wL4 wL4
wL4
v1 ( ) = v2 ( ) :
=
+ C8 ,
C8 = −
2
2
384 192
384
3
wL
v2 ( L) = 0 : C7 =
384
7 wL3
Equation (4): C3 = −
128
Then, Eqs.(1) through (3) yield the results given in the solution of Prob. 10.17.
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________________________________________________________________________
PROBLEMS (10.22 and 10.23) Redo Probs. 10.18 and 10.19, using the multiple-integration
method.
SOLUTION (10.22)
y
M0
A
a
C
L
RA=M0/L
B
b
x
RB= M0/L
Segment AC
EIv1 '''' = 0
EIv1 ''' = C1 = −
M0
L
0
M0
x + C2
L
M
EIv1 ' = − 0 x 2 + C3
2L
M0 3
EIv1 = −
x + C3 x + C4
6L
EIv1 '' = −
(1)
(2)
Segment BC
EIv2 '''' = 0
EIv2 ''' = C5 = −
M0
L
M0
x + C6
EIv2 ''( L) = 0,
L
M
EIv2 ' = − 0 x 2 + M 0 x + C7
2L
M0 3 1
EIv2 = −
x + M 0 x 2 + C7 x + C8
6L
2
EIv2 '' = −
C6 = M 0
(3)
Boundary and continuity conditions:
v1 (0) = 0 : C4 = 0
M 0 L2
+ C7 L + C8 = 0
3
v1 '(a ) = v2 '(a ) : C3 = M 0 a + C7
1
v1 (a ) = v2 (a) : C3 a = M 0 a + C7 a + C8
2
v2 ( L) = 0 :
Continued on next slide
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Solving,
M0L
M 0a2
C3 = −
+ M 0a −
3
2L
2
M L M a
M a2
C7 = − 0 − 0
C8 = 0
3
2L
2
Then, Eqs.(1) through (3) yield the results given in solution of Prob.10.18.
SOLUTION (10.23)
y
P
A
k
a
C
a
B
x
P/2
P/2
Segment AC
0
P
Px
EIv1 ''' =
EIv1 '' =
+ C1
2
2
1 2
EIv1 ' = Px + C2
4
1
EIv1 = Px 3 + C2 x + C3
12
Segment BC
P
EIv2 ''' = −
2
(1)
(2)
1
EIv2 '' = − Px + C4
2
EIv2 ''(2a) = 0 : C4 = Pa,
1
EIv2 '' = − Px + Pa
2
1
EIv2 ' = − Px 2 + Pax + C5
4
1
1
EIv2 = − Px3 + Pax 2 + C5 x + C6
12
2
Boundary & continuity conditions:
P
PEI
v1 (0) = − : C3 = −
2k
2k
C6 2 2
v2 (2a ) = 0 : C5 = − − Pa
2a 3
1
v1 '(a) = v2 '(a) : C2 = Pa 2 + C5
2
1
PEI
v1 (a) = v2 (a) : C6 = Pa 3 −
6
2k
(3)
Continued on next slide
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Solving,
3
PEI
1
PEI
C5 = − Pa 2 +
C2 = − Pa 2 +
4
4ak
4
4ak
Then, Eqs.(1) through (3) yield the results given in solution of Prob.10.19.
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________________________________________________________________________
PROBLEM (10.24) Calculate the deflection at midspan of the beam shown in Fig. P10.17 if the
beam is an S 200 x 34 steel shape (see Table B.9).
Given: w = 15 kN/m , L = 3 m,
E = 210 GPa.
SOLUTION
From Table B.9: I = 27 × 106 mm 4
See solution of Prob. 10.17:
5 wL4
5
15 × 103 (3) 4
vC =
=
768 EI
768 210 ×109 (27 × 10−6 )
= 1.4 × 10−3 m = 1.4 mm ↓
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________________________________________________________________________
PROBLEM (10.25 and 10.26) For the beam and loading shown in Figs. P10.25 and 10.26, using
the double-integration method, determine:
(a) The equation of the deflection curve.
(b) The slope at the end A.
(c) The deflection at the midspan.
SOLUTION (10.25)
w = w0 sin
(a)
y
A
RA
L/2
C
πx
L
L2
0
sin
πx
L
dx =
w0 L
π
RB
x
V
x
B
L/2
RA = RB = w0 ∫
V=
M
w0 L
π
cos
M=
w0 L2
π
πx
2
L
x
sin
πx
L
x
We have
L
πx
EIv '' = w0 ( ) 2 sin
π
L
L
πx
EIv ' = − w0 ( )3 cos
+ C1
π
L
L
πx
EIv = − w0 ( ) 4 sin
+ C1 x + C2
π
L
(1)
(2)
Boundary conditions:
v(0) = 0 : C2 = 0 ,
Equations (1) and (2) become
w L3
πx
v ' = 30 cos
π EI
L
4
wL
πx
v = 40 sin
↓
π EI
L
L
v '( ) = 0 : C1 = 0
2
(3)
(4)
Continued on next slide
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(b)
Make x=0 in Eq. (3):
w0 L3
θA = 3
π EI
(c) Make x=L/2 in Eq. (4):
w L4
vC = 40 ↓
π EI
SOLUTION (10.26)
(a)
x 2 w0 L
x
L
1 w0 3 1
M =−
x + w0 Lx
3 L
4
x/3
2w0
y
A
x
RA=w0L/4
O
V
Due to symmetry only part AC of the beam need be considered.
1
1
EIv '' = − w0 x 3 + w0 Lx
3L
4
1
1
EIv ' = −
w0 x 4 + w0 Lx 2 + C1
12 L
8
1
1
EIv = −
w0 x 5 + w0 Lx3 + C1 x + C2
60 L
24
(1)
(2)
Boundary conditions:
5
L
v '( ) = 0 : C1 = −
w0 L3
2
192
Thus, Eqs.(1) and (2) become
w L3
x
x
v ' = 0 [−80( ) 4 + 120( ) 2 − 25]
960 EI
L
L
4
wL
x
x
x
v = 0 [−16( )5 + 40( )3 − 25( )]
960 EI
L
L
L
v(0) = 0 : C2 = 0,
(3)
(4)
(b) Make x=0 in Eq. (3):
5w0 L3
θA =
192 EI
Continued on next slide
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( c)
Make x=L/2 in Eq. (4):
w0 L4
vC =
↓
120 EI
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________________________________________________________________________
PROBLEMS (10.27 and 10.28) Solve Probs. 10. 25 and 10.26, using the multiple-integration
approach.
SOLUTION (10.27)
EIv '''' = − w0 sin
π 2
πx
π
πx
EIv ''' = w0 ( ) cos
+ C1
L
L
L
πx
EIv '' = w0 ( ) sin
+ C1 x + C2
L
L
π
πx 1
+ C1 x 2 + C2 x + C3
EIv ' = − w0 ( )3 cos
L
L 2
π
πx 1
1
+ C1 x 3 + C2 x 2 + C3 x + C4
EIv = − w0 ( ) 4 sin
2
L
L 6
(1)
(2)
Boundary conditions:
v(0) = 0 : C4 = 0
v ''(0) = 0 : C2 = 0
v ''( L) = 0 : C1 = 0
v( L) = 0 : C3 = 0
Then, Eqs.(1) and (2) yield the results given in solution of Prob.10.25.
SOLUTION (10.28)
∑F = 0: R = R = w L 4
y
A
B
0
Due to symmetry only segment AC need be considered and
2x
w = − w0
L
Thus,
2x
x2
EIv '''' = − w0
EIv ''' = − w0 + C1
L
L
3
x
EIv '' = −
w0 + C1 x + C2
3L
1
x4
EIv ' = −
w0 + C1 x 2 + C2 x + C3
(1)
12 L
2
x5
1
1
EIv = −
w0 + C1 x 3 + C2 x 2 + C3 x + C4
(2)
60 L
6
2
Boundary conditions:
1
1
EIv '''(0) = w0 L : C1 = w0 L
4
4
EIv ''(0) : 0 : C2 = 0,
v(0) = 0 : C4 = 0
Continued on next slide
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5
L
v '( ) = 0 : C3 = −
w0 L3
2
192
Then, Eqs.(1) and (2) yield the results given in solution of Prob.10.26.
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________________________________________________________________________
PROBLEM (*10.29) Find the slope at A of the beam in Fig. P10.26 if the beam is a W 250 x 80
steel shape (see Table B.8).
Given: w = 50 kN/m, L = 4 m,
E = 200 GPa.
*SOLUTION
From Table B.8: I = 126 × 106 mm 4
See solution of Prob.10.26:
5 w0 L3
5
50 × 103 (4)3
θA =
=
192 EI
192 200 ×109 (126 × 10−6 )
= 3.31× 10−3 rad
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________________________________________________________________________
PROBLEM (*10.30 through 10.33) A beam supported and loaded as shown in Figs. P10.30
through P10.33, using a direct-integration method, determine:
(a) The equation of the elastic curve.
(b) The deflection at midspan.
(c) The magnitude and location of the maximum deflection.
(d) The slope at the right end.
SOLUTION (*10.30)
(a)
wo
y
A
RA
xm
RB
B
x
L
x
x2
EIv '''' = − w0
EIv ''' = −
w0 + C1
2L
L
x3
EIv '' = −
w0 + C1 x + C2
6L
1
x4
EIv ' = −
w0 + C1 x 2 + C2 x + C3
24 L
2
5
x
1
1
EIv = −
w0 + C1 x3 + C2 x 2 + C3 x + C4
120 L
6
2
Boundary conditions:
v ''(0) = 0 : C2 = 0
wL
v ''( L) = 0 : C1 = 0
6
v(0) = 0 : C4 = 0
7 w0 L3
v( L) = 0 : C3 = −
360
Equations (1) and (2) become
w0
(−15 x 4 + 30 L2 x 2 − 7 L4 )
v' =
360 EIL
w0 x
(−3 x 4 + 10 L2 x 2 − 7 L4 )
v=
360 EIL
(1)
(2)
(3)
(4)
(b) Make x=L/2 in Eq.(4):
5 w0 L4
vC =
↓
768 EI
Continued on next slide
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Set v ' = 0 in Eq.(3):
( c)
8
= 0.5193L
15
xm = L 1 −
Then Eq.(2) gives
w0 L4
vmax = v( xmax ) = 0.00652
↓
EI
(d) Make x=L in Eq.(1):
w L3
θB = 0
45 EI
SOLUTION (10.31)
(a)
y
L/2
A
xm
M0/L
M0
C
L/2
B
x
M0/L
Due to asymmetric but equal deflection configuration, only segment AC need be
considered.
M0
x
L
M
EIv ' = 0 x 2 + C1
2L
M0 3 1
EIv =
x + C1 x + C2
6L
2
EIv '' =
Boundary conditions:
v(0) = 0 : C2 = 0
M L
L
v( ) = 0 : C1 = − 0
2
12
Then,
M0
(12 x 2 − L2 )
v' =
12 EIL
M0x
(4 x 2 − L2 )
v=
12 EIL
(1)
(2)
Continued on next slide
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(b)
Let x=L/2 in Eq.(2):
vC = 0 as required
(c) Set v ' = 0 in Eq.(1):
L
xm =
= 0.2887 L
12
Then, Eq.(2) gives
M 0 L2
M L2
vmax = v( xmax ) = −
= 0.00802 0 ↓
EI
36 12 EI
or, in segment BC:
M L2
vmax = 0.00802 0 ↑
( xm = 0.7113L )
EI
(d) Make x=0 in Eq.(1):
M L2
θ A = −θ B = 0
24 EI
SOLUTION (10.32)
(a)
y
M0
M =−
A
x
V
M0
x + M0
L
RA=M0/L
M0
x + M0
L
M
EIv ' = − 0 x 2 + M 0 x + C1
2L
M0 3 1
EIv = −
x + M 0 x 2 + C1 x + C2
6L
2
EIv '' = −
(a)
(b)
Boundary conditions:
v(0) = 0 : C2 = 0,
1
v( L) = 0 : C1 = M 0 L
3
Continued on next slide
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Then Eqs.(a) and Eqs.(b) become
M0
(3 x 2 − 6 Lx + 2 L2 )
v' = −
6 EIL
M x
v = − 0 ( x 2 − 3Lx + 2 L2 )
6 EIL
(1)
(2)
(b) Let x=L/2 in Eq.(b):
M 0 L2
vM =
↓
16 EI
(c) For v ' = 0 , Eq.(1) yields
3 x 2 − 6 Lx + 2 L2 = 0 or
Substitute this into Eq.(2):
vmax = v( xm ) = 0.0642
xm = 0.4227 L .
M 0 L2
↓
EI
(d) Making x=L in Eq.(1):
M L
θB = 0
6 EI
SOLUTION (10.33)
y
(a)
w
MA=wL2/8
A
L/2
C
L/2
B
x
RA=wL/2
Segment AC
1
1
1
wLx − wL2 − wx 2
2
8
2
4
w
EIv1 ' = (2 Lx 2 − L2 x − x3 ) + C1
8
3
3
2 2
w 2 Lx L x
x4
EIv1 = (
−
− ) + C1 x + C2
8 3
2
3
EIv1 '' = M 1 =
Segment BC
EIv2 '' = 0
EIv1 ' = C3
EIv = C3 x + C4
(a)
(b)
(c)
Continued on next slide
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Boundary and continuity condition:
v1 (0) = 0 : C2 = 0
v1 '(0) = 0 : C1 = 0
L
L
wL3
v1 '( ) = v2 '( ) : C3 = −
2
2
48
L
L
wL4
v1 ( ) = v2 ( ) : C4 = −
2
2
384
Then Eqs.(b) and (c) become
wx 2
(4 Lx − 3L2 − 2 x 2 )
v1 =
48 EI
wx3
v2 ' = −
48 EI
wx 3
(−8 x + L)
v2 =
384 EI
(1)
(2)
(3)
(b) Let x=L/2 in Eq.(3)
3 wL4
vC =
↓
384 EI
(c) Let x=L in Eq.(3)
7 wL4
vB =
↓
384 EI
(d) We have
θ B = θC =
wL3
48 EI
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________________________________________________________________________
PROBLEM (*10.34 and *10.35) An overhanging beam is loaded as shown in Figs. P10. 34 and
P10.35. Using a direct-integration method, determine:
(a) The equation of the elastic curve.
(b) The deflection at midspan.
(c) The magnitude and location of the maximum deflection.
(d) The slope at the right end.
SOLUTION (*10.34)
(a)
y
w
A
L
3wL/8
B
L/2
C
x
9wL/8
Segment AB
3
1
EIv1 '' = M 1 = wLx − wx 2
8
2
3
1
EIv1 ' = wLx 2 − wx3 + C1
16
6
1
1
EIv = wLx3 − wx 4 + C1 x + C2
16
24
(a)
(b)
Segment BC
3
wx 2 9
EIv2 '' = M 2 = wLx −
+ wL( x − L)
8
2
8
3
3
9
wx
EIv2 ' = wLx 2 −
+ wL( x − L) 2 + C3
16
6 16
3
4
3
wLx wx
EIv2 =
−
+ wL( x − L)3 + C3 x + C4
16
24 16
Boundary & continuity conditions:
wL3
v1 (0) = 0 : C2 = 0,
v1 ( L) = 0 : C1 = −
48
v1 '( L) = v2 '( L) : C1 = C3 , v2 ( L) = 0 : C4 = 0
(c)
(d)
Continued on next slide
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Then, Eqs.(a) through (d) become
w
(9 Lx 2 − 8 x 3 − L3 )
48 EI
wx
(3Lx 2 − 2 x 3 − L3 )
v1 =
48 EI
w
[9 Lx 2 − 8 x 3 − 27( x − L) 2 − L3 ]
v2 ' =
48 EI
w
[3Lx 3 − 2 x 4 − L3 x + 9 L( x − L)3 ]
v2 =
48 EI
v1 ' =
(1)
(2)
(3)
(4)
(b) Let x=L/2 in Eq.(2):
wL4
vM =
↓
192 EI
(c) Let x=3L/2 in Eq.(4):
wL4
wL4
(v2 ) max = vC = −
= 0.0078
↓
128EI
EI
To find (v2 ) max , make v1 ' = 0 in Eq.(1): 9 Lx 2 − 8 x 3 − L3 = 0
Solving, by trial and error: xm = 0.422 L
Then Eq.(2) yields
wL4
(v1 ) max = 0.0054
↓
EI
(d) Let x=3L/2 in Eq.(3):
1 wL3
θC =
12 EI
SOLUTION (*10.35)
(a)
y
C
P
P
P
L/2
A
3P/2
E
L
B
L/2 D
x
3P/2
Due to symmetry only one-half the beam need be considered.
Continued on next slide
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Segment CA
EIv1 '' = M 1 = − Px
1
EIv1 ' = − Px 2 + C1
2
1
EIv1 = − Px 3 + C1 x + C2
6
(a)
(b)
Segment AE
EIv1 '' = M 2 = − Px +
3P
1
3
L
( x − ) = Px − PL
2
2
2
4
1 2 3
Px − PLx + C3
4
4
1
3
EIv2 = Px3 − PLx 2 + C3 x + C4
12
8
EIv2 ' =
(c)
(d)
Boundary & continuity conditions
1 2 3 2
1
v2 '( L) = 0 :
PL − PL + C3 = 0,
C3 = PL2
4
4
2
2
2
2
5
L
L
PL
PL 3PL PL2
,
v1 '( ) = v2 '( ) : −
+ C1 =
−
+
C1 = PL2
2
2
8
16
8
2
16
3
3
3
13PL
L
PL 5 PL
v1 '( ) = 0 :
+
+ C2 = 0,
C2 =
2
48
32
96
3
3
3
L
PL 3PL PL
PL3
v2 ( ) = 0 :
−
+
C4 = 0,
C4 = −
2
96
32
4
6
Equations (a), (b), and (d):
P
(−8 x 2 + 5 L2 )
v1 ' =
16 EI
P
(−16 x 3 + 30 L2 x − 13L3 )
v1 =
96 EI
P
(8 x 3 − 36 Lx 2 + 48 L2 x − 16 L3 )
v2 =
96 EI
(1)
(2)
(3)
(b) Making x=L in Eq.(3):
PL3
vE =
↑
24 EI
Continued on next slide
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( c)
Making x=0 in Eq.(2):
13PL3
vC =
↓
96 EI
Thus, vC = vmax
(d) Making x=0 in Eq.(1):
5 PL2
θC = −θ D =
16 EI
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________________________________________________________________________
PROBLEM (10.36) Calculate the slope at A of the beam shown in Fig. P10.31 if the beam is a W
12 x 72 steel shape (see Table B.8).
Given: M = 400 kips ⋅ in., L = 15 ft,
E = 30 x 106 psi
SOLUTION
From Table B.8, for a W12x72 section: I = 597 in.4
See solution of Prob. 10.31:
M 0 L2
θA =
24 EI
Substitute the given data
400 × 103 (15 × 12) 2
θA =
= 30.24 × 10−3 rad
6
24(30 × 10 )(597)
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________________________________________________________________________
PROBLEM (10.37) A cantilever of variable cross section supports a uniformly distributed load w
(Fig. P10.37). Verify, using the last of Eqs. (10.4), that the expression for the deflection curve is
v=
wL
(− x 3 + 3L2 x − 2 L3 )
Eb0 h3
(P10.37)
Here h represents the depth of the beam and b0 is the width at the fixed end.
SOLUTION
1 x
( b0 )h3
12 L
The last of Eq. (10.4):
(b0 x L)h3 d 2 v
d2
d 2v
d2
(
)
[
] = −W
EI
=
E
dx 2
dx 2
dx 2
dx 2
12
Integrating twice
(b x L)h3 d 2 v
d
[E 0
] = − wx + C1 = −V
dx
dx 2
12
I=
(b0 x L)h3 d 2 v
1
E
= − wx 2 + C1 x + C2 = M
2
12
2
dx
Boundary conditions:
M (0) = 0 : C2 = 0
V (0) = 0 : C1 = 0
Equation (1) becomes then
d 2v
6wL
dv
3wL 2
=−
x
=−
x + C3
2
3
dx
Eb0 h
dx
Eb0 h3
wL 3
v=−
x + C3 x + C4
Eb0 h3
We have
3wL3
2 wL4
v '( L) = 0 : C3 =
,
v ( L ) = 0 : C4 = −
Eb0 h3
Eb0 h3
Substitution of these into Eq (2) results in Eq. (P10.37).
(1)
(2)
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________________________________________________________________________
PROBLEM (10.38) A structural steel shaft of specific weight γ , diameter d, and elastic modulus E,
is supported by roller bearings at its ends A and B (Fig. P10.38). Determine the maximum permissible
length L for this shaft.
Requirement: The deflection of the shaft due to its own weight per unit length w at the midspan is limited
to vmax .
Assumption: Bearings act as simple supports.
Given: vmax =5 mm,
d=75 mm,
g=9.81 kg/s2,
4
vmax =5wL /384EI
γ = 7860 kg/m3
E=200 GPa
(from Table B.4)
(from Table B.14)
SOLUTION
5wL4
vmax =
(by Table B.14)
384 EI
or
384 EIvmax 1 4
)
L=(
5w
Where
π d 4 π (75)4
I=
=
= 1.55(106 ) mm 4
64
64
π × 0.0752
w = (γ g ) A = (7,860 × 9.81)(
)
4
= 340.65 N m
Thus
384(200 ×109 )(1.55 × 10−6 )(0.005) 1 4
L =[
] = 4.32 m
5(340.65)
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________________________________________________________________________
PROBLEM (*10.39) A beam AB supported at the ends as shown in Fig. P10.39 carries a
concentrated load P at the mid span C. Determine the vertical deflection of the beam at the end B.
*SOLUTION
For segment AC
EIv1 '' = Px
1
EIv1 ' = Px 2 + C1
2
1 3
EIv1 = Px + C1 x + C2
6
P
RA=P
B
C
A
Px
x
L/2
Px
PL/2
x
L/2
For segment CB
1
EIv2 '' = PL
2
1
EIv1 ' = PLx + C3
2
1
EIv2 = PLx 2 + C3 x + C4
4
Boundary conditions:
v1 (0) = 0 : C2 = 0
v2 (0) = 0 : C3 = 0
Then
L
L
P L 3
L
P
L
v1 ( ) = v2 ( ) :
( ) + C1 ( ) = PL( ) 2 + C4
2
2
6 2
2
4
4
L
L
P L 2
P L
v1 '( ) = v2 '( ) :
( ) + C1 = − L( )
2
2
2 2
2 2
Solving
3
11
C1 = − PL2
C4 = − PL3
8
48
At B(x=0):
11
11 3
vB = v2 (0) = − PL3 =
PL ↓
48
48
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PROBLEMS (10.1 through 10.4) A cantilever beam is loaded as shown in Figs. P10.1 through
10.4. Using the double-integration method, determine:
(a) The equation of the elastic curve.
(b) The slope at the free end.
(c) The deflection at the free end.
SOLUTION (10.1)
1
(a) EIv " = − wx 2
2
1
EIv ' = − wx 3 + C1
6
1
EIv = − wx 4 + C1 x + C2
24
Boundary conditions:
1
v '( L) = 0 : C1 = wL3
6
1
1
v( L) = 0 : − wL4 + wL4 + C2 ,
24
6
Thus,
w
( x3 − L3 )
v' = −
6 EI
w
( x 4 − 4 L3 x + 3L4 )
v=−
24 EI
(b) x=0 in Eq. (1): θ A =
y
w
A
V
x
1
M = − wx 2
2
1
C2 = − wL4
8
(1)
(2)
wL3
6 EI
(c) x=0 in Eq. (2): v A = −
wL4 wL4
=
↓
8EI 8 EI
SOLUTION (10.2)
M = − P( L − x) = P( x − L)
(a)
y
C
A
2EI
PL
L/2
P
EI
L/2
x
B
P
Segment AC
2 EIv1 '' = P( x − L)
Continued on next slide
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1
P( x − L) 2 + C1
2
1
2 EIv1 = P( x − L)3 + C1 x + C2
6
2 EIv1 ' =
Segment CB
EIv2 '' = P( x − L)
1
EIv2 ' = P( x − L) 2 + C3
2
1
EIv2 = P( x − L)3 + C3 x + C4
6
Boundary conditions:
1
1
v1 '(0) = 0 : C1 = − PL2 ,
v1 (0) = 0 : C2 = PL3
2
6
Thus
P
v1 ' =
( x 2 − 2 xL)
4 EI
P
v1 =
( x 3 − 3 x 2 L)
12 EI
(1)
(2)
Boundary conditions:
L
L
3
1
5
v1 '( ) = v2 '( ) : − PL2 = PL2 + C3 , C3 = − PL2
2
2
16
8
16
L
L
5
1
5
1
v1 ( ) = v2 ( ) : − PL3 = − PL3 − PL3 + C4 , C4 = − PL3
2
2
96
48
32
8
So,
P
5
[( x − L) 2 − L2 ]
2 EI
8
P 1
L3
5
v2 =
[ ( x − L)3 − L2 x + ]
2 EI 3
8
4
v2 ' =
(3)
(4)
(b) x=L in Eq. (3):
θB = −
5 PL2 5 PL2
=
16 EI 16 EI
(c) x=L in Eq. (4):
3 PL3 3 PL3
θB = −
=
↓
16 EI 16 EI
Continued on next slide
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SOLUTION (10.3)
(a)
x2
w0
2L
1
EIv " = −
w0 x 3
6L
1
EIv ' = −
w0 x 4 + C1
24 L
1
EIv = −
w0 x 5 + C1 x + C2
120 L
y
x
w0
L
A
M =−
x/3
Boundary conditions:
1
v '( L) = 0 : C1 =
w0 L3
24
1
1
v( L) = 0 : −
w0 L5 + w0 L4 + C2 = 0
120
24
Therefore
w0
v' =
( x 4 − L4 )
24 EIL
w0
( x 5 − 5L4 x + 4 L5 )
v=−
120 EIL
x
V
C2 =
1
w0 L3
24
x3
w0
6L
(1)
(2)
(b) x=0 in Eq. (1):
w L2
θA = 0
24 EI
(c) x=0 in Eq. (2):
vA = −
w0 L4 w0 L4
=
↓
30 EI 30 EI
SOLUTION (10.4)
y
(a)
MA =
3 2
wa
2
w
A
a
C
a
B
x
RA= wa
Segment AC
Continued on next slide
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3
wa
(2 x − 3a )
M 1 = − wa 2 + wax =
2
2
wa
EIv1 '' =
(2 x − 3a)
2
wa 2
EIv1 ' =
( x − 3ax) + C1
2
wa x 3 3 2
EIv1 =
( − ax ) + C1 x + C2
2 3 2
Boundary conditions:
v1 '(0) = 0 : C1 = 0,
Thus
wax
v1 ' =
( x − 3a)
2 EI
wax 2
v1 =
(2 x − 9a)
12 EI
(0 ≤ x ≤ a )
v1 (0) = 0 : C2 = 0
(1)
(2)
Segment BC
w
w
M 2 = − (2a − x) 2 = (− x 2 + 4ax − 4a 2 )
2
2
w
EIv2 '' = (− x 2 + 4ax − 4a 2 )
2
w x3
EIv2 ' = (− + 2ax 2 − 4a 2 x) + C3
2
3
w x4 2
EIv2 = (− + ax3 − 2a 2 x 2 ) + C3 x + C4
2 12 3
Continuity conditions:
7
1
v1 '(a) = v2 '(a) : − wa 3 = − wa 3 + C3 ,
C3 = wa 3
6
6
7
13
1
v1 (a) = v2 (a) : − wa 4 = − wa 4 + C4 ,
C4 = − wa 4
12
24
24
Thus
w
(−2 x3 + 12ax 2 − 24a 2 x − 2a 3 )
12 EI
w
v2 =
(− x 4 + 8ax3 − 24a 2 x 2 + 4a 3 x − a 4 )
24 EI
v2 ' =
(3)
(4)
Continued on next slide
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(b) x=2a in Eq. (3):
7 wa 3 7 wa 3
θB = −
=
6 EI
6 EI
(c) x=2a in Eq. (4):
41 wa 4 41 wa 4
vB = −
=
↓
21 EI 21 EI
________________________________________________________________________
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PROBLEMS (10.5 and 10.6) Solve Probs. 10.1 and 10.2, using the multiple-integration method.
SOLUTION (10.5)
y
w
A
L
B
x
EIv '''' = − w
EIv ''' = − wx + C1
1
EIv '' = − wx 2 + C1 x + C2
2
1
1
EIv ' = − wx3 + C1 x 2 + C3
6
2
1
1
1
EIv = − wx 4 + C1 x 3 + C2 x 2 + C3 x + C4
24
6
2
(1)
(2)
Boundary conditions:
EIv '''(0) = −VA = 0,
C1 = 0
EIv ''(0) = M A = 0,
EIv '( L) = 0 : C3 = wL3 6
EIv( L) = 0,: C3 = − wL4 8
Then, Eqs.(1) & (2) give the results of the solution of Prob. 10.1.
SOLUTION (10.6)
y
P
EI
MA=PL
2EI
L/2
A
RA=P
C
Segment AC
2 EIv1 ''' = −V = P
2 EIv1 '' = Px + C1
1
2 EIv1 ' = Px 2 + C1 x + C2
2
1
1
2 EIv1 = Px 3 + C1 x 2 + C2 x + C3
6
2
Segment BC
EIv2 ''' = P
EIv2 '' = Px + C4
L/2
x
B
(1)
(2)
Continued on next slide
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1 2
Px + C4 x + C5
2
1
1
EIv2 = Px 3 + C4 x 2 + C5 x + C6
6
2
Boundary conditions:
2 EIv1 ''(0) = − PL : C1 = − PL,
2 EIv1 '(0) = 0 : C2 = 0
EIv1 (0) = 0 : C3 = 0
EIv2 ' =
(3)
(4)
Conditions of continuity:
L
L
PL PL
=
+ C4 ,
C4 = − PL
2 EIv1 ''( ) = EIv2 ''( ) : −
2
2
2
2
L
L
3PL2
3PL2
3
v1 '( ) = v2 '( ) : −
=−
+ C5 ,
C5 = PL2
2
2
16
8
16
3
3
L
L
PL
PL3
5PL
v1 ( ) = v2 ( ) : −
=−
+ C6 ,
C6 = −
2
2
96
96
24
Then, Eqs. (1) through (4) yield the results given in solution of Prob. 10.2.
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________________________________________________________________________
PROBLEM (10.7 and 10.8) Redo Probs. 10.3 and 10.4, using the multiple-integration method.
SOLUTION (10.7)
(x/L)w0
y
w0
A
B
x
L
x
x2
EIv '''' = − w0
EIv ''' = −
w0 + C1
L
2L
x3
EIv '' = −
w0 + C1 x + C2
6L
x4
1
EIv ' = −
w0 + C1 x 2 + C2 x + C3
24 L
2
5
x
1
1
EIv = −
w0 + C1 x 3 + C2 x 2 + C3 x + C4
120 L
6
2
(1)
(2)
Boundary conditions:
EIv '''(0) = 0 : C1 = 0
EIv ''(0) = 0 : C2 = 0,
EIv '( L) = 0 : C3 =
1
w0 L3
24
1
1
1
w0 L4 + w0 L4 + C4 = 0,
C4 = − w0 L4
120 L
24
30
Then, Eqs.(1) and (2) yield the results given in solution of Prob. 10.3.
EIv( L) = 0 : −
SOLUTION (10.8)
y
w
3
MA= wa2
2
A
a
C
a
B
x
RA=wa
Continued on next slide
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Segment AC
EIv1 '''' = 0
EIv1 ''' = C1
EIv1 '' = C1 x + C2
1
EIv1 ' = C1 x 2 + C2 x + C3
2
1
1
EIv1 = C1 x 3 + C2 x 2 + C3 x + C4
6
2
(1)
(2)
Boundary conditions:
EIv1 '''(0) = −VA = wa : C1 = wa
3
3
EIv1 ''(0) = − wa 2 : C2 = − wa 2
2
2
EIv1 '(0) = 0 : C3 = 0,
EIv1 (0) = 0, C4 = 0
Segment BC
EIv2 '''' = − w
EIv2 ''' = − wx + C5
1
EIv2 '' = − wx 2 + C5 x + C6
2
1
1
EIv2 ' = − wx 3 + C5 x 2 + C6 x + C7
6
2
1
1
1
EIv2 = − wx 4 + C5 x 3 + C6 x 2 + C7 x + C8
24
6
2
Continuity conditions:
v1 '''(a) = v2 '''(a ) : C1 = − wa + C5 ,
(3)
(4)
C5 = 2 wa
wa 2
wa 2
=−
+ 2 wa 2 + C6 ,
C6 = −2 wa 2
2
2
wa 3
1
v2 '(a) = v2 '(a) : − wa 3 = −
+ wa 3 − 2wa 3 + C7 ,
C7 = wa 3
6
6
v1 (a) = v2 (a) :
v1 ''(a) = v2 ''(a) : −
−
wa 4 wa 4 wa 4
7
wa 4 = −
+
+
+ C8 ,
12
24
3
6
C8 = −
1
wa 4
24
Then, Eqs.(1) through (4) yield the results given in solution of Prob.10.4.
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________________________________________________________________________
PROBLEM (10.9) Calculate the maximum slope and maximum deflection of the beam in Figure
P10.1. The beam is an S 100 x 14 steel (see Table B.9):
Given: w = 1.2 kN/m, L = 2 m,
E = 200 GPa
SOLUTION
See solution of Prob. 10.1:
wL3
wL4
θA =
vA =
6 EI
8EI
From Table B.9: I = 2.83 × 106 mm 4
Thus,
1.2 ×103 (2)3
= 2.827 × 10−3 rad
θA =
9
−6
6(200 × 10 )(2.83 × 10 )
1.2 × 103 (2) 4
= 4.24 mm ↓
vA =
8(200 × 109 )(2.83 × 10−6 )
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________________________________________________________________________
PROBLEMS (10.10 and 10.11) A cantilever beam is loaded as depicted in Fig. P10.10 and
P10.11. Using the double-integration approach, determine:
(a) The equation of the deflection curve.
(b) The slope at the free end.
(c) The deflection at the free end.
SOLUTION (10.10)
(a)
y
PL
P
A
L/2
C
L/2
B
x
M 1 = PL
(0 ≤ x ≤ L 2)
3
L
M 2 = PL − Px
( ≤ x ≤ L)
2
2
Segment AC
EIv1 '' = PL
EIv1 ' = PLx + C1
1
EIv1 = PLx 2 + C1 x + C2
2
Segment BC
3
EIv2 '' = PL − Px
2
3
1
EIv2 ' = PLx − Px 2 + C3
2
2
3
1
EIv2 = PLx 2 − Px3 + C3 x + C4
4
6
Boundary conditions:
3 2 1 2
v2 '( L) = 0 :
PL − PL + C3 = 0, C3 = − PL2
2
2
3 3 1 3
5
v2 ( L) = 0 :
PL − PL − PL3 + C4 = 0, C4 = PL3
4
6
12
Thus,
P
(3Lx − x 2 − 2 L2 )
v2 ' =
2 EI
P
(9 Lx 2 − 2 x 3 − 12 L2 x + 5 L3 )
v2 =
12 EI
(1)
(2)
Continued on next slide
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Continuity conditions:
1 2
3
L
L
v1 '( ) = v2 '( ) :
PL + C1 = − PL2 ,
2
2
2
8
5
L
L
PL3
,
v1 ( ) = v2 ( ) : − PL3 + C2 = −
2
2
16
12
7
C1 = − PL2
8
19
C2 = − PL2
48
Therefore,
P
(−8 Lx − 7 L2 )
8 EI
P
(24 Lx 2 − 42 L2 x + 19 L3 )
v1 =
48 EI
v1 ' =
(3)
(4)
(b) x=0 in Eq. (3):
7 PL2 7 PL2
θA = −
=
8 EI 8 EI
(c) x=0 in Eq. (4):
19 PL3
vA = −
↑
48 EI
SOLUTION (10.11)
y
(a)
w
MA=2wa2
A
2a
C
a
B
x
RA=2wa
M1 =
1
w(−4a 2 + 4ax − x 2 )
2
M2 = 0
(0 ≤ x ≤ 2a )
L
( ≤ x ≤ L)
2
Continued on next slide
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Segment AC
w
EIv1 '' = (−4a 2 + 4ax − x 2 )
2
w
x3
EIv1 ' = (−4a 2 x + 2ax 2 − ) + C1
2
3
2 3 x4
w
2 2
EIv1 = (−2a x + ax − ) + C1 x + C2
2
3
12
Boundary conditions:
v1 '(0) = 0 : C1 = 0,
v1 (0) = 0 : C2 = 0
Thus,
w
(−12a 2 x + 6ax 2 − x 3 )
v1 ' =
6 EI
w
(−24a 2 x 2 + 8ax 3 − x 4 )
v1 =
24 EI
Segment BC
EIv2 '' = 0,
EIv2 ' = C3 ,
(1)
(2)
EIv2 = C3 x + C4
Continuity conditions:
4
v1 '(2a) = v2 '(2a) : − wa 3 = C3
3
8
v1 (2a ) = v2 (2a ) : − 2wa 4 = − wa 4 + C4 ,
3
2
C4 = − wa 4
3
Therefore,
4 wa 3
3 EI
wa 3
(−4 x + 2a)
v2 =
3EI
v2 ' = −
(3)
(4)
(b) x=3a in Eq. (3):
4 wa 3 4 wa 3
θB = −
=
3 EI
3 EI
(c) x=3a in Eq. (4):
10 wa 4 10 wa 4
vB = −
=
↓
3 EI
3 EI
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________________________________________________________________________
PROBLEM (10.12) Calculate the deflection at the center of the beam shown in Fig. P10.10 if the
beam is a solid rod of diameter d.
Given:
d = 2 in., P = 80 lb,
L = 3 ft,
E = 10 x 106 psi.
SOLUTION
See solution of Prob.10.10:
P
(24 Lx 2 − 42 L2 x + 19 L3 )
v1 =
48 EI
For x=L/2:
1 PL3 1
80(3 ×12)3
= 39.6 ×10−3 in.
=
vC =
12 EI 12 10 ×106 (π × 24 64)
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_______________________________________________________________________
PROBLEMS (10.13 and 10.14) Solve Probs. 10.10 and 10.11, using the multiple-integration
approach.
SOLUTION (10.13)
y
PL
P
A
L/2
C
L/2
B
x
Segment AC
EIv1 '''' = 0
EIv1 ''' = C1
EIv1 '' = C1 x + C2
1
EIv1 ' = C1 x 2 + C2 x + C3
2
1
1
EIv1 = C1 x 3 + C2 x 2 + C3 x + C4
6
2
(1)
(2)
Segment BC
EIv2 '''' = 0
EIv2 ''' = C5
EIv2 '' = C5 x + C6
1
EIv2 ' = C5 x 2 + C6 x + C7
2
1
1
EIv2 = C5 x 3 + C6 x 2 + C7 x + C8
6
2
(3)
(4)
Boundary conditions:
v1 '''(0) = 0 : C1 = 0
v1 ''(0) = PL : C2 = PL
L
v2 '''( ) = − P : C5 = − P
2
3
L
v2 ''( ) = PL : C6 = PL
2
2
1 2 3 2
v2 '( L) = 0 : − PL + PL + C7 = 0,
C7 = − PL2
2
2
3
5
PL 3 3
v2 ( L) = 0 : −
+ PL − PL3 + C8 = 0,
C8 = PL3
6
4
12
Continued on next slide
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Continuity conditions:
3
7
L
L
PL2
v1 '( ) = v2 '( ) :
+ C3 = − PL2 ,
C3 = − PL2
2
2
2
8
8
3
3
3
19
L
L
PL 7 PL
PL
,
v1 ( ) = v2 ( ) :
−
+ C4 =
C4 = − PL3
2
2
8
16
12
48
Then, Eqs.(1) through (4) yield the results given in solution of Prob. 10.10.
SOLUTION (10.14)
y
w
MA=2wa2
A
2a
C
a
B
x
RA=2wa
Segment AC
EIv1 '''' = − w
EIv1 ''' = − wx + C1
1
EIv1 '' = − wx 2 + C1 x + C2
2
1 3 1
EIv1 ' = − wx + C1 x 2 + C2 x + C3
6
2
1
1
1
EIv1 = − wx 4 + C1 x 3 + C2 x 2 + C3 x + C4
24
6
2
Segment BC
EIv2 '''' = 0
EIv2 ''' = C5
EIv2 '' = C5 x + C6
1
EIv2 ' = C5 x 2 + C6 x + C7
2
1
1
EIv2 = C5 x 3 + C6 x 2 + C7 x + C8
6
2
(1)
(2)
(3)
(4)
Continued on next slide
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Boundary conditions:
EIv1 '''(0) = −V = 2aw : C1 = 2aw
EIv1 ''(0) = −2aw2 : C2 = −2aw2
EIv1 '(0) = 0 : C3 = 0
EIv1 (0) = 0 : C4 = 0
Continuity conditions:
v1 '''(2a) = v2 '''(2a ) : C5 = 0
v1 ''(2a) = v2 ''(2a) : C6 = 0
4
v1 '(2a) = v2 '(2a) : − wa 3 = C7
3
8
2
v1 (2a ) = v2 (2a ) : − 2wa 4 = − wa 4 + C8 ,
C8 = wa 4
3
3
Then, Eqs.(1) through (4) yield the results given in solution of Prob. 10.11
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________________________________________________________________________
PROBLEM (*10.15) A cantilever beam is loaded as shown in Fig. P10.15. Using the multiple-integration method, determine:
(a) The equation of the elastic curve.
(b) The deflection at the free end.
(c) The reactions at the fixed support.
*SOLUTION
(a) EIv '''' = − w0 cos
EIv ''' = −(
2 w0 L
πx
2L
πx
+ C1
2L
4 w L2
πx
+ C1 x + C2
EIv '' = ( 02 ) cos
π
2L
8w L3
πx 1
+ C1 x 2 + C2 x + C3
EIv ' = ( 03 ) sin
π
2L 2
4
16w0 L
1
πx 1
+ C1 x 3 + C2 x 2 + C3 x + C4
) cos
EIv = −(
4
π
2L 6
2
π
) sin
Boundary conditions:
16w0 L4
v(0) = 0 : C4 =
4
π
v '(0) = 0 : C3 = 0
2w L
v(0) ''' = 0 : C1 = 0
π
v(0) '' = 0 : C2 = −
2 w0 L2
π
Therefore,
w
πx
− π 3 Lx3 + 3π 3 L2 x 2 − 48 L4 ]
v = − 40 [48L4 cos
3π EI
2L
(b) Let x=L in the above expression:
w
w L4
vB = − 20 (−π 3 L4 + 3π 3 L4 − 48L4 ) = 0.04795 0 ↓
3π EI
EI
(c) Expressions for shear & moment are
2w L
πx
V = 0 (1 − sin )
π
2L
2w0 L
πx
M = 2 (cos
+ π x − π L)
π
2L
Continued on next slide
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These result in, for x=0:
2w L
RA = 0 ↑
π
2(π − 2) w0 L2
MA =
π2
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________________________________________________________________________
PROBLEMS (10.16 through *10.19) A simple beam is loaded as shown in Figs. P10.16
through P10.19. Using the double-integration method, determine:
(a) The equation of the elastic curve.
(b) The slope at the end A.
(c) The deflection at midspan.
SOLUTION (10.16)
(a)
w0
y
L−x
( L − x)2
w0
L
2L
wo
A
L
2
w0L/3
x
C
L
2
M
B
O
V
w0L/6
w0L/6
w0
1
L-x
∑ M = 0 : − M + 6 w L( L − x) − 6 L ( L − x) = 0
o
3
0
or
M=
w0 x 3 w0 x 2 w0 Lx
−
+
6L
2
3
Thus,
w0 x3 w0 x 2 w0 Lx
−
+
6L
2
3
4
3
w x w x w Lx 2
EIv ' = 0 − 0 + 0
+ C1
24 L
6
6
w x 5 w x 4 w Lx 3
EIv = 0 − 0 + 0
+ C1 x + C2
120 L
24
18
EIv '' =
(1)
(2)
Boundary conditions:
v(0) = 0 : C2 = 0
v( L) = 0 :
w0 L4 w0 L4 w0 L4
−
+
+ C1 L,
120
24
18
C1 = −
Equations (1) and (2):
w0
(15 x 4 − 60 Lx 3 + 60 L2 x 2 − 8 L4 )
v' =
360 EIL
w0
(3x 5 − 15Lx 4 + 20 L2 x 3 − 8L4 x)
v=
360 EIL
w0 L3
45
(3)
(4)
Continued on next slide
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(b) Make x=0 in Eq. (3):
w0 L3
w0 L3
θA = −
=
45 EI 45 EI
(c) Make x=L/2 in Eq. (4):
5w L4 5w0 L4
vC = − 0 =
↓
768EI 768EI
SOLUTION (10.17)
y
A
w
L/2
C
L/2
RA=wL/8
1
L
(0 ≤ x ≤ )
M 1 = wLx
8
2
1
1
L 2
M 2 = wLx − w( x − )
8
2
2
5
1 2 1 2
= wLx − wx − wL
8
2
8
B
x
RB=3wL/8
L
( ≤ x ≤ L)
2
(a) Segment AC
1
EIv1 '' = wLx
8
1
EIv1 ' = wLx 2 + C1
16
1
EIv1 =
wLx 3 + C1 x + C2
48
Segment BC
5
1
1
EIv2 '' = wLx − wx 2 − wL2
8
2
8
5
1
1
EIv2 ' = wLx 2 − wx3 − wL2 x + C3
16
6
8
5
1
1
EIv2 =
wLx3 − wx 4 − wL2 x 2 + C3 x + C4
48
24
16
(1)
(2)
(3)
Continued on next slide
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Boundary and continuity conditions:
v1 (0) = 0 : C2 = 0
wL3
L
L
wL3
wL3
C3 = C1 +
v1 '( ) = v2 '( ) :
+ C1 = −
+ C3 ,
48
2
2
64
192
4
4
4
4
L
L
wL C1 L 5wL wL wL C1 L wL4
v1 ( ) = v2 ( ) :
+
=
−
−
+
+
+ C4 ,
2
2
384
2
384 384 64
2
96
wL4
C4 = −
384
4
4
5wL wL wL4
wL3
wL4
C3 =
v2 ( L) = 0 :
−
−
+ C3 L +
= 0,
384
48
24
16
384
3
Equation (a) gives C1 = − 7 wL 384 .
Equations (1) through (3):
w
(24 Lx 2 − 7 L3 )
v1 ' =
384 EI
w
(8 Lx 3 − 7 L3 x)
v1 =
384 EI
w
(40 Lx3 − 16 x 4 − 24 L2 x 2 + L3 x − L4 )
v2 =
384 EI
(a)
(4)
(5)
(b) Make x=L/2 in Eq. (4):
7 wL3
7 wL3
θA = −
=
384 EI 384 EI
(c) Make x=0 in Eq. (5):
5wL4
5wL4
vC = −
=
↓
768EI 768EI
Continued on next slide
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SOLUTION (10.18)
(a)
y
M0
A
a
b
C
B
x
L
RA=M0/L
RB= M0/L
M0
(0 ≤ x ≤ a)
x
L
M
M2 = − 0 x + M0
(0 ≤ x ≤ a )
L
Segment AC
M
EIv1 '' = − 0 x
L
M
EIv1 ' = − 0 x 2 + C1
2L
M
EIv1 = − 0 x 3 + C1 x + C2
6L
M1 = −
(1)
(2)
Segment BC
M
M
EIv2 '' = − 0 x + M 0
EIv2 ' = − 0 x 2 + M 0 x + C3
2L
L
M0 3 M0
EIv2 = −
x +
x + C3 x + C4
6L
2
Boundary conditions:
v1 (0) = 0 : C2 = 0
v2 ( L) = 0 :
M 0 L2
+ C3 L + C4 = 0
3
Continuity conditions:
v1 '(a) = v2 '(a ) : C1 = M 0 a + C3
1
v1 (a ) = v2 (a) : C1a = M 0 a 2 + C3 a + C4
2
Solving Eqs.(a) through (c):
M L
M a2
C1 = − 0 + M 0 a − 0
3
2L
M L M a
M a2
C3 = − 0 − 0
C4 = − 0
3
2L
2
(3)
(a)
(b)
(c)
Continued on next slide
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Then, Eqs.(1) through (3):
M0
(3x 2 + 2 L2 − 6aL + 3a 2 )
v1 ' = −
6 EIL
M0
( x 3 + 2 L2 x − 6aLx + 3a 2 x)
v1 = −
6 EIL
M0
( x3 − 3Lx 2 + 2 L2 x + 3a 2 x − 3L2 a )
v2 = −
6 EIL
(4)
(5)
(6)
(b) Make x=0 in Eq. (4):
M
θ A = − 0 (2 L2 − 6 La + 3a 2 )
6 EIL
(c) Make x=L/2 in Eq. (5):
M 9 L2
v=− 0 (
− 6aL + 3a 2 )
6 EI 4
*SOLUTION (10.19)
(a)
y
k
P
A
a
C
P/2
a
B
x
P/2
1
Px
(0 ≤ x ≤ a)
2
1
1
M 2 = Px − P( x − a) = − Px + Pa
2
2
M1 =
Segment AC
1
EIv1 '' = Px
2
1
EIv1 ' = Px 2 + C1
4
1
EIv1 = Px 3 + C1 x + C2
12
( a ≤ x ≤ 2a )
(1)
(2)
Continued on next slide
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Segment BC
1
EIv2 '' = − Px + Pa
2
1
EIv2 ' = − Px 2 + Pax + C3
4
1
1
EIv2 = − Px3 + Pax 2 + C3 x + C4
12
2
(3)
Boundary and continuity conditions:
P
PEI
v1 (0) = − : C2 = −
2k
2k
3
C4 2 Pa 2
2 Pa
3
C3 = − −
v2 (2a ) = 0 : −
+ 2 Pa + 2aC3 + C4 = 0 ,
2a
3
3
2
2
2
Pa
Pa
Pa
+ C3
C1 =
v1 '(a) = v2 '(a) :
+ C1 = −
+ Pa 2 + C3 ,
2
4
4
Pa 3
PEI
Pa 3 Pa 3
Pa 3
v1 (a) = v2 (a) :
+ C1a −
=−
+
+ C1a −
+ C4 ,
12
2k
12
2
2
Pa 3 PEI
C4 =
−
6
2k
Solve Eqs.(a) and (b):
3
PEI
C3 = − Pa 2 +
4
4ak
2
Pa
PEI
C1 = −
+
4
4ak
Then, Eqs.(1) through (3):
P
EI
( x2 − a2 + )
v1 ' =
4 EI
ak
3EI
6 EI
P
( x3 − 3a 2 x +
)
v1 =
x−
12 EI
ak
k
3EI
6 EI
P
(− x 3 + 6ax 2 − 9a 2 x +
)
v2 =
x + 2a 3 −
12 EI
ak
k
(a)
(b)
(4)
(5)
Continued on next slide
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(b)
Make x=0 in Eq. (4):
P
EI
(−a 2 + )
θA = −
4 EI
ak
(c) Make x=a in Eq. (5):
3EI
P
(2a 3 +
)↓
vC =
12 EI
k
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________________________________________________________________________
PROBLEMS (10.20 and 10.21) Redo Probs. 10.16 and 10.17, using the multiple-integration
method.
SOLUTION (10.20)
y
w0
x
( L − x) = w0 − w0
L
L
wo
A
w0 L
3
x
x
B
L-x
w0 L
6
x
x2
w0 ,
EIv ''' = − w0 x +
w0 + C1
2L
L
1
x3
EIv '' = − w0 x 2 +
w0 + C1 x + C2
2
6L
1
1
x4
3
EIv ' = − w0 x +
w0 + C1 x 2 + C2 x + C3
6
24 L
2
5
1
x
1
1
EIv = − w0 x 4 +
w0 + C1 x3 + C2 x 2 + C3 x + C4
24
120 L
6
2
EIv '''' = − w0 +
(1)
(2)
Boundary conditions:
wL
wL
v '''(0) = −V = 0 : C1 = 0
3
3
v ''(0) = 0 : C2 = 0
v(0) = 0 : C4 = 0
w0 L4 w0 L4 w0 L4
w L3
+
+
+ C3 L = 0,
C3 = − 0
24
120
18
45
Then, Eqs.(1) and (2) yield the results given in solution of Prob.10.16.
v( L) = 0 : −
Continued on next slide
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SOLUTION (10.21)
y
A
wL/8
w
L/2
C
L/2
B
x
3wL/8
Segment AC
EIv1 '''' = 0
EIv1 ''' = C1
EIv1 '' = C1 x + C2
1
EIv1 ' = C1 x 2 + C2 x + C3
2
1
1
EIv1 = C1 x 3 + C2 x 2 + C3 x + C4
6
2
(1)
(2)
Segment BC
EIv2 '''' = − w0
EIv2 ''' = − w0 x + C5
1
EIv2 '' = − w0 x 2 + C5 x + C6
2
1
1
EIv2 ' = − w0 x 3 + C5 x 2 + C6 x + C7
6
2
1
1
1
EIv2 = − w0 x 4 + C5 x 3 + C6 x 2 + C7 x + C8
24
6
2
Boundary and continuity conditions:
EIv1 (0) = 0 : C4 = 0
1
1
EIv1 '''(0) = −V = wL : C1 = wL
8
8
EIv1 ''(0) = 0 : C2 = 0
3
3
EIv2 '''( L) = − wL : − w0 L + C5 = wL,
8
8
1 2
EIv2 ''( L) = 0 : C6 = − wL
8
(3)
5
C5 = wL
8
Continued on next slide
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and
1
L
L
wL3
wL3
v1 '( ) = v2 '( ) :
+ C3 = −
wL3 + C7 ,
C7 = C3 +
(4)
2
2
64
192
48
L
L
wL4 wL4
wL4
v1 ( ) = v2 ( ) :
=
+ C8 ,
C8 = −
2
2
384 192
384
3
wL
v2 ( L) = 0 : C7 =
384
7 wL3
Equation (4): C3 = −
128
Then, Eqs.(1) through (3) yield the results given in the solution of Prob. 10.17.
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________________________________________________________________________
PROBLEMS (10.22 and 10.23) Redo Probs. 10.18 and 10.19, using the multiple-integration
method.
SOLUTION (10.22)
y
M0
A
a
C
L
RA=M0/L
B
b
x
RB= M0/L
Segment AC
EIv1 '''' = 0
EIv1 ''' = C1 = −
M0
L
0
M0
x + C2
L
M
EIv1 ' = − 0 x 2 + C3
2L
M0 3
EIv1 = −
x + C3 x + C4
6L
EIv1 '' = −
(1)
(2)
Segment BC
EIv2 '''' = 0
EIv2 ''' = C5 = −
M0
L
M0
x + C6
EIv2 ''( L) = 0,
L
M
EIv2 ' = − 0 x 2 + M 0 x + C7
2L
M0 3 1
EIv2 = −
x + M 0 x 2 + C7 x + C8
6L
2
EIv2 '' = −
C6 = M 0
(3)
Boundary and continuity conditions:
v1 (0) = 0 : C4 = 0
M 0 L2
+ C7 L + C8 = 0
3
v1 '(a ) = v2 '(a ) : C3 = M 0 a + C7
1
v1 (a ) = v2 (a) : C3 a = M 0 a + C7 a + C8
2
v2 ( L) = 0 :
Continued on next slide
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Solving,
M0L
M 0a2
C3 = −
+ M 0a −
3
2L
2
M L M a
M a2
C7 = − 0 − 0
C8 = 0
3
2L
2
Then, Eqs.(1) through (3) yield the results given in solution of Prob.10.18.
SOLUTION (10.23)
y
P
A
k
a
C
a
B
x
P/2
P/2
Segment AC
0
P
Px
EIv1 ''' =
EIv1 '' =
+ C1
2
2
1 2
EIv1 ' = Px + C2
4
1
EIv1 = Px 3 + C2 x + C3
12
Segment BC
P
EIv2 ''' = −
2
(1)
(2)
1
EIv2 '' = − Px + C4
2
EIv2 ''(2a) = 0 : C4 = Pa,
1
EIv2 '' = − Px + Pa
2
1
EIv2 ' = − Px 2 + Pax + C5
4
1
1
EIv2 = − Px3 + Pax 2 + C5 x + C6
12
2
Boundary & continuity conditions:
P
PEI
v1 (0) = − : C3 = −
2k
2k
C6 2 2
v2 (2a ) = 0 : C5 = − − Pa
2a 3
1
v1 '(a) = v2 '(a) : C2 = Pa 2 + C5
2
1
PEI
v1 (a) = v2 (a) : C6 = Pa 3 −
6
2k
(3)
Continued on next slide
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Solving,
3
PEI
1
PEI
C5 = − Pa 2 +
C2 = − Pa 2 +
4
4ak
4
4ak
Then, Eqs.(1) through (3) yield the results given in solution of Prob.10.19.
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________________________________________________________________________
PROBLEM (10.24) Calculate the deflection at midspan of the beam shown in Fig. P10.17 if the
beam is an S 200 x 34 steel shape (see Table B.9).
Given: w = 15 kN/m , L = 3 m,
E = 210 GPa.
SOLUTION
From Table B.9: I = 27 × 106 mm 4
See solution of Prob. 10.17:
5 wL4
5
15 × 103 (3) 4
vC =
=
768 EI
768 210 ×109 (27 × 10−6 )
= 1.4 × 10−3 m = 1.4 mm ↓
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________________________________________________________________________
PROBLEM (10.25 and 10.26) For the beam and loading shown in Figs. P10.25 and 10.26, using
the double-integration method, determine:
(a) The equation of the deflection curve.
(b) The slope at the end A.
(c) The deflection at the midspan.
SOLUTION (10.25)
w = w0 sin
(a)
y
A
RA
L/2
C
πx
L
L2
0
sin
πx
L
dx =
w0 L
π
RB
x
V
x
B
L/2
RA = RB = w0 ∫
V=
M
w0 L
π
cos
M=
w0 L2
π
πx
2
L
x
sin
πx
L
x
We have
L
πx
EIv '' = w0 ( ) 2 sin
π
L
L
πx
EIv ' = − w0 ( )3 cos
+ C1
π
L
L
πx
EIv = − w0 ( ) 4 sin
+ C1 x + C2
π
L
(1)
(2)
Boundary conditions:
v(0) = 0 : C2 = 0 ,
Equations (1) and (2) become
w L3
πx
v ' = 30 cos
π EI
L
4
wL
πx
v = 40 sin
↓
π EI
L
L
v '( ) = 0 : C1 = 0
2
(3)
(4)
Continued on next slide
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(b)
Make x=0 in Eq. (3):
w0 L3
θA = 3
π EI
(c) Make x=L/2 in Eq. (4):
w L4
vC = 40 ↓
π EI
SOLUTION (10.26)
(a)
x 2 w0 L
x
L
1 w0 3 1
M =−
x + w0 Lx
3 L
4
x/3
2w0
y
A
x
RA=w0L/4
O
V
Due to symmetry only part AC of the beam need be considered.
1
1
EIv '' = − w0 x 3 + w0 Lx
3L
4
1
1
EIv ' = −
w0 x 4 + w0 Lx 2 + C1
12 L
8
1
1
EIv = −
w0 x 5 + w0 Lx3 + C1 x + C2
60 L
24
(1)
(2)
Boundary conditions:
5
L
v '( ) = 0 : C1 = −
w0 L3
2
192
Thus, Eqs.(1) and (2) become
w L3
x
x
v ' = 0 [−80( ) 4 + 120( ) 2 − 25]
960 EI
L
L
4
wL
x
x
x
v = 0 [−16( )5 + 40( )3 − 25( )]
960 EI
L
L
L
v(0) = 0 : C2 = 0,
(3)
(4)
(b) Make x=0 in Eq. (3):
5w0 L3
θA =
192 EI
Continued on next slide
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( c)
Make x=L/2 in Eq. (4):
w0 L4
vC =
↓
120 EI
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________________________________________________________________________
PROBLEMS (10.27 and 10.28) Solve Probs. 10. 25 and 10.26, using the multiple-integration
approach.
SOLUTION (10.27)
EIv '''' = − w0 sin
π 2
πx
π
πx
EIv ''' = w0 ( ) cos
+ C1
L
L
L
πx
EIv '' = w0 ( ) sin
+ C1 x + C2
L
L
π
πx 1
+ C1 x 2 + C2 x + C3
EIv ' = − w0 ( )3 cos
L
L 2
π
πx 1
1
+ C1 x 3 + C2 x 2 + C3 x + C4
EIv = − w0 ( ) 4 sin
2
L
L 6
(1)
(2)
Boundary conditions:
v(0) = 0 : C4 = 0
v ''(0) = 0 : C2 = 0
v ''( L) = 0 : C1 = 0
v( L) = 0 : C3 = 0
Then, Eqs.(1) and (2) yield the results given in solution of Prob.10.25.
SOLUTION (10.28)
∑F = 0: R = R = w L 4
y
A
B
0
Due to symmetry only segment AC need be considered and
2x
w = − w0
L
Thus,
2x
x2
EIv '''' = − w0
EIv ''' = − w0 + C1
L
L
3
x
EIv '' = −
w0 + C1 x + C2
3L
1
x4
EIv ' = −
w0 + C1 x 2 + C2 x + C3
(1)
12 L
2
x5
1
1
EIv = −
w0 + C1 x 3 + C2 x 2 + C3 x + C4
(2)
60 L
6
2
Boundary conditions:
1
1
EIv '''(0) = w0 L : C1 = w0 L
4
4
EIv ''(0) : 0 : C2 = 0,
v(0) = 0 : C4 = 0
Continued on next slide
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5
L
v '( ) = 0 : C3 = −
w0 L3
2
192
Then, Eqs.(1) and (2) yield the results given in solution of Prob.10.26.
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________________________________________________________________________
PROBLEM (*10.29) Find the slope at A of the beam in Fig. P10.26 if the beam is a W 250 x 80
steel shape (see Table B.8).
Given: w = 50 kN/m, L = 4 m,
E = 200 GPa.
*SOLUTION
From Table B.8: I = 126 × 106 mm 4
See solution of Prob.10.26:
5 w0 L3
5
50 × 103 (4)3
θA =
=
192 EI
192 200 ×109 (126 × 10−6 )
= 3.31× 10−3 rad
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________________________________________________________________________
PROBLEM (*10.30 through 10.33) A beam supported and loaded as shown in Figs. P10.30
through P10.33, using a direct-integration method, determine:
(a) The equation of the elastic curve.
(b) The deflection at midspan.
(c) The magnitude and location of the maximum deflection.
(d) The slope at the right end.
SOLUTION (*10.30)
(a)
wo
y
A
RA
xm
RB
B
x
L
x
x2
EIv '''' = − w0
EIv ''' = −
w0 + C1
2L
L
x3
EIv '' = −
w0 + C1 x + C2
6L
1
x4
EIv ' = −
w0 + C1 x 2 + C2 x + C3
24 L
2
5
x
1
1
EIv = −
w0 + C1 x3 + C2 x 2 + C3 x + C4
120 L
6
2
Boundary conditions:
v ''(0) = 0 : C2 = 0
wL
v ''( L) = 0 : C1 = 0
6
v(0) = 0 : C4 = 0
7 w0 L3
v( L) = 0 : C3 = −
360
Equations (1) and (2) become
w0
(−15 x 4 + 30 L2 x 2 − 7 L4 )
v' =
360 EIL
w0 x
(−3 x 4 + 10 L2 x 2 − 7 L4 )
v=
360 EIL
(1)
(2)
(3)
(4)
(b) Make x=L/2 in Eq.(4):
5 w0 L4
vC =
↓
768 EI
Continued on next slide
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Set v ' = 0 in Eq.(3):
( c)
8
= 0.5193L
15
xm = L 1 −
Then Eq.(2) gives
w0 L4
vmax = v( xmax ) = 0.00652
↓
EI
(d) Make x=L in Eq.(1):
w L3
θB = 0
45 EI
SOLUTION (10.31)
(a)
y
L/2
A
xm
M0/L
M0
C
L/2
B
x
M0/L
Due to asymmetric but equal deflection configuration, only segment AC need be
considered.
M0
x
L
M
EIv ' = 0 x 2 + C1
2L
M0 3 1
EIv =
x + C1 x + C2
6L
2
EIv '' =
Boundary conditions:
v(0) = 0 : C2 = 0
M L
L
v( ) = 0 : C1 = − 0
2
12
Then,
M0
(12 x 2 − L2 )
v' =
12 EIL
M0x
(4 x 2 − L2 )
v=
12 EIL
(1)
(2)
Continued on next slide
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(b)
Let x=L/2 in Eq.(2):
vC = 0 as required
(c) Set v ' = 0 in Eq.(1):
L
xm =
= 0.2887 L
12
Then, Eq.(2) gives
M 0 L2
M L2
vmax = v( xmax ) = −
= 0.00802 0 ↓
EI
36 12 EI
or, in segment BC:
M L2
vmax = 0.00802 0 ↑
( xm = 0.7113L )
EI
(d) Make x=0 in Eq.(1):
M L2
θ A = −θ B = 0
24 EI
SOLUTION (10.32)
(a)
y
M0
M =−
A
x
V
M0
x + M0
L
RA=M0/L
M0
x + M0
L
M
EIv ' = − 0 x 2 + M 0 x + C1
2L
M0 3 1
EIv = −
x + M 0 x 2 + C1 x + C2
6L
2
EIv '' = −
(a)
(b)
Boundary conditions:
v(0) = 0 : C2 = 0,
1
v( L) = 0 : C1 = M 0 L
3
Continued on next slide
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Then Eqs.(a) and Eqs.(b) become
M0
(3 x 2 − 6 Lx + 2 L2 )
v' = −
6 EIL
M x
v = − 0 ( x 2 − 3Lx + 2 L2 )
6 EIL
(1)
(2)
(b) Let x=L/2 in Eq.(b):
M 0 L2
vM =
↓
16 EI
(c) For v ' = 0 , Eq.(1) yields
3 x 2 − 6 Lx + 2 L2 = 0 or
Substitute this into Eq.(2):
vmax = v( xm ) = 0.0642
xm = 0.4227 L .
M 0 L2
↓
EI
(d) Making x=L in Eq.(1):
M L
θB = 0
6 EI
SOLUTION (10.33)
y
(a)
w
MA=wL2/8
A
L/2
C
L/2
B
x
RA=wL/2
Segment AC
1
1
1
wLx − wL2 − wx 2
2
8
2
4
w
EIv1 ' = (2 Lx 2 − L2 x − x3 ) + C1
8
3
3
2 2
w 2 Lx L x
x4
EIv1 = (
−
− ) + C1 x + C2
8 3
2
3
EIv1 '' = M 1 =
Segment BC
EIv2 '' = 0
EIv1 ' = C3
EIv = C3 x + C4
(a)
(b)
(c)
Continued on next slide
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Boundary and continuity condition:
v1 (0) = 0 : C2 = 0
v1 '(0) = 0 : C1 = 0
L
L
wL3
v1 '( ) = v2 '( ) : C3 = −
2
2
48
L
L
wL4
v1 ( ) = v2 ( ) : C4 = −
2
2
384
Then Eqs.(b) and (c) become
wx 2
(4 Lx − 3L2 − 2 x 2 )
v1 =
48 EI
wx3
v2 ' = −
48 EI
wx 3
(−8 x + L)
v2 =
384 EI
(1)
(2)
(3)
(b) Let x=L/2 in Eq.(3)
3 wL4
vC =
↓
384 EI
(c) Let x=L in Eq.(3)
7 wL4
vB =
↓
384 EI
(d) We have
θ B = θC =
wL3
48 EI
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________________________________________________________________________
PROBLEM (*10.34 and *10.35) An overhanging beam is loaded as shown in Figs. P10. 34 and
P10.35. Using a direct-integration method, determine:
(a) The equation of the elastic curve.
(b) The deflection at midspan.
(c) The magnitude and location of the maximum deflection.
(d) The slope at the right end.
SOLUTION (*10.34)
(a)
y
w
A
L
3wL/8
B
L/2
C
x
9wL/8
Segment AB
3
1
EIv1 '' = M 1 = wLx − wx 2
8
2
3
1
EIv1 ' = wLx 2 − wx3 + C1
16
6
1
1
EIv = wLx3 − wx 4 + C1 x + C2
16
24
(a)
(b)
Segment BC
3
wx 2 9
EIv2 '' = M 2 = wLx −
+ wL( x − L)
8
2
8
3
3
9
wx
EIv2 ' = wLx 2 −
+ wL( x − L) 2 + C3
16
6 16
3
4
3
wLx wx
EIv2 =
−
+ wL( x − L)3 + C3 x + C4
16
24 16
Boundary & continuity conditions:
wL3
v1 (0) = 0 : C2 = 0,
v1 ( L) = 0 : C1 = −
48
v1 '( L) = v2 '( L) : C1 = C3 , v2 ( L) = 0 : C4 = 0
(c)
(d)
Continued on next slide
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Then, Eqs.(a) through (d) become
w
(9 Lx 2 − 8 x 3 − L3 )
48 EI
wx
(3Lx 2 − 2 x 3 − L3 )
v1 =
48 EI
w
[9 Lx 2 − 8 x 3 − 27( x − L) 2 − L3 ]
v2 ' =
48 EI
w
[3Lx 3 − 2 x 4 − L3 x + 9 L( x − L)3 ]
v2 =
48 EI
v1 ' =
(1)
(2)
(3)
(4)
(b) Let x=L/2 in Eq.(2):
wL4
vM =
↓
192 EI
(c) Let x=3L/2 in Eq.(4):
wL4
wL4
(v2 ) max = vC = −
= 0.0078
↓
128EI
EI
To find (v2 ) max , make v1 ' = 0 in Eq.(1): 9 Lx 2 − 8 x 3 − L3 = 0
Solving, by trial and error: xm = 0.422 L
Then Eq.(2) yields
wL4
(v1 ) max = 0.0054
↓
EI
(d) Let x=3L/2 in Eq.(3):
1 wL3
θC =
12 EI
SOLUTION (*10.35)
(a)
y
C
P
P
P
L/2
A
3P/2
E
L
B
L/2 D
x
3P/2
Due to symmetry only one-half the beam need be considered.
Continued on next slide
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Segment CA
EIv1 '' = M 1 = − Px
1
EIv1 ' = − Px 2 + C1
2
1
EIv1 = − Px 3 + C1 x + C2
6
(a)
(b)
Segment AE
EIv1 '' = M 2 = − Px +
3P
1
3
L
( x − ) = Px − PL
2
2
2
4
1 2 3
Px − PLx + C3
4
4
1
3
EIv2 = Px3 − PLx 2 + C3 x + C4
12
8
EIv2 ' =
(c)
(d)
Boundary & continuity conditions
1 2 3 2
1
v2 '( L) = 0 :
PL − PL + C3 = 0,
C3 = PL2
4
4
2
2
2
2
5
L
L
PL
PL 3PL PL2
,
v1 '( ) = v2 '( ) : −
+ C1 =
−
+
C1 = PL2
2
2
8
16
8
2
16
3
3
3
13PL
L
PL 5 PL
v1 '( ) = 0 :
+
+ C2 = 0,
C2 =
2
48
32
96
3
3
3
L
PL 3PL PL
PL3
v2 ( ) = 0 :
−
+
C4 = 0,
C4 = −
2
96
32
4
6
Equations (a), (b), and (d):
P
(−8 x 2 + 5 L2 )
v1 ' =
16 EI
P
(−16 x 3 + 30 L2 x − 13L3 )
v1 =
96 EI
P
(8 x 3 − 36 Lx 2 + 48 L2 x − 16 L3 )
v2 =
96 EI
(1)
(2)
(3)
(b) Making x=L in Eq.(3):
PL3
vE =
↑
24 EI
Continued on next slide
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( c)
Making x=0 in Eq.(2):
13PL3
vC =
↓
96 EI
Thus, vC = vmax
(d) Making x=0 in Eq.(1):
5 PL2
θC = −θ D =
16 EI
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________________________________________________________________________
PROBLEM (10.36) Calculate the slope at A of the beam shown in Fig. P10.31 if the beam is a W
12 x 72 steel shape (see Table B.8).
Given: M = 400 kips ⋅ in., L = 15 ft,
E = 30 x 106 psi
SOLUTION
From Table B.8, for a W12x72 section: I = 597 in.4
See solution of Prob. 10.31:
M 0 L2
θA =
24 EI
Substitute the given data
400 × 103 (15 × 12) 2
θA =
= 30.24 × 10−3 rad
6
24(30 × 10 )(597)
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________________________________________________________________________
PROBLEM (10.37) A cantilever of variable cross section supports a uniformly distributed load w
(Fig. P10.37). Verify, using the last of Eqs. (10.4), that the expression for the deflection curve is
v=
wL
(− x 3 + 3L2 x − 2 L3 )
Eb0 h3
(P10.37)
Here h represents the depth of the beam and b0 is the width at the fixed end.
SOLUTION
1 x
( b0 )h3
12 L
The last of Eq. (10.4):
(b0 x L)h3 d 2 v
d2
d 2v
d2
(
)
[
] = −W
EI
=
E
dx 2
dx 2
dx 2
dx 2
12
Integrating twice
(b x L)h3 d 2 v
d
[E 0
] = − wx + C1 = −V
dx
dx 2
12
I=
(b0 x L)h3 d 2 v
1
E
= − wx 2 + C1 x + C2 = M
2
12
2
dx
Boundary conditions:
M (0) = 0 : C2 = 0
V (0) = 0 : C1 = 0
Equation (1) becomes then
d 2v
6wL
dv
3wL 2
=−
x
=−
x + C3
2
3
dx
Eb0 h
dx
Eb0 h3
wL 3
v=−
x + C3 x + C4
Eb0 h3
We have
3wL3
2 wL4
v '( L) = 0 : C3 =
,
v ( L ) = 0 : C4 = −
Eb0 h3
Eb0 h3
Substitution of these into Eq (2) results in Eq. (P10.37).
(1)
(2)
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________________________________________________________________________
PROBLEM (10.38) A structural steel shaft of specific weight γ , diameter d, and elastic modulus E,
is supported by roller bearings at its ends A and B (Fig. P10.38). Determine the maximum permissible
length L for this shaft.
Requirement: The deflection of the shaft due to its own weight per unit length w at the midspan is limited
to vmax .
Assumption: Bearings act as simple supports.
Given: vmax =5 mm,
d=75 mm,
g=9.81 kg/s2,
4
vmax =5wL /384EI
γ = 7860 kg/m3
E=200 GPa
(from Table B.4)
(from Table B.14)
SOLUTION
5wL4
vmax =
(by Table B.14)
384 EI
or
384 EIvmax 1 4
)
L=(
5w
Where
π d 4 π (75)4
I=
=
= 1.55(106 ) mm 4
64
64
π × 0.0752
w = (γ g ) A = (7,860 × 9.81)(
)
4
= 340.65 N m
Thus
384(200 ×109 )(1.55 × 10−6 )(0.005) 1 4
L =[
] = 4.32 m
5(340.65)
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________________________________________________________________________
PROBLEM (*10.39) A beam AB supported at the ends as shown in Fig. P10.39 carries a
concentrated load P at the mid span C. Determine the vertical deflection of the beam at the end B.
*SOLUTION
For segment AC
EIv1 '' = Px
1
EIv1 ' = Px 2 + C1
2
1 3
EIv1 = Px + C1 x + C2
6
P
RA=P
B
C
A
Px
x
L/2
Px
PL/2
x
L/2
For segment CB
1
EIv2 '' = PL
2
1
EIv1 ' = PLx + C3
2
1
EIv2 = PLx 2 + C3 x + C4
4
Boundary conditions:
v1 (0) = 0 : C2 = 0
v2 (0) = 0 : C3 = 0
Then
L
L
P L 3
L
P
L
v1 ( ) = v2 ( ) :
( ) + C1 ( ) = PL( ) 2 + C4
2
2
6 2
2
4
4
L
L
P L 2
P L
v1 '( ) = v2 '( ) :
( ) + C1 = − L( )
2
2
2 2
2 2
Solving
3
11
C1 = − PL2
C4 = − PL3
8
48
At B(x=0):
11
11 3
vB = v2 (0) = − PL3 =
PL ↓
48
48
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________________________________________________________________________
PROBLEMS (10.40 through 10.43) A beam is supported and loaded as shown in Figs. P10.40
through P10.43. Determine:
(a) The equation of the elastic curve.
(b) The slope at the free end.
(c) The deflection at the free end.
SOLUTION (10.40)
(a)
y
3
M A = wa 2
2
w
A
a
C
a
B
x
RA=wa
3
1
EIv '' = wax − wa 2 − w < x − a > 2
2
2
1
3
1
EIv ' = wax 2 − wa 2 x − w < x − a >3 +C1
2
2
6
1
3
1
EIv = wax3 − wa 2 x 2 − w < x − a > 4 +C1 x + C2
6
4
24
Boundary conditions:
v(0) = 0 : C2 = 0
v ' (0) = 0 : C1 = 0
Equation (a) and (b) become
wax
1
EIv ' = − w < x − a >3 −
(3a − x)
6
2
wax 2
1
EIv = − w < x − a > 4 −
(9a − 2 x)
24
12
(a)
(b)
(1)
(2)
(b) Make x=2a in Eq.(1):
7 wa 3
θB =
6 EI
(c) Make x=2a in Eq.(2):
41 wa 4
vB =
↓
24 EI
Continued on next slide
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SOLUTION (10.41)
(a)
y
P
PL
A
L/2
C
L/2
B
x
L
>
2
P
L
EIv ' = PLx − < x − > 2 +C1
2
2
1
P
L
EIv = PLx 2 − < x − >3 +C1 x + C2
2
6
2
EIv '' = PL − P < x −
(a)
(b)
Boundary conditions:
PL2
7
v '( L) = 0 : PL2 −
+ C1 ,
C1 = − PL2
8
8
2
3
3
PL PL 7 PL
19
v( L) = 0 :
−
−
+ C2 = 0,
C2 = − PL3
2
48
8
48
Equations (a) and (b) become
P
L
v' =
[8Lx − 4 P < x − > 2 −7 L2 ]
8 EI
2
P
L
v=
[24 Lx 2 − 8P < x − >3 −42 L2 + 19 L3 ]
48 EI
2
(1)
(2)
(b) Make x=2a in Eq.(1):
7 PL2
θA =
8 EI
(c) Make x=0 in Eq.(2):
19 PL3
vB =
↑
48 EI
Continued on next slide
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SOLUTION (10.42)
(a)
y
w
B
MA=2wa2
A
C
2a
x
a
RA=2wa
1
1
EIv '' = 2 wax − 2 wa 2 − wx 2 + w < x − 2a > 2
2
2
1
1
EIv ' = wax 2 − 2wa 2 x − wx 3 + w < x − 2a >3 +C1
6
6
1
1
1
EIv = wax 3 − wa 2 x 2 − wx 4 + w < x − 2a > 4 +C1 x + C2
3
24
24
Boundary conditions:
v '(0) = 0 : C1 = 0
v(0) = 0 : C2 = 0
Then, Eqs. (a) and (b) become
w
v' =
[6ax 2 − 12a 2 x − x 3 + < x − 2a >3 ]
6 EI
w
v=
[8ax 3 − 24a 2 x 2 − x 4 + < x − 2a > 4 ]
24 EI
(a)
(b)
(1)
(2)
(b) Make x=3a in Eq.(1):
4 wa 3
θB =
3 EI
(c) Make x=3a in Eq.(2):
10 wa 4
vB =
↓
3 EI
Continued on next slide
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SOLUTION (10.43)
y
w
A
L
RA
RB
x
C
B
w
M0
Refer to the Fig. P10.43:
1
∑ M A = 0 : − M 0 − 2 wL2 + RB L = 0
M
1
RB = wL + 0
or
L
2
M
1
wL M
∑ Fy = 0 : RA = wL − 2 wL − L0 = 2 − L0
Thus
d 2v
1
1
EI 2 = M ( x) = RA x − wx 2 + RB < x − L > + w < x − L > 2
dx
2
2
1
1
1
dv 1
EI
= RA x 2 − wx3 + RB < x − L > 2 + w < x − L >3 +C1
6
2
6
dx 2
1
1
1
1
EIv = RA x 3 − wx 4 + RB < x − L >3 + w < x − L > 4 +C1 x + C2
6
24
6
24
Boundary conditions:
EIv(0) = 0 : C2 = 0
1
1
EIv( L) = 0 :
RA L3 − wL4 + C1 L = 0
6
24
or
R L2 wL3
wL3 M 0 L
C1 = − A +
=−
+
6
24
24
6
(1)
(2)
Continued on next slide
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Equation (1) and (2) become then
dv
1 wL 2 M 0 2 wx 3 wL M 0
(a)
=θ =
x −
x −
+(
+
[
) < x − L >2
dx
L
L
2 EI 2
3
2
w
wL3 M 0 L
]
+ < x − L >3 −
+
3
12
3
wL M 0
1 wL 3 M 0 x 3 wx 4
v=
x −
−
+(
+
[
) < x − L >3
L
L
6 EI 2
4
2
w
wL3
+ < x − L >4 −
x + M 0 Lx]
4
4
(b)
3L
1 wL3
θC = θ ( ) =
− 5M 0 L)
(
2
6 EI 4
(c) vC = v(
3L
1 9wL4 31
− M 0 L2 )
)=
(
2
24 EI 16
3
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________________________________________________________________________
PROBLEMS (10.44 through 10.47) For the beam loaded and supported as shown in Figs.
P10.44 through P10.47, determine:
(a) The equation of the deflection curve.
(b) The slope at end B for the given numerical values:
P = 4 kips, a = 3 ft, L = 10 ft, E = 10 x 106 psi,
I = 72 in.4 ,
w=2 kips/ft
SOLUTION (10.44)
y
P
B
MA=3Pa
A
x
C a
3a
RA=P
(a)
EIv '' = Px − 3Pa − P < x − 3a >
1
1
EIv ' = Px 2 − 3Pax − P < x − 3a > 2 +C1
2
2
1
3
1
EIv = Px3 − Pax 2 − P < x − 3a >3 +C1 x + C2
6
2
6
Boundary conditions:
v '(0) = 0 : C1 = 0
v(0) = 0 : C2 = 0
Equations (a) and (b) become
P
v' =
[3x 2 − 18ax − 3 < x − 3a > 2 ]
6 EI
P 3
v=
[ x − 9ax 2 − < x − 3a >3 ]
6 EI
(a)
(b)
(1)
(2)
(b) Make x=4a in Eq.(1):
9 Pa 2
θB =
2 EI
9(4 ×103 )(36) 2
=
= 32.4 × 10−3 rad .
2(10 ×106 )(72)
SOLUTION (10.45)
(a)
y
A
P
P
P
C
D
L
B
x
P
Continued on next slide
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EIv '' = Px − P < x − a > − P < x − L + a >
Px 2 P
P
EIv ' =
− < x − a > 2 − < x − L + a > 2 +C1
2
2
2
3
Px P
P
EIv =
− < x − a >3 − < x − L + a >3 +C1 x + C2
6
6
6
Boundary conditions:
L
PL2 P L
Pa
v '( ) = 0 :
− ( − a) + C1 = 0,
C1 =
(a − L)
2
8
2 2
2
v(0) = 0 : C2 = 0
Equations (a) and (b) become
P
v' =
[ x 2 − < x − a > 2 − < x − L + a > 2 + a(a − L)]
2 EI
P 3
v=
[ x − < x − a >3 − < x − L + a >3 +3ax(a − L)]
6 EI
(b)
(a)
(b)
(1)
(2)
Make x=L in Eq.(1):
Pa
θB =
( L − a)
2 EI
4 × 103 (3 × 12)(10 − 3)12
=
= 8.4 × 10−3 rad .
2(10 × 106 )(72)
SOLUTION (10.46)
(a)
y
P
Pa
A
RA =
2a
P
4
C
L
2a
B
x
RB =
3P
4
Px
− P < x − 2a > + Pa < x − 2a > 0
4
Px 2 P
EIv ' =
− < x − 2a > 2 + Pa < x − 2a > +C1
8
2
3
Px P
Pa
EIv =
− < x − 2a > 3 +
< x − 2a > 2 +C1 x + C2
24 6
2
EIv '' =
(a)
(b)
Continued on next slide
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Boundary conditions:
v(0) = 0 : C2 = 0
v(4a ) = 0 :
64 3 8 3
Pa − Pa + 2 Pa 3 + 4aC1 = 0,
24
6
C1 = −
Equations (a) and (b) become
P
v' =
[3x 2 − 12 < x − 2a > 2 +24a < x − 2a > −20a 2 ]
24 EI
P
v=
[ x3 − 4 < x − 2a >3 +12a < x − 2a > 2 −20a 2 x]
24 EI
5Pa 2
6
(1)
(2)
(b) Make 4a in Eq.(1):
7 Pa 2
θB =
6 EI
7(4 ×103 )(3 ×12) 2
=
= 8.4 ×10−3 rad
6(10 × 106 )(72)
SOLUTION (10.47)
A
C
M(x)
L/2
x
RA
2w
dV
L
= w= − 0 < x− >
2
dx
L
w0
dM
L 2
=V = −
< x − > + RA
2
dx
L
w
L
M ( x ) = − 0 < x − > 3 + RA x
3L
2
At x = L,
M ( L) = 0 : 0 = −
w0
L
( L − ) 3 + RA L
3L
2
or
RA =
w0 L
24
Therefore
w
wL
d 2v
L
EI 2 = M ( x) = − 0 < x − >3 + 0 x
dx
3L
2
24
Continued on next slide
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w
wL
dv
L
= − 0 < x − > 4 + 0 x 2 + C1
dx
12 L
2
48
w
wL
L
EIv = − 0 < x − >5 + 0 x 3 + C1 x + C2
60 L
2
144
EI
Boundary conditions:
EIv(0) = 0 : C2 = 0
EIv( L) = 0 : −
(a) v =
(b)
w0 L 5 wL4
( ) +
+ C1 L = 0,
60 L 2
144
C1 = −
37 w0 L3
5760
w0
L
L2 3 37 L4
1
x −
x]
[− < x − >5 +
EIL 60
5
144
5760
dv w0
L
L2
1
37 L4
=
[− < x − > 4 + x 2 −
]
dx EIL 12
5
48
5760
37 w0 L3
θ A = v '(0) =
5760 EI
Substituting the given data
37(2 × 103 )(10 × 12)3
θA =
= 0.03083 rad = 1.77 o
5760(10 ×106 )(72)
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________________________________________________________________________
PROBLEMS (10.48 through *10.51) A simple beam loaded as shown in Figs. P10.48 through
P10.51. Determine:
(a) The equation of the elastic curve.
(b) The angle of rotation at the end A.
(c) The deflection at the midspan.
SOLUTION (10.48)
(a)
w
y
A
L/2
wL/8
C
L/2
B
x
3wL/8
1
w
L
EIv '' = wLx − < x − > 2
8
2
2
1
w
L
EIv ' = wLx 2 − < x − >3 +C1
16
6
2
1
w
L
EIv =
wLx3 −
< x − > 4 +C1 x + C2
48
24
2
Using boundary conditions:
v(0) = 0 : C2 = 0
(a)
(b)
1
7
wL4
wLx 4 −
+ C1 L = 0,
C1 = −
wL3
48
384
384
Equations (a) and (b) become
w
L
v' =
[24 Lx 2 − 64 < x − >3 −7 L3 ]
384 EI
2
w
L
v=
[8Lx 3 − 16 < x − > 4 −7 L3 x]
384 EI
2
(b) Make x=0 in Eq.(1):
7 wL3
θA =
384 EI
v( L) = 0 :
(1)
(2)
(c) Make x=L/2 in Eq.(2):
5 wL4
vC =
↓
768 EI
Continued on next slide
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SOLUTION (10.49)
(a)
y
M0
A
a
C
b
L
M0/L
x
B
M0/L
M0
x − M 0 < x − a >0
L
M
EIv ' = 0 x 2 + M 0 < x − a > +C1
2L
M0 3 M0
EIv =
x +
< x − a > 2 +C1 x + C2
6L
2
Boundary conditions:
EIv '' =
v(0) = 0 : C2 = 0
(a)
(b)
v( L) = 0 : C1 = −
M0
(3b 2 − L2 )
6L
Equations (a) and (b) become
M0
v' =
[−3x 2 − 6 L < x − a > −(3b 2 − L2 )]
6 EIL
M0
v=
[− x3 + 3L < x − a > 2 −(3b 2 − L2 ) x]
6 EIL
(b) Make x=0 in Eq.(1):
M0
θA =
( L2 − 3b 2 )
6 EIL
(1)
(2)
(c) Make x=L/2 in Eq.(2):
M 0 9 L3
3
vM =
− 3aL2 + 3La 2 − b 2 L)
(
6 EIL 8
2
SOLUTION (10.50)
(a)
y
P
A
a
k
P/2
C
a
B
x
P/2
Continued on next slide
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P
x−P < x−a >
2
P
P
EIv ' = x 2 − < x − a > 2 +C1
4
2
P 3 P
EIv = x − < x − a >3 +C1 x + C2
12
6
EIv '' =
(a)
(b)
Boundary conditions:
P
PEI
v(0) = − : C2 = −
2k
2k
3
8Pa Pa 3
PEI
v(2a) = 0 :
−
+ 2C1a −
,
12
6
2k
C1 = −
Pa 2 PEI
+
4
4ak
Equations (a) and (b) become
P
EI
v' =
[ x 2 − 2 < x − a >2 −a 2 + ]
4 EI
ak
6 EI
P
EI
v=
[ x 3 − 2 < x − a >3 +3(− a 2 + ) x −
]
12 EI
ak
k
(1)
(2)
(b) Make x=0 in Eq.(1):
P
EI
θA =
(−a 2 + )
4 EI
ak
(c) Make x=a in Eq.(2):
3EI
P
vC =
(2a 3 +
)↓
12 EI
k
SOLUTION (*10.51)
(a)
2
x
wo
x
w0
L
2w0
y
B
A
L/2
C
L/2
w0L/4
B
w0L/4
(a) Actual load
=
A
C
x
w0L/4
w0L/4 4 w0 ( x − L )
2
L
2w0
(b) Equivalent load
Continued on next slide
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Refer to Fig.(b):
wL
1 2 x2
1
x 1 4w
L
L
M = 0 x− (
w0 ) + [ 0 < x − > 2 ] < x − >3
4
2 L
3 2 L
2
3
2
Thus,
wL
w
2w
L
EIv '' = 0 x − 0 x 3 + 0 < x − >3
4
3L
3L
2
w0 L 2 w0 4 w0
L
EIv ' =
x −
x +
< x − > 4 +C1
8
12 L
6L
2
wL
w
w
L
EIv = 0 x 3 − 0 x 5 + 0 < x − >5 +C1 x + C2
24
60 L
30 L
2
(a)
(b)
Boundary conditions:
v(0) = 0 : C2 = 0
v( L) = 0 : C1 = −
5
w0 L3
192
Equations (a) and (b) become
w0
L
v' =
[24 L2 x 2 − 16 x 4 + 32 < x − > 4 −5 L2 ]
192 EIL
2
w0
L
v=
[8L2 x 3 − 3.2 x 5 + 6.4 < x − > 4 −5L3 x]
192 EIL
2
(1)
(2)
(b) Substitute x=0 into Eq.(1):
5 w0 L2
θA =
192 EI
(c) Substitute x=L/2 into Eq.(2):
w0 L4
vC = vmax =
↓
120 EI
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________________________________________________________________________
PROBLEM (10.52 ) A simple beam AB is subjected to couple M0 at midspan acting as shown in
Fig.P10.52.Determine:
(a) The equation of the deflection curve.
(b) The deflection at midspan.
(c) The magnitude and location of the maximum deflection.
(d) The slope at the free end.
SOLUTION
y
M0
A
L/2
M0/L
L/2
C
L
B
x
M0/L
M0
L
x − M 0 < x − >0
2
L
M
L
EIv ' = 0 x 2 − M 0 < x − > +C1
2L
2
M
M
L
EIv = 0 x 3 − 0 < x − > 2 +C1 x + C2
6L
2
2
EIv '' =
Boundary conditions:
v(0) = 0 : C2 = 0
M L
v( L) = 0 : C1 = − 0
24
Equations (a) and (b) become
M0
L
EIv ' =
[12 x 2 − 24 L < x − > − L2 ]
24 EIL
2
M0
L
EIv =
[4 x 3 − 12 L < x − > 2 − L2 x]
24 EIL
2
(a)
(b)
(1)
(2)
These lead to the results given in solution of Prob. 10.31.
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________________________________________________________________________
PROBLEMS (*10.53 and *10.54) An overhanging beam is supported and loaded as shown in
Fig. P10.53 and P10.54. Determine the equation of the elastic curve and deflection at point A.
SOLUTION (*10.53)
y
wa2
w
C
a
B
A
x
Equivalent loading
w
RA=wa/2
RB
Refer to the above figure:
w
w
wa
a
EIv '' = − ( x + a) 2 + < x > 2 +
< x > + a 2 w < x − >0
2
2
2
2
w
w
wa
a
EIv ' = − ( x + a)3 + < x >3 +
< x > 2 + a 2 w < x − > +C1
6
6
4
2
2
w
w
wa
a w
a
EIv = − ( x + a ) 4 +
< x >4 +
< x >3 +
< x − > 2 +C1 x + C2
24
24
12
2
2
Using boundary conditions:
wa 4
wa 4
v(0) = 0 : −
+ C2 = 0,
C2 =
24
24
4
4
2
3
wa
wa
wa 4
wa 4
v(a) = 0 : − wa 4 +
+
+
+ C1a +
= 0,
C1 = wa 3
3
24
12
8
24
8
Equation (1) becomes
w
a
v=
[−( x + a) 4 + < x > 4 +2a < x >3 +12a 2 < x − > 2 +9a 3 x + a 4 ]
24 EI
2
Let x=-a:
wa 4
vC =
↓
3EI
(1)
SOLUTION (*10.54)
From conditions of equilibrium:
RA = 2 P ↑
RB = 0
Using singularity functions:
M = − Px + 2 P < x − a > − P < x − 2a >
Using singularity functions:
d 2v
EI 2 = − Px + 2 P < x − a > − P < x − 2a >
dx
(1)
Continued on next slide
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dv
1
1
= − Px 2 + P < x − a > 2 − P < x − 2a > 2 +C1
dx
2
2
1
1
1
EIv = − Px 3 + P < x − a >3 − P < x − 2a >3 +C1 x + C2
6
3
6
EI
(2)
(3)
Boundary conditions:
1 3
Pa
6
EIv(3a) = 0 : 3aC1 + C2 = 2 Pa 2
11
3
C1 = Pa 2 , then C2 = − Pa 3
Eq.(5)-Eq.(4):
12
4
Equation (3) is therefore
P
x3 < x − a > < x − 2a >3 11a 2 x 3a 3
−
+
−
v=
[− +
]
EI
6
3
6
12
4
At the free end (x=0), we have
3Pa 3 3Pa 3
vC = −
=
↓
4 EI
4 EI
EIv(a) = 0 : C1a + C2 =
(4)
(5)
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________________________________________________________________________
PROBLEM (10.55) A cantilever beam carries a concentrated force P and moment M = PL as
shown in Fig. P10.55. Determine the deflection at the free end A.
SOLUTION
P
PL
A
L/2
C
L/2 B
=
P
PL
A
L
B
+ A
C
B
(θC ) II
Loading I
Loading II
For each loading, we determine the deflection at A, using Table B.14:
Loading I:
M A L2 PL3
=
(v A ) I =
2 EI
2 EI
Loading II:
P( L 2)3
PL3
=−
(vC ) II = −
3EI
24 EI
L
10 PL3
(v A ) II = (vC ) II + (θC ) II ( ) = −
2
96 EI
3
2
PL
P( L 2) L
10 PL3
=−
−
( )=−
24 EI
2 EI
2
96 EI
Deflection at A:
19 PL3
v A = (v A ) I + (v A ) II =
↑
48 EI
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________________________________________________________________________
PROBLEM (10.56) An overhang beam ABC supports a concentrated load P as shown in Fig 10.56.
what is the deflection at the free end C ?
SOLUTION
P
L/2
A
L
C
B
P/2
=
3P/2
P
P
A
L
+
B
PL/2
A
B
C
θB
The displacement of C is produced by the rotation of B and deflection of the
segment BC as cantilever. Using Table B.14:
L
P( L 2) L P( L 2) PL3
vC = (vC ) M + (vC ) P = θ B ( ) + (vC ) P = −
( )−
=
↓
3EI 2
24 EI
8 EI
2
and
( P L 2) L P( L 2) 2
7 PL2
=
θC = (θC ) M + (θC ) P = −
−
3EI
2 EI
24 EI
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________________________________________________________________________
PROBLEM (10.57) A cantilever AB carries a uniform load of intensity w over the half span CB, as
shown in Fig. P10.57. Find the deflection at the free end B.
SOLUTION
w
w
a
A
C
a
B
=
+
w
Loading I
=
a
C
a
2a
A
Use Table B.14:
w(2a ) 4
wa 3
( vB ) I = −
(θ B ) II = −
8 EI
6 EI
and
(vB ) II = (vC ) II + (θC ) II (a)
A
(vC ) II = −
Loading II
wa 4
8 EI
wa 4 wa 3
7 wa 4
−
(a) =
8EI 6 EI
24 EI
Thus,
vB = (vB ) I + (vB ) II
=−
2wa 4 7 wa 4 41 wa 4
+
=
↓
EI
24 EI 24 EI
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B
________________________________________________________________________
PROBLEMS (10.58 and 10.59) A simply supported beam AB is loaded as shown in Fig. P10.58
and P10.59. Calculate the angle of rotation at end B.
Given: P=4 kips,
a=3 ft, L=10 ft,
E=10x106 psi,
I=72 in.4
SOLUTION (10.58)
P
a
P
a
A
B
D
C
Actual
Loading
=
L
P
P
A
+A
a C
D B
L-a
B
Loading I
Loading II
Use Table B.14: θ B = (θ B ) I + (θ B ) II
where,
Pa( L2 − a 2 )
P( L − a )[ L2 − ( L − a ) 2 ]
(θ B ) I =
(θ B ) II =
6 EIL
6 EIL
P
(aL2 − a 3 + 2 L2 a − 3La 2 + a 3 )
6 EIL
Pa
=
( L − a)
2 EI
4 × 103 (3 × 12)
=
(10 − 3)(12) = 8.4 ×10−3 rad
6
2(10 ×10 )(72)
Thus, θ B =
SOLUTION (10.59)
P
P
B
A
2
C 2a
P
A
C
Loading I
=
Actual
Loading
P
B + A
C
B
Loading II
Continued on next slide
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Use Table B.14:
θ B = (θ B ) I + (θ B ) II
where,
P(4a) 2 Pa 2
(θ B ) I =
=
16 EI
EI
2
Pa(4a)
Pa 2
(θ B ) II =
=
24 EI
6 EI
Thus,
Pa 2 Pa 2 7 Pa 2
=
θB =
+
EI 6 EI 6 EI
7(4 ×103 )(3 ×12) 2
=
= 8.4 ×10−3 rad
6
6(10 ×10 )(72)
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________________________________________________________________________
PROBLEMS (10.60 and *10.61) A beam is loaded and supported as shown in Fig. P10.60 and
P10.61. Determine the deflection at the free end .
SOLUTION (10.60)
Q
P
a
a
A
a
C
B
D
Actual
Loading
=
Q
P
a
a
A
C
+
B
D
A
B
Equivalent Loadings
Use Table B.14
vB = (vB ) P + ( vB ) Q
where,
(vB ) P = (vC ) P + (θC ) P (2a)
=−
Pa 3 Pa 2
4 Pa 3
−
(2a) = −
3EI 2 EI
3 EI
(vB )Q = (vD )Q + (θ D )Q (a)
=−
Q(2a)3 Q(2a) 2
14 Qa 3
−
(a) = −
3EI
2 EI
3 EI
Thus,
2a 3
vB =
(2 P + 7Q) ↓
3EI
SOLUTION (*10.61)
P
M0
B
A
L/2 C L/2
L/2 D
Actual
Loading
=
P
A
C
(θ B ) M
P
M0
a
+
B
A
C
B
M0
Continued on next slide
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Use Table B.14 (cases 7, 9, and 2):
vD = ( vB ) M , P + (vD ) M
where,
L
(vD ) M , P = (θ B ) M , P ( )
2
2
PL L M 0 L L
( )−
( )
=−
16 EI 2
3EI 2
(vD )Q = (vD )Q + (θ D )Q (a)
( vD ) M = −
M 0 ( L 2) 2
2 EI
Hence,
vD =
L2
(3PL − 28M 0 )
96 EI
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________________________________________________________________________
PROBLEM (*10.62) A nonprismatic cantilever beam with flexural rigidity 2EI in part AC and
flexural rigidity EI in part BC carries a uniform load of intensity w (Fig. P10.62). What is the deflection at
the free end ?
*SOLUTION
w
A
L/2
w
A
2EI
C
L/2
=
Actual
Loading
B
wL
2
wL2 + C
8
Equivalent Loadings
C
w
EI
B
Use Table B.14 (cases 3.1, and 2):
17 wL4
w( L 2) 4 ( wL 2)( L 2)3 ( wL2 8)( L 2) 2
=−
vC = −
−
−
768 EI
8(2 EI )
3(2 EI )
2(2 EI )
θC = −
7 wL3
w( L 2)3 ( wL 2)( L 2)2 ( wL2 8)( L 2)
=−
−
−
96 EI
6(2 EI )
2(2 EI )
2 EI
Thus,
L w( L 2) 4
vB = vC + θC ( ) −
2
8( EI )
17 wL4
7 wL4
wL4
=−
−
−
768 EI 192 EI 128EI
17 wL4
=
↓
256 EI
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________________________________________________________________________
PROBLEM (10.63) An S 250 x 38 shape beam (see table B.9) is loaded as shown in Fig. P10.63.
Calculate the slope at B.
Given: a = 2 m,
P = 2Q = 10 kN,
I = 51.6 x 106 m4
E = 200 GPa,
SOLUTION
I = 51.4 × 106 mm 4 (Table B.9).
P
A
Q
a C a D a
B
Actual
Loading
=
Q
P
A
C
B
+
A
D
B
Equivalent Loadings
Use Table B.14 (case 1): θ B = (θ B ) P + (θ B )Q
where,
Pa 2
(θ B ) P = (θC ) P = −
2 EI
Q(2a ) 2
2Qa 2
(θ B )Q = (θ D )Q = −
=−
2 EI
EI
Therefore,
a2
θB =
( P 2 + 4Q)
2 EI
(2) 2 (10 + 20)103
=
= 5.81× 10−3 rad
9
−6
2(200 ×10 )(51.6 × 10 )
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________________________________________________________________________
PROBLEM (10.64) The beam AC in Fig. P10.64 is made of a W 150 x 24 rolled-steel shape (see
Table B.8) and M = 10 kN m, P = 4 kN, L = 3 m, and E = 210 GPa. Calculate the deflection at C.
SOLUTION
Table B.8: I = 13.4 × 106 mm 4
P
A
L/2
C
P
M0
L/2
B
L/2
Actual Loading
M0
A
C
B
Equivalent loading
Use Table B.14 (cases 7 and 9):
vC = (vC ) P + (vC ) M
PL3 M 0 ( L 2) L2
−
( − L2 )
48EI
6 EIL
4
2
L
=
(− PL + 3M 0 )
48EI
=−
Thus,
vC =
(3)2 (−4 × 3 + 3 × 10)103
= 1.2 mm ↑
48(210 ×109 )(13.4 ×10−6 )
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________________________________________________________________________
PROBLEM (10.65) The slope at the wall of the clamped beam shown in Fig. P10.65a is as seen in
Fig. P10.65. Find the reaction on the roller at A.
SOLUTION
y
w
A
1
M = RA x − wx 2
2
V
x
1
2
R
EIv '' = RAA x − wx
2
1
1
EIv ' = RA x 2 − wx3 + C1
2
6
1
1
EIv = RA x 3 − wx 4 + C1 x + C2
6
24
Boundary conditions:
v(0) = 0 : C2 = 0
RA L2 wL3
33
1
wL3
wL3
:
,
−
+ C1 =
C1 =
wL3 − RA L2
192 EI
2
6
192
192
2
3
4
4
3
RA L wL 33wL RA L
25
v( L) = 0 :
−
+
−
=0,
RA =
wL
6
24
192
2
64
v '( L) =
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________________________________________________________________________
PROBLEM (10.66) Before the uniform load of intensity w is applied to the beam shown in Fig.
P10.66, a small gap δB = 1/8 in. exists between the beam AC and the support at B. What is the reaction at
each support developed subsequent to the loading ?
Given: w = 1.2 kips/ft, L=12 ft, EI = 400(106) lb ⋅ in.2
SOLUTION
y
w
MA
δB
B
A
L
RA
C
L/2
RB
Example 10.10 is resolved with the last of Eqs.(c) in page 498 is modified as
v( L) = −δ B .
In so doing
7
K
13
19
K
K
M A = wL2 +
RA =
wL +
RB =
wL −
32
8
24
8L
24
8L
where
24 EI δ B 24(400 × 106 )(0.125)
K=
= 57.87 kip ⋅ in.
=
(12 ×12) 2
L2
Thus,
7 1.2
57.87
M A = ( )(144) 2 +
= 460.8 kip ⋅ in.
32 12
8
13 1.2
57.87
RA = ( )(144) +
= 7.85 kips ↑
24 12
8(144)
19 1.2
57.87
RB = ( )(144) −
= 11.35 kips ↑
24 12
8(144)
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________________________________________________________________________
PROBLEM (10.67) Before linearly varying load of maximum intensity w0 is applied to the beam
shown in Fig. 10.67, a small gap δA= 1/16 in. exists between the beam AB and the support at A.
Determine the reaction at each support developed after the loading.
Assumption: The bearing block is taken to be rigid.
L = 10 ft, E =10 x 10 6 psi,
I = 65 in.4
Given: w0 = 2.4 kips/ft,
SOLUTION
y
δA
w0
L
RA
MB
B
RB
Resolve Example 10.12a with v(0) = −δ A to obtain: C2 = − EI δ A .
Hence,
w0 L2
w0 L
2
RA =
−K
RB = w0 L + K
MB =
+ KL
5
15
10
where
3EI δ A 3(30 ×106 )(65)(0.0625)
K=
=
= 0.2116 kips
L3
(120)3
Substitute the given data:
(2.4 12)120
RA =
− 0.2116 = 2.188 kips ↑
10
2(2.4 12)120
RB =
+ 0.2116 = 9.82 kips ↑
5
(2.4 12)(120) 2
MB =
+ 0.2116(120) = 217.4 kip ⋅ in.
15
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________________________________________________________________________
PROBLEMS (10.68 through 10.71) A fixed-end beam constructed of a W150x30 wide-flange
section (see Table B.8) is loaded as shown in Figs. P10.68 through P10.71. Using a direct-integration
method, determine:
(a) All of the reactions.
(b) The mid span deflection for the given values:
w = w0/2 = 15 kN/m, P = 25 kN, M0 = 10 kN ⋅ m, L = 4 m, E = 200 GPa, I = 17.2x106 mm4
SOLUTION (10.68)
(a)
y
w
MA
A
x
V
1
M = RA x − wx 2 − M A
2
RA
Due to symmetry:
RA = RB =
1
wL ↑
2
MA = MB
We have
1
1
EIv '' = wLx − M A − wx 2
2
2
1
1
EIv ' = wLx 2 − M A x − wx3 + C1
4
6
1
1
1
EIv = wLx3 − M A x 2 − wx 4 + C1 x + C2
12
2
24
Boundary conditions:
v '(0) = 0 : C1 = 0 ,
v(0) = 0 : C2 = 0
1
v( L) = 0 : M A = wL2
12
(b) Then Eq.(a) becomes
wx 2
wx 2
v=
(2 Lx − L2 − x 2 ) = −
( L − x)2
24 EI
24 EI
Make x=L/2:
wL4
vC =
↓
384 EI
Substitute the given data
15 ×103 (4) 4
= 2.91×10−3 m = 2.91 mm ↓
vC =
9
−6
384(200 ×10 )(17.2 ×10 )
Continued on next slide
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SOLUTION (10.69)
(a)
y
MA
M0
A
B
C
L/2
L/2
x
MB
RB
RA
Segment AC
EIv1 '' = RA x − M A ,
EIv1 =
EIv1 ' =
1
RA x 2 − M A x + C1
2
1
1
RA x 3 − M A x 2 + C1 x + C2
6
2
(1)
Segment BC
EIv1 '' = RA x − M A + M 0 ,
RA x 2
EIv2 ' =
− M A x + M 0 x + C3
2
1
1
1
RA x 3 − M A x 2 + M 0 x 2 + C3 x + C4
6
2
2
Boundary & continuity conditions:
v1 '(0) = 0 : C1 = 0
v1 (0) = 0 : C2 = 0
M L
6M A
L
L
L
v1 ( ) = 0 : RA =
v1 '( ) = v2 '( ) : C3 = − 0
2
2
2
2
L
1
1
v2 '( L) = 0 :
RA L2 − M A L + M 0 L − M 0 L = 0
2
2
Solving,
1
3 M0
RA =
M A = M0
↓
4
2 L
Statics:
3 M0
1
RB =
↑
MB = M0
2 L
4
We have, from v( L) = 0 : C4 = M 0 L2 8 .
EIv2 =
(2)
(b) Equation (1) for x=L/2:
L
vC = v1 ( ) = 0
2
Continued on next slide
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SOLUTION (10.70)
(a)
w0
( L − x)
L
y
w0
MA
w0
( L − x) 2
2L
MB
x
A
B
L
RA
MB
M
O
V
L-x
RB
RB
w0
( L − x)3
6L
w
EIv '' = RB L − RB x − M B − 0 ( L3 − 3L2 x + 3Lx 2 − x 3 )
6L
w0 3 3 2 2
1
x4
2
3
( L x − L x − Lx − ) + C1
EIv ' = RB Lx − RB x − M B x −
2
6L
2
4
3 2
2 3
w Lx
1
1
1
L x Lx 4 x5
EIv = RB Lx 2 − RB x 3 − M B x 2 − 0 (
−
−
− ) + C1 x + C2 (1)
2
6
2
6L 2
2
4
20
M = RB ( L − x) − M B −
Boundary conditions:
v '(0) = 0 : C1 = 0
v(0) = 0 :
C2 = 0
2
w0 4 3L4
RB L
L4
4
(L −
v '( L) = 0 : RB L −
− MBL −
+L − )=0
2
6L
2
4
2M B w0 L
RB =
+
12
L
3
2
RB L M B L w0 L5 L5 L5 L5
v( L) = 0 :
−
− ( − + − )=0
L 2 2 4 20
3
2
3M B w0 L
RB =
+
2L
10
2
Equations (2) and (3) yield
w L2
MB = 0
30
Statics:
7
RA =
w0 L ↑
20
RB =
(2)
(3)
3
w0 L ↑
20
MA =
w0 L
20
Continued on next slide
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(b)
Equation (1) becomes
w0
v=
(−3L3 x 2 + 7 L2 x3 − 5Lx 4 + x5 )
120 EIL
Making x=L/2:
w L4
vM = 0
↓
768 EI
30 × 103 (4) 4
= 2.91 mm ↓
=
768(200 ×109 )(17.2 ×10−6 )
SOLUTION (10.71)
(a)
2x
w0
L
y
MA
A
RA
x
Due to symmetry:
RA = RB
w0
C
L/2
B
MB
RB
MA = MB
vC ' = 0
Statics:
1
w0 L ↑
4
Thus, only segment AC need be considered.
1
1
EIv '' = w0 Lx − M A +
w0 x 3
4
3L
1
1
EIv ' = w0 Lx 2 − M A x −
w0 x 4 + C1
8
12 L
1
1
1
EIv =
w0 Lx3 − M A x 2 −
w0 x5 + C1 x + C2
24
2
60 L
RA = RB =
(1)
Continued on next slide
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(b) Boundary conditions:
v '(0) = 0 : C1 = 0
v(0) = 0 :
3
C2 = 0
3
w0 L M A L w0 L
L
v '( ) :
−
−
=0
2
32
2
192
5
5
M A = w0 L2
M B = w0 L2
96
96
Equation (1) becomes
w0
3.2 5
v=
(8Lx 3 − 5L2 x 2 −
x )
192 EI
L
Let x=L/2:
7 w0 L4
vC =
↓
3840 EI
7(30 × 103 )(4) 4
= 4.07 mm
vC =
3840(200 ×109 )(17.2 ×10−6 )
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________________________________________________________________________
PROBLEM (*10.72) Figure P10.72 shows a nonprismatic propped cantilever beam AB with
flexural rigidity EI from A to C and 2EI from C to B. The beam supports a uniform load of intensity w.
Using a direct-integration approach, find :
(a) The support reactions (RA, MA, and RB).
(b) The mid span deflection if w = 60 lb/ft, L = 16 ft, and EI = 200 x 106 lb ⋅ in.2
*SOLUTION
w
(a)
MA
A
L/2
C
L/2
RA
B
RB
w
MA
M
A
C
L/2
RA
x
V
1
M 1 = − M A + RA x − wx 2
2
L
L
M 2 = − M A + RA [ + ( x − )]
2
2
wL L
L
L
1
−
[ + ( x − )] − w( x − ) 2
2 4
2
2
2
Segment AC
1
EIv1 '' = − M A + RA x − wx 2
2
1
1
EIv1 ' = − M A x + RA x 2 − wx 3 + C1
2
6
1
1
1
EIv1 = − M A x 2 + RA x3 − wx 4 + C1 x + C2
2
6
24
(1)
Segment CB
M
M
R L wL2 1
wL
L w
L
)( x − ) − ( x − ) 2
EIv2 '' = 2 = − A + A −
+ ( RA −
2
2
4
16 2
2
2
4
2
2
M x R Lx wL x 1
wL
L
w
L
)( x − ) 2 − ( x − )3 + C3
EIv2 ' = − A + A −
+ ( RA −
2
4
16
4
2
2
12
2
2
2
2 2
M x
R Lx wL x
1
wL
L
w
L
)( x − )3 − ( x − ) 4 + C3 x + C4
EIv2 = − A + A
−
+ ( RA −
4
8
32
12
2
2
48
2
Continued on next slide
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Boundary & continuity conditions:
v1 '(0) = 0 : C1 = 0
v1 (0) = 0 :
C2 = 0
M L R L2 wL3
M L R L2 wL3
L
L
v1 '( ) = v2 '( ) : − A + A −
=− A + A −
+ C3
2
2
2
8
48
4
8
32
M L wL3
C3 = − A +
4
96
2
3
M A L RA L wL4
M A L2 RA L3 wL4
L
L
v1 ( ) = v2 ( ) : −
+
−
=−
+
−
2
2
8
48 384
16
32 128
M L2 wL4
− A +
+ C4
8
192
M L2 R L3
C4 = − A − A
16
96
7M A 7
7
1
7
v2 ( L) = 0 : − M A L2 + RA L3 −
wL4 = 0 , RA =
+ wL
16
8
256
2 L 32
Statics:
9
∑ M = 0 : M = 80 wL
B
2
A
and
RA =
49
wL ↑
80
31
∑ F = 0 : R = 80 wL ↑
y
B
(b) Equation (1) becomes
w0
(−27 L2 x 2 + 49 x3 − 20 x 4 )
v1 =
480 EI
Letting x=L/2:
wL4
vC =
↓
256 EI
Substitute the given data:
60(16) 4
= 0.077 in.
vC =
256(200 ×103 )
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________________________________________________________________________
PROBLEMS (10.73 through 10.76) A propped cantilever beam is loaded as shown in Figs.
P10.73 through P10.76. Using a direct-integration method, obtain:
(a) All of the reactions.
(b) The midspan deflection for the given data:
P = 30 kN,
M0 = 8 kN ⋅ m,
L = 5 m,
w = w0 /2 =20 kN/m,
as required.
EI = 3 MN ⋅ m2
SOLUTION (10.73)
(a)
y
P
B
A
L/2
C
L/2
x
MB
RB
RA
Segment AC
EIv1 '' = RA x
EIv1 =
EIv1 ' =
1
RA x 2 + C1
2
1
RA x 3 + C1 x + C2
6
(1)
Segment CB
L
EIv2 '' = RA x − P( x − )
2
1
P
L
EIv2 ' = RA x 2 − ( x − ) 2 + C3
2
2
2
1
P
L
EIv2 = RA x3 − ( x − )3 + C3 x + C4
6
6
2
Boundary and continuity conditions:
L
L
v1 (0) = 0 : C2 = 0
v1 '( ) = v2 '( ) : C1 = C3
2
2
1
1
v2 '( L) = 0 :
RA L2 − PL2 + C3 = 0
2
8
L
L
v1 ( ) = v2 ( ) : C4 = 0
2
2
1
1
v2 ( L) = 0 :
RA L3 − PL3 + C3 L = 0
6
48
From Eqs.(2) and (3):
5
RA = P ↑
16
and
(2)
(3)
Continued on next slide
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C1 = C3 = −
1
PL2
32
Then
11
∑ F = 0 : R = 16 P ↑
y
B
3
∑ M = 0 : M = 16 PL
A
B
(b) Substitute constants into Eq.(1):
PL3
x
x
v1 =
[5( ) − 3( )]
L
L
96 EI
Make x=L/2 in the above
7 PL3
vC =
↓
768 EI
7(30 × 103 )(5)3
= 11.39 mm ↓
=
768(3 × 106 )
SOLUTION (10.74)
(a)
y
MA
A
B
L/2
RA
C
L/2
x
M0
RB
EIv '' = RB ( L − x) + M 0 = RB L − RB x + M 0
1
EIv ' = RB Lx − RB x 2 + M 0 x + C1
2
1
1
1
EIv = RB Lx 2 − RB x3 + M 0 x 2 + C1 x + C2
2
6
2
(1)
Continued on next slide
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Boundary conditions:
v(0) = 0 : C2 = 0
v '(0) = 0 : C1 = 0
1
1
1
v( L) = 0 :
RB L3 − RB L3 + M 0 L2 = 0 ,
2
6
2
M
3 0
↑
Then, ∑ Fy = 0 : RA =
2 L
1
∑ M y = 0 : M A = 2 M0
RB =
3 M0
↓
2 L
(b) Introduce the constants into Eq.(1):
M0
v=
(− Lx 2 + x3 )
4 EIL
Let x=L/2,
M 0 L2
vC =
↓
32 EI
8 × 103 (5) 2
=
= 2.08 mm ↓
32(3 × 106 )
SOLUTION (10.75)
(a)
w
MA
A
L/2
RA
C
L/2
B
RB
Segment AC
1
EIv1 '' = RA x − M A − wx 2
2
1
1
EIv1 ' = RA x 2 − M A x − wx3 + C1
2
6
1
1
1
EIv1 = RA x 3 − M A x 2 − wx 4 + C1 x + C2
6
2
24
(1)
Segment BC
wL
L
(x − )
2
4
1
wL
L
( x − ) 2 + C3
EIv2 ' = RA x 2 − M A x −
2
4
4
EIv2 '' = RA x − M A −
Continued on next slide
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EIv2 =
1
1
wL
L
( x − )3 + C3 x + C4
RA x 3 − M A x 2 −
6
2
12
4
Boundary & continuity conditions:
v1 '(0) = 0 : C1 = 0
v1 (0) = 0 :
C2 = 0
L
L
wL3
wL3
wL3
v1 '( ) = v2 '( ) : −
=−
+ C3 ,
C3 = −
2
2
48
64
192
4
4
4
L
L
wL
wL wL
wL4
v1 ( ) = v2 ( ) : −
=−
−
+ C4 ,
C4 =
2
2
384
768 384
768
3
2
4
4
4
R L M L 9 wL wL wL
M
15
v2 ( L) = 0 : − A − A −
+
+
= 0, RA = 3 A + wL (b)
6
2
256 192 763
L 64
Statics:
9
∑ M = 0 : M = 128 wL
B
2
A
57
∑ F = 0 : R = 128 wL ↑
y
B
(b) Equation (1) becomes
w
v1 =
= (57 Lx3 − 27 L2 x 2 − 32 x 4 )
768 EI
Make x=L/2 in the above
13wL
vC =
↓
6144 EI
13(20 ×103 )(5) 4
= 8.82 mm ↓
=
6144(3 × 106 )
Continued on next slide
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SOLUTION (10.76)
(a) See Table B.6.
y
w0 x 2
( ) x
3 L
x
w = w0 ( ) 2
L
A
RA
w0
x/4
MB
L
RB
A
x
RA
x
w0 ( ) 2
L
M
O
V
w0 4
x
12 L2
w
1
EIv ' = RA x 2 − 0 2 x5 + C1
2
60 L
w0
1
EIv = RA x3 −
x 6 + C1 x + C2
6
360 L2
EIv '' = M = RA x −
(1)
Boundary conditions:
v(0) = 0 : C2 = 0
w
1
v '( L) = 0 :
RA L2 − 0 L3 + C1 = 0
2
60
w
1
v( L) = 0 :
RA L3 − 0 L4 + C1 L = 0
6
360
w L3
From Eqs.(2) and (3): C1 = − 0
240
and
1
RA =
w0 L ↑
24
Statics:
1
7
RB =
w0 L ↑
∑ Fy = 0 : RA + RB = 3 w0 L ,
24
1
1
L
1
∑ M B = 0 : 24 w0 L2 + M B − 3 w0 L( 4 ) = 0 , M B = 24 w0 L
(2)
(3)
Continued on next slide
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(b)
Equation (1) becomes
w
v = 0 (5 L3 x 3 − 2 x 6 − 3L5 x)
720 L
Making x=L/2:
w L4
vM = 0
↓
960 EI
40 ×103 (5) 4
=
= 8.68 mm
960(3 ×106 )
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________________________________________________________________________
PROBLEMS (10.77 and 10.78) A beam AB made of a W150 x 30 wide-flange section
supported and loaded as shown in Figs. P10.77 and P10.78. Determine:
(a) All of the reactions for this beam.
(b) The deflection at C for the given data:
P = 20 kN, M = 10 kN ⋅ m, L = 5 m, E = 200 GPa, I = 17.2 x 106 mm4
(see Table B.8)
SOLUTION (10.77)
(a)
P
y
A
L/2
RA
C
MB
L/2
B
x
RB
L
>
2
1
P
L
EIv ' = RA x 2 − < x − > 2 +C1
2
2
2
1
P
L
EIv = RA x 3 − < x − >3 +C1 x + C2
6
6
2
EIv '' = RA x − P < x −
Using boundary conditions:
v(0) = 0 : C2 = 0
1
1
v '( L) = 0 : C1 = PL2 − RA L2
8
2
1
1
1
1
v( L) = 0 :
RA L3 − PL3 + PL3 − RA L2 = 0,
6
48
8
2
Then, from statics:
11
∑ Fy = 0 : RB = 16 P ↑
3
∑ M A = 0 : M B = 16 PL
(1)
RA =
5
P↑
16
(b) Equation (1) becomes
P
L
v=
[5 x 3 − 16 < x − > +12 L2 x − 15L2 x]
96 EI
2
Making x=L/2:
7 PL3
vC =
↓
768 EI
7(20 ×103 )(5)3
= 6.62 mm
=
768(200 ×109 )(17.2 ×10−6 )
Continued on next slide
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SOLUTION (10.78)
y
M0
MA
A
L/2
C
L/2
B
x
MB
RA
RB
L 0
EIv '' = − M A + RA x − M 0 < x − >
2
1
L
EIv ' = − M A x + RA x 2 + M 0 < x − > +C1
2
2
2
3
M x
R x M
L
EIv = − A + A + 0 < x − > 2 +C1 x + C2
2
6
2
2
(1)
(2)
Using boundary conditions:
1
1
v '( L) = 0 : − M A L + RA L2 + M 0 L + C1 = 0
2
2
v '(0) = 0 : C1 = 0
v(0) = 0 : C2 = 0 :
1
1
1
v( L) = 0 : − M A L2 + RA L3 + M 0 L2 = 0
2
6
8
Solving
3 M0
1
RA =
↓
M A = M0
2 L
4
Statics:
M
3 M0
RB =
↑
MB = 0
2 L
4
(b) Equation (2) for x=L/2:
M 0 L2 3 M 0 L3
vC =
−
=0
32
12 L 8
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________________________________________________________________________
PROBLEM (10.79) A propped cantilever beam AB supports a uniform load of intensity w over
half-span AC (Fig. P10.79). Determine:
(a) All of the reactions for this beam.
(b) The deflection at C for the given numerical values if w = 10 kN/m, L = 6 m, and EI = 3.44 MN ⋅ m2.
SOLUTION
(a)
y
w
MA
A
L/2
C
L/2
B
x
RB
RA
EIv '''' = − w + w < x −
L 0
>
2
L
> +C1
2
w
L
1
EIv '' = − wx 2 + < x − > 2 +C1 x + C2
2
2
2
w
L
1
1
EIv ' = − wx3 + < x − >3 + C1 x 2 + C2 x + C3
6
6
2
2
1
w
L
1
1
EIv = − wx 4 +
< x − > 4 + C1 x3 + C2 x 2 + C3 x + C4
24
24
2
6
2
EIv ''' = − wx + w < x −
Using boundary conditions:
v(0) = 0 : C4 = 0
v '(0) : C3 = 0
1
1
v ''( L) = 0 : − wL2 + wL2 + C1 L + C2 = 0
2
8
1
1
1
1
v( L) = 0 : − wL4 +
wL4 + C1 L3 + C2 L2 = 0
24
384
6
2
Solving,
C1 =
57
wL
128
RA =
57
wL ↑
128
C2 = −
9
wL2
128
Thus,
Then
∑F = 0:
y
RB =
MA =
9
wL2
128
7
wL ↑
128
Continued on next slide
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(b) Equation (a) is therefore
w
L
v=
[−32 x 4 + 32 < x − > 4 +57 Lx3 − 27 L2 x 2 ]
768 EI
2
Make x=L/2 in the above
13 wL4
vC =
↓
6144 EI
Substituting the given data
13 10 × 103 (6) 4
vC =
= 7.97 mm ↓
6144 3.44(106 )
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________________________________________________________________________
PROBLEM (10.80) A propped cantilever beam AB constructed of a S10 x 35 shape (see Table B.9)
is subjected to a linearly varying load at the half-span as shown in Fig. P10.80. Calculate:
(a) The reaction RA at support A.
(b) The deflection at C.
I = 147 in.4
Given: E = 29 x 106 psi
SOLUTION
6 kip/ft
y
A
RA
x
B
C
4 ft
4 ft
k=1.5
A
B
C
6 kip/ft
k=1.5
Note: forces in kips and lengths in ft.
6 kip ft
Slop of load : k =
= 1.5 kip ft 2
4 ft
Refer to superposed load diagram:
dV
= w( x) = −1.5 x + 6 < x − 4 > 0 +1.5 < x − 4 >
dx
dM
= V = RA − 0.75 x 2 + 6 < x − 4 > +0.75 < x − 4 > 2
dx
d 2v
= M ( x) = RA x − 0.25 x3 + 3 < x − 4 > 2 +0.25 < x − 4 >3
2
dx
dv 1
EI
= RA x 2 − 0.0625 x 4 + < x − 4 >3 +0.0625 < x − 4 > 4 +C1
dx 2
1
EIv = RA x 3 − 0.0125 x 5 + 0.25 < x − 4 > 4 +0.0125 < x − 4 >5 +C1 x + C2
6
(a) EI
Continued on next slide
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Boundary conditions:
EIv(0) = 0 : C2 = 0
1
EIv '(8) = 0 :
RA (8) 2 − 0.625(8) 4 + (4)3 + 0.0625(4) 4 + C1 = 0
2
C1 = 176 − 32 RA
1
EIv(8) = 0 :
RA (8)3 − 0.0125(8)5 + 0.25(4) 4 + 0.0125(4)5
6
+176(8) − 32 RA (8) = 0,
RA = 6.3 kips ↑
Then C1 = 176 − 32(6.3) = −25.6
(b) Deflection at midspan (x=4):
1
EIvC = (6.3)(4)3 − 0.0125(4)5 + 0 + 0 − 25.6(4) = −48 kip ⋅ ft 2
6
We have
EI = (29 × 106 )(147) = 4.263 × 106 kip ⋅ in.2 = 29, 604 kip ⋅ ft 2
It follows that
48
vC = −
= −1.621×10−3 ft = 0.019 in. ↓
29, 604
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________________________________________________________________________
PROBLEMS (10.81 and *10.82) A beam ABC loaded and supported as shown in Figs. P10.81
and P10.82.Determine :
(a) The reactions at supports A and B.
(b) The deflection at the middle of span AB.
SOLUTION (10.81)
(a)
y
P
P/2L
MA
B
A
RA
L
L
C
RB
P 2L
< x − L >2
2
R x2 R
1 P
EIv ' = − M A x + A + B < x − L > 2 − ( ) < x − L >3 +C1
2
2
6 2L
2
3
M x
R x R
1 P
EIv = − A + A + B < x − L >3 − ( ) < x − L > 4 +C1 x + C2
2
6
6
24 2 L
Using boundary conditions:
v '(0) = 0 : C1 = 0
v(0) = 0 : C2 = 0
M
1
1
v( L) = 0 : − M A L2 + RA L3 = 0,
RA = 3 A
2
6
L
Statics:
3
9
13
M A = PL
RA = P ↑
RB = P ↓
8
8
8
EIv '' = − M A + RA x + RB < x − L > −
(1)
(2)
(b) Then make x=L/2 in Eq.(1):
3PL L 2 1 9 P L3
3
( ) +
EIvM = −
=−
PL3
16 4
6 8 8
128
or
3 PL3
vM =
↓
128 EI
SOLUTION (*10.82)
By inspection, the beam is statically indeterminate to the first degree. Due to
symmetry RA = RC and θ B = 0 .
(a) M ( x) = RA x −
wx 2
+ RB < x − L >
2
Continued on next slide
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So
d 2v
wx 2
EI 2 = RA x −
+ RB < x − L >
2
dx
< x − L >2
dv
x 2 wx 3
= RA −
+ RB
+ C1
EI
2
6
2
dx
x 3 wx 4
< x − L >3
+ RB
+ C1 x + C2
EIv = RA −
6
24
6
(1)
Boundary conditions:
EIv(0) = 0 : C2 = 0
L2 wL3
−
+ C1 = 0
2
6
L3 wL4
EIv( L) = 0 : RA −
+ C1 L = 0
6
24
EIv '( L) = 0 : RA
Solving
wL3
3
RA = wL = RC
C1 = −
8
48
From equilibrium
3
5
RB = 2wL − wL = wL
4
4
(b) Equation (1) becomes
R x3 R < x − L >3 wx 4
wL3
−
−
v= A − B
6 EI
6 EI
24 EI 48 EI
At the middle of span AB:
R L3 5wL4
L
wL4
v( ) = A −
=
↓
2
48EI 384 EI 192 EI
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________________________________________________________________________
PROBLEMS (10.83 through 10.86) A beam is loaded and supported as shown in Figs. P10.83
through P10.86. Determine:
(a) The reaction at support A.
(b) The deflection at the midspan.
SOLUTION (10.83)
y
(a)
A
M0
a
C
b
RA
x
MB
B
L
RB
EIv '' = RA x − M 0 < x − a > 0
1
EIv ' = RA x 2 − M 0 < x − a > +C1
2
M
1
EIv = RA x 3 − 0 < x − a > 2 +C1 x + C2
6
2
Using boundary conditions:
v(0) = 0 : C2 = 0
1
v '( L) = 0 :
RA L2 − M 0b + C1 = 0
2
M
1
v( L) = 0 :
RA L3 − 0 b 2 + C1 L = 0
6
2
From Eqs.(2) and (3):
3M 0 2
(L − a2 ) ↑
RA =
2 L3
and C1 = M 0 (b − 2a )b 4 L
(1)
(2)
(3)
(b) Making x=L/2 in Eq.(1)
M0
vM =
[3( L2 − a 2 ) + 12b(b − 2a)]
96 EI
M0
=
( L − a)(5L − 11a)
32 EI
SOLUTION (10.84)
Equivalent loading:
(a)
y
A
RA
w
a
B
C
3a
MB
x
D
w
RB
Continued on next slide
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w
w
< x − a > 2 − < x − 4a > 2
2
2
1
w
w
EIv ' = RA x 2 − < x − a >3 + < x − 4a >3 +C1
2
6
6
1
w
w
EIv = RA x 3 −
< x − a >4 +
< x − 4a > 4 +C1 x + C2
6
24
24
EIv '' = RA x −
Using boundary conditions:
v(0) = 0 : C2 = 0
1
1
1
v '(5a ) = 0 : − RA (5a ) 2 − w(4a )3 + wa 3 + C1 = 0
2
6
6
1
1
1
v(5a ) = 0 :
RA (5a )3 − w(4a ) 4 + wa 4 + 5C1a = 0
6
24
24
(1)
(2)
(3)
Multiply Eq.(2) by 5a & subtract Eq.(3) from it:
RA = 1.005wa ↑
Then, Eq.(b) gives C1 = −2.0625wa 3
(b) Then, substitute x=2.5a into Eq.(1):
EIvM = 2.61719 wa 4 − 0.21094 wa 4 − 5.15625wa 4
or
wa 4
vM = 2.75
↓
EI
SOLUTION (10.85)
(a)
y
w
MA
A
a
RA
C
a
B
MB
RB
w
< x − a >2
2
1
w
EIv ' = − M A x + RA x 2 − < x − a >3 +C1
2
6
1
1
w
EIv = − M A x 2 + RA x 3 −
< x − a > 4 +C1 x + C2
2
6
24
EIv '' = − M A + RA x −
(1)
Continued on next slide
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Using boundary conditions:
v '(0) = 0 : C1 = 0
v(0) = 0 : C2 = 0
RA
wa 3
=0
(2a) 2 −
2
6
1
1
1
1
v( L) = 0 : − wL4 +
wL4 + C1 L + C2 L2 = 0
24
384
6
2
v '(2a) = 0 : − M A (2a ) +
(2)
(3)
Multiply Eq.(2) by a & subtract Eq.(3) from it:
4
1 1
3
(2 − ) RA a 3 − ( − ) wa 4 = 0,
RA = wa
3
6 24
16
Then Eq.(2) gives
5
MA =
wa 2
48
(b) Make x=a in Eq.(1) and substitute the foregoing values:
wa 4
vC =
↓
48EI
SOLUTION (10.86)
(a)
y
MA
w
B
A
L/2
RA
C
L/2
x
RB
w
L
< x − >2
2
2
1
w
L
EIv ' = M A x + RA x 2 − < x − >3 +C1
2
6
2
1
1
w
L
EIv = M A x 2 + RA x 3 −
< x − > 4 +C1 x + C2
2
6
24
2
EIv '' = M A + RA x −
(1)
Boundary conditions:
v '(0) = 0 : C1 = 0
v(0) = 0 : C2 = 0
v( L) = 0 :
wL4
1
1
M A L2 + RA L3 −
=0
2
6
384
(2)
Continued on next slide
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Statics:
1
∑ M = 0 : − M − R L + 8 wL − M = 0
B
A
A
2
0
(3)
Multiply Eq.(3) by L2 2 & add to Eq.(2):
1 1
1
1
1
( − ) RA L3 + ( −
) wL4 − M 0 L = 0
6 2
16 384
2
3M 0
23
RA =
wL −
128
2L
Equation (3) yields
7
1
MA =
wL2 − M 0
128
2
(b) Then, making x=L/2 in Eq.(1):
4
7 wL4 M 0 L2 23wL M 0 L2
EIvC = −
+
+
−
1024
8
6144
32
or
L2
(−19wL2 + 576M 0 )
vC =
6144 EI
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________________________________________________________________________
PROBLEM (10.87) A propped cantilever beam with an overhang is loaded as shown in Fig. P10.87.
Determine:
(a) The reaction at support A.
(b) The deflection at the midspan.
SOLUTION
(a)
w
y
M0
C
L/2
A
L
x
B
RA
MB
RB
L
w
L
> − < x − >2
2
2
2
R
L
w
L
EIv ' = − M 0 x + A < x − > 2 − < x − >3 +C1
2
2
6
2
R
1
L
w
L
EIv = − M 0 x 2 + A < x − >3 − < x − > 4 +C1 x + C2
2
6
2
24
2
EIv '' = − M 0 + RA < x −
(1)
Boundary condition:
3
3
1
1
v '( L) = 0 : C1 = M 0 L − RA L2 + wL3
2
2
2
6
5
1
1
L
v( ) = 0 : C2 = − M 0 L2 + RA L3 − wL4
2
8
4
12
3
3 M0 3
v ( L ) = 0 : RA =
+ wL ↑
2
2 L 8
(b) Make x=L in Eq.(1) and simplify:
5
35
EIvM = − M 0 L2 −
wL4
16
384
or
5 L2
vM =
(24M 0 + 7 wL2 ) ↓
384 EI
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________________________________________________________________________
PROBLEM (10.88) Before the loads are applied, a small gap δB, exists between beam AC and
support B, as seen in Fig. P10.88. Subsequent to loading, the gap closes and reactions develop at each
support. Find these reactions.
SOLUTION
wa
y
a
A
B a
RA
Statics:
w
a
D
RB
C
x
RC
∑ F = 0 : R + R + R = 2wa
5
∑ M = 0 : 3R + 2 R = 2 wa
y
A
C
B
A
C
B
w
< x − 2a > 2
2
1
1
w
EIv ' = RA x 2 + ( RB − wa) < x − a > 2 − < x − 2a > 3 +C1
2
2
6
1
1
w
EIv = RA x3 + ( RB − wa ) < x − a >3 − < x − 2a > 4 +C1 x + C2
6
6
24
Boundary conditions:
v(0) = 0 : C2 = 0
3
4
33
v(3a ) = 0 : C1 = − RA a 2 + RB a 2 − wa 3 = 0
2
9
72
1
3
4
33
v(a) = −δ B :
RA a 3 − RA a 3 − RB a 3 + wa 4 = −δ B EI
6
2
9
72
From Eq.(2):
5
3
RB = wa − RA
4
2
Substitute this into Eq.(3) to obtain:
3 δ B EI 7
RA =
− wa
2 a3
48
(1)
(2)
EIv '' = RA x + ( RB − wa) < x − a > −
(3)
(4)
Equation (4) gives
47
9 δ B EI
RB = − wa −
32
4 a3
Equation (1) yields
65
3 δ B EI
RC = − wa +
96
4 a3
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________________________________________________________________________
PROBLEM (10.89) A cantilever beam AC with an overhang BC supports a uniform load of
intensity w, as shown in Fig. P10.89. Determine the reactions at supports A and B.
SOLUTION
w
MA
B
L
A
=
L/3
Actual
Loading
RB
RA
w
A
L
+
B
A
Equivalent Loadings
L
B
RB
Use Table B.14 (cases 2 and 1):
wL2 2
19 wL4
4
4 2
(vC )W = −
[ L − 4( ) L + 6( L) ] = −
24 EI
3
3
72 EI
3
R L
( vB ) R = − B
3EI
Since
(vB )W + (vB ) R = 0
We have
19
RB =
wL ↑
24
Statics:
13
∑ F = 0 : R = 24 wL ↑
y
A
7
∑ M = 0 : M = 72 wL
A
2
A
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________________________________________________________________________
PROBLEMS (10.90 through 10.93) A beam is supported and loaded as shown in Figs. P10.90
through P10.93. What are the reactions at each support?
SOLUTION (10.90)
P
MB
B
A
C
L/2
=
L/2
RB
RA
P
A
B
L
+
A
RA
B
C
Use Table B.14 (case 1):
v A = (v A ) P + ( v A ) R = 0
P( L 2)3
L R L3
− (θC ) P ( ) + A = 0
3EI
2
3EI
3
3
3
R L
5
PL
PL
RA = P ↑
=−
−
− A = 0,
24 EI 16 EI 3EI
16
=−
Statics:
RB =
11
P↑
16
MB =
3
P
16
SOLUTION (10.91)
M0
MA
A
=
B
L
RB
RA
RB
M0
A
L
B
+
A
B
Continued on next slide
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Use Table B.14 (case 2 and 1):
vB = (vB ) M + ( vB ) R = 0
M 0 L2 RB L3
3 M0
=−
−
= 0, RB =
↓
2 EI
3EI
2 L
Statics:
RA =
3 M0
↑
2 L
MA =
1
M0
2
SOLUTION (10.92)
M0 B
L/2
MA
MB
L/2
C
A
RA
RB
M0
MA
=
MB
A
B
A
B
A
B
Use Table B.14 (cases 11 and 9):
θ A = −(θ A ) M A + (θ A ) M B + (θ A ) M 0 = 0
M AL M B L M 0L
+
+
=0
3EI
6 EI 24 EI
Similarly, θ B = 0 gives
M AL M B L M 0L
−
+
=0
6 EI
3EI 24 EI
=−
(1)
(2)
Solving Eqs.(1) and (2):
MB = M0 4
M A = M0 4
Statics:
3 M0
3 M0
RA =
RB =
↓
↑
2 L
2 L
Continued on next slide
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SOLUTION (10.93)
w
B
MA
A L/2
w
A
C
L/2
=
RB
RA
B
C
+
A
B
RB
Use Table B.14 (case 3 and 1):
w( L 2)3
vB = ( vB ) w + ( vB ) R = 0
(θC ) w = −
6 EI
4
3
3
w( L 2) w( L 2) L RB L
( )+
=−
−
=0
8EI
6 EI
2
3EI
wL4
wL4 RB L3
=−
−
+
=0
128EI 96 EI 3EI
Solving
RB =
7
wL ↑
128
RA =
57
wL ↑
128
Statics:
MA =
9
wL2 ↑
128
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________________________________________________________________________
PROBLEM (10.94) A fixed-end beam AB carries a linearly varying load as shown in Fig. P10.94.
Find the reactions at the ends A and B.
SOLUTION
w0
MA
MB
B
A
=
RB
RA
w0
MB
A
+
B
A
B
+
RB
A
B
Use Table B.14 (case 4, 1, and 2):
vB = ( vB ) w + ( vB ) R + ( vB ) M = 0
w0 L4 RB L3 M B L2
=−
+
−
=0
30 EI 3EI
2 EI
(1)
Similarly,
w0 L3 RB L2 M B L
+
−
=0
24 EI 2 EI
EI
Solving Eqs.(1) and (2):
w0 L2
3
RB =
w0 L ↑
MB =
20
30
θB = −
(2)
Statics:
7
∑ F = 0 : R = 20 w L ↑
y
0
A
1
∑ M = 0 : M = 20 w L
A
B
0
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________________________________________________________________________
PROBLEM (10.95) A nonprismatic cantilever beam propped at end B supports a uniform load of
intensity w, as shown in Fig. P10.95. Determine all of the reactions.
SOLUTION
w
MA
L/2
A
RA
w
A
EI
B =
L/2
C
RB
wL
2
w
2
C wL
+
C
2EI
RB L 2
+
B
C
8
Use Table B14 (cases 1, 2, and 3):
vB = (vB ) w + ( vB ) R = 0
We have
L w( L 2) 4
=0
(vB ) w = (vC ) w + (θC ) w ( ) −
2
8(2 EI )
where
3
3
w( L 2) 4 ( wL 2)( L 2) ( wL2 8)( L 2)
vC = −
−
−
8 EI
3EI
2 EI
4
17 wL
=−
384 EI
2
w( L 2)3 ( wL 2)( L 2) ( wL2 8)( L 2)
θC = −
−
−
6 EI
2 EI
EI
3
7 wL
=−
48 EI
Thus
wL4
17 wL4 7 wL4
−
−
( vB ) w = −
384 EI 96 EI 256 EI
31 wL4
=−
256 EI
2EI
+
B
RB
C
B
2EI
RB
(1)
(2)
Continued on next slide
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Similarly,
3
( RB L 2)( L 2) 2 RB ( L 2)
5 RB L3
(vC ) R =
+
=
2 EI
3EI
48 EI
2
( R L 2)( L 2) RB ( L 2)
3 RB L2
+
=
(θC ) R = B
2 EI
8 EI
EI
Thus,
3
L R ( L 2)
(vB ) R = (vC ) R + (θC )( ) + B
2
3(2 EI )
3
RB L 5 3 1
5 RB L3
=
( + + )=
EI 8 6 48 16 EI
(3)
Then, Eq.(1) leads to
31
RB = wL ↑
80
Statics:
49
∑ F = 0 : R = 80 wL ↑
y
A
9
∑ M = 0 : M = 80 wL
B
2
A
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________________________________________________________________________
PROBLEM (10.96 and 10.97) A beam is supported and loaded as shown in Figs. P10.96 and
P10.97. Determine all of the reactions.
SOLUTION (10.96)
w0
MB =
B
L
A
MA
RB
RA
w0
A
MB
C
B
+
A
B
2w0
+
RB
A
B
w0
A
A
C
B
C
2w0
From symmetry: RA = RB
Statics: RA = RB =
B
+
(θC ) w =
2w( L 2)3
24 EI
MA = MB
1
wL ↑
4
Use Table B14 (cases 4.1 and 2):
vB = ( vB ) w + ( vB ) R + ( vB ) M = 0
Here
4
3
(2w0 ) L4 2 w0 ( L 2) 2w0 ( L 2) L
( vB ) w = −
( )
+
+
30 EI
30 EI
24 EI
2
4
11 wL
=−
192 EI
(vB ) R = RB L3 3EI
(vB ) M = − M B L3 2 EI
Thus,
11 w0 L4 RB L3 M B L2
vB = −
+
−
=0
192 EI
3EI
2 EI
Solving
5
5
M B = w0 L2
M A = w0 L2
96
96
Continued on next slide
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SOLUTION (10.97)
w
C
M0
A
L/2
B
L
RA
MB
=
RB
w
M0
A
B
+
B
RA
Using Table B.14 (cases 2, 3 and 1):
M L2 wL4 RA L3
vA = − 0 −
+
=0
2 EI 8 EI 3EI
or
3 M0 3
RA =
+ wL ↑
2 L 8
Then, the reactions RB and M B are found from equilibrium.
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________________________________________________________________________
PROBLEMS (10.98 through 10.101) A beam is supported and loaded as shown in Fig. P10.98
and P10.101. What are the reactions at each support?
SOLUTION (10.98)
w
A
2a
a
C
RB
RC
B
RA
Actual
= Loading
w
+
B
A
C
B
A
RB
Equivalent loadings
C
Using Table B.14 (cases 10 and 8):
w(2a)
( vB ) w = −
[(20)3 − 2(3a )(2a) 2 + (3a)3 ]
24 EI
11 wa 4
=−
12 EI
( vB ) R =
RB a(2a)
4 wa 4
[(2a ) 2 − (3a ) 2 + a 2 ] =
6(3a) EI
9 EI
Boundary condition,
vB = (vB ) w + (vB ) R = 0 , gives:
33
RB = wa ↑
16
Statics:
13
1
RA = wa ↑
RC = wa ↑
16
8
SOLUTION (10.99)
M0
L
A
L
RB
RA
B
C
Actual
Loading
=
RC
M0
+
RB
A
C
A
Equivalent loading
B
C
Continued on next slide
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Use Table B.14 (cases 7 and 9):
M0L
R (2 L)3
vB = − B
−
[ L2 − (2 L) 2 ] = 0
48 EI
6(2 L) EI
M
1
= − RB + 0 = 0
6
4L
or
3 M0
RB =
↓
2 L
Statics:
1M
∑ M A = 0 : RC = 4 L0 ↑
5M
∑ Fy = 0 : RA = 4 L0 ↑
SOLUTION (10.100)
w
A
L B
RA
A
B
L
Actual
= Loading
C
RB
RC
RC
w
C
+
A
B
C
Equivalent loadings
Using Table B.14 (case 10, 1, and 9):
vC = (vC ) w + (vC ) R = 0
where,
wL3
L4
(vC ) w = (θ B ) w ( L) = −
( L) =
24 EI
24 EI
3
R L ( R L) L
R L3
(vC ) R = (θ B ) R ( L) + C = C
( L) + C
3EI
3EI
3EI
3
2 RC L
=
3 EI
Equation (1) yields
1
RC = wL ↓
16
(1)
Continued on next slide
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Statics:
5
RB = wL ↑
8
RA =
7
wL ↑
16
SOLUTION (10.101)
w
MA
A
L
k
B
RA
Actual
= Loading
RB
w
A
B
+
B
A
RB
Equivalent loading
Using Table B.14 (case 3 and 1):
vB = ( vB ) w + ( vB ) R
=−
wL4 RB L3
+
8EI 3EI
(1)
Since
RB
(2)
k
Equations (1) and (2) give
3w L B
↑
RB =
3EI
1+ 3
kL
(Interestingly, for a rigid spring ( k → ∞ ): RB = 3wL 8 and for a free end
( k = 0 ): RB = 0. )
Statics:
RA = wL − RB ↑
1
M A = wL2 − RB L
2
where RB is given above.
vB = −
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________________________________________________________________________
PROBLEMS (10.102 and 10.103) The compound beam AHB shown in Figs. P10.102 and
10.103 has fixed supports at ends A and B and consists of two members joined by pin connection at H.
Determine all reactions of the beam due to the concentrated load P.
SOLUTION (10.102)
Use Table B.14 (case 1):
RH
H
L/3
A
H
RH
vH = −
+ H
B
2L/3
RH ( L 3)3
R L3
=− H
3EI
81EI
θC
(1)
B
C
P( L 3) 2
PL2
θC =
=
2 EI
18 EI
3
R (2 L 3) P( L 3)3 PL2 L
vH = H
−
−
( )
3EI
3EI
18EI 3
8 RH L3
5 PL3
=
−
81 EI
162 EI
(2)
From Eqs.(1) and (2):
5
RH = P
18
Then
A
MA
5
P↑
18
5
MA =
PL
54
5
P
18
L/3
RA =
H
RA
5
P
18
P
C
MB
B
RB
13
P↑
18
8
MB =
PL
54
RB =
Continued on next slide
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SOLUTION (10.103)
Use Table B.14 (case 1):
P
A
MA
Loading I
( vH ) L =
( P − RH ) L3
3EI
MB Loading II
( vH ) R =
RH a 3
3EI
H
L
RH
RA
RH
H
a
B
RB
Since (vH ) L = (vH ) R :
( P − RH ) L3 RH a 3
=
,
3EI
3EI
or
PL3
RH = 3
L + a3
Statics:
Loading I:
PL3 − RH L3 = RH a 3
Pa 3
∑ Fy = 0 : RA = L3 + a3 ↑
Pa 3 L
∑ M = 0 : M = (P − P ) = L + a
A
A
H
3
3
Loading II:
Pa 3
↑
L3 + a 3
PL3 a
M
=
M
=
0
:
∑ B
B
L3 + a 3
∑ Fy = 0 : RB =
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________________________________________________________________________
PROBLEM (10.104) A cantilever beam AB and a simple beam CD are supported and loaded as
shown in Fig. P10.104. A roller fits snugly between the two beam at point B. Determine:
(a) The force F transmitted between the beams at B.
(b) The reactions RC and RD for the lower beam.
Assumption: The flexural rigidity EI is constant for both beams.
SOLUTION
(a) Refer to Table B.14.
F
∑M = 0: R = 2
F
C
L/2
L/2
B v
RC
C
B
RD
P
B
A
D
vB
=
P
vB '
+
vB ''
F
(b) Substitute F = 2 P 3 into Eq. (1):
D
(1)
PL3
↓
48 EI
P( L 2)3
PL3
vB ' =
=
↓
3EI
24 EI
F ( L 2)3
FL3
vB '' =
=
↑
3EI
24 EI
Condition of compatibility at B:
vB = vB '+ vB ' ↓
vB =
FL3
PL3
FL3
=
−
48EI 24 EI 24 EI
2
F= P
or
3
1
RD = P = RC
3
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________________________________________________________________________
PROBLEM (10.105) A cantilever beam AB supported by a rod AC at its free end A, carries a
uniform load of intensity w (Fig. P10.105). Determine the force T in the rod.
Assumptions: IE is the flexural rigidity of the beam and EA is the axial rigidity of the rod.
Before the load is applied, the rod is free of force.
SOLUTION
Refer to Table B.14.
w
B
(v A ) w
A
wL4
(v A )W =
↓
8EI
L
T
TL3
(v A )T =
↑
3EI
(v A )T
δACompatibility condition at A:
T
δ A = (vA )W + (vA )T ;
Solving,
T=
δA =
Ta
AE
Ta wL4 TL3
=
−
AE 8EI 3EI
3wAL4
8 AL3 + 24aI
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________________________________________________________________________
PROBLEM (10.106) A beam ABC with simple supports at A and B carries a uniform load of
intensity w over the overhang BC as shown in Fig. P10.106 and discussed in Example 10.19. The beam is
constructed of a W 10 x 33 shape (see Table B.8). Calculate:
(a) The deflection at point C.
(b) The slope at point A.
(c) The maximum deflection between supports.
Given: w = 6 kips/ft, L = 6 ft,
E = 30 x 106 psi
SOLUTION
Table B.8: I = 170 in.4
(a) Use Eq.(10.53):
vC =
11 wL4 11(6 12)(103 )(6 × 12) 4
= 0.075 in. ↓
=
384 EI
384(30 ×106 )(170)
(b) Apply Eq.(10.54):
wL3
(6 12)(103 )(6 × 12)3
θA =
= 0.762 × 10−3 rad
=
6
48 EI
48(30 × 10 )(170)
(c) From Eq.(10.55):
vD =
wL4
(6 12)(103 )(6 ×12) 4
=
= 0.021 in. ↑
72 3EI 72 3(30 × 106 )(170)
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________________________________________________________________________
PROBLEM (10.107) A cantilever beam AB supports a couple moment M0 at the free end A as
shown in Fig. 10.107. Determine the vertical displacements of points A and C for this beam.
SOLUTION
The M EI diagram and elastic curve of the beam is as shown below.
M/EI
L/2
L/2
A
A1
A1
L/4
A
tA/B
C
tC/B
We have A2 = 2 A1 . Thus
vA = t A B
B x
L
= A2 ( )
-M0
2
M
L
= (− 0 )( L)( )
EI
2
2
M 0 L M 0 L2
B
=−
=
↓
2 EI
2 EI
L at A M L L
Tangent
vC = tC B = A1 ( ) = (− 0 )( )( )
4
EI 2 4
2
2
M L M L
=− 0 = 0 ↓
8EI
8 EI
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________________________________________________________________________
PROBLEM (10.108 through 10.111) A cantilever beam is loaded as shown in Figs. P10.108
through P10.111.Determine:
(a) The slope at the free end.
(b) The deflection at the free end.
SOLUTION (10.108)
We have θ A = 0 and v A = 0 .
y
MA =
2 2
wL
9
w
A 2L/3
2
RA = wL
3
M
A
EI
2
− wL2
9
C
B
L/3
C
B
x
Parabolic
Spandrel
x
Tangent at A
A
C
vB
θB
3
3 2L L
x = b=
=
4
4 3
2
1 1 2L 2 2
1 bh
4wL3
=−
wL = −
A=−
EI 3 3 9
EI 3
81EI
(a) Equation (10.48) becomes
θ B A = θ B − 0 = A , or
θB =
4wL3
81EI
(b) Equation (10.50):
L
4wL3 5 L
10 wL4
vB = t B A = − A( x + ) = −
( )=
↓
3
81EI 6
243 EI
Continued on next slide
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SOLUTION (10.109)
P
2P
C
A L/2
M
EI
L/2
B
x1
x
A1
-PL
M
EI
x2
A2
x
C
-PL
Tangent at A
A
tB/A
tC/A
(a) θ B = θ B / A = A1 + A2 = −
θA = 0
vA = 0
1 PL2 1 PL2 3 PL2
−
=
2 EI 4 EI
4 EI
PL2 2 L PL2 5 L
13 PL3
( )−
( )=
↓
(b) vB = t BA = A1 x1 + A2 x2 = −
2 EI 3
4 EI 6
24 EI
SOLUTION (10.110)
w0
A
L
B
w0L2/6
w0L2/6
x
Cubic
spandrel
M
EI
A
w0 L2
−
6
x=
B
vB = 0
θB = 0
Tangent at A
tAB
n +1
4
b= L
5
n+2
A
w0 L2
w0 L3
bh
1
( L)(
)=−
A=
=
4 EI 4 EI
6
24 EI
Continued on next slide
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w0 L3
θA =
(a) θ B / A = θ B − θ A = A ,
24 EI
3
w L4
w L 4L
(b) t BA = v A = Ax = − 0 ( ) = 0 ↓
24 EI 5
30 EI
0
SOLUTION (10.111)
a
A
M EI
Pa
2
Pa/2
A1
P
a
C
B
3a 2
x
A2
-
4a 3
Tangent at A
A
tB/A
1 Pa 2
2 EI
Pa 2
A2 = −2
EI
A1 =
vA = 0
θA = 0
(a) θ B / A = θ B − θ A0 = A1 − A2
Pa 2 2 Pa 2 3 Pa 2
−
=
2 EI
2 EI
EI
3a
4a
(b) t B / A = vB = A1 ( ) + A2 ( )
2
3
2
Pa 3a 2 Pa 2 2
23 Pa 3
( )−
( ⋅ 2a ) =
or
vB =
↓
2 EI 2
12 EI
EI 3
θB =
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________________________________________________________________________
PROBLEMS (10. 112 and 10.113) A cantilever beam supports a partial uniform load of intensity
w as shown in Figs. P10.112 and P10.113. Determine:
(a) The angle of rotation at free end B.
(b) The deflection at the free end B.
SOLUTION (10.112)
w
B
A
EI
L/2
3EI
L/2
M
x
−3wL2 8
x1
M
EI
− wL2 8EI
− wL2 24 EI
A1
A2
x
A3
− 8wLEI
2
Parabolic
Spandrel
x3
x2
Tangent at A
A
vB
B
1 wL2 L
wL3
=−
2 24 EI 2
96 EI
2
1 wL L
wL3
A2 = −
=−
2 8EI 2
32 EI
2
wL3
1 wL L
A3 = −
=−
3 8EI 2
48EI
A1 = −
θB
L L 2L
+ =
2 6
3
L L 5L
x2 = + =
2 3 6
3 L 3L
x3 = ( ) =
4 2
8
x1 =
(a) θ B / A = θ B − θ A0 = A1 + A2 + A3
θB = −
wL3
wL3
(1 + 3 + 2) =
96 EI
16 EI
Continued on next slide
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(b) t B / A = vB
=−
wL3 2 L wL3 5L wL3 3L
−
−
96 EI 3 32 EI 6 48EI 8
or
vB = −
47 wL4
wL4
↓
(8 + 30 + 9) =
1152 EI
1152 EI
SOLUTION (10.113)
w
a
A
a
C
M
EI
x2
x1
− wa 2 2 EI A2
−3 wa 2 2 EI
B
a
D
A3
x
Spandrel
Parabolic
A1
x3
Tangent at A
vB
A
D
B
θB
bh
1 wa 2
wa 3
A1 =
=− a
=−
3EI
3 2 EI
6 EI
3
wa
1
1 wa 3
A2 = −
A3 = − a( wa 2 ) = −
2 EI
2
2 EI
x1 = a +
3a 7 a
=
4
4
x2 =
5a
2
x3 =
8a
3
(a) θ B / A = θ B − θ A0 = A1 + A2 + A3
θB = −
wa 3
7 wa 3
(2 + 6 + 6) =
12 EI
6 EI
Continued on next slide
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(b) t B / A = vB = A1 x1 + A2 x2 + A3 x3
vB = −
wa 4
69 wa 4 23 wa 4
(7 + 30 + 32) = −
=
↓
24 EI
24 EI
8 EI
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________________________________________________________________________
PROBLEM (10.114) Calculate the maximum slope and maximum deflection of the cantilever beam
in Fig. P10.108 for an S 6 x 12.5 shape (see Table B.9).
Given: w = 6 kips/ft, L = 8 ft, E = 30 x 106 psi.
SOLUTION
Table B.9:
I = 22.1 in.4
See solution of Prob.10.110:
w0 L3
24 EI
(6 12)(103 )(8 × 12)3
=
= 27.8 × 10−3 rad
6
24(30 × 10 )(22.1)
θ max = θ A =
Similarly,
w0 L4
↓
30 EI
(6 12)(103 )(8 × 12) 4
=
= 2.135 in. ↓
30(30 × 106 )(22.1)
vmax = v A =
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________________________________________________________________________
PROBLEM (10.115) Compute the slope and deflection at point D of the cantilever beam seen in
Fig. P10.113 for a W 460 x 74 shape beam (see Table B.8).
Given: w = 150 kN/m,
a = 1 m,
E = 200 GPa.
SOLUTION
Table B.8: I = 333 × 106 mm 4
3
wa
A1 = −
6 EI
3
wa
A2 = −
2 EI
See solution of Prob.10.113:
wa 3
A3 = −
2 EI
The distances from point D to the centroid of each area are
x1 = 3a 4
x1 = 3a 2
x1 = 5a 3
Thus,
7 wa 3
7(150 × 103 )(1)3
=
θD = θB =
6 EI
6(200 × 109 )(370 × 10−6 )
= 2.36 × 10−3 rad
t D / A = vD = A1 x1 + A2 x2 + A3 x3
or
wa 4
41 wa 4
(3 + 18 + 20) =
vD = −
↓
24 EI
24 EI
Substituting the given data:
41
150 × 103 (1) 4
= 3.85 mm ↓
vD =
24 (200 ×109 )(333 × 10−6 )
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________________________________________________________________________
PROBLEMS (10.116 through 10.119) A simple beam is loaded as shown in Figs. P10.116
through P10.119. Determine:
(a) The slope at point C.
(b) The deflection at point C.
SOLUTION (10.116)
P
B
2L/3
L/3
RB=P/3
RA=2P/3
M
EI
A1
2 PL
9 EI
A2
x
x1
y
A
tangent at A
x2
θA
C’’
x
vC
tC/A C’
1 L 2 PL
PL2
A1 =
=
2 3 9 EI 27 EI
1 2 L 2 PL 2 PL2
A2 =
=
2 3 9 EI 27 EI
Thus,
C
tB/A
2L 1 L 7
+
= L
3 33 9
2 2L 4
x2 =
= L
3 3 9
x1 =
PL3
5 PL3
(7 + 8) =
t B A = A1 x1 + A2 x2 =
243EI
81 EI
2
t
5 PL
θA = − B A = −
81 EI
L
PL2
5 PL2
2 PL2
(a) θC / A = A1 ,
θC = θC / A + θ A =
−
=
27 EI 81 EI
81 EI
Continued on next slide
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PL3
1L
)=
(b) tC / A = CC ' = A1 (
33
243EI
and
1
vC = tC / A − C ' C '' = tC / A − t B / A
3
3
3
PL
5 PL
4 PL3
=
−
=
↓
243EI 243 EI
243 EI
SOLUTION (10.117)
y
M0
A
3EI
L/2
M0/L
EI
x
L/2
C
B
M0/L
M
M0
M0 / 2
x
M/EI
x1
M0/3EI
M0/6EI
A1
A2 A3
x
x3
y
A
x2
θA
C’’
C vC
C’
tangent at A
1 M0 L M0L
=
2 3EI 2 12 EI
1 M0 L M0L
A2 =
=
2 6 EI 2 24 EI
1 M0 L M0L
A3 =
=
2 3EI 2 12 EI
A1 =
B
tC/A
tB/A
L 2 L 5L
+
=
2 32 6
L L 2L
x2 = + =
2 6
3
2L L
x3 =
=
32 3
x1 =
Continued on next slide
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We have
t B / A = A1 x1 + A2 x2 + A3 x3
=
M 0 L2
1 M 0 L2
(5 + 2 + 2) =
72 EI
8 EI
t
1 M0L
θ A = − BA = −
8 EI
L
M0L M0L 1 MoL
+
=
12 EI 24 EI 8 EI
1 M0L 1 M0L
θC = θC / A + θC =
−
=0
8 EI
8 EI
(a) θC / A = A1 + A2 =
2L
1L
) + A2 (
)
32
32
M L2
5 M 0 L2
= 0 (4 + 1) =
144 EI
144 EI
1
5 M 0 L2 1 M 0 L2
1 M 0 L2
vC = tC / A − t B / A =
=
↓
−
2
144 EI
16 EI
36 EI
(b) tC / A = A1 (
SOLUTION (10.118)
w
A
B
a
3wa/4
wa/4
M/EI
x1
wa2/2EI
wa 2
4 EI
A1
A2
-wa2/2EI
y
A
tA/B
a
C
x2
B
Parabolic
spandrel
C’’
C vC
C’
tC/B
θB
B
x
x
tangent at B
Continued on next slide
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1 wa 2
1 wa 3
2a
(2a) =
A1 =
x1 =
2 2 EI
2 EI
3
2
bh
1 wa
a
A2 =
x2 =
=− a
3
3 2 EI
4
t A / B = A1 x1 + A2 x2
7 M 0 L4
wL4
(8 − 1) =
=
24 EI
24 EI
1
7 wa 3
θ B = − tB / A =
2a
48 EI
1 wa 2
1 wa 3
(a ) =
(a) θC / B =
2 4 EI
8 EI
θC = −θC / B + θ B
=−
(b) tC / B =
1 wa 3 7 wa 3
1 wa 3
+
=
8 EI 48 EI
48 EI
wa 4
1 wa 3 a
( )=
8 EI 3
24 EI
1
7 wL4
5 wL4
wa 4
vC = tC / B − t B / A =
=
↓
−
2
24 EI 48 EI
48 EI
Continued on next slide
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SOLUTION (10.119)
M0
B
A
C L/4
3L/4
M0/L
M/EI
M0/L
x2
M0/4EI
A2
x
A1
-3M0/4EI
x1
tangent at B
tA/B
A
vC
C tC/B
C’
C’’
θB
B
x
9M 0 L
3L 2 L
1 3M 0 3L
A1 = −
=
=−
4 3 2
2 4 EI 4
32 EI
3L 1 L 5 L
1 M0 L M0L
x2 =
A2 =
+
=
=
4 34 6
2 4 EI 4 32 EI
x1 =
t B / A = A1 x1 + A2 x2
M 0 L2
11 M 0 L2
(5 − 27) = −
=
192 EI
96 EI
M
L
t AB
11 0
θB =
=−
96 EI
L
(a) θC / B = A2 =
M0L
32 EI
θC = θ B − A2 = −
11 M 0 L M 0 L
7 M0L
−
=
96 EI
32 EI 48 EI
M L2
1L
)= 0
34
384 EI
M L2
1
11 M 0 L2
1 M 0 L2
vC = tC / B − t A / B = 0 +
=
↑
4
384 EI 384 EI
32 EI
(b) tC / A = A2 (
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________________________________________________________________________
PROBLEMS (10.120 and 10.121) A simple beam with two different moments of inertia supports
loading as shown in Figs. P10.120 and P10.121.
SOLUTION (10.120)
P
A
EI
3EI
a
a
P/2
M/EI
Pa/2E
A2
B
a
a
C
P/2
x2
A1
Pa/6EI
Pa/3EI
x
x1
A3
x3
y
B
A
vC
θA
x
tC/A = vC
Pa 2
1 Pa
(a) =
2 2 EI
4 EI
Pa 2
2a 5a
1 Pa
(a) =
x2 = a +
A2 =
=
3
3
2 6 EI
12 EI
2
a 3a
Pa
x3 = a + =
A3 =
2 2
6 EI
x1 =
2
a
3
A1 =
(a) θC = 0 from symmetry about C
(b) θC / A = 0 − θ A , θ A = −θC / A , θ C / A = A1 + A2 + A3
Thus,
Pa 2
θ A = − A1 + A2 + A3 = −
(3 + 1 + 2)
12 EI
Pa 2
=
2 EI
Pa 3
vC = tC / A = A1 x1 + A2 x2 + A3 x3 =
(6 + 5 + 9)
36 EI
5 Pa 3
=
↓
9 EI
Continued on next slide
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SOLUTION (10.121)
w
A
EI
wa
B
C
2EI
a
wa
a
M
C
A
x
B
wa2/2
M/EI
4a/3
wa2/EI
2a/3
wa2/2EI
A1
A3
-wa2/2EI
-wa2/4EI
x
A4
Spandrel
parabola
y
A
A2
Spandrel
parabola
3a/4
5a/4
C’’ vC
θA
C
tC/A
C’
tangent at3 A
1 wa 2
1 wa
(a) =
2 EI
2 EI
2
wa 3
1 wa
(a) = −
A3 = −
3 2 EI
6 EI
A1 =
x
B
tB/A
wa 3
1 wa 2 a
( )=
2 EI 2
4 EI
2
wa 3
1 wa
(a ) = −
A4 = −
3 4 EI
12 EI
A2 =
Thus,
4a
2a
5a
3a
) + A2 ( ) + A3 ( ) + A4 ( )
3
3
4
4
3
1
9 wa
θ A = − tB / A = −
2a
32 EI
4
wa
9 wa 4
(32 + 8 − 10 − 3) =
=
48EI
16 EI
t B / A = A1 (
1 wa 3 1 wa 3
−
2 EI 6 3EI
wa 3 9wa 3
5 wa 3
=
θC = θC / A + θ A =
−
3EI 32 EI 96 EI
(a) θC / A = A1 + A3 =
Continued on next slide
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a
a
(b) tC / A = A1 ( ) + A3 ( )
3
4
4
4
wa
wa
1 wa 4
=
−
=
6 EI 24 EI 8 EI
5 wa 4
1
wa 4 9wa 4
vC = tC / A − t B / A =
=
↓
−
2
8 EI 32 EI 32 EI
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________________________________________________________________________
PROBLEM (10.122) A simply supported beam carries a concentrated load P as shown in Fig.
P10.116. Calculate the deflection at point D located L/4 from support A.
I = 20 in.4
Given: P = 4 kips,
L = 9 ft,
E = 10 x 106 psi,
SOLUTION
See solution of Prob.10.116.
2PL/9EI
M/EI
PL/6EI
A3
A
D
x
L/4
L/3
1
1L
)
vD = − t BA + A3 (
4
34
5PL3 1 PL L L
71 PL3
=−
+
=
↓
81× 4 2 6 EI 4 12 5184 EI
Substituting the given data:
71(4 × 103 )(9 × 12)3
vD =
= 0.345 in. ↓
5184(10 × 106 )(20)
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________________________________________________________________________
PROBLEM (10.123) For the W 610 x 125 shape (see table B.8) beam and loading shown in Fig.
P10.118, calculate slope at point D located a/2 from support B.
Given: w = 120 kN/m, a = 4 m,
E = 200 GPa
SOLUTION
See solution of Prob.10.118.
M/EI
wa2/2EI
wa2/8EI
A4
D
B
x
a/2
I = 985 × 106 mm 4 (Table B.8)
7 wa 3
48 EI
Thus,
θB =
θD = θD / B + θB
7 wa 3 wa 3 17 wa 3
+
=
48 EI 32 EI 96 EI
17(120 × 103 )(4)3
=
= 6.91× 10−3 rad
9
−6
96(200 × 10 )(985 ×10 )
θD =
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________________________________________________________________________
PROBLEM (10.124) A 4-m-long simple beam is subjected to a moment of M = 50 kN ⋅ m (see
Fig. P10.119). Determine the value of EI vD at point D located L/8 from B.
SOLUTION
See solution of Prob.10.119.
M/EI
L/4
L/8
L/24
M0/4EI
M0/8EI
D
A3
B
x
1
L
vD = − t A / B + A3 ( )
8
24
2
11 M 0 L 1 1 M 0 L L
45 M 0 L2
) +
( )( ) =
↑
= −( −
96 EI 8 2 8EI 8 24
3072 EI
Substituting the given data:
45
(50 ×103 )(4) 2 = 11.72 × 103 lb ⋅ in.3
EIvD =
3072
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________________________________________________________________________
PROBLEMS (10.125 through *10.128) An overhanging beam is loaded as shown in Fig.
P10.125 through P10.128. Determine:
(a) The slope at B.
(b) The deflection at point C.
SOLUTION (10.125)
A
2a
M0/2a
M0
B a
C
M0/2a
a/2
M/EI
A2
A1
M0/EI
x
5a/3
C
y
A
θA
tB/A
vC
C’’
x
tC/A C’C’’
C’
M a
M a
1 M0
(2a) = 0
A1 =
A2 = 0
2 EI
EI
EI
M 0 a 5a M 0 a a 13 M 0 a 2
( )+
( )=
tC / A =
6 EI
EI 3
EI 2
2
M a 1
t
M a
2 M 0a
,
t B / A = 0 ( 2a ) =
θA = − B/ A = − 0
3 EI
2a
3EI
EI 3
M 0a
EI
M a M a 4 M 0a
θB = 0 + 0 =
3EI 3 EI
EI
3
(b) We have C ' C " = t B / A = M 0 a 2 EI . Thus,
2
M a2
7 M 0a2
vC = tC / A − C ' C " = 0 (13 − 6) =
↑
6 EI
6 EI
(a) θ B − θ A = A1 =
Continued on next slide
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SOLUTION (10.126)
Beam is symmetric about midspan D, θ D = 0 .
w
w
A
B
2EI
EI
a
a
a
a
wa
wa
M/EI
3a/4
a/2
A1 A
2
2
-wa /4EI
2
-wa /2EI
(a) θ B =
x
Spandrel
a/4 parabola
tB/D
D
A
wa 2
wa 3
a
( )=−
4 EI
4 EI
2
wa 3
1 wa
(a ) = −
A2 = −
3 2 EI
6 EI
θ B = θ B / D = A1
A1 = −
EI
D
tC/D
B
C vC
wa 3
4 EI
a
wa 4
(b) t B / D = A1 ( ) = −
2
8EI
a
wa 4
3a
tC / D = A1 (a + ) + A2 ( ) = −
2
4
2 EI
4
4
4
wa
wa
3 wa
vC = tC / D − t B / D = −
=
↓
+
2 EI 8EI 8 EI
Continued on next slide
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SOLUTION (10.127)
P
P
A
L/2
D
B L/2 C
L/2
RA=0
RB=2P
M/EI
L/6
D
A
A3
-PL/2EI
L/6
M/EI
B
A1
D
A
x
A2
L/3
y
A
PL/EI
- PL/EI
B’
θA
tA/B
D
B
vC
tC/A x
PL2
1 PL L PL2
1 PL C
( L) = −
A1 =
A2 = −
=
2 EI 2 4 EI
2 EI
2 EI
2
PL
L
PL3
1 PL L
A3 = −
t B / A = A3 ( ) = −
=−
2 2 EI 2
8EI
6
48 EI
2
t
PL
θ A = − BA =
L 48EI
AD is a straight line.
(a) θ B / A = θ B − θ A = − A3
θ B = − A3 + θ A =
PL2 PL2
7 PL2
−
=
48EI 8EI 48 EI
L
L
PL3
PL3
t
A
A
(
)
(
)
=
+
=
(1 − 4) = −
(b) C / A
1
2
6
3
24 EI
8 EI
3
3
3
PL
PL
3 PL3
vC = tC / A − t B / A = −
=
↓
+
2
8EI 32 EI 32 EI
Continued on next slide
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SOLUTION (*10.128)
w
wa
a
A
3a
B
7wa/6
C
17wa/6
M/EI
17wa2/2EI
A2
A1
-4wa2/EI
B A5
x
A3
-wa2/EI
3a/2
2a
M/EI
9a/4
A A
4
2
-9wa /2
x
B
θA
Spandrel
parabola
A
tangent at A
B
C’’
tB/A C
vC
C’C
C’
wa 2
1 17 wa 2
51wa 3
3wa 3
)3a =
(3a) = −
A1 = (
A2 = −
2 2 EI
4 EI
EI
EI
2
3
2
1 3wa
9wa
1 9wa
9wa 3
(3a) = −
(3a) = −
A3 = −
A4 = −
2 EI
2 EI
3 2 EI
2 EI
2
3
wa
1 wa
(a) = −
A5 = −
2 EI
2 EI
3a
9a
t B / A = A1 (2a) + A2 ( ) + A3 (2a ) + A4 ( )
2
4
4
4
4
4
51wa 9 wa 9 wa 81wa 15wa 4
=
−
−
−
=
2 EI
2 EI
8 EI
8EI
EI
3
1
5 wa
θ A = − tB / A =
3a
8 EI
(a) θ B / A = θ B − θ A = A1 + A2 + A3 + A4 =
θB =
3wa 3
4 EI
3wa 3 5wa 3 1 wa 3
−
=
4 EI
8 EI 8 EI
Continued on next slide
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5a
13a
2a
) + A3 (3a) + A4 (
) + A5 ( )
2
4
3
4
4
wa 153 15 27 117 1
55 wa
(
=
− − −
− )=
2 2
8 3
24 EI
EI 4
Then,
4
vC = −C ' C "+ tC / A = − t B / A + tC / A
3
4
4 15wa
55wa 4
5 wa 4
)+
=
↓
=− (
3 8 EI
24 EI
24 EI
(b) tC / A = A1 (3a) + A2 (
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________________________________________________________________________
PROBLEMS (10.129 and 10.130) A beam ABC with an overhang BC supports a lod as shown in
Figs. P10.129 and P130. Determine:
(a) The angle of rotation at support B.
(b) The deflection at the free end C.
SOLUTION (10.129)
P
A
P/3
C
a
B
3a
4P/3
M/EI
2a
2a/3
x
A2
A1
-Pa/EI
θA
4tA/B/3
tB/A
A
tC/A
vC
B
C
2
Pa 2
1 Pa
3 Pa
1 Pa
(3a) = −
(a ) = −
A2 = −
2 EI
2 EI
2 EI
2 EI
2
3
3
Pa
2a
3Pa
10 Pa
tC / A = A1 (2a ) + A2 ( ) = −
−
=−
3
3EI
3 EI
EI
2
3Pa
t B / A = A1 (a) = −
2 EI
1
Pa 2
θ A = − tB / A =
3a
2 EI
A1 = −
(a) θ A / B = θ A − θ B = A1
θ B = θ A − A1 =
1 Pa 2 3 Pa 2 Pa 2
−
=
2 EI 2 EI
EI
4
10 Pa 3 4 3Pa 3
)
− (−
(b) vC = tC / A − t B / A = −
3
3 EI 3
2 EI
4 Pa 3
=
↓
3 EI
Continued on next slide
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SOLUTION (10.130)
w
A
2a
a
B
C
M/EI
Parabola
wa2/2EI
A1
A2
x
A3
Spandrel
parabola
-wa2/2EI
y
tangent at B
tB/A
θB B θ
A
B
C’’
C’
C
vC
tC/B
wa 2 2 wa 3
2
A1 = − (2a )
=
3
2 EI 3 EI
wa 3
wa 3
1 wa 2
1 wa 2
(2a) = −
(a ) = −
A2 = −
A3 = −
2 2 EI
2 EI
3 2 EI
6 EI
4
4
4a
2 wa 4 wa
t A / B = A1 (a) + A2 ( ) = −
−
=0
3
3EI
6 EI
(a) θ B =
1
t A / B = 0 Thus C ' C " = 0 .
2a
wa 4
3a
(b) tC / B = A3 ( ) = −
4
8 EI
4
wa
vC = tC / B =
↓
8EI
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________________________________________________________________________
PROBLEM (10.131) The 16-ft-long overhanging beam AC seen in Fig. P10.128 supports the
loading shown, where w = 400 lb/ft and EI = 12 x 106 lb in.2. Compute the value of angle of rotation θ A
at point A.
SOLUTION
See solution of Prob.10.128.
5 wa 3
θA =
8 EI
Substituting the given data
5 400 (4 ×12)3
)
θA = (
= 0.192 rad = 11o
6
8 12 12 × 10
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________________________________________________________________________
PROBLEM (10.132) A 6-m-long overhanging beam AC supports a uniform load of intensity w
(see Fig. P10.130). Calculate the slope at point C.
Given: w = 40 kN/m, E = 70 GPa, I = 20 x 106 mm4
SOLUTION
See solution of Prob.10.130.
wa 3
A3 = −
θA = 0
6 EI
Thus,
θC / B = θC − θ B 0 = A3 =
wa 3
6 EI
Substituting the given values:
40 × 103 (2)3
θC =
= 38.1× 10−3 rad
9
−6
6(70 ×10 )(20 ×10 )
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________________________________________________________________________
PROBLEM (10.133) The 12-ft overhanging beam AC seen in Fig. P10.125 supports a moment
of M0 = -500 kip ⋅ in. Compute the maximum value of EI vmax between supports A and B.
SOLUTION
See solution of Prob.10.125:
M/EI
M0x/2aEI
M0/EI
A
A
2a
x
B
y
C
x
x
tA/D
A
θA = −
vma B
M 0a
3EI
θD / A = θD0 − θ A = A =
(1)
1 M0 2
x
2 2aEI
(2)
Equations (1) and (2) give
2
x=
a
3
Then,
M x3
2x
4 M 0a2
t A / D = A( ) = − 0 = −
3
6aEI
9 3 EI
Hence,
4 M 0a2
vmax = v A / D =
↓
9 3 EI
and
4
[500 ×103 (4 × 12)2 ] = 295.6 ×106 lb ⋅ in.3
EIvmax =
9 3
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________________________________________________________________________
PROBLEMS (10.134 through 10.137) A beam is supported and loaded as shown in Figs.
P10.134 and P10.137. Determine the maximum deflection between the supports A and B.
SOLUTION (10.134)
P
P
B
A
2a
P/4
C
P/4
M/EI
8a/3
A2
Pa/4EI
E
A1
- Pa/2EI
y
tA/O
a D a
x
a
3
A1 = −
A2 =
θA
x
O
1 Pa 8a
2 Pa 2
( )=−
2 2 EI 3
3 EI
Pa 2
1 Pa 4a
( )=
2 4 EI 3
6 EI
tB/A
vmax
E
B
x
4 a 1 2 a 8a
1
4a
+ ( + )] + A2 [ (a + ]
3 3 3
3
3
3
3
2
3
44 Pa
7 Pa
3 Pa
=−
+
=−
27 EI 54 EI
2 EI
2
1
3 Pa
θ A = − tB / A =
4a
8 EI
3Pa 2 1 Pa x
) x,
θO / A = θO − θ A :
= (
x=a 3
8EI
2 2 EI 2a
1 Pa a 2 ⋅ 3 2
3 Pa 3
] (a 3) =
Thus,
vmax = vO / A = [
↑
2 2 EI 2a 3
4 EI
t B / A = A1[
Continued on next slide
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SOLUTION (10.135)
w
A
a
wa
C
a
B
a
D
5wa/6
7wa/6
M/EI
5wa
x
6 EI
2
5wa /2EI
A1
-wa2/2EI A2
wa(x-a) B
x
M/EI
-wa2/2EI
y
A
tA/B
A1 =
A3
C
x
vmax
θB B
tB/O
tangent at O
2
at15
B wa 2
1 5watangent
(3a) =
2 2 EI
4 EI
2
1 2wa
2wa 3
A2 = −
(2a ) = −
EI
2 EI
2
3
wa
1 wa
A3 = −
(a) = −
3 2 EI
6 EI
a
2a
57 wa 4
t A / B = A1 (a) + A2 ( ) + A3 ( ) =
3
4
24 EI
3
1
57 wa
θB = tA/ B =
3a
72 EI
1 5
1
θ B / O = θ A − θO 0 = [ wax 2 − wa ( x − a)2 ]
EI
2 8
Thus,
57 3 5
1
wa = wax 3 − wa ( x − a ) 2 ,
72
12
2
x = 1.472a
Then,
vmax = vO = t B / O =
1 5
2x
1
2
( wax 2 ) −
( wa )( x − a )[ ( x − a ) + a ]
2 EI 6
3 2 EI
3
Making x = 1.472a into the above
wa 4
vmax = vO = 0.74
↓
EI
Continued on next slide
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SOLUTION (10.136)
w0
A
a
C
B
a
w0a/2
w0a/2
M/EI
w0a2/2EI
2a/3
A1
A
Cubic
4a/5
A2
C
-w0a2/6EI
x
y
B
A
tA/C
vmax
C
x
tangent at C
w0 a 3
1 w0 a 2
A1 =
(a) =
2 2 EI
4 EI
2
w a3
1 w0 a
A2 = −
(a) = − 0
4 6 EI
24 EI
w a 4 w a 4 2w a 4
2a
4a
t B / A = A1 ( ) − A2 ( ) = 0 − 0 = 0
3
5
6 EI 30 EI 15 EI
2 w0 a 4
vmax = t AC =
↓
15 EI
Continued on next slide
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SOLUTION (10.137)
PL P
B
A
C
L/2
L/2
P/2
3P/2
M/EI
3PL/4EI
3Px/2EI
A1
-PL/4EI
A2
y
tA/D
A
θA
x
B
x
vmax
D
B
x
tB/A
tangent at A
− PL2
1 3PL L
3 PL2
1 PL L
A2 = −
( )=
( )=
2 4 EI 2 16 EI
2 4 EI 2 16 EI
3
L L
L
PL
PL3
5 PL3
t B / A = A1 ( + ) + A2 ( ) =
−
=
2 6
3
8EI 48EI 48 EI
1
5 PL2
θ A = − tB / A = −
L
48 EI
0
θ A/ D = θ A − θD
Also
3Px
3Px 2
θA = −
( x) = −
2 EI
2 EI
5
Equations (1) and (2) yield x =
L
6
Then,
1 3P( 5 L 6) L 5 2 5L
vmax = t A / D = [
L]
2
2 EI
6
3 6
3
3
PL
5 5 PL
=
= 0.0259
↓
432 EI
EI
A1 =
(1)
(2)
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________________________________________________________________________
PROBLEM (*10.138 and 10.139) A simple beam AB with two different moments of inertia
carries a load as shown in Figs. P10.138 and P10.139. What is the largest deflection between the supports A
and B?
SOLUTION (*10.138)
P
EI
4EI
A
L/2
C
L/2
B
P/2
P/2
M/EI
PL/4EI
Px/2EI
PL/16EI
A
A2
A1
C D
x
θA
A
B
vmax
tA/B
D
x
θB B
tB/D
tangent at D
tangent at B
A1 =
1 PL L
1 PL2
( )=
2 16 EI 2 64 EI
A2 =
1 PL L
1 PL2
( )=
2 4 EI 2 16 EI
L
2L
3 PL3
t A / B = A1 ( ) + A2 ( ) =
3
3
64 EI
2
1
3 PL
θB = tA/ B =
L
64 EI
θB / D = θB − θD0
Also
θB =
Px 2
1 Px
( x) =
2 2 EI
4 EI
and
3 PL2 Px 2
=
,
64 EI 4 EI
x=
3L
4
Therefore
Px 3
1 Px 2 2 x
( )=
2 2 EI 3
6 EI
3
P 3 3L
3 PL3
=
=
↓
6 EI 64
128 EI
vmax = t B / D =
Continued on next slide
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SOLUTION (10.139)
A
M0
2EI
L/2
EI
B
L/2
C
M0/L
M0/L
x
M0/4EI
A1
M/EI
L/3
x
A2
M0x/4EIL
tangent at B
tA/B
A
D
tangent at D
vmax
tB/D
θB
B
M L
1 M0 L M0L
1 M0 L
( )=
( )=− 0
A2 = −
2 4 EI 2 16 EI
2 2 EI 2
8EI
2L
1 M0L
L
t B / A = A1 ( ) + A2 ( ) = −
3
3
16 EI
M
1
θ B = t AB = 0
L
16 EI
1 M0x
θ B / D = θB − θD0 = −
( x)
2 EIL
L
From Eqs. (1) and (2): x =
2 2
We have
M x3
1 M 0 x2 2
( x) = − 0
tB / D = −
2 EIL 3
3EIL
Hence,
M 0 L2
vmax = vD =
↑
48 2 EI
A1 =
(1)
(2)
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________________________________________________________________________
PROBLEM (10.140) For the overhanging S 4 x 7.7 rolled-shape (see Table B.9) beam seen in Fig.
P10.126, calculate the slope at end C.
Given: w = 200 lb/ft, a = 3 ft,
E = 30 x 106 psi
SOLUTION
Table B.9: I = 6.08 in.4 See solution of Prob. 10.126.
wa 3
wa 3
A1 = −
A2 = −
4 EI
6 EI
θC / D = θC − θ D = A1 + A2
Thus
wa 3
5 wa 3
θC = −
(3 + 2) =
12 EI
12 EI
5 (200 12)(3 × 12)3
=
= 1.776 ×10−3 rad
12 30 × 106 (6.08)
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________________________________________________________________________
PROBLEM (10.141) The 4-m-long simple beam shown in Fig. P10.136 supports linearly varying
load of maximum intensity w0 = 60 kN/m. Compute the value of EI θ A at the support A.
SOLUTION
See solution of Prob.10.136:
w a3
w a3
A1 = 0
A2 = − 0
4 EI
24 EI
So,
θ A / C = θ A − θC 0 = A1 + A2 , or
θA =
w0 a 3
5 w0 a 3
(6 − 1) =
24 EI
24 EI
and
5(60 × 103 )(2)3
EIθ A =
= 100 ×103 N ⋅ m 2
24
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________________________________________________________________________
PROBLEM (*10.142) Determine the free-end deflection v A of the cantilever beam loaded as
shown in Fig. P10.142.
Assumption: The beam has a rectangular cross section of constant width b.
*SOLUTION
y
b
L
P
h0
2h0
A
h
x
B
M/EI
x
dx
x
A
x
h
x
h = h0 (1 + ) = 0 ( L + x)
L
L
I
1
I A = bh03
I = A3 ( L + x)3
L
12
v A = t A / B = Ax = ∫
=−
-PL/EIB
M
− PxL3
=
EI EI A (1 + x)3
Px 2 L3 dx
EI A (1 + x)3
PL3
x 2 dx
EI A ∫ ( L + x)3
L
PL3
L2
2L
=−
−
ln( L + x) +
EI A
L + x 2( L + x) 0
=−
5
PL3
PL3
PL3
= 0.06815
↓
(ln 2 − ) = −0.06815
EI A
EI A
EI A
8
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________________________________________________________________________
PROBLEM (10.143) Redo Prob. 10.142 for a beam of solid circular cross section and diameters h0
and 2h0 at the ends A and B, respectively.
SOLUTION
I=
π h04
64
x
I = I A (I + )
L
y
L
P
h0
2h0
A
B
x
h
Referring to solution of Prob.10.38, we now write.
M
− Px
=
EI EI (1 + x ) 4
A
L
and
v A = t AB = Ax = ∫
Mx
P L x 2 dx
dx = −
dx
EI
EI A ∫0 (1 + x )3
L
L
PL3 −1
2L
L2
=−
+
−
EI A 1 + x 2( w + x x ) 2( L + x)
L
L
0
3
3
1 1 1
1
PL3
PL
PL
=
↓
[− + − + 1 − 1 + ] =
=−
3 24 EI 24 EI A
EI A 2 4 24
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________________________________________________________________________
PROBLEM (*10.144) For the beam of variable cross section illustrated in Fig. P10.144. Determine
the maximum deflection vC .
*SOLUTION
Beam is symmetric about C.
P
A
L/2
B
L/2
C
P/2
P/2
PL/4
M
A
B
C
x
y
B
A
vC
tA/C
x
C
1
1
x
b0 h3 , I = bx h3 = I o ( I + ) ,
L
12
12
1 3
1
I C = ( b0 )h3 = b0 h3
12 2
8
I0 =
M=
P
x
2
M
PL L 2 x 2
tA/ C = ∫
xdx =
dx
2 EI 0 ∫0 ( L + x)
EI
=
PL 2 ⎧ 3 3 ⎫
PL
[ L ⎨ln( ) − ⎬] =
(0.03L2 )
2 EI 0
⎩ 2 8 ⎭ 2 EI 0
and
vC = t AC =
0.18 PL3 0.0225 PL3
=
↓
Eb0 h3
EI C
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________________________________________________________________________
PROBLEM (10.145) In the fixed beam AB shown in Fig. P10.145, the left-hand support has settled
a distance δ0 below the right-hand support. Determine the support reactions.
SOLUTION
Consider RA and M A as redudants.
MB
B
MA
δ0
A
L
RB
RA
MA
A
x
B
RA
M/EI
L/2
MA/EI
A1
x
A2
-RAL/EI
2L/3
M AL
1 RA L2
A2 = −
2 EI
EI
0
0
θ B / A = θ B − θ A = A1 + A2
A1 =
Thus,
2 M A = RA L
(1)
2
3
M L R L
2
L
t A / B = −δ 0 = A1 ( ) + A2 ( L) = A − A
2
3
2 EI
3EI
or
6 EI δ 0
−2 RA L + 3M A = −
L2
From Eqs. (1) and (2):
12 EI δ 0
↓
RA = −
L3
MA = −
(2)
6 EI δ 0
L2
Continued on next slide
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Symmetry dictates that
RB = − RA ↑
MB = MA
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________________________________________________________________________
PROBLEMS (10.146 through 10.149) A beam is supported and loaded as shown in Figs.
P10.146 through P10.149. Determine the reactions at the support.
SOLUTION (10.146)
Let M A be redundant. Because of symmetry:
wL
MA = MB
RA = RB =
↑
and
2
w
MA
RA
C
L/2
MB
L/2
RB
2
wL /16EI
M/EI
Parabola
A1
A
x
A2
-MA/EI
L/2
A1 =
Thus
wL3
2 wL2
L
=
3 8EI 12 EI
M AL
EI
3
M L
wL
θ A / B = 0 = A1 + A2 =
− A
12 EI
EI
A2 = −
θ A / B = θ B 0 − θ A0 = 0
or
MA =
wL2
12
MB =
wL2
12
Continued on next slide
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SOLUTION (10.147)
Let RA and M A as redundants.
M0
A
MA
RA
L/2
C
B
RB
L/3
M/EI
A
MB
L/2
A1
B
A2
C
RAL/EI
A3
x
-MA/EI
-M0/EI
L/4
RA L2
1 RA L
A1 =
( L) =
2 EI
2 EI
M0L
A3 = −
2 EI
A2 = −
M AL
EI
M L
RA L2
θ A = 0 = A1 + A2 + A3 =
− M AL − o
2
2
or
RA L − 2 M A = M 0
We have vB = 0 :
L
L
L
t BA = 0 : A1 ( ) + A2 ( ) + A3 ( ) = 0
3
2
4
or
3
RA L − 3M A = M 0
4
Solving Eqs.(1) and (2):
1
3 M0
M A = M0
↑
RA =
4
2 L
Statics:
RB = − RA
MB = MA
(1)
(2)
Continued on next slide
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SOLUTION (10.148)
Let M A be redundant.
tangent at B
P
θA = 0
RB
C
A1 =
B
A3
C
a
MA
B
2a
A
RA
M
MA
A
4a/3
A1
A2
x
1
M A (2a ) = M A a
2
1
A2 = − Pa (2a) = − Pa 2
2
1 2
A3 = − Pa
2
-Pa
2a/3
vB = t B / A = 0 =
1
4a
2a
[ A1 ( ) + A2 ( )]
3
3
EI
or
MA =
1
Pa
2
Statics:
RB =
7
P↑
4
RA =
3
↓
4
SOLUTION (10.149)
Consider RA and M A as redundant.
θ B / A = θ B 0 − θ A0 = 0
tB / A = 0
Continued on next slide
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P
MA
A
a
RA
y
RB
L
tangent at A
A
MA
MB
B
b
C
x
B
RA
L/2
L/3
M
A
RAL
A1
A2
C
A3
x
-MA
-Pb
b/3
EIθ B / A = A1 + A2 + A3 = 0
or
RA L2 − 2M A L − Pb 2 = 0
L
L
b
EIt B / A = A1 ( ) + A2 ( ) + A3 ( ) = 0
3
2
3
or
RA L3 − 3M A L2 − Pb3 = 0
Solving Eqs.(1) and (2)
Pb 2
Pab 2
RA = 3 (2a + L) ↑
MA = 2
L
L
Statics:
Pa 2b
0
:
M
=
M
=
∑ B
B
L2
Pa 2b
0
:
(2a + b) ↑
F
=
R
=
P
−
∑ y
B
L2
1
1
RA L( L) = RA L2
2
2
A2 = − M A L
1
1
A3 = − Pb(b) = − Pb 2
2
2
A1 =
(1)
(2)
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________________________________________________________________________
PROBLEMS (10.150 through 10.153) Determine all of the reactions for the propped cantilever
beam AB shown in Figs. P10.150 through P10.153.
SOLUTION (10.150)
Let M A be redundant.
w
tangent at A
MA
A
B
L
RA
RB
V
wL2/8
wL/2
x
-wL/2
M
2 L wL2 wL2
=
3 8
12
1
1
A2 = − M A ( L) = − M A L
2
2
L/2
wL2/8
A1 =
Parabola
A1
x
A2
-MA
3L/2
L
wL4 M A L2
2L
EIt B / A = 0 = A1 ( ) + A2 ( ) =
−
2
3
24
3
or
1
M A = wL2
8
5
3
Statics: RA = wL ↑
RB = wL ↑
8
8
SOLUTION (10.151)
Assume RA as redundant
tangent at A
MA
A
RA
B
L
RB
M
L/2
M0
-RBL
M0
A1
A2
2L/3
x
A1 = M 0 L
1
1
A2 = − RB L( L) = − RB L2
2
2
Continued on next slide
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2L 1
1
L
EIt B / A = 0 = A1 ( ) + A2 ( ) = RB L − M 0
2
3
3
2
3 M0
or
RB =
↓
2 L
Statics:
3M
∑ Fy = 0 : RA = 2 L0 ↑
1
∑ M A = 0 : M A = 2 M0
SOLUTION (10.152)
Consider RA as redundant.
w0
tangent at B
A
RA
M
MB
B
L
RB
2L/3
RAL
A1
x
A2
4L/3
Cubic
EIt A / B = 0 = A1 (
RA =
or
-w0L2/6
1
1
RA L( L) = RA L2
2
2
2
w L3
1 w0 L
( L) = − 0
A2 = −
4 6
24
A1 =
2L
4L 1
1
) + A2 ( ) = RA − w0 L
3
3
3
30
1
w0 L ↑
10
Statics:
∑F = 0:
y
∑M = 0:
B
2
w0 L ↑
5
1
M B = w0 L2
15
RB =
Continued on next slide
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SOLUTION (10.153)
M/EI
Choose RA as redundant.
2L/3
RAL
A1
A
C
A2
L/2
3L/8
Tangent at A
A
B
M2
Parabolic
Spandrel
x
wL2
1 L
M 2 = − w( ) 2 = −
2 2
8
R L2
1 R L
A1 = ( A )( L) = A
2 EI
2 EI
2
wL3
1 wL L
A2 = (−
)( ) = −
3 8EI 2
48EI
B
Requirement:
2L
L 3L
) + A2 ( + )
3
2 8
3
4
R L
7 wL
= A −
3EI 384 EI
t A B = 0 = A1 (
Solving
RA =
7 wL
↑
128
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________________________________________________________________________
PROBLEM (10.154) A propped cantilever beam AB is subjected to loads as shown in Fig. P10.154.
Calculate the reactions at each support.
SOLUTION
5 kN/m
10 kN
MB
A
B
C
Select RA as redundant.
Draw
M EI diagrams (see Table A.4).
10 RA
50 RA
1 10 RA
)(10) =
A1 = (
2 EI
EI
1 80
320
A2 = − ( )(8) = −
x
2 EI
EI
1 90
180
A3 = − ( )(6) = −
-80
3 EI
EI
Place reference tangent at B:
x θ B = 0. Also
2
2
t A B = A1[ (10)] + A2 [ (8) + 2]
-90
3
3
1
+ A3 [10 − (6)] = 0
4
RB
RA
M/EI
A1
A2
M/EI
A3
Parabolic
Spandrel
Then
tA B =
50 RA 20 320 28 180
( )−
( )−
(8.5) = 0
EI 3
EI 3
EI
or
RA = 13.55 kN ↑
Equilibrium:
RB = 10 + 30 − 13.55 = 26.45 kN ↑
M B = 34.5 kN ⋅ m
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________________________________________________________________________
PROBLEMS (10.155 and 10.156) A beam AB is supported and loaded as shown in Figs.
P10.155 and P10.156. Determine the reactions at supports A and B.
SOLUTION (10.155)
Consider RA as redundant.
P
tangent at A
A
L/2
B
L/2
C
MB
RA
vA = t A / B = 0
θB = 0
RB
M
RAL
2L/3
A1
A
C
L/2
EIt A / B = 0 = A1 (
B
x
A2
L/3
1
1
A1 = ( RA L) L = RA L2
2
2
1 PL L
1
A2 = −
( ) = − PL2
2 2 2
8
-PL/2
2L
5L
) + A2 ( )
3
6
or
RA =
5
P↑
16
RB =
11
P↑
16
Statics:
MB =
3
PL
16
Continued on next slide
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SOLUTION (10.156)
Consider RA and M A as redundants.
w0
Tangent at B
MA
A
L
RA
M
MB
B
RB
2L/3
RAL
L/2
A1
x
A2
-MA
Cubic
4L/5
1
1
RA L( L) = RA L2
2
2
A2 = − M A L
A1 =
A3 = −
A3
-w0L2/6
w L2
1 w0 L2
( L) = − 0
4 6 EI
24 EI
θ A / B = θ B 0 − θ A0 = 0 = A1 + A2 + A3
or
1
1
RA L =
w0 L2
2
24
2L
4L
L
EIt AB = 0 = A1 ( ) + A2 ( ) + A3 ( )
3
2
5
or
2
1
− M A + RA L = w0 L2
3
15
From Eqs.(1) and (2):
1
3
M A = w0 L2
RA =
w0 L ↑
30
20
Statics:
7
1
RB =
w0 L ↑
MB =
w0 L2
20
20
−M A +
(1)
(2)
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________________________________________________________________________
PROBLEM (10.157) A fixed-guided roller beam AB carries a uniform load of intensity w as
shown in Fig. P10.157. Determine the reactions RA , MA , and MB for the beam.
SOLUTION
Equilibrium:
w
L
MA
MB
B
RA
L
A
A
y
tangent at A
A
∑ M = 0 : M + M − wL( 2 ) = 0 (1)
∑ F = 0 : R = wL
B
A
Since both tangent at A and tangent
at B are horizontal, θ B A = 0 .
B
Thus
θB A = 0 = (
tangent at B
M/EI
A1
O
-wL2/2
A2
L/3
Parabolic
spandrel
or
MB
B
MB
1 wL2
)( L) + (−
)( L)
EI
3 2 EI
x
wL2
6
Equation (1) gives then
wL2
MA =
3
MB =
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________________________________________________________________________
PROBLEM (10.158) A continuous beam ABC of two spans, carries moments M0 at each end A
and C as shown in Fig. P10.158. Determine the reactions RA , RB , and RC for the beam.
SOLUTION
Select RA as redundant.
Requirement:
tA B = 0
L/2
M/EI
A1
A
=
M0
x
B
A2
M0
2L
L 1 R L
( L)( ) + (− A )( L)( )
2 2
3
EI
EI
or
M 0 L2 RA L3
0=
−
2 EI
3EI
2L/3
Solving
L
RA =
3M 0
↓
2L
Equilibrium:
M0
A
B
L
L
3M 0
∑M = 0:
RC
or
C
RB
3M0/2EI
M0
B
RC =
3M 0
( L) − RC ( L) = 0
2L
3M 0
↓
2L
3M 0
∑ F = 0 : R − 2L − 2L = 0
y
or
B
RB =
3M 0
↑
L
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________________________________________________________________________
PROBLEMS (10.159 and *10.160) A two-span continuous beam ABC is loaded as shown in
Figs. P10.159 and P10.160. Find the reactions RA , RB , and RC for the beam.
SOLUTION (10.159)
Consider RB as redundant.
P
-tC/B= tA/B
P
B
a
EI
RA
tangent at B
RB
M
C
a
A 2EI 2a
tC/B
RC
a
A1
A
A1
-Pa/2
M
C
x
4a/3
B
A
A2
A2
C
1 Pa
1
(2a) = Pa 2
2 2
2
1
A2 = M B (2a) = M B a
2
2
2 EIt A / B = − A1 (a) − A2 ( ⋅ 2a)
3
2
EIt B / C = − A1 (a) − A2 ( ⋅ 2a )
3
A1 =
x
-MB
We have
t A / B = −tC / B : −
from which
Pa 3 2
Pa 3 4M B a 2
− M B a 2 = −(
−
)
4
3
2
3
1
M B = Pa
8
Segment BC:
M B = RC (2a) − Pa
Continued on next slide
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Thus,
Pa
= RC (2a ) − Pa ,
8
Segment AB:
Pa
−
= − RA (2a ) + Pa ,
8
1
Statics:
RB = P ↑
8
−
RC =
7
P↑
16
RA =
9
P↓
16
w
tC/B
A
tA/B
M
L
tangent at B
RA
B L/2
C
RC
RB
L/2
wL2/8
A1
L/3
A2
2L/3
A3
x
-RCL/2
Continued on next slide
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SOLUTION (*10.160)
Assume RC as redundant.
wL3
2 wL2
( L) =
3 8
12
R L2
1 RC L
( L) = − C
A2 = −
2 2
4
RC L2
1 RC L L
( )=−
A3 = −
2 2 2
8
A1 =
L
wL4 RC L3
2L
EIt A / B = A1 ( ) − A2 ( ) =
−
2
3
24
6
3
R L
L
EItC / B = A3 ( ) = − C
3
24
We have
R L3 wL4 RC L3
2tC / B = −t A / B : C =
−
,
12
24
6
Statics:
3
5
RB = wL ↑
RA = wL ↑
4
12
RC =
1
wL ↓
6
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________________________________________________________________________
PROBLEM (*10.161) Before the load is applied to the beam shown in Fig. P10.161, a small gap δC
exists between beam AB and the support at C. Following the application of load, the gab closes and
reactions develop at each support. Determine the reactions for the beam.
*SOLUTION
Consider RC as redundant.
A
RA
M
B
2a
2a
RC
RB
4a/3
4RBa
2RBa
A1
x
A2
-
P
δc
Pa
a
M
1
A1 = (4 RB a)4a = 8 RB a 2
2
1
9
A2 = − (3Pa )3a = − Pa 2
2
2
1
A3 = (2 RC a )2a = 2 RC a 2
2
2RCa
A
A3
x
C
2a/3
y
δc
x
A
2a
tA/B=2( tC/B+ δc)
B
tC/B
tangent at B
4a
2a 32
9
4
) + A2 (a ) + A3 ( ) =
RB a 3 − Pa 3 + RC a 3
3
3
3
2
3
1
1
1
a 4
EItC / B = [ (2 RB a)2a] − [ ( Pa)a] = RB a 3 − Pa 3
6
2
2
3 3
EIt A / B = A1 (
We have 2(tCB + δ C ) = t AB :
2δ EI
4
25
= 8RB + RC − P = C3
a
3
6
1
3
∑ M A = 0 : RB + 2 RC = 4 P
3
3 δ C EI
9
3 δ EI
From Eqs.(1) and (2): RC = P −
RB = P + C 3
3
8
4 a
16
8 a
Then, ∑ Fy = 0 :
RA =
(1)
(2)
1
3 δ EI
P+ C3
16
8 a
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PROBLEM (11.1) Determine the critical load P for the system shown in Fig. P11.1. The rigid
bar AB is supported by a linear spring of stiffness k at A and a torsional spring of stiffness k t at B.
SOLUTION
P
A
θL
F = kθ L
∑ M = 0 : −k θ + Pθ L − kθ L = 0
θ
or
L
M t = ktθ
2
t
B
Pcr = kL +
kt
L
B
________________________________________________________________________
PROBLEM (11.2) Redo Prob. 11.1 for the case in which the bar is free at A.
SOLUTION
P
A
θL
∑ M = 0 : −k θ + Pθ L = 0
B
L
θ
M t = ktθ
t
or
Pcr =
kt
L
B
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PROBLEM (11.1) Determine the critical load P for the system shown in Fig. P11.1. The rigid
bar AB is supported by a linear spring of stiffness k at A and a torsional spring of stiffness k t at B.
SOLUTION
P
A
θL
F = kθ L
∑ M = 0 : −k θ + Pθ L − kθ L = 0
θ
or
L
M t = ktθ
2
t
B
Pcr = kL +
kt
L
B
________________________________________________________________________
PROBLEM (11.2) Redo Prob. 11.1 for the case in which the bar is free at A.
SOLUTION
P
A
θL
∑ M = 0 : −k θ + Pθ L = 0
B
L
θ
M t = ktθ
t
or
Pcr =
kt
L
B
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PROBLEMS (11.3 and 11.4) Determine the critical load P cr for the rigid bar-spring
system shown in Figs. P11.3 and P11.4, where k represents the linear spring constant.
SOLUTION (11.3)
P
Bar AB
∑ M B = 0 : PLθ − LR = 0 ,
R
A
θ
R = Pθ
L
Bar BC
∑ MC = 0 :
B
R
Lθ
B
kLθ
L
C
R
− Pθ L − RL + k
kLθ
4
From Eqs.(1) and (2):
1
Pcr = kL
8
or
(1)
Lθ L
=0
2 2
R = − Pθ +
(2)
SOLUTION (11.4)
P
R
A
Ra − Pθ a = 0, R = Pθ
a
B
aθ
a
Bar AB
∑MB = 0:
F = kaθ
θ
C
(1)
System ABC
∑ M C = 0 : R(2a) − kθ a(a) = 0
R=
1
kaθ
2
From Eqs.(1) and (2): Pcr =
(2)
1
ka
2
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PROBLEMS (11.5 and 11.6) A rigid bar-spring assembly is shown in Figs. P11.5 and
P11.6, where k t represents the torsional spring constant. What is the critical load P cr for the
system?
SOLUTION (11.5)
P
P
RC
A
A
θ
a
a
aθ
B
B
M = 2θ kt
a
System ABC
∑ M C = 0 : RC = 0
Bar AB
∑MB = 0:
k
2θ kt − Paθ = 0 , Pcr = 2 t
a
θ
C
SOLUTION (11.6)
Pcr
FA
Bars AC and CB
We have M B = ktθ
A
θ
M B = 0 : ktθ − 2 LFA = 0,
C
θ
B
FB
FA =
ktθ
2L
Bar AC:
M C = 0 : Pcr ( Lθ ) − LFA = 0
or
k
F
Pcr = A = t
θ 2L
MB=ktθ
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PROBLEM (11.7) Solve Prob. 11.4, assuming that the two rigid bars AB and BC are
replaced by a single rigid bar AC of length 2a and that the support at A is removed.
SOLUTION
2aθ
P
a
aθ
a
θ
FB = kaθ
A
C
∑M = 0:
C
− P(2aθ ) + kaθ (a ) = 0
From which
1
Pcr = ka
2
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PROBLEM (11.8) Calculate the critical load of a pinned-end wooden yardstick (E = 1.6 x 106
psi) of 1/8-in. by 1 1/8-in. rectangular cross section.
SOLUTION
1
1.125
(0.125)3 = 183.15 × 10−6 in.4
I min = bh3 =
2
12
Thus,
π 2 EI π 2 (1.6 ×106 )(183.15 ×10−6 )
Pcr = 2 =
= 2.23 lb
L
(3 × 12) 2
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PROBLEM (11.9) The jib crane shown in Fig. P11.9 is of capacity W = 20 kN. For α =
30 o and a factor of safety of n s = 3, calculate the required minimum cross-sectional area for
circular steel bar AB (E = 200 GPa).
SOLUTION
FBC
20
∑ F = 0 : F = sin 30 = 40 kN
∑ F = 0 : F = 40 cos 30 = 34.64 kN
30o
B
FAB
y
BC
x
AB
o
o
W=20 kN
Pcr = FAB =
So,
I=
π c4
π 2 EI
,
f s L2
I=
34.64 ×103 (3)(2.5) 2
= 0.329 × 10−6 m 4
2
9
π (200 ×10 )
= 0.329 ×10−6 ,
c = 25.44 mm
4
A = π c 2 = π (25.44) 2 = 2033 mm 2
Justification of the formula used:
r = I A = 12.7 mm
L r = 2500 12.7 = 197 O.K.
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PROBLEM (11.10) Redo Prob. 11.9 for W = 30 kN and α = 40°.
SOLUTION
FBC
30
∑ F = 0 : F = sin 40 = 46.67 kN
∑ F = 0 : P = F = 46.67 cos 40
o
40
y
BC
x
cr
o
o
FAB
B
AB
= 35.75 kN
W=30 kN
Pcr = FAB =
π 2 EI
,
f s L2
I=
35.75 ×103 (3)(2.5) 2
= 0.34 ×10−6 m 4
π 2 (200 ×109 )
Therefore,
I=
π c4
4
= 0.34 × 10−6 ,
c = 25.6 mm
and
A = π (25.6)2 = 2066 mm 2
Justification of the formula used:
r = I A = 12.8 mm
L r = 2500 12.8 = 195 O.K.
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PROBLEM (11.11) Two equal round bars are supported in line with one another and are
separated by a small gap (Fig. P11.11). Determine the temperature rise necessary to cause them
to:
(a) Just touch.
(b) Buckle elastically.
Given: a = 12 x 10 -6 /°C, L = 1 m, d = 25 mm, δ = 1.2 mm
SOLUTION
(a) α L(ΔT ) =
δ
ΔT =
,
δ
2α L
2
Substituting the given data:
1.2 × 10−3
ΔT =
= 50o C
−6
2(12 × 10 )1
(b) A =
π
(0.025) 2 = 490.9 ×10−6 m 2 I =
4
We have
α L(ΔT ) −
π
64
(0.025) 4 = 19.17 ×10−9 m 4
PL δ
=
AE 2
Thus,
ΔT =
δ
π 2I
+
2α L 4 Aα L2
or
π 2 (19.17 × 10−9 )
= 50 + 8 = 58o C
4(490.9 ×10−6 )(12 × 10−6 )(1) 2
Justification of the formula used:
r = I A = d 4 = 6.25 mm
L r = 1000 6.25 = 160 O.K.
ΔT = 50 +
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PROBLEM (11.12) Redo Prob. 11.11 for the case in which δ = 0.5 mm and L = 0.6 m.
SOLUTION
See solution of Prob. 11.11:
A = 490.9 × 10−6 m 2
I = 19.17 × 10−9 m 4
Hence,
L r = 600 6.25 = 96
and Euler’s formula applies.
(a) ΔT =
0.5 × 10−3
δ
=
= 34.7 o C
−6
2α L 2(12 ×10 )(0.6)
(b) ΔT =
δ
π 2I
+
2α L 4 Aα L2
= 34.7 +
π 2 (19.17 × 10−9 )
−6
−6
4(490.9 ×10 )(12 ×10 )(0.6)
2
r = 6.25 mm
= 57 o C
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PROBLEM (11.13) A 60-mm outer-diameter and 50-mm inner-diameter steel pipe (E = 20
GPa) 2 m long acts as a spreader in the assembly of Fig. P11.13. Determine for a factor of safety
n s = 2, the value of F that will cause buckling of the pipe.
SOLUTION
Pcr
1
2
2
F
I=
4
(c 4 − b 4 ) =
Pcr
F
1
π
1
2
π
4
F 2
=
Pcr 1
(304 − 254 ) = 32.94 × 104 mm 4
Pcr π 2 EI π 2 (200 × 109 )(0.3294 ×10−6 )
=
=
= 81.28 kN
ns
ns L2
2(2) 2
F = 2(81.28) = 162.6 kN
Justification of the formula used:
A = π (c 2 − b 2 ) = π (302 − 252 ) = 863.94 mm 2
r=
329, 400
I
=
= 19.5 mm
A
863.94
and
L 2000
=
= 102.6 O.K.
r 19.5
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PROBLEM (*11.14) A jib crane is made of a rectangular steel bar AB with a width b by depth
h cross section (Fig. P11.14). Determine the buckling load of the column.
Given: b = 2 in., h = 3 in., E = 30 x 106 psi, σy = 36 ksi
*SOLUTION
Refer to Example 11.3. We have axial force in bar AB is P=4W/3.
1
1
I y = (3)(2)3 = 2 in.4 , I z = (2)(3)3 = 4.5 in.4 , A = (2)(3) = 6 in.2
12
12
Iy
I
ry =
= 0.577 in.
rz = z = 0.866 in.
A
A
Buckling in xy-plane ( Le = L ).
96
L
=
= 110.9
rz 0.866
Pcr =
or
π 2 EA
( L rz )
2
=
π 2 (30 ×106 )6
(110.9)
2
= 144.4(103 ) =
4W
3
W = 106.3 kips
Buckling in xz-plane
Line of action of the compressive force again passes through A and thus Le = L .
So,
L
96
=
= 166.4
ry 0.577
Hence
π 2 EA π 2 (30 ×106 )(6) 4W
=
=
Pcr =
( L ry ) 2
(166.4) 2
3
Solving
W = 48 kips
The column will buckle when W > 48 kips .
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PROBLEM (11.15) Two aluminum alloy 2014-T6 bars are used as compression members,
each with an effective length of L e , have the cross sections shown in Fig. P11.15. Calculate:
(a) The wall thickness of the hollow square bar so that the bars have the same cross-sectional
area.
(b) The critical load of each bar.
Given: L e = 8 ft, E = 10.6 x 10 6 psi (from Table B.4)
SOLUTION
(a) Same area:
π
4
(d o2 − di2 ) = bo2 − bi2
bi2 = bo2 −
π
4
(d o2 − di2 ) = 22 −
π
4
(22 − 1.52 )
or
1
t = (bo − bi ) = 0.19 in.
2
bi = 1.62 in.
(b) Circular bar
I=
π
64
(d o4 − di4 ) =
Pcr =
π 2 EI
L2e
π
64
=
(24 − 1.54 ) = 0.537 in.4
π 2 (10.6 ×106 )(0.537)
(8 ×12) 2
= 6.1 kips
Square bar
1
1
I = (bo4 − bi4 ) = (24 − 1.624 ) = 0.759 in.4
12
12
2
π EI π 2 (10.6 ×106 )(0.759)
= 8.62 kips
Pcr = 2 =
Le
(8 ×12) 2
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PROBLEM (11.16) A control linkage consists of a high-strength ASTM-A242 rod AB of
diameter d and a pivot arm CD (Fig. P11.16). The load is transmitted to the rod through a pin
A. What is the largest value of F that can be applied on the basis of the buckling strength of the
rod and a safety factor of n s ?
Given: a = 160 mm,
E = 200 GPa,
b = 40 mm, L = 0.6 m, d = 10 mm,
σy = 345 MPa
n s = 1.7
(from Table B.4)
SOLUTION
I=
πd4
=
π (10) 4
= 490.874 mm 4
64
64
2
πd
π (10) 2
=
= 78.54 mm 2
A=
4
4
2
π EI π 2 (200 ×109 )(490.874 ×10−12 )
= 2.692 kN
PA = Pcr = 2 =
(0.6) 2
L
The corresponding stress is
P
2, 692
σ cr = cr =
= 34.4 MPa
A 78.54 × 10−6
which is well below the yield stress of 345 MPa. We have
∑ M C = 0 : (a + b) F = bPcr
or
200 F = 40(2, 692), F = 538.4 N
Therefore
F 538.4
Fall = =
= 317 N
ns
1.7
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PROBLEM (11.17) A round steel column of length L is fixed at its base and pinned at its
top end supports a maximum permissible load P (Fig. P11.17). Calculate the diameter d of the
member based on a safety factor of n s with respect to buckling.
Given: L = 50 in., P = 4.5 kips,
E = 30 x 10 6 psi.
n s = 3,
σy = 55 ksi
SOLUTION
Pcr = ns P = 3(4.5) = 13.5 kN ,
I=
π
d4
A=
πd
2
r=
Le = 0.7 L = 0.7(50) = 35 in.
d
4
64
4
Equation (11.5) gives
64 Pcr L2e 64(13.5 ×103 )(35) 2
d4 =
,
=
π 3E
π 3 (30 ×106 )
Hence
Le
35
=
= 135.5
r 1.033 4
Equation (11.10):
d = 1.033 in.
L
30 × 103
E
( e )c = π
=π
= 73.4 < 135.5
r
σy
55
Euler formula applies, Hence
d = 1.033 in.
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PROBLEM (11.18) Redo Prob. 11.17 for the case in which the length of the column is L =
20 in. and the maximum load permissible load equals P = 30 kips.
SOLUTION
Refer to Solution of Prob. 11.17. Now we have Le = 0.7(20) = 14 in.
and Pcr = 3(30) = 90 kips . Equation (11.5):
64 Pcr L2e 64(90 ×103 )(14) 2
,
d =
=
π 3E
π 3 (30 × 106 )
4
d = 1.05 in.
and
Le
14
=
= 53.5 < 73.4
r 1.05 4
Euler formula does not apply.
Apply Johnson formula, Eq. (11.14):
2
Pc σ y Le 1 2
90
55(14) 2 1 2
]
+
d = 2(
) = 2[
+
π × 55 π 2 (30 × 103 )
πσ y π 2 E
or
d = 1.493 in.
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PROBLEM (11.19) Two C 6 x 10.5 steel channels (see Table B.9) are bolted back to back
as shown in Fig. P11.19, and used as a pinned-end column 8 ft long. Using E = 30 x 10 6 psi,
determine the buckling load for the two members.
SOLUTION
Y
C 6 × 10.5 :
y
A1
A1
z
z
Y
I z = 15.2 in.4
I y = 0.865 in.4
Since I y < I z , buckling occurs about Y-Y
axis
A1 = 3.09 in.2
d=0.5 in.
IY = 2( I y + A1d 2 ) = 2[0.865 + 3.09(0.52 )] = 3.275 in.4
96
L
3.275
=
= 131.9
= 0.728 in.
rY 0.728
2 × 3.09
Using Euler formula
π 2 EA π 2 (30 ×106 )(30.9 × 2)
= 105.2 kips
Pcr =
=
( L r )2
(131.9) 2
rY =
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PROBLEM (11.20) Two L 3 x 3 x ¾ steel angles (see Table B.11) are bolted back to
back as depicted in Fig. P11.20, and used as a fixed-pinned end column 12 ft long. What is the
y
buckling load for he two members?
Use E = 29 x 10 6 psi.
SOLUTION
L3 × 3 × 1 4 :
Y
A1 = 1.44 in.2
z
z
A1
Y
IY > I z = 1.24 in.2
rz = 0.93 in.
A1
Buckling will occur about z-z axis:
Le = 0.7 L = (0.7)(12 × 12) = 100.8 in.
Le r = 100.8 0.93 = 108.4
Thus
π 2 EA π 2 (29 ×106 )(1.44 × 2)
= 70.2 kips
Pcr =
=
( L r )2
(108.4) 2
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PROBLEM (11.21) An S 310 x 74 steel section (see Table B.9) is used as a column of actual
length L. Based upon a factor of safety n s , determine the critical load P cr for each of the following
conditions:
(a) Pinned-pinned
(b) Fixed-pinned.
(c) Fixed-free.
Given: L = 4 m, E = 200 GPa,
ns = 2
SOLUTION
y
z
S 310 × 74 :
Buckling occurs about y-y axis:
I y = 6.35 × 10−6 m 4 < I z
z
ry = 26.2 mm
y
(a) Le r = 4000 26.2 = 152.7
Pcr =
π 2 EI
=
ns L2e
π 2 (200 ×109 )(6.53 ×10−6 )
2(4) 2
= 402.8 kN
(b) Le = 0.7 L = 2.8 m
Pcr =
π 2 (200 ×109 )(6.53 × 10−6 )
2(2.8) 2
= 822 kN
(c) Le = 2 L = 8 m
Pcr =
π 2 (200 ×109 )(6.53 × 10−6 )
2(8) 2
= 101 kN
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PROBLEM (11.22) Redo Prob. 11.21 for a W 250 x 80 steel column of actual length L = 9
m (see Table B.9).
SOLUTION
S 250 × 80 :
I y = 42.8 × 10−6 m 4 < I z
y
Buckling occurs about y-y axis:
ry = 65 mm
z
z
y
(a) Le = L = 9 m ( Le ry = 9 0.065 = 138.5)
Pcr =
π 2 EI
ns L2e
=
π 2 (200 ×109 )(42.8 ×10−6 )
2(9) 2
= 521.5 kN
(b) Le = 0.7 L = 6.3 m
Pcr =
π 2 (200 ×109 )(42.8 ×10−6 )
2(6.3) 2
= 1064 kN
(c) Le = 2 L = 18 m
Pcr =
π 2 (200 ×109 )(42.8 ×10−6 )
2(18) 2
= 130.4 kN
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PROBLEMS (11.23 and 11.24) Given on a factor of safety n s = 2, determine the largest
load F that may be applied to the structure shown in Figs. P11.23 and P11.24.
Assumption: Each column is a 50-mm-diameter steel bar having E = 210 GPa.
SOLUTION (11.23)
F
2
4
4
1
3
A
F
B
3
C
FAB
2
1
FBC
Joint B
∑ Fx = 0 : FBC = 0.6 5FAB
2
FBC + 0.8 FAB = F
5
∑F = 0:
y
Solving,
FAB = 0.5F
We have
I=
π
FBC = 0.671F
(0.05) 4 = 306.8 ×10−9 m 4
64
Bar AB ( L r = 200 )
0.5F = Pcr =
π 2 EI
=
ns L2
π 2 (210 ×109 )(306.8 ×10−9 )
2(2.5) 2
r = I A = d 4 = 12.5 mm
,
F = 101.7 kN
Bar BC ( L r = 178.9 )
0.671F =
π 2 (210 ×109 )(306.8 ×10−9 )
2( 5) 2
= 63.59 ,
F = 94.77 kN
Therefore,
Fall = 94.77 kN
Continued on next slide
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SOLUTION (11.24)
I=
π
(0.05) 4 = 306.8 ×10−9
64
L
I d 50
r=
= 160
= =
= 12.5 mm
r
A 4 4
We have
π 2 EI π 2 (210 ×109 )(306.8 ×10−9 )
= 79.5 kN
=
Pcr =
2(2) 2
ns L2
Thus,
1.2
Fall =
Pcr = 2(79.5) = 110.4 kN
0.6
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PROBLEMS (11.25 and 11.26 ) Knowing that the factor of safety n s = 1.5, calculate the
maximum load F that can be applied to the structure shown in Figs. P11.25 and P11.26.
Assumption: Each column is of 1 5/8 in.-diameter steel bar with E = 10 x 10 6 psi.
SOLUTION (11.25)
I=
π
(1.625) 4 = 0.342 in.4
64
r = I A = d 4 = 1.625 4 = 0.406 mm
13
5
F
FBC = F
12
12
Bar AB ( L r = 78 0.406 = 192 )
FAB =
F
B
FBC
12
Pcr =
5
FAB
=
π 2 EI
ns L2
π 2 (10 ×106 )(0.342)
1.5(78) 2
= 3.7 kips
and
13
F = 3.7,
12
F = 3.41 kips
Bar BC ( L r = 36 0.342 = 105 )
Pcr =
π 2 (10 ×106 )(0.342)
1.5(36) 2
= 17.36 kips
and
5
F = 17.36,
12
F = 41.66 kips
Hence
Fall = 3.41 kips
Continued on next slide
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SOLUTION (11.26)
LBC = 42 + 22 = 4.472 ft = 53.66 in.
I=
π
64
d4 =
π
64
(1.625) 4 = 0.342 in.4 ,
r=
d 1.625
=
= 0.406 in.
4
4
LAB = 22 + 22 = 2.828 ft = 33.94 in.
Bar BC ( ( L r = 53.66 0.406) = 132.2 . Thus
( FBC ) all =
Pcr π 2 (10 × 106 )(0.342)
=
= 7.82 kips
1.5(53.66) 2
ns
Bar AB
Pcr π 2 (10 ×106 )(0.342)
( FAB ) all =
=
= 19.53 kips
1.5(33.94) 2
ns
Joint B
F
1
1
FAB
1
2
∑ F = 0 : 2 F − 5 F = 0, F = 0.791F
x
B
AB
1
1
FBC
BC
AB
1
∑ F = 0 : 2 F + 5 F − F = 0, F = 1.061F
y
2
BC
AB
BC
AB
Solving
F = 1.061FAB
F = 1.341FBC
The allowable value for F:
F < 1.341(19.53) = 26.2 kips
F < 1.341(7.82) = 9.4 kips
Thus
Fall = 10.5 kips
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________________________________________________________________________
PROBLEM (*11.27) The truss ABC shown in Fig. P11.27 supports a load P = 2 kN.
Determine the factor of safety n s with respect to buckling of the members.
Assumption: Each member is a circular steel pipe having E = 210 GPa.
*SOLUTION
1
∑ F = 2 F − 1 = 0,
x
1.732 kN
AB
1
∑ F = 2 (1.414) + F − 1.732 = 0
B
y
1 kN
BC
FBC = 0.732 kN
1
1
FAB
FAB = 1.414 kN
FBC
( FAB )cr =
Member AB
π d 4 π (20) 4
I AB =
=
= 7.854 × 103 mm 4
64
64
LAB = 2.828 m
π 2 EI AB
=
π 2 (210 ×109 )(7.854 ×10−9 )
L2AB
(2.828) 2
(F )
2.035
ns = AB cr =
= 1.44
1.414
FAB
= 2.035 kN
Member BC
I BC =
πd4
64
( FBC )cr =
=
π (15)4
64
π 2 EI BC
=
= 2.485 ×103 mm 4
LBC = 2 m
π 2 (210 ×109 )(2.485 ×10−9 )
(2) 2
L2BC
(F )
1.288
ns = BC cr =
= 1.76
0.732
FBC
The smallest governs: ns = 1.44
= 1.288 kN
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PROBLEM (*11.28) A column AB of length L and flexural rigidity EI has a support at its
base that does not permit rotation but allows horizontal displacement and is pinned at its top as
shown in Fig. P11.28. Determine the critical load P cr .
*SOLUTION
Differential equation for a free-body of a segment of x is given by Eq.(11.1):
d 2v
EI 2 + Pv = 0
dx
or
d 2v
P
+ p 2v = 0
p=
2
dx
EI
General solution is
v = A sin px + B cos px
Boundary conditions :
v(0) = 0 : 0 + 0 + B = 0,
B=0
v '( L) = 0 : AP cos pL = 0,
Ap = 0
Since A ≠ 0 :
P
π
P
L = n( )
cos
L = 0 or
EI
EI
2
2
P
π
For n = 1,
= 2
EI 4 L
Therefore
π 2 EI
Pcr =
4 L2
(1)
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PROBLEM (11.29) A fixed-ended long column of rectangular cross section 1 in.by 3 in.
is made of aluminum alloy. Determine the minimum length L c that the column may have and the
corresponding critical load P cr .
Given: E = 10 x 10 6 psi, σp = 18 ksi
SOLUTION
1
(3)(1)3 = 0.25 in.4 ,
A = 1× 3 = 3 in.2
12
rmin = 0.25 3 = 286.68 × 10−3 in.
Equation (11.10):
I min =
Le
10 ×106
π 2E
=π
= 74.048
( )C =
18 × 103
r
σ pl
or
Le = 21.376 in. ,
LC =
Le
= 42.752 in.
0.5
Thus,
Pcr =
or
π 2 EI
L2e
=
π 2 (10 ×106 )(0.25)
(21.376) 2
= 54 kips
Pcr = σ p A = 18(3) = 54 kips
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________________________________________________________________________
PROBLEM (11.30) A rectangular aluminum (E = 70 GPa) tube of uniform thickness t =
20 mm (Fig. P11.30) serves as a 6-m-long column fixed at both ends. Calculate the critical
stress in the column.
SOLUTION
1
(200 × 1003 − 160 × 603 ) = 13.79 × 106 mm 4
12
A = 200 × 100 − 160 × 60 = 10.4 × 103 mm3
I min =
rmin = I min A = 36.4 mm
Le = 0.5L = 3 m
Le r = 3, 000 36.4 = 82.4
Hence,
π 2E
π 2 (70 ×109 )
=
= 101.8 MPa
σ cr =
( Le r ) 2
(82.4) 2
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PROBLEM (11.31) Repeat Prob. 11.30, assuming that the column is pinned at one end and
fixed at the other.
SOLUTION
Le = 0.7 L = 42 m . From solution of Prob.11.30: rmin = 36.4 mm .
We now have
Le r = 4200 36.4 = 115.4
Hence,
π 2 E π 2 (70 ×109 )
=
= 51.9 MPa
σ cr =
( Le r )
(115.4) 2
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PROBLEM (11.32) Brace BD of the structure shown in Fig. P11.32 is a steel rod (E = 200
GPa and σy = 220 MPa) of square cross section (40 mm on a side). Determine the safety factor
n s against failure by buckling.
SOLUTION
10 kN
B
2m
A
1.4 m
x
1
1
FBD
1
∑ M = 0 : − 10(3.4) + 2 F (2) = 0 ,
FBD = 24.04 kN
I = b 4 12 = 404 12 = 213.3 × 103 mm 4 ,
A = 40 × 40 = 1.6 × 103 mm 2
A
r=
BD
I
= 11.5 mm
A
L 2000
=
= 173.9
r 11.5
So,
σ cr =
π 2E
=
( L r )2
π 2 (200 ×109 )
(173.9) 2
= 65.27 MPa
We have
σ cr < σ pl
(solution is valid)
( FBD )cr = 65.27 ×10 (1.6 × 10−3 ) = 104.43 kN
and
(F )
104.43
ns = BD cr =
= 4.3
FBD
24.04
6
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PROBLEM (11.33) A 6-ft-long column of 4-in. by 2-in. rectangular cross section fits
between a rigid ceiling and a rigid floor. Calculate the change in temperature that - will cause the
column to buckle.
Given: E = 30 x 10 6 psi, α = 6.7 x 10 -6 /°F, σy = 34 ksi
Assumption: The ends are taken to be rounded .
SOLUTION
Assume that σ y = σ cr . So
π 2 (30 ×106 )
L
= 93.3 ,
( )c =
I min = 4(2)3 12 = 2.667 in.4
34 ×103
r
A = 2 × 4 = 8 in.2 ,
r = I min A = 0.578 in.
and
L r = 6 ×12 0.578 = 124.6
Thus,
σ = α E (ΔT ) =
π 2E
(L r)
Substituting the given data:
ΔT =
,
2
π2
−6
6.7 ×10 (124.6)
2
ΔT =
π2
α ( L r )2
= 94.9o F
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PROBLEM (11.34) The jib crane of Fig. P11.34a is constructed of a 35-mm-diameter rod
BC and a steel tube AB. The approximate stress-strain curve for both members is shown in
P11.34b. Determine the value of W at which failure occurs.
Given: The steel tube has the following properties:
A = 400 mm2 , I = 8x10 3 mm 4 , E = 200 GPa.
SOLUTION
FBC = 2W
C
Rod
d=35 m
FAB = W
1
Rod BC
1
45o
2.5 m
A
B
W
π
2W = 5 × 106 ( × 0.0352 ) = 4,811
4
or
W = 3.4 kN
Bar AB
Py = 8 × 106 (400 ×10−6 ) = 3.2 kN
W = Pcr =
π 2 (200 ×109 )(8 × 10−9 )
(2.5) 2
= 2.53 kN
and
2530
= 6.325 MPa < 8 MPa
400 ×10−6
Conclusion: bar AB fails as a column, and Wall = 2.53 kN
σ cr =
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PROBLEM (*11.35) A 60-in. long, 2 in.-diameter round column is fixed at its base and
pinned at its top. Calculate the axial buckling load P cr , if the stress-strain diagram for the material
is approximated as depicted in Fig. P11.35.
*SOLUTION
A=
πd2
4
I=
πd4
64
r=
Le = 0.7 L = 0.7(60) = 42 in.
I d 2
= = = 0.5 in.
A 4 4
(see Fig. 11.3c),
Le 42
=
= 84
r 0.5
Equation (11.12):
π 2 Et
π 2 Et
=
= 141.7(10−6 ) Et
(1)
σ cr =
2
2
( Le r )
(84)
From stress-strain diagram, we have
15 ×103
(45 − 15)103
Et = E =
= 15 × 106 psi
Et =
= 5 × 106 psi
0.001
0.007 − 0.001
Equation (1) gives
σ cr = 141.7(10−6 )(15 ×106 ) = 2.13 ksi < 15 ksi (O.K.)
σ cr = 141.7(10−6 )(5 × 106 ) = 0.709 ksi < 15 ksi
(not valid)
Thus
Pcr = σ cr A = 2.13[π (1) 2 ] = 6.69 kips
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________________________________________________________________________
PROBLEM (11.36) A 0.5-m-long, 50-mm by 75-mm rectangular column is pinned at both
ends. If the stress-strain curve for the material is approximated as shown in Fig.P11.36, what is
the buckling load P cr ?
SOLUTION
A = 50 × 75 = 3.75 × 103 mm 2
r= I A=
b
2 3
L r = 500 14.4 = 34.7 (inelastic buckling)
Use Eq. (11.12):
π 2 Et
π 2 Et
=
= 0.0082 Et
σ cr =
( L r ) 2 (34.7) 2
Elastic range:
σ cr = 0.0082(20 ×109 ) = 164 MPa
Figure shows that E is not valid above 40 MPa.
=
50
= 14.4 mm
2 3
(1)
Second range:
σ cr = 0.0082(20 ×109 ) = 82 MPa
Et1 is not valid above 60 MPa.
Third range:
σ cr = 0.0082(7.5 ×109 ) = 61.5 MPa
Et 2 is valid between 60 MPa and 75 MPa. Thus,
Pcr = 61.5 ×106 (3.75 ×10−3 ) = 231 kN
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________________________________________________________________________
PROBLEM (*11.37) Determine the value of w 0 that may cause the pinned column AB
in the structure of Fig. P11.37 to buckle. Assume that the column is constructed of a material for
which the stress-strain curve is as shown in Fig. P11.36. Use L = 0.5 m.
SOLUTION
w0
∑M = 0:
C
P=
C
L/2
P
L/2
2
w0 L
3
(1)
r = b 2 3 = 50 2 3 = 14.4 mm
L r = 500 14.4 = 34.7 (inelastic buckling)
See solution of Prob. 11.36: σ cr = 61.5 MPa
Thus, Pcr = σ cr A = 61.5 ×106 (0.05 × 0.05) = 154 kN
From Eqs. (1) and (2):
2
154 × 103 = w0 L ,
w0 = 462 kN m
3
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________________________________________________________________________
PROBLEM (11.38) Redo Prob. 11.36, given that one end of the column is fixed and the
other is pinned.
SOLUTION
Le = 0.7 L = 350 mm
Referring to solution of Prob. 11.36:
Le 350
=
= 24.3 (inelastic buckling)
r 14.4
π 2 Et
π 2 Et
=
= 0.0167 Et
σ cr =
( Le r ) 2 (24.3) 2
Elastic range:
σ cr = 0.0167(20 ×109 ) = 334 MPa
E is not valid above 40 MPa
(1)
Second range:
σ cr = 0.0167(10 ×109 ) = 167 MPa
Et1 is not valid above 60 MPa.
Third range:
σ cr = 0.0167(7.5 ×109 ) = 125 MPa
Et 2 does not apply above 75 MPa.
Fourth range:
σ cr = 0.0167(5 ×109 ) = 83.5 MPa
Et 3 is valid between 75 MPa and 85 MPa. Thus,
Pcr = 83.5 × 106 (3.75 ×10−3 ) = 313 kN
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________________________________________________________________________
PROBLEM (11.39) A structural steel column is fixed at its base and pinned at its top. The
column has a rectangular cross section of width b = 40 mm and depth h = 12 mm. Calculate the
buckling
load P cr for the following lengths:
(a) L =200 mm.
(b) L = 600 mm.
SOLUTION
From Table B.4:
E = 200 GPa
σ y = 250 MPa
The properties of area are
A = bh = (40)(10) = 400 mm 2 ,
r=
I
5760
=
= 3.795 mm ,
400
A
I=
1 3 1
bh = (40)(12)3 = 5760 mm 4
12
12
L
E
200 ×103
=π
= 88.9
( e )c = π
250
r
σy
(a) From Fig. 11.3c, Le = 0.7 L = 0.7(200) = 140 mm . Hence
Le
140
=
= 36.9
r 3.795
Since 36.9 < 88.9 , John Formula should be used. Thus:
σ y ( Le r ) 2
]
Pcr = Aσ y [1 −
4π 2 E
250(36.9) 2
] = 95.71 kN
= (400)(250)[1 − 2
4π × 200 ×103
(b) Now we have
Le 0.7(600)
=
= 110.7 > 88.9
3.795
r
Euler formula should used. So
π 2 EI π 2 (200 ×109 )(5760 ×10−12 )
Pcr = 2 =
= 64.5 kN
L2
(20) 2
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________________________________________________________________________
PROBLEM (*11.40) Determine the critical load P cr for an aluminum alloy 6061-T6 pipe
with pinned ends. The column has outside diameter d 0 = 75 mm, inside diameter d i = 72 mm,
and length L = 800 mm.
*SOLUTION
E = 70 GPa
A=
I=
π
4
π
σ y = 260 MPa
π
2
2
(d o2 − di2 ) =
(d o4 − di4 ) =
4
π
(from Table B.4)
(75 − 72 ) = 346.361 mm 2
(754 − 724 ) = 233.99(103 ) mm 2
64
64
I
233.99(103 )
r=
=
= 26 mm
346.36
A
So,
L 800
=
= 30.8
26
r
Equation (11.10):
70 ×103
L
E
=π
=π
= 51.5
260
σy
r
Since 30.8 < 51.5 , Johnson formula should be used.
σ ( L r )2
]
Pcr = Aσ y [1 − y 2
4π E
260(30.8) 2
] = 82 kN
= (346.361)(260)[1 − 2
4π (70 × 103 )
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________________________________________________________________________
PROBLEM (11.41) A structural steel column with pinned ends has a cross-sectional area of
width b = 3 in. by depth h = 4 in. Determine the shortest length L min that the column may have
using Euler’s formula.
Given: σy = 36 ksi, E = 29 x 10 6 psi (from Table B.4)
SOLUTION
The smallest moment of inertia is
1
1
I min = hb3 = (4)(3)3 = 9 in.4
12
12
and
I
9
rmin = min =
= 0.866 in.
A
3× 4
Euler’s formula (11.9) with σ y = σ cr :
L
Lmin
π 2 E π 2 (29 ×103 )
=
= 7,905.519 ,
( min ) 2 −
= 89.2
36
σ cr
r
r
Hence, Lmin = (89.2)(0.866) = 77.2 in.
Comment: If this column is 77.2 or more in length, it will be buckle.
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________________________________________________________________________
PROBLEM (11.42) Verify the specific Johnson’s formulas, Eqs. (11.13a) and (11.13b), for
the intermediate sizes of columns with round and rectangular cross sections, respectively.
SOLUTION
A solid circular section of diameter d:
A=
π
d2
I=
π
d4
4
64
Using Eq. (11.13a), we have
4 Pcr
1 σ y 4 Le 2
(
)
=
σ
−
y
πd2
E 2π d
Solving,
σ y2 L2e
P
d = 2( cr + 2 )1 2
πσ y π E
r=
I d
=
A 4
Q.E.D.
A rectangular section of height h and width b:
1 3
h2
2
A = bh
I = bh
r =
12
12
By Eq. (11.13a), we obtain
Pcr
1 σ 12 Le 2
)
=σy − ( y
bh
E 2π h
from which
Pcr
Q.E.D.
b=
3L2eσ y
hσ y (1 − 2 2 )
π Eh
It is assumed that h ≤ b .
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________________________________________________________________________
PROBLEM (11.43) An ASTM A-47 cast-iron pipe of outer diameter d and wall thickness t
is to serve as a pinned-end column of length L and is to support an eccentric load of P with a
safety factor n s (Fig. P11.43). Calculate the largest allowable eccentricity e.
Given: d = 10 in., t = ¾ in., L = 15 ft, P = 160 kips, n s = 2, E = 24 x 10 6 psi,
σy = 33 ksi
SOLUTION
P = ns (160) = 2(160) = 320 kips ,
I=
π
A=
π
4
(102 − 8.52 ) = 21.79 in.2
(104 − 8.54 ) = 234.6 in.4
64
From Example A.4:
1
1
r=
D2 + d 2 =
102 + 8.52 = 3.28 in.
4
4
We have
320
P
=
= 14.68 ksi
A 21.79
5e
ec
=
= 0.4648e
2
(3.28) 2
r
L 180
=
= 54.88
r 3.28
Substitute these into Eq. (11.20b):
⎡
⎛
14.68 × 103 ⎞ ⎤
33 = 14.68 ⎢1 + 0.4648e sec ⎜ 27.44
⎟⎥
6
⎜
⎟
24
10
×
⎢⎣
⎝
⎠ ⎥⎦
from which
e = 2.09 in.
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PROBLEM (11.44) An aluminum alloy 2014-T6 pipe of outside diameter d o , inner
diameter d i , and length L, both ends pinned, carries a compressive load P with eccentricity e
(Fig. P11.43). What is the maximum stress in the pipe?
Given: d o = 3 in., d i = 2.9 in., L = 35 in., e = 0.2 in., P = 1.5 kips
E = 10.6 x 10 6 ksi, σy = 60 ksi (from Table B.4)
SOLUTION
A=
I=
π
4
π
(d o2 − di2 ) =
(d o4 − di4 ) =
π
4
π
(32 − 2.92 ) = 0.463 in.4
(34 − 2.94 ) = 0.504 in.4 ,
64
64
6
σ y = 60 ksi
E = 10.6 ×10 psi
We have
1.5
P
=
= 3.24 ksi
A 0.463
ec (0.2)(1.5)
=
= 0.276
r2
(1.043) 2
Equation (11.15b) is thus
⎛ L
P ⎡ ec
σ max = ⎢1 + 2 sec ⎜⎜
A ⎣⎢ r
⎝ 2r
r=
I
0.504
=
= 1.043 in.
0.463
A
(from Table B.4)
35
L
=
= 32.14
r 1.089
P ⎞⎤
⎟⎥
AE ⎟⎠ ⎦⎥
⎡
⎛
⎞⎤
3.5
= (3.24) ⎢1 + 0.276sec ⎜⎜16.07
⎥
3 ⎟
(0.463)(10.6 × 10 ) ⎟⎠ ⎥⎦
⎢⎣
⎝
= 4.22 ksi
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PROBLEMS (*11.45) A structural ASTM-A36 steel tube has length L, outside diameter d o ,
thickness t, and both ends pinned. A compressive load P acts with an eccentricity e as shown
in Fig. P11.43. Calculate:
(a) The maximum stress in the tube.
(b) The allowable load based on a safety factor n with respect to material yielding.
Given: d = 70 mm, t = 5 mm, L = 2.5 m, e = 32.5 mm, P = 10 kN, n s = 3,
E = 210 GPa, σy = 250 MPa (from Table B.4)
*SOLUTION
di = d o − 2t = 70 − 2(5) = 60 mm
A=
I=
π
4
π
64
r=
(d o2 − di2 ) =
(d o4 − di4 ) =
π
4
(702 − 602 ) = 1021 mm 4
π
64
(704 − 604 ) = 542.4 × 103 mm 4
L 2.5 × 103
=
= 108.5
28.05
r
542.4(10)3
I
=
= 23.05 mm
1021
A
(a) Use Eq. (11.15b)
σ max =
⎛ L
P ⎡ ec
⎢1 + 2 sec ⎜⎜
A ⎣⎢ r
⎝ 2r
P ⎞⎤
⎟⎥
AE ⎟⎠ ⎦⎥
where
10 ×103
P
=
= 9.804 MPa ,
A 1.02 × 10−3
ec (32.5)(35)
=
= 2.141
r2
(23.05) 2
Thus
⎡
⎛
⎢⎣
= 32.3 MPa
⎜
⎝
σ max = (9.804) ⎢1 + (2.141) sec ⎜ 54.25
⎞⎤
10 × 103
⎥
⎟
(1021)(10−6 )(210 ×109 ) ⎠⎟ ⎥
⎦
(1)
(b) Apply Eq.(11.16)
σy =
⎛ L
Py ⎡ ec
⎢1 + 2 sec ⎜⎜
A ⎢⎣ r
⎝ 2r
P ⎞⎤
⎟⎥
AE ⎟⎠ ⎥⎦
⎡
⎛
⎞⎤
Py
⎢
1
(2.141)
sec
54.25
+
⎜
⎟⎥
3 ⎟
⎜
1021(10−16 ) ⎢
(1021)(210
10
)
×
⎝
⎠ ⎥⎦
⎣
Solving numerically: Py = 57.92 kN . Equation (11.22) gives then
Py
250(104 ) =
Pall =
Py
ns
=
(2)
57.92
= 19.31 kN
3
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PROBLEM (11.46) Redo Prob. 11.45 for the case in which the tube fixed at its base and
free at its top, and has a factor of safety of n s = 2 (Fig. P11.43).
SOLUTION
Refer to Solution of Prob. 11.45. Here
Le = 2 L = 2(2.5) = 5 m (See Fig. 11.3a)
Le 5 × 103
=
= 217
r
23.05
Le
= 108.5
2r
(a) Equation (1) yields
⎡
⎛
σ max = (9.794) ⎢1 + (2.141) sec ⎜108.5
⎜
⎢⎣
⎝
= 38.2 kN
10 ×103
(1021)(210 × 103 )
⎞⎤
⎟⎥
⎟
⎠ ⎥⎦
(b) Equation (2) gives
⎡
⎛
⎞⎤
Py
⎢
1
(2.141)
sec
108.5
+
⎜
⎟⎥
3 ⎟
⎜
1021(10−16 ) ⎢
(1021)(210
10
)
×
⎝
⎠ ⎥⎦
⎣
Solving numerically, Py = 29.94 . Equation (11.22):
Py
250(106 ) =
Pall =
Py
ns
=
29.94
= 14.79 kN
2
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PROBLEM (11.47) Rework Prob. 11.45 for tube with fixed-pinned ends and a factor of
safety of n s = 1.5.
SOLUTION
Refer to Solution of Prob. 11.45. We now have
Le = 0.7 L = 0.7(2.5) = 1.75 m (See Fig. 11.3c)
Le 1.75 ×103
=
= 75.92
r
23.05
Le
= 37.94
2r
(a) Equation (1) becomes
⎡
⎛
σ max = (9.794) ⎢1 + (2.141) sec ⎜ 37.96
⎜
⎢⎣
⎝
10 × 103
(1021)(210 × 103 )
⎞⎤
⎟ ⎥ = 31.5 MPa
⎟
⎠ ⎥⎦
(b) Equation (2) Leads to
⎡
⎛
⎞⎤
Py
⎢
1
(2.141)
sec
37.96
+
⎜
⎟⎥
3 ⎟
⎜
1021(10−16 ) ⎢
(1021)(210
10
)
×
⎝
⎠ ⎥⎦
⎣
Solving numerically, Py = 68.1 kN . Equation (11.22) results in
Py
250(106 ) =
Pall =
Py
ns
=
68.1
= 45.4 kN
1.5
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PROBLEM (11.48) A pin-ended rod AB of diameter d carries an eccentrically applied
load of P as shown in Fig. P11.48. If the maximum deflection at the midlength is v max ,
calculate:
(a) The eccentricity e.
(b) The maximum stress in the rod.
Given: d = 40 mm, P = 60 kN, v max = 0.6 mm, E = 200 GPa
SOLUTION
P
A = π (20) 2 = 1256.64 mm 2
A
I=
40 mm
vmax
π
4
Pcr =
0.7 m
B
e
Figure (a)
(20) 4 = 125.66 ×103 mm 4
π 2 EI
L2
π 2 (200 × 109 )(125.66 ×10−9 )
=
(0.7) 2
= 506.2 kN
P
(a) Using Eq. (11.18):
⎡ ⎛π
60 ⎞ ⎤
o
0.6 ×10−3 = e ⎢sec ⎜⎜
⎟⎟ − 1⎥ = e ⎡⎣sec(31 ) − 1⎤⎦ ,
⎣⎢ ⎝ 2 506.2 ⎠ ⎦⎥
e = 3.6 mm
(b) Referring to Fig. (a):
M = P (vmax + e) = 60(0.6 + 3.6) = 252 N ⋅ m
P Mc
Hence, σ max = +
A I
60 ×103
252(20)10−3
= 87.9 MPa
=
+
1256.64 ×10−6 125.66 ×10−9
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PROBLEM (11.49) Resolve Prob. 11.48, assuming that ends B and A of column AB are
fixed and free, respectively.
SOLUTION
Le = 2 L = 1.4 m . Refer to solution of Prob. 11.48
Pcr =
π 2 EI
L2e
=
π 2 (200 ×109 )(125.66 ×10−9 )
(1.4) 2
= 126.6 kN
(a) Equation (11.18):
⎡ ⎛π
60 ⎞ ⎤
o
0.6 ×10−3 = e ⎢sec ⎜⎜
⎟⎟ − 1⎥ = e[sec(62 ) − 1] ,
⎢⎣ ⎝ 2 126.6 ⎠ ⎥⎦
e = 0.53 mm
(b) M = P(vmax + e) = 60(0.6 + 0.53) = 67.8 N ⋅ m . Therefore,
P Mc
σ max = +
A I
67.8(20 × 10−3 )
= 47.75 + 10.8 = 58.6 MPa
= 47.75 +
125.66 ×10−9
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PROBLEM (11.50) A 50-mm-diameter steel pinned-end column is loaded as shown in Fig.
P11.48. For a factor of safety of n s with respect to yielding, use Fig. 11.16a to determine
allowable load P all , for eccentricities of:
(a) e = 5 mm.
(b) e = 3 mm.
Given: E = 200 GPa, σy = 275 MPa,
ns = 3
SOLUTION
I=
π
4
(25) 4 = 306.8 × 103 mm 4 ,
r = d 4 = 50 4 = 12.5 mm
A = π (25) 2 = 1963.5 mm 2
(a) e=5 mm
ec 5 × 25
=
= 0.8
r 2 (12.5) 2
From Fig. 11.16a: Py A = 133 MPa. So,
Py = 133 ×106 (1963.5 ×10−6 ) = 261 kN
Pall =
Py
ns
=
261
= 87 kN
3
(b) e=3 mm
ec 3 × 25
=
= 0.48
r 2 (12.5) 2
Figure 11.16a:
Py A = 163 MPa. Thus,
Py = 163 ×103 (1963.5 × 10−6 ) = 320 kN
Pall =
320
= 106.7 kN
3
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PROBLEM (11.51) Redo Prob. 11.50, this time applying Eq. (11.21).
SOLUTION
See solution of Prob. 11.50:
(a) A = 1963.5 × 10−6 m 2
Le
ec
= 0.8
= 56
2
r
r
Equation (11.21) becomes
⎛
⎞⎤
Py ⎡
Py
275 × 106 = ⎢1 + 0.8sec ⎜ 28
⎟⎥
⎜
200 × 109 A ⎟⎠ ⎥
A⎢
⎝
⎣
⎦
Solving by trial and error: Py A = 133 MPa
Then
Py = 133 ×106 (1963.5 × 10−6 ) = 261 kN
So,
Pall = 261 3 = 87 kN
(b)
Le
ec
= 0.48
= 56
2
r
r
Equation (11.21) becomes
⎛
⎞⎤
Py ⎡
Py
275 ×106 = ⎢1 + 0.48sec ⎜ 28
⎟⎥
9
⎜
⎟
200
10
×
A⎢
A
⎝
⎠ ⎥⎦
⎣
Solving by trial and error: Py A = 163 MPa
Py = 163 ×106 (1963.5 × 10−6 ) = 320 kN
and
Pall = 320 3 = 106.7 kN
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PROBLEM (11.52) Use Fig. 11.16a to determine the maximum load P that can be applied
at both ends of an S 310 x 52 steel link (see Table B.9), given an eccentricity e from the center of
the section measured along the web as shown in Fig. P11.52. Check the value found by
applying Eq. (11.21).
Assumption: σy = 250 MPa,
E = 200 GPa,
L =5m
SOLUTION
From Table B.9, for S310x52:
A = 6.64 × 103 mm 2
I = 95.3 ×106 mm 4
S = 625 ×103 mm3
r = 120 mm
We have
ec eA 0.06 × 6.64 × 10−3
=
=
= 0.64 ,
r2 S
625 ×10−6
So,
Py A = 155 MPa (Fig. 11.16a)
Le
5
=
= 41.7
r 0.12
or
Py = 155 ×106 (6.64 × 10−3 ) = 1029 kN
Check : Equation (11.21) becomes
⎛
⎞⎤
Py ⎡
Py
⎥
275 ×106 = ⎢1 + 0.64sec ⎜ 20.85
⎟
9
⎜
⎟⎥
200
10
×
A⎢
A
⎝
⎠⎦
⎣
Solving by trial and error: Py A = 155.5 MPa
Then
Py = 155.5 ×106 (6.64 × 10−3 ) = 1032 kN
It differs 3 kN from that of approximate solution.
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PROBLEM (11.53) An aluminum hollow box column of length L is fixed at its base and
free at its top as shown in Fig. P11.53. If an eccentric load P acts at the middle of side AB (that
is, e = 50 mm) of the free end, calculate the largest stress σmax in the column.
Given: L = 2 m, E = 70 GPa, P = 150 kN, e = 50 mm
SOLUTION
A
100 mmC
20 mm
Le = 2 L = 4 m
A = 200 × 100 − 160 × 60
= 10.4 × 103 mm 2
P
200 mm
D
B
P
L
1
(200 ×1003 − 160 × 603 )
12
= 13.79 × 106 mm 4
I=
r = I A = 36.4 mm
e = c = 50 mm
Thus,
ec 50 × 50
=
= 1.89
r 2 (36.4) 2
Le
= 54.95
2r
Use Eq.(13.20b) with L = Le :
⎛
⎞⎤
150 ×103 ⎡
150 ×103
⎢
+
σ max =
1
1.89sec
54.95
⎜
⎟ ⎥ = 53.1 MPa
⎜
10.4 × 10−3 ⎢
70 × 109 (10.4 × 10−3 ) ⎟⎠ ⎥
⎝
⎣
⎦
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PROBLEM (11.54) Rework Prob. 11.53, assuming that the load acts at the middle of side
AC (that is, e = 100 mm).
SOLUTION
C
200 mm
P
A
D
100 mm
B
20 mm
P
Le = 2 L = 4 m
A = 200 × 100 − 160 × 60
= 10.4 × 103 mm 2
1
(100 × 2003 − 60 ×1603 )
12
= 46.19 × 106 mm 4
I=
e = c = 100 mm
r = I A = 66.6 mm
Therefore,
ec 100 × 100
=
= 2.25
r2
(66.6) 2
Le
= 30.03
2r
Apply Eq.(11.20b) with L = Le :
σ max =
⎛
⎞⎤
15 × 103 ⎡
150 ×103
⎢
+
1
2.25sec
30.03
⎜
⎟ ⎥ = 5.01 MPa
−3 ⎟
9
⎜
×
10.4 × 10−3 ⎢
70
10
(10.4
×
10
)
⎝
⎠ ⎥⎦
⎣
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PROBLEM (*11.55) An aluminum alloy 2014-T6 hollow box column of square cross section and
length L is fixed at its base and free at its top as shown in Fig. P11.55. The column supports a
compressive load P that acts with an eccentricity of e. Determine:
(a) The horizontal deflection of the top of the column.
(b) The maximum stress in the column.
Given: a = 5 in., b = 4.2 in.,
e = 2.3 in.,
L = 6 ft,
P = 40 kips, E = 10.6 x 106 psi,
σy = 60 ksi (from Table B.4)
*SOLUTION
We have
A = a 2 − b 2 = 52 − 4.22 = 7.36 in.2
1
1
I = (a 4 − b 4 ) = (54 − 4.24 ) = 26.153 in.4
12
12
I
26.153
r=
=
= 1.885 in.
7.36
A
a
c = = 2.5 in.
Le = 2 L = 2(6 × 12) = 144 in.
2
Hence
P
40
ec (2.3)(2.5)
=
= 5.435 ksi
=
= 1.618 ,
2
2
A 7.86
r
(1.885)
Pcr =
π 2 EI
=
π 2 (10.6 ×106 )(26.153)
L2e
(144) 2
40
P
=
= 0.303
Pcr 131.9
= 131.9 kips
(a) Apply Eq. (11.18):
π P
vmax = e[sec(
) − 1]
2 Pcr
= (2.3)[sec(
π
2
0.303 − 1)] = 2.32 in.
(b) Use Eq.(11.20a)
π
P
ec
σ max = [1 + 2 sec(
2
A
r
P
)]
Pcr
= (5.435)[1 + (1.618) sec(
π
2
0.303)] = 18.99 ksi
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PROBLEM (*11.56) The compressive load P acts with an eccentricity e on the structural steel
tubular column shown in Fig. P11.55. If the horizontal deflection at the top due to loading is vmax ,
calculate:
(a) The eccentricity e.
(b) The maximum stress in the column.
Given: a = 100 mm,
b = 80 mm,
L =1.8 m,
P = 200 kN, vmax = 20 mm, E = 200 GPa
σy = 250 MPa (from Table B.4)
*SOLUTION
A = a 2 − b 2 = 1002 − 802 = 3600 mm 2
1
1
I = (a 4 − b 4 ) = (1004 − 804 ) = 4.92 × 106 mm 4
12
12
Le = 2( L) = 2(1.8) = 3.6 m
c = 50 mm
Pcr =
π 2 EI
=
π 2 (200 ×109 )(4.92 × 10−6 )
L2e
200
P
=
= 0.267
Pcr 749.4
(3.6) 2
= 749.4 kN
(a) Equation (11.18):
π P
vmax = e[sec(
) − 1]
2 Pcr
20 = e[sec(
π
2
0.267) − 1] ,
e = 19.6 mm
(b) M max = P(e + vmax ) = 200(19.6 + 20)10−3 = 7.92 kN ⋅ m
σ max =
P M max c 200 × 103 7.92 × 103 (0.05)
+
=
+
= 86 MPa
A
I
36 ×10−3
4.92(10−6 )
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PROBLEM (11.57) An S 6 x 12.5 steel column (see Table B.9) with pinned ends carries an axial
load P = 20 kips. What is the longest allowable column length?
Given: E = 29 x 10 6 psi, σy = 36 ksi.
SOLUTION
S6x12.5 : A = 3.67 in.2 ,
rmin = ry = 0.705 in.
σ all = 20 × 103 3.67 = 5.45 ksi
(1)
2π 2 (29 × 106 )
= 126.1
36 × 103
Assuming L r ≥ Cc , Use Eq. (11.25) and (1):
Cc =
σ all =
π 2 (29 ×106 )
2
= 5.45 × 103 ,
L
= 165.4
r
1.92( L r )
Our assumption was correct. Thus,
L
L
L = 116.7 in.
=
= 165.4 ,
ry 0.705
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PROBLEM (11.58) Redo Prob. 11.57 for a W 6 x 20 steel column (see Table B.8).
SOLUTION
W6x16: A = 4.74 in.2
rmin = ry = 0.967 in.
σ all =
P 20 × 103
=
= 4.219 ksi
A
4.74
Cc =
2π 2 (29 × 106 )
= 126.1
36 × 103
(1)
Assuming L r ≥ Cc , use Eq. (11.25) and (1):
σ all =
π 2 (29 ×106 )
2
= 4.219 ×103 ,
L
= 188
r
1.92( L r )
Our assumption was correct. Hence,
L = (188)ry = (188)(0.967) = 181.8 in.
or
L = 181.8 in.
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PROBLEM (11.59) Determine the allowable axial load P for a W 250 x 80 steel column
(see Table B.8) for each of three effective lengths:
(a) L e = 2 m.
(b) L e = 5 m.
(c) L e = 10 m.
Given: E = 210 GPa, σy = 250 MPa.
SOLUTION
WF250x80: A = 10.2 × 103 mm 2
Cc =
2π 2 E
σ yp
=
rmin = ry = 65 mm
2π 2 (210 × 109 )
= 128.8
250 × 103
(a) Le = 2 m
Le ry = 2000 65 = 30.77
Since L ry < Cc , use Eq. (11.24):
5 3 30.77 1 30.77 3
ns = +
− (
) = 1.75
3 8 128.8 8 128.8
250 ×106
(30.77) 2
[1 −
] = 138.8 MPa
σ all =
1.75
2(128.8) 2
Thus,
Pall = 138.8 × 106 (10.2 ×10−3 ) = 1416 kN
(b) Le = 5 m
Le ry = 5000 65 = 76.92
Since L ry < Cc , use Eq. (11.24):
5 3 76.96 1 76.96 3
ns = +
− (
) = 1.87
3 8 128.8 8 128.8
250 ×106
(76.96) 2
[1 −
] = 109.8 MPa
σ all =
1.87
2(128.8) 2
Therefore,
Pall = 109.8 ×106 (10.2 × 10−3 ) = 1120 kN
Continued on next slide
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(c) Le = 10 m
Le ry = 10 ×103 65 = 153.85 > Cc
Use Eq. (11.25):
π 2 (250 ×109 )
σ all =
= 45.61 MPa
1.92(153.85) 2
Thus,
Pall = 45.61× 106 (10.2 × 10−3 ) = 465.2 kN
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PROBLEM (*11.60) Determine the allowable axial load P for a W 14 x 145 steel
column (see Table B.8) with built-in ends and a length of 36 ft, braced at midpoint C (Fig.
P11.60).
Given: E = 30 x 10 6 psi,
σy = 40 ksi
*SOLUTION
P
W14x145:
A = 42.7 in.4 ,
18 ft
P
18 ft
=
Le=0.7L
r = 3.98 in.
2π 2 (30 × 106 )
= 121.7
40 × 103
Le 151.2
=
= 37.99
r
3.98
Cc =
Since L ry < Cc , use Eq. (11.24):
5 3 37.99 1 37.99 3
ns = +
− (
) = 1.78
3 8 121.7 8 121.7
40 ×103
1 37.99 3
σ all =
[1 − (
) ] = 22.13 ksi
1.78
2 121.7
So,
Pall = 22.13(42.7) = 945 kips
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PROBLEM (11.61) Redo Prob. 11.60 for an S 20 x 75 steel column (.see Table B.9).
SOLUTION
r = 1.16 in.
S20x75 : A = 22.1 in.2
From solution of Prob. 11.60:
Cc = 121.7
L2 = 151.2 in.
We now have
Le r = 151.2 1.16 = 130.3 > Cc
Apply Eq. (11.25)
π 2 (30 ×106 )
σ all =
= 9.08 ksi
1.92(130.3) 2
and
Pall = 9.08 × 22.1 = 201 kips
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PROBLEM (11.62) Rework Prob. 11.59 for a steel pipe column of outer diameter 160 mm
and inner diameter 140 mm.
SOLUTION
A=
π
4
(1602 − 1402 ) = 4712.4 mm 2
From Example A.4:
1
1
r=
D2 + d 2 =
1602 + 1402 = 53.2 mm
4
4
From solution of Prob. 11.59: Cc = 128.8
(a) Le = 2 m
Le r = 37.59 < Cc
Equation (11.24):
5 3 37.59 1 37.59 3
ns = +
− (
) = 1.77
3 8 128.8 8 128.8
250 × 106
1 37.59 2
σ all =
[1 − (
) ] = 135.2 MPa
1.77
2 128.8
Thus,
Pall = 135.2 × 106 (4712.4 × 10−6 ) = 637 kN
(b) Le = 5 m
Le r =
5 × 103
= 93.98 < Cc
53.2
Use Eq. (11.24):
5 3 93.98 1 98.98 3
ns = +
− (
) = 1.89
3 8 128.8 8 128.8
250 ×106
1 93.98 2
σ all =
[1 − (
) ] = 97.1 MPa
1.89
2 128.8
Hence,
Pall = 97.1× 106 (4712.4 × 10−6 ) = 457 kN
Continued on next slide
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(c) Le = 10 m
10 × 103
= 187.97 > Cc
53.2
Use Eq. (11.25):
π 2 (210 ×109 )
σ all =
= 30.55 MPa
1.92(187.97) 2
So,
Pall = 30.55 × 106 (4712.4 × 10−6 ) = 144 kN
Le r =
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PROBLEM (11.63) Determine the smallest allowable diameter d of a steel rod which may
be used to support an axial load P = 40 kips for L e = 12 ft.
Given: E = 29 x 10 6 psi, σy = 36 ksi
SOLUTION
A =πd2 4
r = I A = d 4,
Cc =
2π 2 (29 × 106 )
= 126.1
36 × 103
Le = 12 ft = 144 in. Assume Le r > Cc
and use Eq. (11.25):
P π 2 (29 ×106 )
40 ×103
π 2 (29 × 106 )
;
=
σ all = =
A 1.92( Le r ) 2
π d 2 4 1.92(144 × 4 d ) 2
from which
1.92(160)(144 × 4) 2
,
d = 3.26 in.
d4 =
π 3 (29 ×103 )
Slenderness ratio is
Le
144
=
= 176.7 > 126.1
r 3.26 4
Our assumption was correct and
d = 3.26 in.
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PROBLEM (11.64) Select the lightest wide-flange steel shape (see Table B.7) to support an
axial load of P = 1150 kN on an effective length of Le =4 m.
Given: E = 200 GPa, σy = 250 MPa .
SOLUTION
See Example 11.12: Cc = 126 .
1150 ×103
P
A=
= 115 MPa,
= 0.01 m 2
6
A
115 ×10
Try (Table B.8): W250x80.
We have
L rmin = 4 × 103 65 = 61.54 < Cc
Use Eq. (11.24):
5 3 61.54 1 61.54 3
ns = +
− (
) = 1.84
3 8 126 8 126
250 × 106
1 61.54 2
σ all =
[1 − (
) ] = 119.7 MPa
1.84
2 126
Pall = 119.7 × 106 (10.2 × 10−3 )
= 1221 kN > 1150 kN O.K.
Try:
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PROBLEM (11.65) A square aluminum alloy 6061-T6 column of length L = 0.6 m is
pinned at both ends (Fig. P11.65). What is the smallest allowable width a if the member is to
support an axial load P = 250 kN?
SOLUTION
A = a2
a
r= I A=
a
2 3
I = a 12
L
0.6
2.078
=
=
r a 2 3
a
4
a
Assume: 9.5 < Le r < 66.
Using Eq. (11.28):
250 × 103
2.078
σ all =
= [140 − 0.87(
)]106
2
a
a
a 2 − 12.91×10−3 a − 1.786 ×10−3 = 0 ,
a = 49.21 mm .
or
So,
r = a 2 3 = 49.21 2 3 = 14.2 mm
L r = 600 14.2 = 42.3 < 66
Our assumption was correct. Use a = 49.2 mm
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PROBLEM (11.66) Redo Prob. 11.65, assuming that the column is pinned at one end and
fixed at the other.
SOLUTION
Le r = 0.42 (a 2 3) = 1.455 a
Le = 0.7 L = 420 mm ,
Assume: 9.5 < Le r < 66. Equation (13.21):
σ all =
250 ×103
1.455
)]106
= [140 − 0.87(
2
a
a
or
a 2 − 9.04 ×10−3 a − 1.786 ×10−3 = 0
Solving: a = 47.02 mm. So,
r = a 2 3 = 13.57 mm
Le r = 420 13.57 = 30.95 < 66
Our assumption was correct. Use a = 47 mm
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PROBLEM (11.67) A rectangular aluminum alloy 6061-T6 tube of the cross section
shown in Fig. P11.67 is used as a column of 5-m effective length. What is the largest permissible
axial load P?
SOLUTION
A = 200 × 100 − 160 × 60 = 10.4 × 103 mm 2
1
I min = (200 × 1003 − 160 × 603 ) = 13.79 × 106 mm 4
12
Le rmin = 5 × 103 36.4 = 137.4
rmin = I min A = 36.4 mm ,
Use Eq. (11.29):
350 ×109
σ all =
= 18.54 MPa
(137.4) 2
Pall = 18.54 × 106 (10.4 ×10−3 ) = 192.8 kN
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________________________________________________________________________
PROBLEM (11.68) Determine the allowable axial load P for an aluminum alloy 6061-T6
pipe of outer diameter 10 in. and inner diameter 8 ½ in. The pipe is used as a column of effective
length L e = 12 ft.
SOLUTION
A=
π
(102 − 8.52 ) = 21.79 in.2
Le = 12 ft = 144 in.
4
From Example A.4:
1
1
r=
D2 + d 2 =
102 + 8.52 = 3.28 in.
4
4
Le r = 144 3.28 = 43.89
Use Eq. (11.28):
σ all = [20.2 − 0.126(43.89)]103 = 14.67 ksi
Hence
Pall = 14.67(21.79) = 320 kips
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PROBLEM (11.69) Redo Prob. 11.68, given that the effective length of the column is
increased to L e = 20 ft.
SOLUTION
See solution of Prob. 11.68:
A = 21.79 in.2
r = 3.28 in.
Thus,
Le r = 20 × 12 3.28 = 73.17
Use Eq. (11.29):
51, 000 ×103
σ all =
= 9.53 ksi
(73.17) 2
and
Pall = 9.53(21.79) = 207.6 kips
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PROBLEM (11.70) A square fixed-pinned end aluminum alloy 6061-T6 column of width
a supports a compression load P. Determine the permissible length of the bar.
Given: a = 2.2 in., P = 40 kips
SOLUTION
A = a2
I=
a4
12
r=
2.2
I
a
=
=
= 0.6351 in.
A 2 3 2 3
P 40(103 )
=
= 8.26 ksi
A (2.2) 2
Assume L r ≥ 66 and use Eq. (11.29)
σ all =
σ all =
51000(103 )
= 8.26(103 )
2
(L r)
Solving
L
= 78.58 > 66
r
(O.K.)
Hence
L = 78.57(0.6351) = 49.9 in.
________________________________________________________________________
PROBLEM (11.71) A solid round aluminum strut (alloy 6061-T6) of length L and pinned
ends supports an axial load P. Calculate the minimum permissible diameter d of the member.
Given: L = 400 mm, P = 60 kN
SOLUTION
A=
πd2
4
I=
πd4
64
r=
I d
=
A 4
L
≤ 66 and use Eq.(11.28).
r
0.6
60(103 )
σ all = [140 − 0.87(
)] =
d 4
πd2 4
2 0.0764
,
140d 2 − 1.74d − 0.0764 = 0
140 − (0.87) =
2
d
d
Assume 9.5 <
or
1.74 ± (1.74) 2 + 4(140)(0.0764)
d=
= 0.0304 m = 30.4 mm
2(140)
Since
d 30.4
=
= 7.6 mm
4
4
Assumption is O.K.
r=
L 400
=
= 52.6 < 66
r 7.6
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PROBLEM (11.72) A timber column is fixed at its base and free at its top as shown in Fig.
P11.72. What is the largest allowable axial load P that it can support?
Given: L = 5 ft, E = 1.8 x 10 6 psi
SOLUTION
For the situation described Le = 2 L (see Fig. 11.3a) and d=3 in. So,
Le 2(5 × 12)
=
= 40
d
3
Since 40 > 26, use Eq. (11.32).
0.3(1.8 × 106 )
σ all =
= 0.338 ksi
(40) 2
and
Pall = σ all A = (0.338)(3 × 4) = 4.05 kips
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PROBLEM (11.73) Rework Prob. 11.72 for the case in which the length of the column is L = 3 ft.
SOLUTION
We now have Le = 2 L = 2(3 × 12) = 72 in. and d=3 in.
Le 72
=
= 24
d
3
Since 24 < 26, apply Eq. (11.31).
1 24
σ all = 1.2[1 − ( ) 2 ] = 0.859 ksi
3 26
Thus
Pall = σ all A = (0.859)(3 × 4) = 10.31 kips
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PROBLEM (11.74) Select the minimum width a of a square timber column of effective
length L e = 4 m if it must support an axial load P = 200 kN.
Given:
E = 12 GPa, σall = 9 MPa
SOLUTION
d=a
Use Eq. (11.32):
P
0.3E
σ all = =
A ( Le d ) 2
or
200 ×103 0.3(12 ×109 )
=
a2
(4 a ) 2
Solving,
a 4 = 16 1.8 ×104 ,
a = 172.7 mm
Check :
200 × 103
σ=
= 6.71 MPa < 9 MPa
(0.1727) 2
Use
a = 173 mm
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PROBLEM (11.75) Determine the allowable axial load P for a built-up timber column of the cross
section shown in Fig. P13.75. The effective length is Le = 15 ft.
Given: E = 1.6 x 10 6 psi, σall = 2 ksi
SOLUTION
A = 102 − 7 2 = 51 in.2
Le = 15 ×12 = 180 in. ,
Use Eq. (11.32) :
0.3(1.6 × 106 )
σ all =
= 1.48 ksi < 2 ksi
(18) 2
Thus,
Pall = 1.48(51) = 75.5 kips
d = 10 in.2
Le d = 18
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PROBLEM (11.76) Calculate the allowable axial load P for a 100-mm by 200-mm
rectangular timber column if the effective length is:
(a) L e = 2 m.
(b) L e = 4 m.
Given: E = 11 GPa, σall = 9 MPa
SOLUTION
A = 100 × 200 = 20 × 103 mm 2 ,
(a) Le = 2 m
d = 100 mm
Le d = 20
Equation (11.32) :
0.3(11× 109 )
σ all =
= 8.25 MPa < 9 MPa
(20) 2
Hence,
Pall = 8.25 ×106 (20 × 10−3 ) = 165 kN
(b) Le = 4 m
Le d = 40
Equation (11.32) :
0.3(11×109 )
σ all =
= 2.06 MPa < 9MPa
(40) 2
And Pall = 2.06 × 106 (20 ×10−3 ) = 41.2 kN
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PROBLEM (*11.77) Determine the largest permissible length L for a timber column
shown in Fig. P11.77.
Given: E = 10.4 GPa, P = 15 kN
Assumption: The board is taken to be pin-supported at its ends.
*SOLUTION
Assume that Eq.(11.32) applies.
P
0.3E
σ all =
=
Le = L
2
( Le d )
A
or
0.3(10.4)(109 )
15(103 )
=
( L 0.035) 2
(0.035)(0.14)
Solving
L = 1.117 m = 1,117 mm
We have
L 1,117
=
= 31.9
d
35
Check: 26 ≤ L d ≤ 50, the solution is valid.
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PROBLEM (11.78) Redo Prob. 11.77 if the column is fixed at its base and free at its top.
SOLUTION
We now have Le = 2 L (see Fig. 11.3a). Refer to Solution of Prob. 11.77.
P
0.3E
σ all =
=
2
A
( Le d )
or
0.3(10.4)(109 )
15(103 )
=
,
L = 0.559 m = 559 mm
(2 L 0.035) 2
(0.035)(0.14)
So,
Le 2(559)
=
= 31.9
d
35
Check: 26 ≤ L d ≤ 50, the solution is valid.
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PROBLEM (*11.79) An aluminum alloy 6061-T6 compression member AB is fixed at its
base and free at its top as shown in Fig. P11.79. Using the interaction formula with (σall ) b = 140
MPa, Calculate the largest permissible eccentric load P.
*SOLUTION
We have Le = 2 L
(from Fig. 11.3a).
A = bh = 50(100) = 5000 mm 2
The smallest moment of inertia is
1
1
I = bh3 = (100)(50)3 = 1.042 ×106 mm 4
12
12
1.042 × 103
I
r=
=
= 14.44
5
A
The largest slenderness ratio is thus
Le 2(1800)
=
= 249.3
r
14.44
Assume Eq.(11.29) applies (249.3 > 66).
350 ×103
350(109 )
6
=
= 5.63 MPa
(σ all )c =
(10
)
( Le r ) 2
(249.3)3
Also
P
P
=
= 200 P
A 5 × 10−3
Mc
Pec
P (0.025)(0.05)
= 3
=
= 300 P
I
bh 12 (0.05)(0.1)3 12
Equation (11.34) becomes
M I
P A
+ c ≤ 1;
(σ all )c (σ all )b
or
P = 26.5 kN
200 P
300 P
+
≤1
6
5.63 ×10 140 × 106
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PROBLEM (11.80) A column of L e = 4-m effective length and 100-mm-square cross
section is made of timber for which E w = 12 GPa and (σall ) b = 10 MPa for compression parallel
the grain. Using the interaction method, calculate the maximum load P that may be safely
supported with an eccentricity of e = 40 mm.
SOLUTION
(σ all )c =
0.3E
0.3(12 ×109 )
= 2.25 MPa
=
( Le d ) 2
(4 0.1) 2
A = 100 × 100 = 1× 104 mm 2
(σ all )b = 10 MPa ,
1
I = (100) 4 = 8.333 × 106 mm 4 ,
c = 50 mm
M = 0.04 P
12
Substitute these into Eq.(11.34):
P 1×10−2 0.04 P(0.05) 8.333 ×10−6
+
≤1
2.25 × 106
10 ×106
or
(0.444 P + 0.24 P)10−4 ≤ 1 ,
P ≤ 14.62 kN
So,
Pmax = 14.62 kN
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PROBLEM (11.81) Redo Prob. 11.80 for the case in which the column is made of steel.
Given: (σall )b = 160 MPa
σy = 250 MPa
E = 200 GPa
SOLUTION
From solution of Prob. 11.80 :
I = 8.333 × 106 mm 4
c = 50 mm ,
Therefore,
r=
A = 1× 104 mm 2
8.333 × 10−6
= 28.9 mm ,
1×10−2
M = 0.04 P
Le r = 4 × 103 28.9 = 138.4
2π 2 (200 × 109 )
= 125.7
250 × 106
Thus, Le r > Cc and we use Eq. (11.25) :
Cc =
(σ all )c =
π 2E
1.92( Le r )
=
2
π 2 (200 ×109 )
1.92(138.4) 2
= 53.67 MPa
Then
P A Mc I
+
≤1
(σ all )c (σ all )
Becomes
P 1×10−2 0.04 P(0.05) 8.333 ×10−6
+
≤ 1,
53.67 × 106
160 ×106
Therefore
Pmax = 297 kN
P ≤ 297 kN
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PROBLEM (11.82) Solve Prob.11.80 for the case in which the column is made of
aluminum alloy 6061-T6 and the allowable stress in bending is 150 MPa.
SOLUTION
E = 72 GPa
6071-T6 (Table B.4) : σ y = 410 MPa
See solution of Prob. 11.81:
Le r = 138.4
I = 8.333 ×106 mm 4 ,
A = 1×104 mm 2
Equation (11.29) gives
350 ×109
(σ all )c =
= 18.27 MPa
(138.4) 2
Then,
P A
Mc I
+
≤1
(σ all )c (σ all )b
P 1× 10−2 0.04 P(0.05) 8.333 ×10−6
+
≤ 1,
18.27 ×106
150 × 106
P ≤ 141.4 kN
and
Pmax = 141.4 kN
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PROBLEM (*11.83) A load of P = 10 kips is applied with an eccentricity of e = 1 ½ in. to
a timber post (Fig. P11.83). Let E = 1.7 x 10 6 psi , ( σall ) b = 1.2 ksi for compression parallel
to the grain, and L e = 5 ft. Employing the interaction method, determine smallest allowable
diameter d of the post.
*SOLUTION
Le = 0.7 L = 0.7(5 × 12) = 42 in. ,
d = 2c
Assume Le d ≤ 50 and use Eq. (11.32) :
(σ all )c =
0.3(1.7 ×106 )
= 1.156 × 103 c 2
2
(42 2c)
Then,
P A
Mc I
+
≤1
(σ all )c (σ all )b
10 × 103 π c 2 1.5 ×10 × 103 c (π c 4 4)
+
≤ 1;
1.156 ×103 c 2
1.2 ×103
2.754 15.92
+ 3 ≤1
c4
c
Solving by trial and error : c = 2.56 in.
d = 5.12 in.
Check :
Le d = 42 5.12 = 8.2 < 50 O.K.
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PROBLEM (11.84) An S 4 x 7.7 rolled steel (see Table B.8) column of L e = 8 ft
effective length supports an eccentric load P applied e = 1 ¾ in. from the center of the section
measured along the web (Fig. P11.84). Using the interaction formula, determine the maximum
safe value of the load P.
Given: E = 30 x 10 6 psi, σy = 36 ksi, ( σall ) b = 25 ksi
SOLUTION
Cc =
y
1 3 4 in.
P
2π 2 E
σy
2π 2 (30 × 106 )
= 128.3
36 × 103
A = 2.26 in.2
S z = 3.04 in.3
rmin = ry = 0.581 in.
c = 2 in.
=
z
S 4x7.7
I z = 6.08 in.4
We have
Le r = 96 0.581 = 165.2 > Cc
Equation (11.25):
(σ all )c =
π 2E
1.92( Le r )
=
2
π 2 (30 ×106 )
1.92(165.2) 2
= 5.65 ksi
Then, we use
P A
Mc I
+
≤1
(σ all )c (σ all )b
Substituting the numerical values:
1.75 P(2) 6.08
P 2.26
+
≤1
6
18.27 ×10
25 × 103
P (0.078 + 0.023)10−3 ≤ 1 ,
P ≤ 9.9 kips
Therefore,
Pmax = 9.9 kips
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PROBLEM (11.85) An aluminum alloy 6061-T6 rod of diameter d supports a load P with
an eccentricity e (see Fig. P11.83). Employing the interaction method, with (σall ) b = 22 ksi,
determine the longest length L that may be used.
Given: d = 2.5 in., P = 8 kips, e = 1in. ( σall ) b = 22 ksi
SOLUTION
M = 8 ×1 = 8 kip ⋅ in. ,
A = π (1.25) 2 = 4.91 in.2
I=
π
4
(1.25) 4 = 1.971 in.4
r = I A = d 4 = 0.625 in.
Then
P A
Mc I
+
≤1
(σ all )c (σ all )b
8 × 103 4.91 8 × 103 (1.25) 1.917
+
≤1
(σ all )c
22 × 103
or
1.629 ×103
(σ all )c ≥ 2.14 ksi
+ 0.237 ≤ 1 ,
(σ all )c
Assume Le r > 66. Equation (11.29):
51, 000
51, 000
ksi;
,
Le r = 154.4 > 66
(σ all )c ≤
2.14 ≤
2
( Le r )
( Le r ) 2
For r = 0.625 in.
Le r = Le 0.625 = 154.4,
and
L
L = e == 137.9 in.
0.7
Le = 96.5 in.
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PROBLEM (11.86) Redo Prob. 11.85 for the case in which the diameter of the column is
decreased to d = 2 in.
SOLUTION
M = 8 ×1 = 8 kip ⋅ in. ,
A = π (1) 2 = π in.2
I=
π
4
(1) 4 =
π
4
in.4
r = I A = d 4 = 0.5 in.
Then
P A
Mc I
+
≤1
(σ all )c (σ all )b
8 × 103 π 8 × 103 (1) (π 4)
+
≤1
(σ all )c
22 × 103
2.547 ×103
+ 0.463 ≤ 1 ,
(σ all )c
(σ all )c ≥ 4.74 ksi
Assume Le r > 66. Equation (11.29):
51, 000
≥ 4.74 ,
Le r = 103.7 > 66
(σ all )c =
( Le r ) 2
For r = 0.5 in. :
Le r = Le 0.5 = 103.7,
and
L
L = e = 74.1 in.
0.7
O.K.
Le = 51.85 in.
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PROBLEM (*11.87) A W 410 x 85 steel column ( Table B. 8) of L e =5-m effective length
supports an eccentric load of P = 400 kN at a distance e from the center of the section measured
along the web (Fig. P11.87). Employ the interaction formula to determine the maximum allowable
value of e.
Given: ( σall ) b = 140 MPa, E = 210 MPa
*SOLUTION
A = 2.26 in.2
I z = 316 ×106 mm 4
y
417 mm
z
P
A = 10.8 × 103 mm 2
c = 417 2 = 208.5 mm
rz = 171 mm
rmin = 40.6 mm
Le rmin = 123.2
e
W 410x85
Equation (11.25):
(σ all )c =
π 2E
1.92( Le rmin ) 2
Interaction formula:
P A
Mc I
+
≤1;
(σ all )c (σ all )b
=
π 2 (210 ×109 )
1.92(123.2) 2
= 71.12 MPa
( P A)ec rz2
P A
+
≤1
7.12 × 106
140 × 106
400 × 103
1
e(0.2085)
+
[
] ≤1
−3
6
10.8 ×10 (10 ) 71.12 (0.17) 2 (140)
e(0.2085)
≤ 0.127 − 0.0141
(140)(0.171) 2
or
e≤
Thus,
0.0129(140)(0.171) 2
≤ 0.253 m
0.2085
e = 253 mm
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PROBLEM (12.1)
A stepped bar of diameters d1 and d2 and length L is subjected to tensile
force P (Fig. P12.1). Calculate the largest amount of strain energy that can be stored in the bar without
causing any yielding.
Given: d1 = 20 mm, d2 = 30 mm, L = 1.2 m,
σ y = 220 MPa, E = 200 GPa
SOLUTION
20 mm
30 mm
A
C
B
0.8 m
P
0.4 m
Segment BC
A=
π
4
(0.02) 2 = 314 ×10−6 m 2
P = σ yp A = 220 × 106 (314 × 10−6 ) = 69.08 kN
U BC =
(69.08 ×103 ) 2 (0.4)
P2 L
=
= 15.2 J
2 AE 2(314 ×10−6 )(200 ×109 )
Segment AB
π
(0.03) 2 = 706.5 × 10−6 m 2
4
(69.08 × 103 ) 2 (0.8)
P2 L
U AB =
=
= 13.51 J
2 AE 2(706.5 × 10−6 )(200 ×109 )
P = 69.08 kN
A=
Total energy
U = 13.51 + 15.2 = 28.71 J
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________________________________________________________________________
PROBLEM (12.2) Redo Prob. 12.1 for d1 = d2 = 25 mm.
SOLUTION
25 mm
A
A=
π
4
1.2 m
C
P
(0.025) 2 = 490.6 ×10−6 m 2
P = σ y A = 220 ×106 (490.6 × 10−6 ) = 107.9 kN
Total Energy:
P 2 L σ y PL 220 ×106 (107.9 ×103 )1.2
U=
=
=
= 71.2 J
2 AE
2E
2(200 × 109 )
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PROBLEM (12.3) A cold-rolled bronze rod of cross-sectional area 12 in.2 is subjected to axial
compression loading. If the allowable axial compressive stress is 14 ksi and E = 16 x 106 psi, calculate
the minimum length of the rod required for a strain energy of U = 1.3 in.-kips to be stored in the rod.
SOLUTION
2
ALmin
P 2 L σ all
=
2 AE
2E
Solving
2 EU 2(16 × 106 )(1.3 × 103 )
Lmin =
=
2
12(14 × 103 ) 2
Aσ all
= 17.687 in.
U=
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PROBLEM (12.4) An aluminum bar AB is of yield strength 38 ksi and modulus of elasticity E =
10 x 106 psi (see Fig. P12.4). A strain energy of U =150 in.-lb must be stored in the bar when axial load
P is applied. Calculate diameter d so that the factor of safety with respect to yielding is ns = 4.
SOLUTION
U y = 4U = 4(150) = 600 in. ⋅ lb
σ y2
(38 × 103 ) 2
= 72.2 in. ⋅ lb in.3
6
2 E 2(10 ×10 )
Uy
U y = (U 0 ) y AL or A =
(U o ) y L
Substituting the given data
600
πd2
A=
=
,
d = 0.383 in.
72.2(6 × 12)
4
(U 0 ) y =
=
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PROBLEM (12.5) A tubular aluminum alloy shaft AB (G = 4 x 106 psi) is fixed at the end B and
carries two torques at A and C as shown in Fig. P12.5. What is the strain energy stored in the shaft?
SOLUTION
Using the method of sections:
TAB = 400 lb ⋅ in.
The polar moment of inertia is
J=
π
TBC = 100 lb ⋅ in.
(34 − 24 ) = 6.381 in.4
32
The strain energy is then
T 2L
1
2
2
U =∑
=
LAB + TBC
LBC )
(TAB
2GJ 2GJ
1
=
[(400) 2 (25) + (100) 2 (10)]
6
2(4 ×10 )(6.381)
U = 80.3(10−3 ) in. − lb
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PROBLEM (12.6) The stepped bar seen in Fig. P12.6 consists of a square cross section with sides a
and a circular cross section of diameter d. The bar is subjected to a tensile force P. Determine the ratio of
d/a so that the strain energy in both parts is the same.
SOLUTION
Axial strain energy
P2 L
UC =
=
2 AE
Use Eq. (12. 2), circular Part:
2P2 L
P2 L
=
πd2
π d 2E
2(
)E
4
Square Part:
US =
P 2 L P 2 ( L 2) P 2 L
=
= 2
2 AE
2a 2 E
4a E
Requirement::
UC = U S ;
P2 L
2P2 L
=
π d 2 E 4a 2 E
d
8
=
π
a
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PROBLEM (*12.7) A solid tapered bar AB of length L is fixed at one end and subjected to .force
P at the other end (Fig. Pl 2.7). Verify that the strain energy of the bar is
U =
P2 L
π Ed 2
(P12.7)
where d is the diameter at the free end.
*SOLUTION
dx
x
ρ
d
2d
A
ρ=
d 2
x
L
O
P
B
L
L
dV = Adx = πρ 2 dx
or
π d 2 x2
d 2 2
dV = π (
x) dx =
dx
4 L2
L
Thus,
σ=
P
P
4 L2 P
=
=
A (π dx 2 4 L2 ) π d 2 x 2
We have
U0 =
σ2
2E
=
8 L4 P 2
π 2d 4 x4 E
U = ∫ U 0 dV = C ∫
V
2L
L
Let
C=
2P 2 L2
πd2
2L
1
dx
=C −
2
x
xL
2
⎡ 1 1⎤ P L
= C ⎢−
+ ⎥=
2
⎣ 2 L L ⎦ π Ed
Q.E.D
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PROBLEM (12.8) A stepped shaft ABC (G = 42 GPa) is subjected to torques as shown in Fig.
P12.8. For TB = 5 kN ⋅ m and TC = 2 kN ⋅ m, what is the strain energy of the shaft?
SOLUTION
TAB = 3 kN ⋅ m
TBC = 2 kN ⋅ m
Segment AB
J AB =
π
(60) 4 = 127.17 × 10−8 m 4
32
T 2L
(3 × 103 ) 2 (0.6)
U AB =
= 50.55 J
=
2GJ 2(42 ×109 )(127.17 × 10−8 )
Segment BC
J BC =
π
(40) 4 = 25.12 × 10−8 m 4
32
T 2L
(2 × 103 ) 2 (0.4)
U BC =
= 75.83 J
=
2GJ 2(42 × 109 )(25.12 × 10−8 )
Total strain energy
U = 50.55 + 75.83 = 126.4 J
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PROBLEM (12.9) Rework Prob. 12. 8 for the case in which the segment AB of the shaft is hollow,
the inside diameter is 30 mm, and TB = 1 kN ⋅ m and TC = 4 kN ⋅ m.
B
SOLUTION
TAB = 3 kN ⋅ m
TBC = 4 kN ⋅ m
40 mm
30 mm
A
B TB=1 kN ⋅ m
60 mm
0.6 m
Segment AB
J AB =
π
C TC=4 kN ⋅ m
0.4 m
(604 − 304 ) = 119.22 ×10−8 m 4
32
(3 × 103 ) 2 (0.6)
T 2L
U AB =
=
= 53.92 J
2GJ 2(42 × 109 )(119.22 ×10−8 )
Segment BC
J BC =
U BC =
π
32
(40) 4 = 25.12 × 10−8 m 4
(4 ×103 ) 2 (0.4)
= 303.31 J
2(42 ×109 )(25.12 × 10−8 )
Total strain energy
U = 53.92 + 303.31 = 357.2 J
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PROBLEM (12.10) The steel shaft ABC (G = 80 GPa) of Fig. P12.8 is subjected to torques TB = 0
and TC . Calculate the strain energy of the shaft when the maximum shearing stress is 50 MPa.
SOLUTION
Segment BC
J BC =
π
(40) 4 = 25.12 × 10−8 m 4
32
−8
J
6 25.12 × 10
= 628 N ⋅ m
T = τ max = 50 × 10
0.02
c
(628) 2 (0.4)
T 2L
U BC =
=
= 3.925 J
2GJ 2(80 × 109 )(25.12 × 10−8 )
Segment AB
J AB =
U AB =
π
32
(60) 4 = 127.17 × 10−8 m 4
(628) 2 (0.6)
= 1.163 J
2(80 ×109 )(127.17 ×10−8 )
Total strain energy
U = 3.925 + 1.163 = 5.088 J
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PROBLEM (12.11) Determine the strain energy of the aluminum shaft (G = 28 GPa) shown in Fig.
P12.8 when TB = 0 and the angle of twist φ is 0.05 rad.
J AB =
J BC =
π
32
π
32
(60) 4 = 127.17 × 10−8 m 4
(40) 4 = 25.12 × 10−8 m 4
TC2
T 2L
0.6
0.4
1
[
] −8 = 36.86 ×10−6 TC2
=
+
9
2GJ 2(28 × 10 ) 127.17 25.12 10
TC
0.6
0.4
1
TL
=
+
φC = ∑
[
] −8 = 73.72 × 10−6 TC
9
GJ 28 ×10 127.17 25.12 10
or
φ (106 )
TC = C
73.72
U =∑
Then
U=
36.86 ×10−6 (φC2 ×1012 )
= 6782φC2 = 67.82(0.05) 2 = 16.96 J
(73.72) 2
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PROBLEMS (12.12 through *12.15) A beam is supported and loaded as shown in Figs. P12.12
through P12.15. Determine the strain energy of the beam caused by bending.
SOLUTION (12.12)
x
w
1
M x = − wx 2
2
x
A
L
B
1 L 1 2 2
1 w2 L5
M 2 dx
(− wx ) dx =
U =∫
=
2 EI
2 EI ∫0 2
40 EI
SOLUTION (12.13)
P
P
B
A
a C
2a
MAC=Px
MCD=Pa
P
P
M
D a
Pa
x
Segment AC
2
a M dx
P2 a 2
P 2 a3
U AB = ∫
=
x
dx
=
0 2 EI
2 EI ∫0
6 EI
Segment CD
2 2
3a P a
P 2 a3
U CD = ∫
dx =
0
2 EI
EI
By symmetry: U AC = U BD .
Total strain energy :
Pa 3 P 2 a 3 4 P 2 a 3
+
=
U =2
EI
6 EI
3 EI
Continued on next slide
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SOLUTION (12.14)
w
B
A
wL
2
L
wL
2
x
wL
wx 2 1
x−
= w( Lx − x 2 )
2
2
2
2
L M
1 L w2
( Lx − x 2 ) 2 dx
U =∫
dx =
∫
0 2 EI
0
2 EI
4
2
L
w
( L2 x 2 − 2 Lx 3 + x 4 )dx
=
∫
0
8EI
M=
After integration
L
w2 L2 x3 2 Lx 4 x 5
1 w2 L5
=
−
+
U=
8 EI 3
4
5 0 240 EI
SOLUTION (*12.15)
u
A
L
aP
L
P
B
a
C
x
(− Pu ) 2 du P 2 a 3
=
0
2 EI
6 EI
a
(− Px) 2
L
P2a2 L
L
U AB = ∫
dx =
0
2 EI
6 EI
U BC = ∫
a
Total strain energy
P 2 a3 P 2 a 2 L P 2a 2
U=
+
=
( L + a)
6 EI
6 EI
6 EI
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________________________________________________________________________
PROBLEMS (12.16 through 12.19) A beam is supported and loaded as shown in Figs. P12.16
through P12.19. What is the strain energy of the beam caused by bending?
SOLUTION (12.16)
A
C
2a
M0
3a
a
B
M0
3a
x
M0
x
3a
M
M BC = 0 u
3a
M AC = −
u
M0
U = U AC + U BC
M
M u
(− 0 x) 2 dx
( 0 ) 2 du
a
2a
3a
=∫
+ ∫ 3a
0
0
2 EI
2 EI
2
3
3
M0
M 02 a
8a a
(
)
=
+
=
2 EI (9a 2 ) 3
3
6 EI
SOLUTION (12.17)
Part AC:
U AC = ∫
12
0
=
M=
1
Px
2
P
C
A
L/2
M2
dx
2 EI
P/2
B
L/2
P/2
3 L2
P2 L 2 2
P2 x
P 2 L3
=
=
x
dx
192 EI
8EI ∫0
8 EI 3 0
By summery U AC = U BC and total energy is
U AB = U AC + U BC =
P 2 L3
96 EI
Continued on next slide
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SOLUTION (12.18)
u
A
L B
M0
L
x
a C
M0
L
M0
M BC = M 0
M
M AB = 0 x
L
U = U AB + U BC
=
2
⎤
a
1 ⎡ L⎛ M0 ⎞
x
dF + ∫ M 02 dx ⎥
⎢ ∫0 ⎜
⎟
0
2 EI ⎢⎣ ⎝ L ⎠
⎥⎦
=
M 02 1 L3
M2
L
( 2 + a) = 0 (a + )
3
2 EI L 3
2 EI
SOLUTION (12.19)
The free-body diagrams
of beam AB and the portion
AO is shown in Fig. (a).
The condition ∑ M 0 = 0 gives
w0 L
2
w0
A
w0 L
6
L/3
2L/3
w
M = 0 ( L2 x − x 3 )
Figure (a)
6L
A
Therefore
w0 L
2
L M dx
6
U =∫
0 2 EI
L w
1
=
{∫ [ 0 ( L2 x − x 3 )]2 dx}
2 EI 0 6 L
L
w02
[( L4 x 2 + x 6 − 2 L2 x 4 )dx]
=
2 ∫0
72 EIL
or
w2 L5
U= 0
945EI
B
w0 L
3
1
2
xw
( L0 ) x x/3 xwL
0
M
O
x
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PROBLEMS (12.20 and 12.21) A beam of rectangular cross section is supported and loaded as
shown in Figs. 12.12 and 12.13. Determine the strain energy of the beam caused by shear deformation.
SOLUTION (12.20)
x
w
b
h
We have α =
L
αV 2
0
2 AG
Us = ∫
A
L
6
5
Vx = − wx
dx =
6 1 L
(− wx) 2 dx
5 2 AG ∫0
B
L
3 w2 x 3
1 w2 L3
=
=
5 AG 3 0 5 AG
SOLUTION (12.21)
We have
VAC = VBD = P
VCD = 0
α =6 5
Thus
a
6 P2a
α V2
P 2 dx
Us = ∫
dx = 2 ∫ α
= 2( )
0
2 AG
2 AG
5 2 AG
2
6Pa
=
5 AG
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PROBLEMS (12.22 and 12.23) A beam of rectangular cross section is supported and loaded as
shown in Figs. 12.14 and 12.15. What is the strain energy of the beam caused by shear deformation?
SOLUTION (12.22)
w
wL
2
wL
2
x
wL
6
+ wx
α=
2
5
2
2
L αV
3w L
L
( x − ) 2 dx
Us = ∫
dx =
∫
0 2 AG
5 AG 0
2
V =−
L
3w2 x3 x 2 L L2 x
1 w2 L3
=
−
+
=
5 AG 3
2
4 0 20 AG
SOLUTION (12.23)
P
α=
B
A
a
L
aP
L
x
C
VAB = −
x’
P
V
6
5
x
a
P
L
VBC = P
-Pa/L
L
αV 2
0
2 AG
Us = ∫
dx =
a
⎤
3 ⎡ L P2a2
dx + ∫ P 2 dx '⎥
⎢
2
∫
0
5 AG ⎣ 0 L
⎦
3P 2 a
=
( L + a)
5 AGL
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PROBLEM (*12.24) A simple beam AB of length L and rectangular cross section is subjected to
a moment M0 at point A as shown in Fig. P12.24. Determine the strain energy of the beam caused by both
bending and shear.
*SOLUTION
Load, shear, and moment diagram is shown in Fig. (a).
For bending
M0
M2
dx '
0 2 EI
M 02 L 2
M 02 L
=
x ' dx ' =
6 EI
2 EIL2 ∫0
Ub = ∫
L
a
M0 A
L
L
B
M0
L
x’
x
V
x
For shear
L
αV 2
0
2 AG
Us = ∫
-M0/L
dx
where
α=
6
5
A = bh
V =−
M0
L
Therefore
3M 02 L
3 M 02
Us =
dx =
5bhGL2 ∫0
5 GbhL
M
M0
x
Figure (a)
Total energy, letting I = bh3 12 :
U s + Ub =
2M 02
3 Eh 2
+
(1
)
Ebh3
10 GL2
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PROBLEM (12.25) A truss ABC consisting of two members supports a vertical load P at joint B as
shown in Fig. P12.25. Each member has axial rigidity AE. Determine the strain energy of the truss.
SOLUTION
Apply the method of joints at pin B. Use cos 60o = 3 2 and sin 60o = 1 2
3
2
FBC − P = 0,
FBC =
2
3
1
1
∑ Fx = 0 : 2 FBC = FAB FAB = 3 P
We have
LAB = L
LBC = 2L
Therefore
F 2L
1
2
2
U =∑
=
LAB +FBC
LBC )
( FAB
2 AE 2 AE
1 P2
4P2
[ ( L) +
(2 L)]
=
2 AE 3
3
P2 L
or
U = 1.5
AE
∑F = 0:
y
P
FAB
B
60o
FBC
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PROBLEM (*12.26) Figure P12.26 depicts a workpiece clamped to a milling machine table by a
bolt of diameter d tightened to a tension of T. The link AB has a rectangular cross section of width b
and depth h. Calculate:
(a) The strain energy in bolt due to axial load.
(b) The strain energy in link caused by bending.
Given: b = ½ in., h = 5/16 in., d = 1/4 in., T = 120 lb, E = 29 x 106 psi
Assumptions: Bolt and link are made of steel. The effect of the hole in bending of the link is disregarded.
*SOLUTION
(a) Axial strain energy in bolt.
P2 L T 2 L
Ub =
=
2 AE 2 AE
(120) 2 (1.1875)
=
= 6 × 10−3 in. ⋅ lb
π
2
6
2[ (0.250) (29 ×10 )]
4
(b) Bending strain energy in link.
I = bh3 12 = (0.5)(0.3125)3 12 = 1.2716 × 10−3 in.4
M 2 dx
0 2 EI
1
2
1
=
{[ ∫ (80 x) 2 dx + ∫ (40 x ') 2 dx '}
0
2 EI 0
2 in.
2144
1 in.
=
EI
A
Substituting the data,
x
2144
120 lb
80 lb
Ub =
(29 ×106 )(1.2716 × 10−3 )
= 58.1× 10−3 in. ⋅ lb
Ub = ∫
L
B
x’
40 lb
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PROBLEM (12.27) At a point in a structure subjected to plane stress, the stresses have the
magnitudes and direction shown acting on an element in Fig. P12.27. Determine the strain-energy density
and its components at the point.
Given: E = 200 GPa, ν = 1/3
SOLUTION
Refer to Fig. P12.27.
We have
200 × 109
G=
= 75 GPa
2(1 + 1 3)
and
7+2
7+2 2
) + (5) 2 ;
σ 1,2 =
± (
2
2
or
σ 1 = 10.09 MPa
σ 2 = −1.09 MPa
Equations(12.17), (12.19), and (12.20) become:
1
1 − 2ν
U0 =
U 0v =
(σ 12 + σ 22 − 2νσ 1σ 2 )
(σ 1 + σ 2 ) 2
2E
6E
1
1
U 0d =
[(σ 1 − σ 2 ) 2 + σ 12 + σ 22 ] =
(σ 12 + σ 22 − σ 1σ 2 )
12G
6G
Substitute the given data:
1012
1
U0 =
[10.092 + 1.092 + 2( )(10.9 ×1.09)] = 276 J m3
9
2(200 ×10 )
3
U 0v =
1 − 2(1 3)1012
(10.09 − 1.09) 2 = 23 J m3
9
6(200 ×10 )
1012
[10.092 + (−1.09) 2 + 10.9 × 1.09] = 253 J m3
9
6(75 × 10 )
Check:
U 0 = U 0 v + U 0 d = 276 J m3
U 0d =
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PROBLEM (12.28) Calculate the strain-energy density and its components at the point depicted in
Fig. P12.28 in a loaded member.
Given: E = 30 x 106 psi , ν = 0.3
SOLUTION
From Fig. P12.28: σ 1 = 300 psi and σ 2 = −700 psi
We have
30 × 106
G=
= 11.5 ×106 psi
2(1 + 0.3)
Apply Eqs.(12.17), (12.19), and (12.20) with σ 3 = 0 :
104
[32 + 7 2 + 2(0.3)(21)] = 118 × 10−4 in. ⋅ lb in.3
2(30 ×106 )
1 − 2(0.3) 4
U 0v =
10 (3 − 7) 2 = 3.5 ×10−4 in. ⋅ lb in.3
6
6(30 × 10 )
U0 =
104
(32 + 7 2 + 3 × 7) = 114.5 × 10−4 in. ⋅ lb in.3
6(11.5 × 106 )
Check:
U 0 = U 0 v + U 0 d = 118 × 10−4 in. ⋅ lb in.3
U 0d =
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PROBLEM (12.29) An element in three-dimensional state of stress is subjected to stresses as shown
in Fig. P12.29. Determine the strain-energy density and its components at the element.
Given: E = 70 GPa , ν = 1/3
SOLUTION
G = E 2(1 + ν ) = 70 2.5 = 28 GPa
Fig. P12.29: σ 1 = 40 MPa.
2
−50 − 30
⎛ −50 − 30 ⎞
2
σ 2,3 =
± ⎜
⎟ + (20)
2
2
⎝
⎠
or
σ 2 = −17.6 MPa
σ 3 = −62.4 MPa
Equation (12.17) yields
1012
U0 =
{402 + 17.62 + 62.42 − 2(0.25)
2 × 70 ×109
×[40(−17.6) + (−17.6)(−62.4) + 40(−62.4)]}
3
= 48.96 kJ m
Equation (12.19) gives
(1 − 2 4)1012
U 0v =
[40 − 17.6 − 62.4]2 = 1.905 kJ m3
9
6(70 ×10 )
Equation (12.20):
1012
U 0d =
[(40 + 17.6) 2 + (−17.6 + 62.4) 2 + 102.42 ] = 47.05 kJ m3
9
12(28 × 10 )
Check:
U 0 = U 0 v + U 0 d = 48.96 kJ m3
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PROBLEM (12.30) An element is subjected to equal tension and compression: pure shear (Fig.
P12.30). Compute the strain-energy densities for the states of stress shown in the figure. Then, equating
the results found, verify that G = E/2(1 + ν ).
σ 2 = −σ 1
SOLUTION
Refer to P12.30, Equation (12.17) gives
σ2
1
U0 =
(σ 12 + σ 22 − 2νσ 1σ 2 ) = 1 (1 +ν )
2E
E
τ max = σ 1
U0 =
2
τ max
2G
=
σ 12
2G
σ1
(1)
(2)
Equating σ 12 from Eqs. (1) and (2), and simplifying the result:
G = E 2(1 + ν )
Q.E.D.
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PROBLEM (12.31) Determine the strain-energy density and its components for the triaxial state of
stress shown in Fig. P12.31.
Given: σ 3 = 30 MPa, σ 2 = 40 MPa,
σ 1 = 100 MPa,
E = 110 GPa,
ν
= 0.3
SOLUTION
G=
110
= 42.3 GPa
2(1 + 0.3)
Equation (12.17):
1012
U0 =
[302 + 402 + 1002 − 2 × 0.3(30 × 40 + 40 × 100 + 30 × 100)]
9
2(110 × 10 )
= 34.46 kJ m3
Equation (12.19) yields
(1 − 2 × 0.3)1012
U 0v =
(30 + 40 + 100) 2 = 17.52 kJ m3
9
6(110 ×10 )
Equation (12.20):
1012
U 0d =
(102 + 602 + 702 ) = 16.94 kJ m3
9
12(42.3 × 10 )
Check:
U 0 = U 0 v + U 0 d = 34.46 kJ m3
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PROBLEM (12.32) Use the work-energy method to determine the deflection at end C of the bar
ABC seen in Fig. P12.32.
SOLUTION
Equation (12.4) for n=2:
22 + 2 P 2 L
P2 L
=
U2 =
3(22 ) (2 AE ) 4 AE
Equation (12.23):
W = U2 =
P2 L
1
Pδ C =
2
4 AE
or
δC =
PL
2 AE
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PROBLEM (12.33) Calculate the angle of twist for shaft ABC (G = 28 GPa) due to TC = 1.2 kN ⋅ m
applied at the free end as shown in Fig. P12.33. Employ the work-energy method.
SOLUTION
See solution of Prob. 12.11:
U = 36.86 ×10−6 TC2
Tφ
1
Equation (12.25): U = TCφC ; 36.86 ×10−6 TC2 = C C
2
2
−6
−6
3
φC = 73.72 ×10 TC = 73.72 ×10 (1.2 × 10 )
or
= 88.46 ×10−3 rad = 5.1o
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PROBLEM (12.34) A beam is supported and loaded as shown in Fig. P12.34. Using the workenergy method, determine the deflection at point C (or D) owing to bending and shear.
SOLUTION
See solution of Probs. 12.13 and 12.21:
4 P 2 a3
6 P2a
Ub =
Us =
3 EI
5 AG
vC = vD (due to symmetry)
Thus,
4 P 2 a3 6 P 2 a 1
1
U=
+
= P (vCb + vCs ) = PvC
3 EI
5 AG 2
2
2a
3
or
vC = 2 Pa(
+
)
3EI 5GA
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PROBLEM (12.35) A beam is supported and loaded as shown in Fig. P12.34. Using the workenergy method, determine the deflection at point C (or D) owing to bending and shear.
SOLUTION
See solution of Probs. 12.15 and 11.23:
P2a2
3P 2 a
Ub =
Us =
(a + L)
( a + L)
6 EI
5 AGL
Thus,
P2a2
3P 2 a
1
U=
(a + L) +
(a + L) = PvC
2
6 EI
5 AG
3
a
or
vC = 2 Pa(a + L)[
+
]
6 EI 5GAL
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PROBLEM (12.36) A beam is supported and loaded as shown in Fig. P12.36. Employ the work method to determine the slope at point C caused by bending.
SOLUTION
See solution of Prob.12.16:
M 2a
U= 0
6 EI
Thus, Eq. (12.26):
M 02 a 1
M a
= M 0θC , θC = 0
6 EI 2
3EI
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PROBLEM (12.37) A beam is supported and loaded as shown in Fig. P12.37. Use the work method
to find the slope at point C caused by bending.
SOLUTION
From solution of Prob. 12.18:
M2 L
U = 0 ( + a)
2 EI 3
Thus, Eq. (12.26) with
W=
1
M 0θC
2
yields
θC =
M0
( L + 3a )
3EI
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PROBLEMS (12.38 and 12.39) A beam is supported and loaded as shown in Fig. P12.38 and Fig.
P12.39. Employ work-energy approach to determine the slope at point C due to the moment M0 .
SOLUTION (12.38)
M0
A
x
C
RA
RB
L/3
2L/3
Reactions are noted in Fig. (a).
B
x’
L
Part AC. M = −
M0x
L
Figure (a)
U AC = ∫
2L 3
0
4M 02 L
M 02 2 L 3 2
M2
=
x
dx
dx =
81EI
2 EI
2 EIL2 ∫0
M0x '
L
2
L3 M
M 02 L
M 02 L 3 2
=
'
U BC = ∫
dx ' =
x
dx
0
2 EI
2 EIL2 ∫0
162 EI
2
M L
Total strain energy: U = U AC + U BC = o
18 EI
M L
1
Therefore
M 0θC = U , θC = 0
2
9 EI
Part BC M = −
SOLUTION (12.39)
M0
B
C
x
A
M0/L
x’
Reactions are shown in Fig. (a).
M0/L
M
Part AC. M = M 0
M0
L2
x
U AC = ∫
0
M 02 L
M2
dx =
2 EI
4 EI
Figure (a)
x
M0
L
2
L M
M 02 L
M 02 L 2
=
x
dx
U AB = ∫
dx =
0 2 EI
2 EIL2 ∫0
6 EI
Part AB. We have M =
Continued on next slide
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Total strain energy:
Hence
5 M o2 L
12 EI
5 M 02 L
θC =
6 EI
U = U AC + U AB =
1
M 0θC = U ,
2
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PROBLEMS (*12.40 and 12.41) Apply the work-energy method to determine the slope at point
C of the beam supported and loaded as shown in Fig. P12.40 and Fig. P12.41.
SOLUTION (*12.40)
M0
A
3EI
C
B
EI
M0/L
x
Bending moment : M = −
x’
M0/L
M0
x
L
Part AC.
U AC = ∫
M
L2
0
x
=
- M0
M 02 x 2
dx
3(2 EI )
M 02 L 2 2
1 M 02 L
=
x
dx
144 EI
6 EIL2 ∫0
Part BC.
M 02
M 02
M 02 L 2
L 3
7 M 02 L
3
U BC = ∫
dx ' =
x ' dx ' =
[L − ( ) ] =
L 2 2 EI
2 EIL2 ∫L 2
6 EIL2
2
48 EI
2
11 M o L
Total strain energy: U = U AC + U BC =
72 EI
Hence
11 M 0 L
1
M 0θ B = U , θ B =
36 EI
2
L
SOLUTION (12.41)
P
PL
Bending moment : M = − Px '
B
A
EI
1.5 EI C
Part AC.
P
M
x
-PL
M2
P2 L 2
dx ' =
x ' dx '
L 2 3EI
3EI ∫L 2
7 P 2 L3
P2 3 L 3
=
[L − ( ) ] =
72 EI
9 EI
2
U AC = ∫
x’
L
1 P 2 L3
M2
P2 L 2 2
'
'
dx ' =
x
dx
=
0
2 EI
2 EI ∫0
48 EI
2 3
17 P L
Total strain energy: U = U AC + U BC =
144 EI
So,
1
17 PL3
PvB = U , vB =
2
72 EI
Part BC.
U BC = ∫
L2
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________________________________________________________________________
PROBLEM (12.42) Using the work-energy method, determine the deflection due to bending at point
B of the cantilever beam shown in Fig. 12.42.
Given:
P = 120 N, E = 200 GPa, h = 2b = 60 mm.
SOLUTION
y
P
60 mm
x
A
O
B
L=0.6 m
x
z= L
2bx
x
3b
b
O
z
b = 30 mm
M = − P ( x − 0.3)
2b
z=
x
L
2(30)
=
x
600
x
=
10
0.3 m
0.6 m
1 x
(0.06)3 = 1.8(10−9 ) x m 4
12 10
2
0.9 ( x − 0.3)
0.9
0.09
M 2 dx
dx = 20∫ ( x − 0.6 +
)dx
= 20∫
U =∫
0.3
0.3
x
x
2 EI
Iz =
0.9
x2
= 20
− 0.6 x + 0.09 ln x = 20[0.36 − 0.36 + 0.099] = 1.98 J
2
0.3
Thus,
1
120
U = PvB ;
vB , vB = 33 mm ↓
1.98 =
2
2
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________________________________________________________________________
PROBLEMS (12.43 and *12.44) For the truss and loading shown in Fig. P12.43 and Fig.
P12.44, obtain the vertical deflection at point C. Use the work-energy method and assume that all
members are of equal axial rigidity AE.
SOLUTION (12.43)
P
C
LAC = LBC =
4
2L
3
A
P
2
L/2
5
L
6
3
L/2
B
P
2
Apply the method of joints at C and B:
5
3
FAC = FCB = P
FAB = P
8
8
Therefore,
25 5
9
19 P 2 L
F 2 L P2 L
=
U =∑
[2( ⋅ ) + ] =
48 AE
2 AE 2 AE 64 6 64
Hence
19 P 2 L 1
19 PL
↓
U=
= Pδ C , δ C =
48 AE 2
24 AE
SOLUTION (*12.44)
Using the method of joint of joints at C, B, D, and A:
FAB = P
FBC = 2 P
FCD = − P
FBD = − P
FAD = 2 P
FAE = 0
FDE = −2P
We have
LAD = LBC = 2 L
Thus,
F 2 L P2 L
=
U =∑
(1 + 2 2 + 1 + 1 + 2 2 + 0 + 4)
2 AE 2 AE
2P2 L
=
(1.75 + 2)
AE
Hence,
2P2 L
1
U=
(1.75 + 2) = Pδ C
AE
2
Continued on next slide
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Solving
δC =
PL
4 PL
↓
(1.75 + 2) ↓ = 12.675
AE
AE
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________________________________________________________________________
PROBLEM (12.45) For the beam loaded as shown in Fig. P12.45, using Castigliano’s theorem,
determine:
(a) The vertical deflection vB of the free end.
(b) The slope θ B of the free end.
SOLUTION
x
L
M = − Px − C
∂M ∂P = − x
P
C
∂M ∂C = −1
(a) In this case C=0.
L M ∂M
1 L
PL3
=
↓
−
x
dx
(
)(
)
vB = ∫
dx =
Px
−
0 EI ∂P
3EI
EI ∫0
(b) For θ B C=0:
1 L
M ∂M
PL2
θB = ∫
(− Px)(−1)dx =
dx =
0 EI ∂C
2 EI
EI ∫0
L
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________________________________________________________________________
PROBLEM (12.46) A cantilever beam is loaded as shown in Fig. P12.46. Apply Castigliano’s
theorem to determine the deflection v A of the free end.
SOLUTION
w
Q
A
a
Segment AC
M 1 = −Qx
C
∂M 1 ∂Q = − x
Segment BC
M 2 = −Qx −
B
2a
W
( x − a)2
2
∂M 2
= −x
∂Q
Thus,
3 a M ∂M
M 1 ∂M 1
2
2
dx
dx + ∫
0 EI ∂Q
a EI ∂Q
Let Q=0. Hence
w 3a
w 3a 3
2
(
)
( x − 2ax 2 + a 2 x)dx
vA =
x
−
a
xdx
=
∫
∫
a
a
2 EI
2 EI
vA = ∫
a
3a
w x 4 2ax3 a 2 x 2
=
−
+
2 EI 4
3
2 a
=
10 wa 4
↓
3 EI
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________________________________________________________________________
PROBLEM (12.47) A cantilever beam is loaded as shown in Fig. P12.46. Apply Castigliano’s
theorem to determine the deflection vB of the free end.
SOLUTION
x
P
x
C
b
a
A
Q
B
L
Segment BC
M 1 = −Qx
∂M 1 ∂Q = − x
Segment AC
M 2 = −Qx − P( x − b),
So,
vB = ∫
∂M 2
= −x
∂Q
M ∂M
dx
EI ∂Q
Let Q=0:
1 L
− P( x − b)(− x)dx
EI ∫b
1 1
Pb 2 2
[ P( L3 − b3 ) −
( L − b )]
=
2
EI 3
vB =
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________________________________________________________________________
PROBLEM (12.48) A pin-ended truss in which all members have the same axial rigidity AE is
loaded as shown in Fig. P12.48. Using Castigliano’s theorem, determine the vertical displacement of
point D.
SOLUTION
B
L
L
A
D
LAB = LCD = 2 L
W
LBC = LAC = LBD = L
L
C
The method of joints(or sections) is applied:
FAB = 2W
FBC = −W
FCD = − 2W
FBD = W
FAC = −W
Thus,
∂F
1
δV =
Fi i Li
∑
∂W
AE
WL
WL
=
[ 2( 2) 2 + 1 + 1 + 1 + 2( 2) 2] = 8.657
↓
AE
AE
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________________________________________________________________________
PROBLEM (12.49) For the pin-ended truss loaded as shown in Fig. P12.48, determine the
horizontal deflection of point D. Employ Castigliano's theorem.
Assumption: All members have the same axial rigidity AE.
SOLUTION
D
B
Q
L
W
L
A
L
LBC = LAC = LBD = L
C
By the method of joints (or sections) :
FAB = 2(W + Q)
FBC = −W − Q
FBD = W + Q
LAB = LCD = 2 L
FCD = − 2W
FAC = −W
Therefore,
∂F
∂FAB
1
= 2
Fi i Li
∑
∂Q
∂Q
AE
∂FBC ∂FBD
∂FCD ∂FAC
=
=1
=
=0
∂Q
∂Q
∂Q
∂Q
Let Q=0:
WL
WL
δH =
(2 2 + 2) = 4.828
→
AE
AE
δH =
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________________________________________________________________________
PROBLEM (12.50) A truss ABC of three steel members supports a vertical load P at C as
shown in Fig. P12.50. The cross-sectional areas of members are AAB , ABC , and AAC . What is the
vertical displacement δ v of joint C? Apply Castigliano’s theorem.
Given:
AAB = 0.002 m,
ABC = 0.003 m,
AAC = 0.0026 m,
P = 25 kN,
E = 210 GPa
SOLUTION
F ∂F
F ∂F
F ∂F
∂U 1
= ∑ ( AB AB + BC BC + AC AC )
∂P E
AAB ∂P
ABC ∂P
AAC ∂P
Joint C
δv =
P
5.4
2.25
We have
C
LAC = 5.42 + 2.252 = 5.85 m
2.25
FAC
LBC = 32 + 2.252 = 3.75 m
LAB = 2.4 m
3
FBC
5.4
(1)
3
∑ F = 0 : − 5.85 F + 3.75 F = 0
x
AC
BC
2.25
2.25
∑ F = 0 : − 5.85 F + 3.75 F = 0
y
AC
Solving,
FAC = 3.25 P
BC
FAC = 3.75 P
Joint B
FBC
2.25
B
3
FAB
3
∑ F = 0 : F + 3.75 F = 0
x
AB
BC
FAB = 3P
RB
Equation (1) is therefore
3
3.75
3.25
P
P
[( )(3) + (
)(3.75) + (
)(3.25)] = 13.25(103 )
δv =
−3
3
2.6
E (10 ) 2
E
The value of displacement at C is
25(103 )
δ v = 13.25(103 )
= 1.577 mm ↓
210(109 )
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________________________________________________________________________
PROBLEMS (12.51 and 12.52) For the beam and loading shown in Fig. P12.51 and P12.52, use
Castigliano's theorem to determine the slope θ B at point B.
SOLUTION (12.51)
x
w
C
A
B
L
1
M = −C − wx 2
∂M ∂C = −1
2
1 L
M ∂M
wx 2
+
dx =
C
θB = ∫
(
)dx
2
EI ∂C
EI ∫0
Letting C=0:
θB =
1 L wx 2
wL3
=
dx
6 EI
EI ∫0 2
SOLUTION (12.52)
y
A
P
P
a D
2
P+C/4a
Segment AD
M1 = ( P +
C
)x
4a
E a
C
B
P-C/4a
∂M 1 ∂M 2 ∂M 3
x
=
=
=
4a
∂C
∂C
∂C
Segment DE
M 2 = (P +
C
) x − P( x − a)
4a
Segment EB
M 3 = (P +
C
) x − P( x − a) − P( x − 3a)
4a
Let C = 0 :
∂M i
1
θB =
Mi
dx
∫
EI
∂C
2
3a
4a
4 a Px
x
x
x
1 a
=
dx]
[ ∫ Px( )dx + ∫ Pa( )dx + ∫ 4 Pa( )dx − ∫
3a
3 a 4a
a
4a
4a
4a
EI 0
Continued on next slide
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Integrating
Pa 3 a 3 9a 3 a 3 64a 3 36a 3 64a 3 27 a 3
θB =
− +
−
+
−
[ +
]
4aEI 3
2
2
2
2
3
3
3 Pa 2
=
2 EI
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________________________________________________________________________
PROBLEMS (12.53 through 12.55) For the beam and loading shown in Figs. P12.53 through
P12.55, employ Castigliano's theorem to determine:
(a) The deflection at point C.
(b) The slope at point C.
SOLUTION (12.53)
x
x’
A
Pa
L
+
B
L
C
L
P
a C C
Segment AB
M1 = − P
a
C
x− x
L
L
Segment BC
M 2 = − Px '− C
(a) For this case C=0:
vC =
1 L Pax
1 a
ax
− Px '(− x ')dx '
(−
)(− )dx +
∫
EI 0
L
L
EI ∫0
L
a
Pa 2 x 3
P x '3
Pa 2
=
+
=
( L + a) ↓
EIL2 3 0 EI 3 0 3EI
(b)
∂M 1
x
=− ,
L
∂C
∂M 2
= −1.
∂C
For C=0, we have :
1 L Pax
1 a
x
θC =
− Px '(− x ')dx '
(−
)(− )dx +
∫
EI 0
L
L
EI ∫0
L
a
Pa x 3
P x '2
=
+
EIL2 3 0 EI 2 0
=
Pa
(2 L + 3a)
6 EI
Continued on next slide
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SOLUTION (12.54)
x
M0
Q
A
M0
3a
− Q3
2a
C
x’
a
B
M0
3a
+ 23Q
Segment AC
M1 = −
M0
Q
x+ x
3a
3
Segment CB
M2 = −
M 0 x ' 2Q
+
x'
3a
3
(a) We have
∂M 1 x
∂M 2 2 x '
Q=0
=
=
3
∂Q 3
∂Q
Thus,
1 2a M 0 x x
1 a M 0 x ' 2x '
vC = −
dx +
dx '
∫
EI 0 3a 3
EI ∫0 3a 3
2 M 0a2
=
↑
9 EI
(b) In this case Q=0:
∂M 1
∂M 2 x '
x
=− ,
=
3a
∂M 0
∂M 0 3a
So,
1 2a M 0 x x
1 a M0x ' x '
θC =
dx +
dx '
∫
EI 0 3a 3a
EI ∫0 3a 3a
M a
= 0
3EI
Continued on next slide
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SOLUTION (12.55)
x
x’
A
M0
L
+
Qa
L
L
Q
M0
a C
B
M0
L
− Q ( LL+ a )
Segment AB
M1 = −
M0
Qa
x−
x
L
L
Segment BC
M 2 = M 0 − Qx '
(a) We have
∂M 1
∂M 2
ax
=−
= − x ' , and Q = 0 .
L
∂Q
∂Q
Hence,
1 L M 0 x ax
1 a
(− )dx +
M 0 (− x ')dx '
vC =
∫
EI 0 L
L
EI ∫0
M a2
= 0 (2 L + 3a) ↑
6 EI
(b) Now Q=0:
∂M 1 x
∂M 2
= ,
= 1.
∂M 0 L
∂M 0
Thus,
a
1 L M0x x
dx + ∫ M 0 (1)dx '
θC =
∫
0
0
EI
L L
M
= 0 ( L + 3a)
3EI
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________________________________________________________________________
PROBLEM (12.56) For the truss loaded as shown in Fig. P12.56, determine the horizontal deflection at point C. Use Castigliano's theorem.
SOLUTION
P
C
5L/6
5L/6
2L/3
A
P
2
We have
∂FAB 1
=
2
∂Q
Let Q=0:
Q
L
∂FAC 5
=
6
∂Q
B
P
2
Apply the method of joint at
C & B:
5
5
FBC = − Q − P
6
8
5
5
FAC = Q − P
6
8
1
3
FBC = Q + P
2
8
∂FBC
5
=−
∂Q
6
∂F
1
Fi Li i
∑
∂Q
AE
1 3P L
5P 5L 5
5P 5L 5
[
=
+ (− )
+ (− ) (− )]
AE 8 2
8 6 6
8 6
6
3 PL
=
→
16 AE
δH =
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________________________________________________________________________
PROBLEM (12.57) For the truss loaded as shown in Fig. P12.56, determine the horizontal deflection at point C. Use Castigliano's theorem.
SOLUTION
P+Q
A
Joint C
B
RA
FBC
2L
2L
L
D
L
RE
E
Q
1
FCD
C
C
P
P
FBC = 2 P
FCD = − P
Joint B
B
FAB
1
FBD
FBC
FAD
Joint D
FAD = 2 P + 2Q
FDE = −2 P − Q
Joint A
1
FBC = P
2
1
FAB = −
FBC = − P
2
FAB =
FDE
FAE
1
1
P
D
Q
P+Q
A
2P
P
P
FAE = P + Q −
1
FAD
2
FAD
Continued on next slide
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We have
∂FAD
∂FDE
∂FAE
= 2,
= −1,
= 0,
Q=0
∂Q
∂Q
∂Q
and other derivatives are zero. Thus,
∂F
1
Fi Li i
δD =
∑
∂Q
AE
1
PL
[ 2 P( 2 L) 2 + (−2 P) L(−1)] =
(2 2 + 2)
=
AE
AE
PL
= 4.828
↓
AE
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________________________________________________________________________
PROBLEMS (12.58 through 12.61) A beam is loaded and supported as shown in Figs. P12.58
through P12.61. Apply Castigliano's theorem to determine the deflection at point C.
SOLUTION (12.58)
y
P
x
2a
A
C
a B
M0 =
3
Pa
2
Segment CB
M1 = M 0
Segment AC
M 2 = M 0 − P( x − a)
We have:
Hence,
∂M 1 ∂P = 0
∂M 2 ∂P = −( x − a)
3a
3a
EIvC = − ∫ M 0 ( x − a)dx + P ∫ ( x 2 − 2ax + a 2 )dx
a
a
3a
2
3a
2
x
x
= M 0 − + ax + P
− ax 2 + a 2 x
2
3
a
a
3Pa 3 1
1
Pa 3
[− (9 − 1) + 2] + Pa 3 [ (27 − 1) − 8 + 2] = −
2
2
3
3
3
Pa
vC =
↑
3EI
=
or
SOLUTION (12.59)
Q
A
RA=Q/2
a
x
x”
w
C a
x’
B
a D
RB=2wa+Q/2
Continued on next slide
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Segment AC
∂M 1 x
1
M 1 = Qx
=
2
∂Q 2
Let Q=0:
∂M 1
1 a
M1
dx = 0
∫
0
EI
∂Q
Segment CB
1
1
M 2 = Q(a + x ') − Qx '− wx '2
2
2
Q
1
= (a − x ') − wx '2
2
2
∂M 2 1
= (a − x ')
∂Q 2
Let Q=0:
∂M 2
1
1 a wx '2 1
wa 4
=
−
−
'
(
)
(
')
M
dx
a
x
dx
'
=
−
2
∂Q
EI ∫
EI ∫0
48EI
2 2
Segment BD
∂M 3
1
M 3 = − wx ''2
=0
2
∂Q
Therefore
∂M i
wa 4
↑
vC = ∫ M i
dx =
∂Q
48EI
Continued on next slide
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SOLUTION (12.60)
Add force Q at point C as shown
in Fig. (a).
2
L M
U =∫
dx
0 2 EI
∂U
1 L ∂M
=
δC =
M
dx
∂Q EI ∫0
∂Q
Q
P
x
C
B
L/2
L/2
Figure (a)
∂M
=0
∂Q
L
M = − Px − Q( x − ),
2
M = − Px,
Part AC.
Part CB.
∂M
L
= −( x − )
∂Q
2
Let Q=0:
L
1 L2
1 L
(− Px)(0)dx +
(− Px)[−( x − )]dx
∫
∫
L
0
2
EI
EI
2
P L 2 L
P 1 3 1 L 3 L 1 2 L 1 L 2
( x − x)dx =
[ L − ( ) −( ) L +( ) ( ) ]
=
2
3 2
2 2
2 2 2
EI ∫L 2
EI 3
3
5 PL
=
48 EI
δC =
SOLUTION (12.61)
Q
L/2
A
wL
8
C
+ Q2
wL Q
+ )x
8
2
L/2
x’
x
Segment AC
M1 = (
w
B
3 wL
8
+ Q2
∂M 1 x
=
∂Q 2
Segment CB
M2 = (
wx '2
3wL Q
+ ) x '−
8
2
2
∂M 2 x '
=
∂Q
2
Let Q=0:
L 2 3wLx '
wLx x
wx '2 x '
) dx '
dx + ∫ (
−
0
0
8 2
8
2 2
2
2
L 2 wLx
L 2 3wLx '
wx '3
wL4
)dx ' =
dx + w∫ (
−
= w∫
0
0
16
16
4
96
Continued on next slide
EIvC = ∫
L2
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or
wL4
vC =
↓
96 EI
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________________________________________________________________________
PROBLEM (12.62) Employ Castigliano's theorem to determine the slope at point B of the curved
bar of Fig. P12.62.
SOLUTION
A
M θ = − PR sin θ + C
∂M θ
=1
∂C
ds = Rdθ
R
dθ
θ
B
P
C
Let C=0:
EIθ B = ∫ M θ
=∫
π 2
0
∂M θ
ds
∂C
π 2
− PR sin θ (1) Rdθ = PR 2 cos θ 0
= PR 2 (0 − 1) = PR 2
or
θB =
PR 2
EI
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________________________________________________________________________
PROBLEM (12.63) Use Castigliano's theorem to determine the vertical component of the deflection
at point B of the curved bar of Fig. P12.62.
SOLUTION
A
ds = Rdθ
M θ = QR(1 − cos θ ) − PR cos θ
∂M θ
= R(1 − cos θ )
∂θ
R
dθ
θ
B
Q
Let C=0:
∂M θ
Rdθ :
∂Q
EI δV = ∫ M θ
EI δV = ∫
π 2
0
P
(− PR sin θ R(1 − cos θ ) Rdθ
= − PR 3 ∫
π 2
0
sin θ (1 − cos θ )dθ
π 2
1
PR 3
2
= − PR − cos θ + cos θ
=−
2
2
0
3
from which
δV =
PR 3
↑
2 EI
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________________________________________________________________________
PROBLEM (12.64) A uniform bar of flexural rigidity EI is fixed at one end and loaded at the other
end as shown in Fig. P12.64. Use Castigliano's theorem to obtain:
(a) The vertical deflection at point D.
(b) The slope at point D.
SOLUTION
x
Pa-C
2a
B
A
a
P
a
D
C
x
C
x
P
(a) C=0.
Segment CD
∂M 1
= −x
∂P
M 1 = − Px
Segment BC
∂M 2
=a
∂P
M 2 = Pa
Segment AB
M 3 = Px − Pa
Thus
EI δ D = ∫ M i
∂M 2
= x−a
∂P
∂M i
dx
∂P
a
a
2a
0
0
0
= P ∫ x 2 dx + Pa 2 ∫ dx + P ∫ ( x 2 − 2ax + a 2 )dx
3
2a
1
x
= Pa ( + 1) + P
− ax 2 + a 2 x
3
3
0
4 3
8
= Pa + Pa 3 ( − 4 + 2)
3
3
3
2 Pa
↓
δD =
EI
3
or
Continued on next slide
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(b)
M 1 = −C − Px
∂M 1 ∂C = −1
M 2 = Pa + C
∂M 2 ∂C = 1
M 3 = Px − Pa − C
∂M 3 ∂C = −1
Let C=0:
2a
a
a
∂M i
3
dx = P ∫ xdx + Pa ∫ dx − P ∫ ( x − a) dx = Pa 2
0
0
0
2
∂C
2
3 Pa
θD =
2 EI
EIθ D = ∫ M i
or
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________________________________________________________________________
PROBLEMS (*12.65 and *12.66) A load P is supported at joint B of a structure consisting of
three bars of equal axial rigidity AE (Fig. P12.65 and P12.66). Use Castigliano's theorem to determine the
force in each bar.
SOLUTION (*12.65)
Consider RD as redundant.
C
FBC
∂FBD
=1
∂RD
3
∑F = 0:
D
4
A
FBD =RD
3
3 ft
4.5 ft
B
P
4
y
B
FAB
FBC = 1.25 P − 1.25RD
P
∑ F = 0 : F = −0.75P + 0.75R
x
AB
D
where
∂FBC ∂RD = −1.25
Thus,
∂FAB ∂RD = 0.75
∂Fi
=0
∂RD
= (1.25 P − 1.25 RD )(7.5)(−1.25)
+(−0.75 P + 0.75 RD )(4.5)(0.75) + RD (3)1 = 0
= −14.25P + 17.25 RD = 0
AEδ D = ∑ Fi Li
or
RD = 0.826 P
Then
FBD = 0.826 P
FAB = −0.131P
FBC = 0.22 P
Continued on next slide
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SOLUTION (*12.66)
RD =FAB=R
D
C
vD=0
R
FBC
0.5L
0.6L
B
3
P
4
5
4
3
0.8L
0.5L
B
P
A
Figure (a)
FAB
Figure (b)
Structure is statically indeterminate to the 1st degree. We have
δ D = ∂U ∂R = 0. That is
δD =
∂F
∂F
∂F
1
[ FAB LAB AB + FBC LBC BC + FBD LBD BD ] = 0
AE
∂R
∂R
∂R
Joint B.
Equilibrium results in:
FAB = 0.8R − 0.8P
FBC = 0.6 P − 0.6 R
FBD = R
(1)
(2)
Equation (1) is thus
δ D = (0.8R − 0.8P )(0.8L)(0.8) + (0.6 P − 0.6 R)(0.6 L)(−0.6)
+ R(0.5L)(1) = 0
Solving
R = 0.593P = FBD
Equations (2) give then
FAB = −0.32 P
FBC = 0.244 P
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________________________________________________________________________
PROBLEM (12.67) Apply Castigliano's theorem to determine the deflection of point C of the
stepped cantilever beam shown in Fig. P12.67.
SOLUTION
2EI1
x
A
M = − Px
Thus,
a
B
EI1
a
P
C
∂M ∂P = − x
vC =
1 a
1 2a
(
)(
)
(− Px)(− x)dx
−
Px
−
x
dx
+
2 EI1 ∫a
EI1 ∫0
2a
Pa 3
P x3
=
+
3EI1 2 EI1 3 a
=
3 Pa 3
↓
2 EI1
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________________________________________________________________________
PROBLEM (12.68) Determine the deflection of free end B of the uniformly loaded cantilever beam
shown in Fig. P12.68. Apply the unit-load method.
SOLUTION
x
w
1 lb
x
A
L
B
1
M = − wx 2
2
A
L
B
m = −x
So,
Mm
1 L
dx =
(− wx 2 )(− x)dx
∫
0 EI
0
2 EI
vB = ∫
L
L
w x4
wL4
=
=
↓
2 EI 4 0 8EI
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________________________________________________________________________
PROBLEM (12.69) For the beam loaded as shown in Fig. P12.69, applying the unit-load method,
determine:
(a) The vertical deflection vB of the free end.
(b)
The slope θ A of the free end.
SOLUTION
1 lb
P
A
x
B
M = − Px
A
x
m = −x
B
1 lb ⋅ in.
A
x
B
m ' = −1
Mm
1 L
dx =
(− Px)(− x)dx
0 EI
EI ∫0
(a) vB = ∫
L
L
P x3
PL3
=
=
↓
EI 3 0 3EI
Mm '
1 L
dx =
(− Px)(−1)dx
0 EI
EI ∫0
PL2
=
2 EI
(b) θ B = ∫
L
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________________________________________________________________________
PROBLEM (12.70) Determine the deflection of point C of the uniformly loaded beam shown in
Fig. P12.70. Use the unit load method.
SOLUTION
w
1 lb
A
wL
2
B
L/3 C
M=
2L/3
1
1
wLx − wx 2
2
2
wL
2
A
B
L/3 C
2L/3
2
lb
3
2
lb
3
2x
3
2x
L
− (x − )
Segment BC: m2 =
3
3
Segment AC: m1 =
L Mm
Mm1
2
dx + ∫
dx
0
L 3 EI
EI
1 L 3 wLx wx 2 2 x
1 L 3 wLx wx 2 2 x
L
(
)
(
)[ − x + ]dx
dx
=
−
+
−
∫
∫
2
2 3
2
2
3
3
EI 0
EI 0
4
11 wL
=
↓
927 EI
vC = ∫
L3
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________________________________________________________________________
PROBLEM (12.71) Determine the midspan deflection of a cantilever beam loaded as shown in Fig.
P12.71. Employ the unit-load method.
SOLUTION
x
w0
L
wo
A
1N
B
L
A
L/2
C
L/2
B
x
w0 x 3
1 w0 x x
M =− x
=−
2 L 3
6L
Segment AC: m1 = x
Segment BC: m2 = 0
Therefore,
1 L Mm
dx
EI ∫0 EI
w0 x 3
1 L 2 w0 x 3
1 L
(−
)(0) +
( x)(−
)dx
=
6L
6L
EI ∫0
EI ∫L / 2
31 w0 L4
=
↓
960 EI
vC =
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________________________________________________________________________
PROBLEM (12.72) Apply the uni