Chapter 1 • Introduction - Solution Manual Fundamentals of Machine Elements 2nd
Chapter 1 • Introduction 1.1 Design transport containers for milk in 1 gallon and 1 liter sizes. Notes: There are a number of design considerations that can be included in this problem. Some
examples are considered here, but the student should not be restricted in the types of solutions pursued. This type of problem is an excellent opportunity for brainstorming, where a group is encouraged to provide all possible answers to the problem regardless of feasibility. For example, the materials considered could include wood barrels, ceramic jugs or foil or plastic bags, as in intraveneous fluid containers. Solution: The transport containers for milk should be easy to handle and store as well as being easy to pour from without spilling. The material should be recyclable or be able to be burned leaving harmless combustion products. The obvious material choices are paper or plastic materials, although glass bottles and metal cans can be considered as possibilities. For a one liter container the standard type Pure Pac or Tetra Pac Bric made of plastic carton is a good design.
A one gallon container has to be compact to be easily stored in a refrigerator. The width, depth, and height should be similar in size, which would be about 156 mm x 156 mm x 156 mm. That size is too large to grip with one hand. Therefore, a one gallon container is preferably made of a plastic with an integral handle.
1.2 Design a simple coat hanger to be used by dry cleaners. It should be able to accommodate
both a winter coat and a pair of trousers. It should function well but also be as inexpensive as possible. Notes: See the notes to problem 1. Solution:
If they are not hung simultaneously, a steel wire frame of triangular shape with a hook at the top could be used. The horizontal portion could be covered with a high friction coating or tape or glue to keep the trousers from sliding off.
For coats, a sturdier frame is needed, requiring either a larger gage of wire or pursuit of another alternative such as wood or plastic hangers. th
1.3 Journal bearings on train boxcars in the early 19 century used a “stink additive” in their
lubricant. If the bearings got too hot, they would attain a noticeable odor, and an oiler would give the bearing a squirt of lubricant at the next train stop. What design philosophy does this illustrate? Explain. Solution:
The most obvious design philosophy is the Doctrine of Manifest Danger. This philosophy suggests that systems be designed so that a failure can be detected before it occurs. In this case, a dry bearing becomes hot, and the bad smell indicates a loss of lubricant before the bearing failure is catastrophic.
It should be noted that other answers are possible. For example, the principal of uniform safety suggests that the level of safety of all components be kept at the same level. If failure of one component, in this case a bearing, is imminent, then extra maintenance on this bearing is justified by the principal of uniform safety.
1.4 A safety chain is used in automotive towing applications. Which design philosophy is
being incorporated into the towing system design? Solution:
Since the safety chain is used in addition to the towbar, it is an example of redundancy, and it is a passive system. Furthermore, it is an example of a fail-safe design, because if the towbar fails, the towed vehicle will follow the car and will not swerve into other lanes and collide with other vehicles.
1.5 A hand-held drilling machine has a bearing to take up radial and thrust load from the
drill. Depending on the number of hours the drill is expected to be used before it is scrapped, different bearing arrangements will be chosen. A rubber bushing has a 50-hr life. A small ball bearing has a 300-hr life. A two bearing combination of a ball bearing and a cylindrical roller bearing has a 10,000 hr life. The cost ratios for the bearing arrangements are 1:5:20. What is the optimum bearing type for a simple drill, a semiprofessional drill, and a professional drill?
Notes: This problem is fairly subjective, but there should be a realization that there is an increase in expected life for a professional drill versus a simple drill. In fact, for certain users, a 50 hour life is probably sufficient. Solution: Notice that the cost per time is highest for the rubber bushing and lowest for the ball bearing cylindrical bearing combination. However, the absolute cost is higher for the longer lasting components, so the expected life must be assessed. Rough estimates based on expected lives are as follows.
a) Simple drill - use rubber bushingbearings.
b) Semi-professional drill - either two ball bearings or a ball and cylindrical roller bearing combination.
c) Professional drill - use a ball and cylindrical roller bearing combination.
1.6 Using the hand-held drill described in Problem 1.5, if the solution with the small ball
bearing was chosen for a semiprofessional drill, the bearing life could be estimated to be 300 hr until the first spall forms in a race. The time from first spall to when the whole rolling-contact surface is covered with spalls is 200hr, and the time from then until a ball cracks is 100 hr. What is the bearing life a) If high precision is required?
b) If vibrations are irrelevent?
c) If an accident can happen when a ball breaks? Notes: This problem is fairly subjective, but there should be a realization that there is a decrease in expected life for the high precision drill, etc.
Solution: a) For high precision, no spall should be allowed so the life is 300 hrs.
b) If vibrations are irrelevent, the drill can be used until a ball breaks, or 600hours. c) Depending on the severity of the accident, it is possible that an entirely different bearing should be used. However, safety allowances of 300-500 hours may be reasonable.
1.7 The dimensions of skis used for downhill competition need to be determined. The
maximum force transmitted from one foot to the ski is 2500N, but the snow conditions are not known in advance, so the bending moment acting on the skis is not known. The weight of the skier and equipment is 100 kg. Estimate the safety factor needed. Notes: There are two different approaches, that of section 1.5.1.1, and that of a worst case scenario.
Solution: Using the approach of section 1.5.1.1, the largest safety factor from Table 1.1 is n =3.95, sx and the largest value of n sy from Table 1.2 is 1.6. Therefore, the largest safety factor, if there is poor control of materials and loading, and if the consequences are sever, is given by Equation 1.2 as:
n =n n =(3.95)(1.6)=6.32
s sx syAnother approach is to perform a worst case scenario analysis. If the weight of skier and equipment is 100 kg and the maximum load on the ski is 2500N, then the centrigugal acceleration 2 is 23 m/s . If the speed of the skier is 35m/s (126km/hr), the radius from the ski track will be 53 m. If the speed is 20 m/s, the radius is 17.4m. This means that no overstressing should ever occur due to bending of the ski, so the safety factor could be chosen just above one.
1.8 A crane has a loading hook that is hanging on a steel wire. The allowable normal tensile
stress in the wire gives an allowable force of 100,000N. Find the safety factor that should be used.
a) If the wire material is not controlled, the load can cause impact, and fastening the hook in the wire causes stress concentrations. (If the wire breaks, people can be seriously hurt and expensive equipment can be destroyed.)
b) If the wire material is extremely well controlled, no impact loads are applied, and the hook is fastened in the wire without stress concentrations. (If the wire breaks, no people or expensive equipment can be damaged.) Notes: Equation (1.2) is needed to solve this problem, with data obtained from Tables 1.1 and 1.2.
Solution:
a). If the quality of materials is poor (A=p), the control over the load to the part is poor (B=p), and the accuracy of the stress analysis is poor (C=p), then nsx=3.95 as given in Table 1.1. If the economic impact and danger are both very serious (D=E=vs), then from Table 1.2, n sy =1.6. Therefore, the required safety factor is obtained from Equation (1.2) as
n n
s =n sx sy =(3.95)(1.6)=6.3
b) If the quality of materials, the control over the load to the part, and the accuracy of the stress analysis are all very good (A=B=C=vg), then n sx =1.1 as given in Table 1.1. If the economic impact and danger are both not serious (D=E=ns), then from Table 1.2, n sy =1.0. Therefore, the required safety factor is obtained from Equation (1.2) as
n s =n sx n sy =(1.1)(1.0)=1.1
1.9 Give three examples of fail-safe and fail-unsafe products.
Notes: This is an open-ended problem, and many different solutions are possible. Solution: Examples of fail safe products are: 1) electric motors, which fail to move and cause no hazards when they fail. 2) Bowling pins, because of their distance from bowlers, do not present any hazards when they shatter. 3) Furniture upholstery, which causes aesthetic objection long before structural failure occurs. Examples of fail unsafe products include: 1) Bungee cords, as in bungee cord jumping from bridges and the like; 2) parachutes, which cause a fatal injury in the vast majority of failures, and 3) A fan on an exhaust hood over a fume producing tank, such as an electroplating tank. Failure of the forced air circulation system exposes plant personnel to harmful fumes.
1.10 An acid container will damage the environment and the people around if it leaks. The
cost of the container is proportional to the container wall thickness. The safety can be increased either by making the container wall thicker or by mounting a reserve tray under the container to collect the leaking acid. The reserve tray costs 10% of the thick-walled container cost. Which is less costly, to increase the wall thickness or to mount a reserve tray under the container? Notes: This problem uses the approach described by Equation (1.2) and Tables 1.1 and 1.2.
Solution: If the safety factor is unchanged, then for a thicker container wall, we see that with danger to personnel rated very serious and economic impact very serious (D=E=vs), then n sy =1.6. If a reserve tray is used, the required safety factor for the container (D=E=ns) is n =1.0. Therefore, if sy
A is the cost of the container with thicker walls, then the cost of the container with a reserve tray
is
A ( )
- 0.1 A = 0.725A
1.6 Therefore, with the same safety factor, it is less costly to have the reserve tray.
1.11 The unit for dynamic viscosity in the SI system is newton-seconds per square meter, or 2
pascal-seconds (N-s/m = Pa-s). How can that unit be rewritten by using the basic relationships described by Newton’s law for force and acceleration? Notes: The only units needed are kg, m, and s. Solution: By definition, a Newton is a kilogram-meter per square second. Therefore, the unit for dynamic viscosity can be written as:
Ns kgm s kg
= =
2
2
2 sm m s m
1.12 The unit for dynamic viscosity in Problem 1.11 is newton-second per square meter 2
(Ns/m ) and the kinematic viscosity is defined as the viscosity divided by the fluid density. Find at least one unit for kinematic viscosity. Notes: The only units needed are m and s. Solution: The units for kinematic viscosity can be written as:
Ns kgm
sm2
2
2 Nsm m
m s
η = η = = = = k kg kg kg s
ρ
3 m
1.13 A square surface has sides 1m long. The sides can be split into decimeters, centimeters,
or millimeters, where 1m=10dm, 1dm=10cm, and 1cm=10mm. How many millimeters, centimeters, and decimeters equal 1m? Also, how many square millimeters, square centimeters and square decimeters equal one square meter? Notes: This is fairly straightforward; Table 1.3(b) is useful.
Solution: From Table 1.3(b), one meter equals 10 dm, 100 cm and 1000 mm. Therefore, a square meter is: 2 2 2 1m =(1000mm) =1 million (mm) . 2 2 2
1m =(100cm) =10,000 (cm) 2 2 2 3 1m =(10dm) =100(dm)
1.14 A volume is 1 tera (mm ) large. Calculate how long the sides of a cube must be to contain
that volume. 12 Notes: From Table 1.3 (b), tera is 1 x 10 .Solution: 3 12 3 A tera (mm ) is 10 (mm ). Therefore, this volume is
3
1m
12
3
3
10 mm = 1000m
1000mm A cube with this volume has sides with length of 10m. 8
1.15 A ray of light travels at a speed of 300,000 km/s=3 x 10 m/s. How fast will it travel in 1
ps, 1 ns, and 1µs? Notes: An interesting assignment is to have students submit string of the required length for the picosecond and nanosecond cases.
Solution:
8 − 12 −
4 x = v ∆ t =
3 × 10 m / s 10 s = 3 × 10 m = 0.3mm
( ) ( )
In one nanosecond, light travels:
8 −
9 x = v ∆ t =
3 × 10 m / s 10 s = 0.3 m = 300mm
( ) ( )
In one microsecond, light travels:
8
6 − x v t m s
= ∆ =
3 × 10 / s 10 = 300m
( ) ( )
1.16 Two smooth flat surfaces are separated by a 10-µm thick lubricant film. The viscosity of
2the lubricant is 0.100 Pa-s. One surface has an area of 1 dm and slides over the plane surface with a velocity of 1 km/hr. Determine the friction force due to shearing of the lubricant film. Assume the friction force is the viscosity times the surface area times velocity of the moving surfaces and divided by the lubricant film thickness.
Notes This problem is simple if consistent units are used. Solution: -5 2 Rewriting terms into consistent units, 10µm=10 m, 0.100Pas=0.100Ns/m , 2 2 2
1dm =(0.1m) =0.01m , and
km
1000m 1 = = 0.2778m/ s
hr
3600s Therefore the friction force is given by:
Ns
2
0.1 0.001m 0.2778m/ s
( )
2
( )
Au b m
F = η = =
27.78N
−
5 h
10 m
1.17 A firefighter sprays water on a house. The nozzle diameter is small relative to the hose
diameter, so the force on the nozzle from the water is
dm a
F v = dt
where v is the water velocity, and dm a /dt is the water mass flow per unit time. Calculate the force the firefighter needs to hold the nozzle if the water mass flow rate is 3 tons/hr and the water velocity is 100km/hr. Notes: A ton can be misperceived; in metric units, a ton is 1000kg, while in English units it is 2000 lbf. Since the units in this problem are metric, the metric interpretation is used. This problem is not difficult as long as consistent units are used. Solution: The velocity can be written as:
km 1 0 0 , 0 0 0m
100 = = 27.78m / s
hr 3600s
The mass flow rate can be rewritten as:
dm
3tons 3000kg
a = = = 0.8333kg/ s dt hr
3600s Therefore, the force needed is
F = v dm a dt
a
2154.45N 1867.19N
= 2010.82 ± 143.6 N =
2 ( )
1.4363m / s
( )
P = m a a = 1400kg ± 100kg
, the force is given by
Because slush and mud adhered to the bottom of the car, the weight was estimated to be 1400±100kg. Calculate the force driving the car and indicate the accuracy. Solution: Since the force is given by P=m a
= 27.78m / s ( )
1.19 During an acceleration test of a car, the acceleration was measured to be 1.4363 m/s 2 .
W tot =1346kg+65.55kg+75.23kg+88.66kg+91.32kg+1.349kg±0.5kg=1668.109kg±0.5kg Therefore the total weight is 1668kg.
88.66kg, and 91.32 kg. It is raining and the additional weight due to the water on the car is 1.349 kg. Calculate the total weight of the car, including the weight of the passengers and the weight of the water, using four significant figures. Solution: The weight is
1.18 The mass of a car is 1346 kg. The four passengers in the car weigh 65.55 kg, 75.23 kg,
23.15 N
( ) =
0.8333kg / s
Chapter 2 • Load, Stress and Strain The stepped shaft A-B-C shown in sketch a is loaded with the forces P
1 and/or P 2 . Note
2.1
that P 1 gives a tensile stress σ in B-C and σ /4 in A-B and that P 2 gives a bending stress σ at B and 1.5 σ at A. What is the critical section a) If only P 1 is applied?
b) If only P 2 is applied?
c) If both P 1 and P 2 are applied? Notes: This solution will ignore advanced concepts such as stress concentrations that would exist at the fillet, since this material is covered later in the text. Similarly, the location in the cross section where the maximum stress occurs could be identified with the information in Chapter 4, but will be left for the student. Solution: If only P 1 is applied, the critical section is the length B-C, since the tensile stress is highest in this section. If only P 2 is applied, the largest stress occurs at A. If both P 1 and P 2 are applied, the maximum stress in section A-B occurs at point A and is
σ = σ /4+1.5 σ =1.75 σ
AThe maximum stress in section B-C occurs at point B and is
σ = σ σ =2 σ + B Therefore, the critical section is B-C, in particular point B.
The stepped shaft in Problem 2.1 (sketch a) has loads P and P . Find the load
2.2 1 2
classification if P 1 ’s variation is sinuous and P 2 is the load from a weight
a) If only P 1 is applied.
b) If only P 2 is applied.
c) If both P 1 and P 2 are applied. Notes: This solution uses the terminology on pages 30-32. Solution:
If only P is applied, then the load is cyclic with respect to time (see page 30), and is a 1 normal load (see page 31).
If only P 2 is applied, the load is sustained with respect to time, and is both a shear load and a bending load. However, for long columns, bending is usually more important than shear, so it is reasonable to classify P2 as a bending load.
If both P 1 and P 2 are applied, then the load is cyclic with respect to time (although the mean stress is not zero), and the loading is combined (see page 32).
2.3 The sign convention for bending moments can be derived from the differential equation
for the elastic line:
EIy
”=-M Where
E
=modulus of elasticity, Pa 4 I =area moment of inertia, m
M =bending moment, N-m 2
2 -1
y "=d y /dx =second derivative of y with respect to x, m y=deflection, m This gives a positive (tensile) stress in the fibers on the beam in the positive (y) direction. Find how the slope of the beam changes with increasing x coordinate for positive and negative bending.
Notes: The slope of the beam is the first derivative of the deflection, so one must integrate the given equation to obtain the solution. Solution:
The slope of the beam is given by θ =dy/dx. Integrating the given equation,
dy Mx
- = θ = − C
dx EI
Where C is a constant of integration, and reflects the effects of boundary conditions. If we ignore
C , we can see that if the bending moment is negative, the slope of the beam is positive for
positive x and negative for negative x. If the moment is positive, the slope is negative for positive x and positive for negative x.
A bar hangs freely from a frictionless hinge. A horizontal force P is applied at the bottom
2.4
of the bar until it inclines 30° from the vertical direction. Calculate the horizontal and vertical components of the force on the hinge if the acceleration due to gravity is g, the bar has a constant cross section along its length, and the mass is m a . Notes: This problem is straightforward once a free body diagram is drawn. Force and moment equilibrium is then sufficient to solve the problem.
Solution: The free body diagram of the bar is as shown to the right. Since the bar cross section is constant, we can put the center of gravity at the geometric center of the bar. The reactions at the hinge have been drawn in red. Notice the absence of a moment reaction, since the hinge was expressly described as frictionless (this is a common assumption for hinges).
Most of the time, it saves time to perform moment equilibrium first, before applying force equilibrium. This is true here as well. Therefore, summing moments about the hinge point, Equation (2.2) gives:
l
2P
M m g Pl g
= = sin30 ° − cos30 ° ; m =
a a∑
2 tan30 ° Taking equilibrium in the x-direction, Equation (2.1) gives:
gm
P R P P = = − ; R = = t a n 3 0 ° x x x
- a
∑
2 Similarly, in the y-direction,
P = = R − gm ; R = gm y y a y a
∑
2.5 Sketch b shows the forces acting on a rectangle. Is the rectangle in equilibrium? Notes: To satisfy equilibrium, Equations (2.3) must be satisfied.
Solution: Taking the sum of forces in the x-direction,
F x =30N-20N-10N=0 Σ
Therefore, equilibrium in the x-direction is satisfied. Taking equilibrium in the y-direction gives
F =-18N+20N-20N+18N=0 Σ y
Therefore, equilibrium in the y-direction is satisfied. Taking moments about point A,
M A =(20N)(7cm)+(10N)(5cm)-18N(7cm)=64Ncm Σ Therefore, moment equilibrium is not satisfied, and the rectangle is not in equilibrium.
2.6 Sketch c shows the forces acting on a triangle. Is the triangle in equilibrium? Notes: To satisfy equilibrium, Equations (2.3) must be satisfied.
Solution: Taking the sum of forces in the x-direction,
F x =65.37N-40N(cos10°)-30N(cos30°)=0.00 Σ
Therefore, equilibrium in the x-direction is satisfied. Taking equilibrium in the y-direction gives
F y =39.76N-17.81N-40N(sin10°)-30N(sin30°)=0.00 Σ
Therefore, equilibrium in the y-direction is satisfied. Taking moments about point A,
M A =(39.76N)(2.5cm)-40N(cos10°)(5cm)+30N(sin30°)(6.5cm)=0.00Ncm Σ Therefore, the triangle is in equilibrium.
2.7 Sketch d shows a cube with side lengths a and eight forces acting at the corners. Is the
cube in equilibrium? Notes: For three dimensional equilibrium, Equations (2.1) and (2.2) must be satisfied.
Solution: Taking the sum of forces in the x-direction,
F x =P+P-P-P=0 Σ
Therefore, equilibrium in the x-direction is satisfied. Taking equilibrium in the y-direction gives
F y =P+P-P-P=0 Σ
Therefore, equilibrium in the y-direction is satisfied. There are no forces in the z-direction, so equilibrium is automatically satisfied. Taking moments about the z-axis,
M oz =Pa+Pa-Pa-Pa=0 Σ
Therefore, equilibrium about the z-axis is satisfied. Taking moments about the x-axis,
M oz =Pa-Pa=0 Σ
Therefore, equilibrium about the x-axis is satisfied. Taking moments about the y-axis,
M oy =Pa-Pa=0 Σ Therefore, the cube is in equilibrium.
2.8 A downhill skier stands on a slope with a 5° inclination. The coefficient of friction
between the skis and the snow is 0.10 when the ski is stationary and 0.07 when the ski starts to slide. Is the skier in static equilibrium when he is stationary and when he slides down the slope? Notes: For equilibrium, the force Equations in (2.3) needs to be satisfied. Solution: A free-body diagram of the skier is shown to the right. The skier’s weight is m a g , and the slope has a normal force as a reaction and a friction force which develops tangent to the slope. Summing forces in the direction normal to the slope yields:
F =N-m g cos5°; N=m g cos5° Σ a a
Taking forces tangent to the slope,
F =m a g sin5°-µN=m a g sin5°-µm a g cos5°; µ=tan5°=0.08749
ΣIf the skier is stationary, the coefficient of friction is 0.10, and there is enough friction to keep the skier stationary. If the skier is moving, then the coefficient of friction is 0.07, so the friction force developed is not enough to maintain equilibrium. The skier will accelerate.
Given the components shown in sketches e and f, draw the free-body diagram of each
2.9 component and calculate the forces.
Notes: For equilibrium, the force Equations in (2.3) needs to be satisfied. Assume there is no friction in sketch (e). In sketch f, moment equilibrium shows that the only possible force in each member is an axial force. Solution: The free body diagram for sketch e is shown to the left. Flat surfaces give normal reaction forces as shown. From Equation (2.3), and summing forces in the y-direction,
F y =0=-W+N 1 cos30°+N 2 cos60°; N 1 =W/cos30°-N 2 (cos60°/cos30°) Σ
Summing forces in the x-direction,
F x =0=N 1 sin30°-N 2 sin60°=(W/cos30°-N 2 (cos60°/cos30°))sin30°-N 2 sin60°; N 2 =0.5W Σ
Substituting into the x-direction equilibrium equation gives N 1 =0.866W. The free body diagram for sketch f is shown to the right. For equilibrium in the x-direction,
F =0=P-P sin30°; P =2P=2.4kN Σ x 2 2 Taking equilibrium in the y-direciton.
F y =-P 1 +P 2 cos30°=-P 1 +2Pcos30°; P 1 =1.732P=2.08kN Σ A 5-m-long bar is loaded as shown in sketch g. The bar cross section is constant along its
2.10 length. Draw the shear and moment diagrams and locate the critical section.
Notes: In such problems, one must first obtain the reactions before generating the shear and moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment diagrams are obtained through the method of sections. Solution: The reaction forces have been added to the figure in red. Summing moments about the origin gives
M o =0=10kN(2m)-A y (3m)+3.2kN(5m);A y =12kN
ΣSumming forces in the y-direction yields
F y =0=-O y +10kN-A y +3.2kN; O y =10kN-12kN+3.2kN=1.2kN Σ The method of sections described on page 39 is used to obtain the shear and moment diagrams.
This proceeds as follows:
a) 0 ≤ x <2m:
F
y =0=-O y +V; V=O y =1.2kN ΣM =0=M-V(x); M=Vx=(1.2kN)x
Σ O
Note: at x=2m, M=2.4kNm
b) 2m ≤ x <3m:
F
y =0=-O y +10kN+V;V=O y -10kN=-8.8kN Σ
M
O =0=M-V(x)-10kN(2m); M=20kNm-(8.8kN)x Σc)3m ≤ x :
F
y =0=-O y +10kN-12kN+V;V=3.2kN ΣM =0=M-V(x)-10kN(2m)+12kN(3m);
Σ O
M=-16kNm+3.2x The shear and moment diagrams are plotted as follows: From these shear and moment diagrams, it can be seen that the critical section is at point A.
2.11 A beam is supported by a wire in one end and by a pin in the other end. The beam is
loaded by a force parallel with the wire. What is the direction of the reaction force in the pin? Notes: This merely requires an understanding of the equations of equilibrium. One of the fundamental notions is that “you can’t push on a rope”, so that the wire can only support an axial, tensile load. Solution: Force equilibrium gives that the force in the pin has to be parallel with the wire since the wire can only transmit an axial force. Therefore, the force in the pin has to be in the same direction as in the wire.
Sketch h shows a simple bridge. The midpoint of the bridge is held up by a wire leading
2.12
to the top hinge of the two equal beams. When a truck runs over the bridge, the force in the wire is measured to be 500,000 N. Determine the forces in the beams and the horizontal force component in the middle of the bridge.
Notes: This problem can be easily solved by drawing a free body diagram of one of the beams and then applying the equations of equilibrium. Solution: A free body diagram of the left beam is shown to the right. It is assumed that each beam takes one-half the load, as shown. From moment equilibrium, about the pin, it can be seen that
M =0=(250,000N)(lcos45°)-P (lsin45°); P =250,000N
Σ H HTherefore, the load carried by the beam is
2
2
= 250kN 250kN = 353.6kN beam ( ) ( )
+
P
Taking force equilibrium in the horizontal direction gives
F h =0=R x -P H ; R x =P H =250kN Σ This is the horizontal force component in the middle of the bridge.
2.13 Sketch i shows a 0.05-m-diameter steel shaft supported by self-aligning bearings (which
can provide radial but not bending loads on the shaft). A gear causes each force to be applied as shown. All length dimensions are in meters. The shaft can be considered weightless. Determine the forces at A and B and the maximum bending stress. Draw shear and moment diagrams. Notes: In such problems, one must first obtain the reactions before generating the shear and moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment diagrams are obtained through direct integration, as suggested by Equations (2.4) and (2.5). The maximum stress is given by σ =Mc/I where I is the moment of inertia, given in Table 4.1 on page 148.
Solution: Taking moment equilibrium about point A yields:
M A =0=(1kN)(0.2m)-B y (0.5m)-(0.5kN)(0.64m); B y =-0.24kN=-240N Σ
Vertical force equilibrium gives A : y
F y =0=A y -1kN+B y +0.5kN; A y =1kN-B y -0.5kN=0.74kN=740N
ΣThe shear and moment diagrams are constructed directly from the applied load, using Equations (2.4) and (2.5). Since the loads are simple, this approach allows simple integration. The diagrams are as follows: As can be seen, the maximum moment is at x=0.2 and has the value of
M max =(740N)(0.2m)=148Nm. This is the area under the shear force curve, indicated in red.
The moment of inertia of a round cross section is (see Table 4.1 on page 148, or the inside back cover):
4
4 d
0.05m
( )
7
4 −
I m
= π = π = 3.068 ×10
64
64 The maximum bending stress in the shaft is:
Mc
148Nm 0.025m
( ) ( )
σ = = =12.06MPa
7
4 −
I m
3.068 ×
10
2.14 Sketch j shows a simply supported bar loaded with a force P at a position one-third of the
length from one of the support points. Determine the shear force and bending moment in the bar. Also, draw the shear and moment diagrams.
Notes: In such problems, one must first obtain the reactions before generating the shear and moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment diagrams are obtained through direct integration, as suggested by Equations (2.4) and (2.5).
Solution: Moment equilibrium about point A yields:
M A =0=P(l/3)-B y (l); B y =P/3 Σ
Vertical force equilibrium gives:
F =0=A -P+B ; A =P-B =P-P/3=2P/3 Σ y y y y y The shear and moment diagrams are obtained through direct integration and are shown below.
Sketch k shows a simply supported bar loaded by two equally large forces P at a distance
2.15 l
/4 from its ends. Determine the shear force and bending moment in the bar, and find the critical section with the largest bending moment. Also, draw the shear and moment diagrams.
Notes: In such problems, one must first obtain the reactions before generating the shear and moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment diagrams are obtained through direct integration, as suggested by Equations (2.4) and (2.5).
Solution: Moment equilibrium about point A yields:
M A =0=P(l/4)+P(3l/4)-B y (l); B y =P Σ
Vertical force equilibrium gives
F y =0=A y -P-P+B y ; A y =P Σ The shear and moment diagrams are obtained through direct integration and are shown below.
The largest bending moment occurs between the two applied forces and has the magnitude of M =Pl/4.
2.16 The bar shown in sketch l is loaded by the force P. Determine the shear force and bending moment in the bar. Also, draw the shear and moment diagrams.
Notes: In such problems, one must first obtain the reactions before generating the shear and moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment diagrams are obtained through direct integration, as suggested by Equations (2.4) and (2.5).
Solution: Moment equilibrium about point A yields:
M A =0=P l +B y l ; B y =-P Σ
Vertical force equilibrium gives:
F =0=-P+A +B ; A =2P
Σ y y y y The shear and moment diagrams are obtained through direct integration and are shown below.Sketch m shows a simply supported bar with a constant load per unit length w imposed
2.17 over its entire length. Determine the shear force and bending moment as functions of x.
Draw a graph of these functions. Also, find the critical section with the largest bending moment.
Notes: In such problems, one must first obtain the reactions before generating the shear and moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment diagrams are obtained through direct integration, as suggested by Equations (2.4) and (2.5).
Solution: For the purposes of equilibrium, the distributed load can be replaced by a point load of magnitude
w l
at the midpoint of the span. Therefore, moment equilibrium about point A yields:
l w lM = = w l B l ; B =
A y y
∑
−
2
2 Force equilibrium in the vertical direction gives:
w l
- F = = A − w l B ; A =
y y y y ∑
2 The shear and moment diagrams are obtained through direct integration and are shown below. It can be seen that the maximum moment occurs at mid-span. The magnitude of this force is the area under either triangle on the shear curve, as suggested by Equation (2.5). This gives
2 w l l w l
1
o
M= max
=
2
2
2
8 Sketch n shows a simply supported beam loaded with a ramp function over its entire
2.18
length, the largest value being 2P/l. Determine the shear force and the bending moment and the critical section with the largest bending moment. Also, draw the shear and moment diagrams.
Notes: One must first obtain the reactions before generating the shear and moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment diagrams are obtained through the method of sections.
Solution: The loading is described by
x
2P
2Px
q x = −
( )
2 l l = − l
In analyzing equilibrium, the loading is replaced by a force of magnitude P at a location of
x
=2l/3. Therefore, moment equilibrium about point A gives:
2l
2P
M = = P B l ; B =
A y ( ) y
∑
−
3
3 Vertical force equilibrium gives:
P
+
F = = A − P B ; A = P − B =
y y y y y
∑3 Taking a section at a location x in the span,
2 P Px
1
2Px
F A x
V = = − − ; V = − y y ( )
∑
2
2
2 l 3 l
1
2Px 2
= = − ; ( )
- M x x Vx M
A ∑
2
2 3
l
3
2
3
P Px Px Px
2Px
x
M = − = − +
2 2
2
3 l
3 3l 3l
The shear and moment diagrams are shown below:
The maximum moment occurs when V=0, which occurs at:
2 P Px l
= − ; x =
2
3
3
l
Substituting this into the moment equation gives
3
P l P l Pl Pl
2Pl
M = − = − = max
2
3 3 3 3 3 9 3 9 3 3l
Draw the shear and moment diagrams, and determine the critical section for the loading
2.19
conditions shown in sketch o. Neglect the weight of the bar. Let w =100kN/m, P=5kN, 4 M =2x10 N-m, and l=300mm.
Notes: One must first obtain the reactions before generating the shear and moment diagrams. Also, it is useful to consider moment equilibrium first. In this problem, the shear and moment diagrams are obtained through singularity functions. Table 2.2 on page 43 is useful for writing the load distribution in terms of singularity functions.
Solution: For equilibrium reasons, the distributed loads can be replaced by point loads as shown below: Taking moment equilibrium about point A gives:
0.200m
M B
= = 5kN 10kN 0.15m 5kN 0.233 m − 0.2m 5kN 0.3m 20kNm + + A ( ) ( ) ( ) ) y ( ) (
- +
∑