Score:
1.
0/100
0
Points
%
Award: 0 out of 15.00 points
Three wooden planks are fastened together by a series of bolts to form a column. The diameter of each bolt is 12-mm and the inner diameter of each
washer is 16-mm, which is slightly larger than the diameter of the holes in the planks. Determine the smallest allowable outer diameter d of the washers,
knowing that the average normal stress in the bolts is 32 MPa and that the bearing stress between the washers and the planks must not exceed 8.5 MPa.
The smallest allowable outer diameter d of the washers is
17
mm.
References
Numeric Response
Difficulty: Medium
Three wooden planks are fastened together by a series of bolts to form a column. The diameter of each bolt is 12-mm and the inner diameter of each
washer is 16-mm, which is slightly larger than the diameter of the holes in the planks. Determine the smallest allowable outer diameter d of the washers,
knowing that the average normal stress in the bolts is 32 MPa and that the bearing stress between the washers and the planks must not exceed 8.5 MPa.
The smallest allowable outer diameter d of the washers is
Explanation:
Bolt:
A Bolt
πd
=
2
π(0.012
=
4
Tensile force in bolt:
σ
=
P
=
=
m)
2
4
=
1.13097
× 10
−4
m
P
A
σA
(32 × 10
6
Pa) (1.13097
× 10
−4
2
m )
3
= 3.6191 × 10 N
Bearing area for washer:
Aw
=
and
Aw
=
π
4
2
2
(d o
− d )
i
P
σ BRG
Therefore, equating the two expressions for Aw gives
π
4
2
(d o
2
do
=
2
− di )
4P
πσ BRG
P
=
σ BRG
2
+ di
3
4(3.6191 × 10
2
do
2
do
=
π(8.5 × 10
=
7.9812
6
× 10
do
=
28.251
do
=
28.3 mm
N)
+ (0.016
Pa)
−4
× 10
−3
m
m
2
m)
2
2
28.3 ± 2% mm.
2.
Award: 0 out of 20.00 points
A load P is applied to a steel rod, whose flange diameter is 40 mm, supported as shown by an aluminum plate into which a 12-mm-diameter hole has been
drilled. Knowing that the shearing stress must not exceed 177 MPa in the steel rod and 70 MPa in the aluminum plate, determine the largest load P that can
be applied to the rod.
The largest load P that can be applied to the rod is
kN.
6720
References
Numeric Response
Difficulty: Medium
A load P is applied to a steel rod, whose flange diameter is 40 mm, supported as shown by an aluminum plate into which a 12-mm-diameter hole has been
drilled. Knowing that the shearing stress must not exceed 177 MPa in the steel rod and 70 MPa in the aluminum plate, determine the largest load P that
can be applied to the rod.
The largest load P that can be applied to the rod is
66.7 ± 2% kN.
Explanation:
For steel:
A1
=
τ1
=
P
=
πdt
=
π (0.012
m) (0.010
m)
=
376.99
× 10
−6
m
2
P
A1
A 1 τ1
=
376.99
(
× 10
−6
2
m ) (177 × 10
6
Pa)
=
66.727
× 10
3
N
For aluminum:
A2
=
τ2
=
P
=
π dt
=
π (0.040
m) (0.008 m)
=
1.00531 × 10
−3
m
2
P
A2
A 2 τ2
=
(1.00531 × 10
−3
2
70
m )(
Limiting value of P is the smaller value, so
P = 66.7 kN
× 10
6
Pa)
=
70.372
× 10
3
N
A 6-mm-diameter pin is used at connection C of the pedal shown. It is given that P = 750 N.
References
Section Break
3.
Difficulty: Medium
Award: 0 out of 6.00 points
Determine the average shearing stress in the pin.
The average shearing stress in the pin is
MPa.
References
Numeric Response
Difficulty: Medium
Determine the average shearing stress in the pin.
The average shearing stress in the pin is
34.4836 ± 2% MPa.
Explanation:
Since BCD is a three-force member, the reaction at C is directed toward E, the intersection of the lines of action of the other two forces.
Using geometry,
−
−−−−−−−−−−−−−−−−−−
2
C E = √ (300 mm)
2
+ (125 mm)
= 325 mm
Using the free-body diagram of BCD,
+↑ΣFy = 0 :
1
τpin =
AP
325 mm
1
C
2
125 mm
=
C
2
π
4
d
2
=
C − P = 0 ⇒ C = 2.6P = 2.6 (750 N) = 1950 N
2C
πd
2
=
2(1950 N)
π(6×10
−3
2
m)
= 34.4836 × 10
6
Pa = 34.4836 MPa
4.
Award: 0 out of 6.00 points
Determine the nominal bearing stress in the pedal at C.
The nominal bearing stress in the pedal at C is
MPa.
References
Numeric Response
Difficulty: Medium
Determine the nominal bearing stress in the pedal at C.
The nominal bearing stress in the pedal at C is
36.1111 ± 2% MPa.
Explanation:
Since BCD is a three-force member, the reaction at C is directed toward E, the intersection of the lines of action of the other two forces.
Using geometry,
−
−−−−−−−−−−−−−−−−−−
2
C E = √ (300 mm)
2
+ (125 mm)
= 325 mm
Using the free-body diagram of BCD,
+↑ΣFy = 0 :
σb =
C
Ab
=
C
dt
125 mm
325 mm
C − P = 0 ⇒ C = 2.6P = 2.6 (750 N) = 1950 N
(1950 N)
=
(6×10
−3
m)(9×10
−3
= 36.1111 × 10
m)
6
Pa = 36.1111 MPa
5.
Award: 0 out of 6.00 points
Determine the nominal bearing stress in each support bracket at C.
The nominal bearing stress in each support bracket at C is
MPa.
References
Numeric Response
Difficulty: Medium
Determine the nominal bearing stress in each support bracket at C.
The nominal bearing stress in each support bracket at C is
32.5 ± 2% MPa.
Explanation:
Since BCD is a three-force member, the reaction at C is directed toward E, the intersection of the lines of action of the other two forces.
Using geometry,
−
−−−−−−−−−−−−−−−−−−
2
C E = √ (300 mm)
2
+ (125 mm)
= 325 mm
Using the free-body diagram of BCD,
+↑ΣFy = 0 :
125 mm
325 mm
C − P = 0 ⇒ C = 2.6P = 2.6 (750 N) = 1950 N
1
C
σb =
2
Ab
=
C
2dtb
(1950 N)
=
2(6×10
−3
m)(5×10
−3
= 32.5000 × 10
m)
6
Pa = 32.5000 MPa
6.
Award: 0 out of 16.00 points
Member ABC, which is supported by a pin and bracket at C and a cable BD, was designed to support the 19-kN load P as shown. Knowing that the ultimate
load for cable BD is 100 kN, determine the factor of safety with respect to cable failure.
The factor of safety with respect to cable failure is
.
References
Numeric Response
Difficulty: Medium
Member ABC, which is supported by a pin and bracket at C and a cable BD, was designed to support the 19-kN load P as shown. Knowing that the ultimate
load for cable BD is 100 kN, determine the factor of safety with respect to cable failure.
The factor of safety with respect to cable failure is
2.9024 ± 2% .
Explanation:
Use member ABC as a free body and note that member BD is a two-force member.
+ ↺ ΣMC = 0 :
(P cos 40°)(1.2 m) + (P sin 40°)(0.6 m) − (FBD cos 30°)(0.6 m) − (FBD sin 30°)(0.4 m) = 0 ⇒ 1.304926P − 0.719615
FBD = 1.813366P = (1.813366)(19 × 10
F.S. =
FU
F BD
=
100×10
3
N
34.4540×10
3
= 2.9024
N
3
N) = 34.4540 × 10
3
N
7.
Award: 0 out of 16.00 points
Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that the average normal stress must not
exceed 150 MPa in either rod, determine the smallest allowable values of the diameters d1 and d2. Take P = 40 kN.
The smallest allowable value of the diameter d1 is
The smallest allowable value of the diameter d2 is
mm.
mm.
References
Numeric Response
Difficulty: Easy
Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that the average normal stress must not
exceed 150 MPa in either rod, determine the smallest allowable values of the diameters d1 and d2. Take P = 40 kN.
The smallest allowable value of the diameter d1 is
The smallest allowable value of the diameter d2 is
18.4 ± 2% mm.
42.2 ± 2% mm.
Explanation:
Rod AB:
Force: P = 40 × 103 N
Stress: σ AB = 150 × 106 Pa
Area: A = π4 d12
σ AB
A
π
4
=
2
d1
2
d1
=
d1
=
P
=
A
P
σ AB
=
P
σ AB
(4)(40 × 10
4P
=
πσ AB
π(150 × 10
18.426
× 10
18.4 mm
Rod BC:
Force: P = 40
d1
3
N)
=
Pa)
339.53
× 10
−6
m
2
m
=
Stress: σ BC
Area: A =
σ BC
2
−3
6
=
P
A
4P
π
4
d
3
N
−
(2) (125 × 10
− 150 × 10
6
3
N)
=
210
–
Pa
2
2
=
4P
πd
2
2
(4)(-210 × 10
d2
=
d2
=
42.22
d2
=
42.2 mm
πσ BC
× 10
=
=
π(−150 × 10
× 10
−3
m
6
3
N)
Pa)
=
1.78
× 10
−3
m
2
× 10
3
N
8.
Award: 0 out of 15.00 points
Two wooden members are joined by plywood splice plates that are fully glued on the contact surfaces. Knowing that the clearance between the ends of the
members is 6 mm and that the ultimate shearing stress in the glued joint is 2.5 MPa, determine the length L for which the factor of safety is 2.75 for the
loading shown. Take P = 15 kN.
The length L is
mm.
References
Numeric Response
Difficulty: Easy
Two wooden members are joined by plywood splice plates that are fully glued on the contact surfaces. Knowing that the clearance between the ends of the
members is 6 mm and that the ultimate shearing stress in the glued joint is 2.5 MPa, determine the length L for which the factor of safety is 2.75 for the
loading shown. Take P = 15 kN.
The length L is
138 ± 2% mm.
Explanation:
τall
2.5
=
MPa
=
2.75
0.90909
MPa
On one face of the upper contact surface,
A
L − 0.006
=
m
2
(0.125
m)
Since there are 2 contact surfaces,
τall
=
P
2A
0.90909 × 10
L
=
0.138 m
L = 138 mm
6
Pa
=
15 × 10
(L − 0.006
3
N
m)(0.125 m)