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PROBLEM 12.61
For the beam and loading shown, design the cross section of
the beam, knowing that the grade of timber used has an
allowable normal stress of 12 MPa.
SOLUTION
0: 2.4 (0.6)(3.6)(3) 0
B
MA=−+ =
2.7 kNA=
0: (1.8)(3.6)(3) 2.4 0
A
MB=− +=
8.1 kNB=
Shear:
2.7 kN
2.7 (2.4)(3) 4.5 kN
4.5 8.1 3.6 kN
3.6 (1.2)(3) 0
A
B
B
C
V
V
V
V
−
+
=
=−=−
=−+ =
=−=
Locate point D where
0.V=
2.4 7.2 6.48
2.7 4.5
0.9 m 2.4 1.5 m
dd
d
dd
−
= =
= −=
Areas of the shear diagram:
A to D:
1(0.9)(2.7) 1.215 kN m
2
V dx
= = ⋅
∫
D to B:
1(1.5)( 4.5) 3.375 kN m
2
V dx
= −=− ⋅
∫
B to C:
1(1.2)(3.6) 2.16 kN m
2
V dx
= = ⋅
∫
Bending moments:
0
0 1.215 1.215 kN m
1.215 3.375 2.16 kN m
2.16 2.16 0
A
D
B
C
M
M
M
M
=
=+= ⋅
=−=− ⋅
=−+ =
Maximum
3
| | 2.16 kN m 2.16 10 N mM= ⋅= × ⋅
6
all 12 MPa 12 10 Pa
σ
= = ×
363 3 3
min 6
all
| | 2.16 10 180 10 m 180 10 mm
12 10
M
S
σ
−
×
== =×=×
×
For rectangular section,
2 23
11
(150) 180 10
66
S bh b= = = ×
3
2
(6)(180 10 )
150
b×
=
48.0 mm
b=
consent of McGraw-Hill Education.
PROBLEM 12.62
For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of
1750 psi.
SOLUTION
PROBLEM 12.63
Knowing that the allowable normal stress for the steel used is 24 ksi, select
consent of McGraw-Hill Education.
PROBLEM 12.64
Knowing that the allowable stress for the steel used is 24 ksi, select the
consent of McGraw-Hill Education.
PROBLEM 12.65
Knowing that the allowable stress for the steel used is 160 MPa, select the
consent of McGraw-Hill Education.
PROBLEM 12.66
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical wide-flange beam to support the loading
consent of McGraw-Hill Education.
PROBLEM 12.67
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical S-shape beam to support the loading shown.
SOLUTION
PROBLEM 12.68
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical S-shape beam to support the loading shown.
SOLUTION
0: 0.8 (0.4)(2.4)(100) (1.6)(80) 0
B
MA∑= − − =
280 kNA= ↓
0: 0.8 (1.2)(2.4)(100) (2.4)(80) 0
A
MB∑= − − =
600 kNB= ↑
Shear:
280 kN
A
V= −
280 (0.8)(100) 360 kN
B
V−=−− =−
360 600 240 kN
B
V
+
=−+ =
240 (1.6)(100) 80 kN
C
V=−=
Areas under shear diagram:
A to B:
1(0.8)( 280 360) 256 kN m
2−− =− ⋅
B to C:
1(1.6)(240 80) 256 kN m
2+= ⋅
Bending moments:
0
A
M=
0 256 256 kN m
256 256 0
B
C
M
M
=−=− ⋅
=−+ =
Maximum
3
256 kN m 256 10 N mM= ⋅= × ⋅
6
all 160 MPa 160 10 Pa
σ
= = ×
333 3 3
min 6
all
256 10 1.6 10 m 1600 10 mm
160 10
M
S
σ
−
×
== =×=×
×
Shape
S (103 mm3)
S510 98.2×
1950
S460 104×
1685
Lightest S-section:
S510 98.2×
consent of McGraw-Hill Education.
PROBLEM 12.61
For the beam and loading shown, design the cross section of
the beam, knowing that the grade of timber used has an
allowable normal stress of 12 MPa.
SOLUTION
0: 2.4 (0.6)(3.6)(3) 0
B
MA=−+ =
2.7 kNA=
0: (1.8)(3.6)(3) 2.4 0
A
MB=− +=
8.1 kNB=
Shear:
2.7 kN
2.7 (2.4)(3) 4.5 kN
4.5 8.1 3.6 kN
3.6 (1.2)(3) 0
A
B
B
C
V
V
V
V
−
+
=
=−=−
=−+ =
=−=
Locate point D where
0.V=
2.4 7.2 6.48
2.7 4.5
0.9 m 2.4 1.5 m
dd
d
dd
−
= =
= −=
Areas of the shear diagram:
A to D:
1(0.9)(2.7) 1.215 kN m
2
V dx
= = ⋅
∫
D to B:
1(1.5)( 4.5) 3.375 kN m
2
V dx
= −=− ⋅
∫
B to C:
1(1.2)(3.6) 2.16 kN m
2
V dx
= = ⋅
∫
Bending moments:
0
0 1.215 1.215 kN m
1.215 3.375 2.16 kN m
2.16 2.16 0
A
D
B
C
M
M
M
M
=
=+= ⋅
=−=− ⋅
=−+ =
Maximum
3
| | 2.16 kN m 2.16 10 N mM= ⋅= × ⋅
6
all 12 MPa 12 10 Pa
σ
= = ×
363 3 3
min 6
all
| | 2.16 10 180 10 m 180 10 mm
12 10
M
S
σ
−
×
== =×=×
×
For rectangular section,
2 23
11
(150) 180 10
66
S bh b= = = ×
3
2
(6)(180 10 )
150
b×
=
48.0 mm
b=
consent of McGraw-Hill Education.
PROBLEM 12.62
For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of
1750 psi.
SOLUTION
PROBLEM 12.63
Knowing that the allowable normal stress for the steel used is 24 ksi, select
consent of McGraw-Hill Education.
PROBLEM 12.64
Knowing that the allowable stress for the steel used is 24 ksi, select the
consent of McGraw-Hill Education.
PROBLEM 12.65
Knowing that the allowable stress for the steel used is 160 MPa, select the
consent of McGraw-Hill Education.
PROBLEM 12.66
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical wide-flange beam to support the loading
consent of McGraw-Hill Education.
PROBLEM 12.67
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical S-shape beam to support the loading shown.
SOLUTION
PROBLEM 12.68
Knowing that the allowable normal stress for the steel used is 160 MPa,
select the most economical S-shape beam to support the loading shown.
SOLUTION
0: 0.8 (0.4)(2.4)(100) (1.6)(80) 0
B
MA∑= − − =
280 kNA= ↓
0: 0.8 (1.2)(2.4)(100) (2.4)(80) 0
A
MB∑= − − =
600 kNB= ↑
Shear:
280 kN
A
V= −
280 (0.8)(100) 360 kN
B
V−=−− =−
360 600 240 kN
B
V
+
=−+ =
240 (1.6)(100) 80 kN
C
V=−=
Areas under shear diagram:
A to B:
1(0.8)( 280 360) 256 kN m
2−− =− ⋅
B to C:
1(1.6)(240 80) 256 kN m
2+= ⋅
Bending moments:
0
A
M=
0 256 256 kN m
256 256 0
B
C
M
M
=−=− ⋅
=−+ =
Maximum
3
256 kN m 256 10 N mM= ⋅= × ⋅
6
all 160 MPa 160 10 Pa
σ
= = ×
333 3 3
min 6
all
256 10 1.6 10 m 1600 10 mm
160 10
M
S
σ
−
×
== =×=×
×
Shape
S (103 mm3)
S510 98.2×
1950
S460 104×
1685
Lightest S-section:
S510 98.2×
consent of McGraw-Hill Education.
