mechanical design an integrated approach Part 4 pptx

supported beam is subjected to a concentrated load at a distance a from the left end as shown in Figure 4.6.. 4.5 BEAM DEFLECTIONS BY SUPERPOSITION The elastic deflections and slopes of

sion of load, shear, or moment can be formulated and individual preference The approach

to solving the deflection problem beginning with Eq (4.16c) or (4.16b) is known as the multiple-integration method When Eq, (4.16a) is used, because two integrations are required to obtain the 0, this is called the double-integration method The constants of the integration are evaluated using the specified conditions on the ends of the beam, that is, the boundary conditions Frequently encountered conditions that may apply at the ends (x = a) of a beam are shown in Figure 4.5 We see from the figure that the force (static) vari- ables M, V, and the geometric (kinematic) variables v, 6 are 0 for common situations

If the beam has a cross-sectional width b that is large compared to the depth h (i.c.,

b > h), the beam is stiffer, and the deflection, is less than that determined by Eqs (4.16) for narrow beams The large cross-sectional width prevents the lateral expansion and con- traction of the material, and the deflection is thereby reduced, as shown in Section 4.9 An improved value for the deflection uv of wide beams is obtained by multiplying the result given by the equation for a narrow beam by (1 — v), where v is Poisson’s ratio

EXAMPLE 4.5 Finding Beam Deflections by the Double-Integration Method

A simply supported beam is subjected to a concentrated load at a distance a from the left end as

shown in Figure 4.6 Develop

(a) The expressions for the elastic curve

(b):: The deflection at point C for the case in which a = b = L/2

Double integrations of these equations give the results

For segment AC For segment CB EH(= 2 Eng = x — pea)

En = Fee be Ett = Fox = 2Œ =4)? +ó Elyse barte Ely = Te 2a) tox ten From the boundary and the continuity conditions, we easily obtain

ves epee EO Be) ea @sxsl)

Then, through the use of Eq (4.15) the slopes for the two parts of the beam can readily be found

Force P acts at the middle of the beam span (a = b = L/2) and hence Eqs (4.17) result in

4.5 BEAM DEFLECTIONS BY SUPERPOSITION

The elastic deflections (and slopes) of beams subjected to simple loads have been solved and are readily available (see Tables A.9 and A.10) In practice, for combined load config- urations,-the method of superposition may be applied to simplify the analysis and design

The method is valid whenever displacements are linearly proportional to the applied loads

‘This is the case if Hooke’s law holds for the material and deflections are small

To demonstrate the method, consider the beam of Figure 4.7a, replaced by the beams depicted in Figures 4.7b and 4.7c At point C, the beam undergoes deflections (v)p and (v)y, due to P and M, respectively Hence, the deflection uc due to combined loading is

ve = (uc)p + (vc) From the cases 1 and 2 of Table A.9, we have

_ 5PLỀ ML2

Similarly, the deflection and the angle of rotation at any point of the beam can be found by

the foregoing procedure

The method of superposition can be effectively applied to obtain deflections or reactions for statically indeterminate beams In these problems, the redundant reactions are considered unknown loads and the corresponding supports are removed or modified accordingly, Next, superposition is employed: The load diagrams are drawn and expressions are written for the deflections produced by the individual loads (both known and unknown);

the redundant reactions are computed by satisfying the geometric boundary conditions Fol- lowing this, all other reactions can be found from equations of static equilibrium

The steps described in the preceding paragraph can be made clearer though the illus- tration of a beam statically indeterminate to the first degree (Figure 4.8a) Reaction Rg is selected as redundant and treated as unknown load by eliminating the support at B

Decomposition of the loads is shown in Figures 4.8b and 4.8c Deflections due to w and the redundant Rg are (see cases 5 and 8 of Table A.9):

(up) so 5w1 ; (op) Velen Ral?

7”2AET pie 6EI

From geometry of the original beam,

Figure 4.8 Deflections of a two-span continuous bear,

The remaining reactions are Ra = R, = 3wL/8, as determined by applying the equa-

tions of equilibrium Having the reactions available, deflection can be obtained using the

method discussed in the preceding section

Determination: of Interaction Force between Two Beams Placed EXAMPLE 4.6 One on Top of the Other

Two simply supported beams are situated one‘on top of the other as shown in Figure 4.9 The top

beam:is: subjected to.a' uniformly: distributed load of intensity w Find (a), The interaction force R'at midspan C acting upward on the beam AB and acting downward

on Đeam ĐE, : (b)- The maxinum miortenf añd deflection in beam AB

Eigure 4.9 Example 4.6:

Solution: Both beams are statically indeterminate to the first degree We select R as redundant and

treat it'as an unknown load Considering the two beams in turn and using the data in Table A.9, the deflections at the center are as follows For beam AB, due to load w and owing to R,

A schematic representation of the frame of a winch crane

is shown in Figure 1.5 Determine the deflection under the

load using the method of superposition

Given: ‘The geometry and loading of the critical portion

of the frame are known from Case Study 3-1

Data:

E = 200 GPa, = 2.01(10~°) m*

Assumptions: The loading is static Deflection due to

transverse shear is neglected

Solution: See Figure 3.31 and Table A.9

When the load P and the weight w of the cantilever

depicted in the figure act alone, displacement at D (from

WINCH CRANE FRAME DEFLECTION ANALYSIS

cases | and 3 of Table A.9) are PL? /3EJ and wlth /SEI, respectively It follows that, the deflection vp at the free

end owing to the combined loading is

3E1 SEI

Substituting the given numerical values into the pre-

ceding expression, we have

= —8.6 mm Here minus sign means a downward displacement

Comment: Since vp </A/2, the magnitude of the

deflection obtained is well within the acceptable range (see Section 3.7)

Case Study 4-2 | Bout CUTTER DEFLECTION ANALYSIS

Members 2 and 3 of the bolt cutter shown in Figure 3.32 are critically stressed Determine the deflections employ- ing the superposition method

Given: The dimensions (in inches) and loading are known from Case Study 3-2 The parts are made of AISI

1080 HR steel having E = 30 x 10° psi

Assumptions: The loading is static The member 2

can be approximated as a simple beam with an overhang

Solution: See Figures 3.32 (repeated here), 4.10 and Table A.9

Member 3 The elongation of this tensile link

(Figure 3.32a) is obtained from Eq (4.1) So, due to

{a) @)

Figure 3.32 (repeated) Some free-body diagrams of bolt cutter shown in Figure 1.6:

(a) link 3; (b) jaw 2 (continued)

Case Study (CONCLUDED)

Đ - ` The angle @, at the support A (from case 7 of Table A.9) is

128(1.25)

= =569(0759 in, 2($) (4) G0 x 109) 6.900") in Member 2 This jaw is loaded as shown in Fig- ure 3.32b The deflection of point D is made up of two parts: a displacement v, owing to bending of part DA act-

ing as a cantilever beam and a displacement v, caused by

the rotation of the beam axis at A (Figure 4.10)

The deflection v; at D (by case | of Table A.9) is

where M = Qa The displacement v2 of point D, due to

only the rotation at A, is equal to @4@, or

Comment: Only very small deflections are allowed in

members 2 and 3 to guarantee the proper cutting stroke,

and the values found are acceptable

of the deflection v and the properties of the area of the bending moment diagram Usually it

gives more rapid solution than integration methods when the deflection and slope at only one point of the beam are required The moment-area method is particularly effective in the

analysis of beams of variable cross sections with uniform or concentrated loading (3-5]

MOMENT-AREA THEOREMS

Two theorems form the basis of the moment-area approach These principles are developed

by considering a segment AB of the deflection curve of a beam under an arbitrary loading

The sketches of the W/E/ diagram and greatly exaggerated deflection curve are shown in

(a) M/E! diagram; (b) elastic curve

Figure 4.11, Here M is the bending moment in the beam and £/ represents the flexural

rigidity The changes in the angle dO of the tangents at the ends of an element of length dx and the bending moment are connected through Eqs (4.14) and (4.15):

This is called the first moment-area theorem: The change in angle 024 between the tangents

to the elastic curve at two points A and B equals the area of the M/EJ diagram between those points Note that the angle 6g4 and the area of the M/EJ diagrams have the same sign That means a positive (negative) area corresponds to a counterclockwise (clockwise) rotation of the tangent to the elastic curve as we proceed in the x direction Hence, 0g,

shown in Figure 4.11b is positive

Inasmuch as the deflection of a beam are taken to be small, we see from Figure 4.11b that the vertical distance dt due to the effect of curvature of an element of length dx equals

x d@, where dO is defined by Eq (a) Therefore, vertical distance AA’, the tangential devi- ation tag of point A from the tangent at B is

Be Máy _ “ae [aca ot diagram between Aand 2] T1 (4.24)

it is obvious whether the beam deflects upward or downward and whether the slope is clockwise or counterclockwise, When this is the case, it is not necessary to follow the sign conventions described for the moment-area method: We calculate the absolute values and find the directions by inspection

APPLICATION OF THE MOMENT-AREA METHOD

Determination of beam deflections by moment-area theorems is fairly routine They are equally applicable for rigid frames In continuous beams, the two sides of a joint are 180° to one another, whereas in rigid frames the sides of a joint often are at 90° to one an- other Our discussion is limited to beam problems A correctly constructed M/EI diagram and a sketch of the elastic curve are always necessary Table A.3 may be used to obtain the

areas and centroidal distances of common shapes The slopes of points on the beam with

respect to one another can be found from Eq (4.23) and the deflection, using Eq (4.24) or (4.25), The moment-area procedure is readily used for beams in which the direction of the tangent to the elastic curve at one or more points is known (e.g., cantilevered beams) For computational simplicity, often M/EJ diagrams are drawn and the formulations made in terms of the quantity E/; that is, numerical values of EJ may be substituted in the final step of the solution

For a statically determinate beam with various loads or an indeterminate beam, the displacements determined by the moment-area method are usually best found by super- position This requires a series of diagrams indicating the moment due to each load or reaction drawn on a separate sketch In this manner, calculations can be simplified, because the areas of the separate M/EJ diagrams may be simple geometric forms When treating statically indeterminate problems, each additional compatibility condition is expressed by

a moment-area equation to supplement the equations of statics

= EXAMPLE 4.7 Shaft: Deflection by the Moment-Aiea Method

A-simple shaft carries its own weight of intensity w, as depicted in Figure 4.12a Determine the slopes

at the ends and center deflection:

Assumption: Bearings act as simple supports

Solution: : Inasmuch as the flexural rigidity EJ is constant, the M/E diagram has the same para-

bolic shape as the bending-moment diagram (Figure 4.12b), where the area properties are taken from

Table'A:3:.Thé elastic curve is depicted in Figure 4.12c, with the tangent drawn at A

©) Figure 4.12 Example 4.7

The shaft and loading aré symmetric about the cénter C; hence, the tangent to the elastic curve

at C is horizontal: @¢ = 0 Therefore, 6¢4 = 0-64 or 04 = ~Oc, and

The minus sign means that the end A of the beam rotates clockwise, as stiown in the figure

Through the use of the second moment-area theorem, Eq (4.25),

The minus sign indicates that the deflection is downward

Comment: Alternatively, the moment of area A, about point A, Eq (4.24), readily gives the nu-

merical value of Umax -

169

Solution: The M/EJ diagram is divided conveniently into its component parts, as shown in Figure 4.13b:

Pape 8EI Age >, 16B! Ỷ 38m SEI

The elastic curve is in Figure 4.13c Inasmuch as 04 = 0 and vq = 0, we have Oc = Oc4, 0g = Opa,

ĐC = te4, and vg = tag

Applying the first moment-area theorem,

5PL?

Op = Art Ag + Ag = ~T6ET (4.28a)

The minus sign means that the rotations are clockwise From the second moment-area theorem,

Reactions of a Propped up Cantilever by the Moment-Area Method

‘A propped cantilevered beam is loaded by a concentrated force P acting at the position shown in Eieure 4144 Determine the reactional forces and moments at the ends of the beam 8

Solution: The reactions indicated in Figure 4.14a shaws that the beam is statically indeterminate

to the first degree We select Rg as a redundant (or unknown) load and remove support B (Fig-

uré 4.14b) The corresponding M/EJ diagram is in Figure 4.14c, with the component areas

A moving body striking a structure delivers a suddenly applied dynamic force that is called

an impact or shock load Details concerning the material behavior under dynamic loading are presented in Section 2.9 and Chapter 8 Although the impact load causes elastic mem- bers to vibrate until equilibrium is reestablished, our concern here is with only the influence

of impact or shock force on the maximum stress and deformation within the member

Note that the design of engineering structures subject to suddenly applied loads is complicated by a number of factors, and theoretical considerations generally serve only qualitatively to guide the design [6-8] In Sections 4.8 and 4.9, typical impact problems are analyzed using the energy method of the mechanics of materials theory together with the following common assumptions:

1 The displacement is proportional to the loads

2 The material behaves elastically, and a static stress-strain diagram is also valid under impact

3 The inertia of the member resisting impact may be neglected

No energy is dissipated because of local deformation at the point of impact or at the

supports

Obviously, the energy approach leads to an approximate value for impact loading It presupposes that the stresses throughout the impacted member reach peak values at the same time In a more exact method, the stress at any position is treated as a function of time, and waves of stress are found to sweep through the elastic material at a propagation rate This wave method gives higher stresses than the energy method However, the former

is more complicated than the latter and not discussed in this text The reader is directed to

references for further information [9-12]

4.8 LONGITUDINAL AND BENDING IMPACT

Here, we determine the stress and deflection caused by linear or longitudinal and bending impact loads In machinery, the longitudinal impact may take place in linkages, hammer-type power tools, coupling-connected cars, hoisting rope, and helical springs Examples of bending

impact are found in shafts and structural members, such as beams, plates, shells, and vessels

FREELY FALLING WEIGHT

Consider the free-standing spring of Figure 4.15a, on which is dropped a body of mass m from a height 4 Inasmuch as the velocity is 0 initially and 0 again at the instant of maximum deflection of the spring (8nax), the change in kinetic energy of the system is 0 Therefore, the work done by gravity on the body as it falls is equal to the resisting work done by the spring:

WÁ + 8imas) = ts max (4.30)

CHAPTER4 ® DEFLECTION AND IMPACT

The deflection corresponding to a static force equal to the weight of the body is simply

W/k This is called the static deflection, 5 The general expression of maximum dynamic deflection is, using Eq (4.30),

HorIZONTALLY MOVING WEIGHT

An analysis similar to the preceding one may be used to develop expressions for the case

of a mass (m = W/g) in horizontal motion with a velocity v, stopped by an elastic body,

In this case, kinetic energy 2, = mu? /2 replaces W(h + Smax), the work done by W, in

173

When the body hits the end of a prismatic bar of length L and axial rigidity AE (Fig- ure 4.15b), we have k = AE/L and hence 6, = mgL/AE Equations (4.36) are therefore

tend to promote brittle fracture This point has been treated in Section 2.9

Impact Loading ona Rod

Solution: The cross-sectional area of the rod A = z(l /23?/4 = 1/16 in? pa RR

(a) For the rod with the washer, the static deflection is pe

Comments: The differencẻ in stress for the preceding two solutions is large This suggests the

need for flexible’ systems for withstanding impact loads Interestingly, bolts subjected to dynamic

Toads, stich as those used to attach the ends to the tube in pneumatic cylinders, are often designed with long’ grips (see Section 15.9) to take advantage of the more favorable stress conditions

EXAMPLE 4.11

Impact Loading on a Beam

Aweight W is dropped a height A, striking at midspan a simply supported steel beam of length L The

beam is of rectangular cross section of width b and depth d (Figure 4.17) Calculate the maximum

deflection and maxiraum stress for these two cases:

(a) The beam is rigidly supported at each end, (b) The beam is supported at each end by springs

Figure 4.17) Example 4.11

176 PART! ® FUNDAMENTALS

Given: W = 100N2A= 150 mm, L =2m, b= 30mm, andd = 60 mm

: Assumptions: Modulus of elasticity 2 = 200 GPa and spring rate k = 200 kN/m

Solution: | We have Moax = WL/4 at point C and 7 = bd? /12 The maximum deflection, due to a

static: load, is (ftom case Sof Table A.9):

(b) The static deflection of the beam due to its own bending and the deformation of the

springs is

50

Sq = 91 + 208” 9.404 mm The impact factor is then

2(0.15) K=l+jt+ 0:404(10-3) 28.27

Hence, Bmax == 28.27(0.404) = 11.42 mm Ginax == 28.27(2.778) == 78.53 MPa

Comments: Comparing the results, we observe that dynamic loading considerably increases de- flection and stress in a beam Also noted is a reduction in stress with increased flexibility, owing to the spring added to the supports However, the values calculated are probably somewhat higher than

the actual quantities, because of our simplifying assumptions 3 and 4

Advantage will be taken of the analogy between linear and torsional systems to readily

write the final relationships

Consider a circular prismatic shaft of flexural rigidity GJ, length L, fixed at one end and subjected to a suddenly applied torque T at the other end (Figure 4.18) The shaft stiff- ness, from Eq (4.10), is k = GJ/L, where J = zZ4/32 and d is the diameter The maxi- mum dynamic angie of twist (in rad), from Eq (4.36a), is

Here A represents the cross-sectional area of the shaft

Recall from Section 1.11 that, for a rotating wheel of constant thickness, the kinetic energy is expressed in the form

(4.41)

Figure 4.18 Bar subjected to impact torsion

In the foregoing, we have

7 = mass moment of inertia (N-s?-m or Ib- ?-in.)

@ = angular velocity (rad/s)

m= mass (kg or lb - s2ñn.)

b = radius

t = thickness

p = mass density (kg/m? or Ib- s*/in.*)

As before, W and g are the weight and acceleration of gravity, respectively A detailed treat- ment of stress and displacement in disk flywheels is given in Section 16.5

Note that in the case of wheel of variable thickness, the mass moment of inertia may conveniently be obtained from the expression

I= mr? (4.43)

The quantity r is called the radius of gyration for the mass It is hypothetical distance from the wheel center at which the entire mass could be concentrated and still have the same moment of inertia as the original mass

Impact Loading ‘on‘a Shaft

EXAMPLE 4.12

A shaft of diameter d and length L has a flywheel (radius of gyration r, weight W, modulus of rigid-

ity G, yield strength in shear S,,) at oné end and runs at a speed of n If the shaft is instantly stopped

at the other end, determine (a).: The maximum shaft angle of twist

{b) The maximuin shear stress

Given: 4= 3 in, E = 2.5 ft, W = 1201b, = 10 in., n = 150 rpm

Assumption: The shaft is made of ASTM-A24? steel So, by Table B.1, G = 11.5 x 10° psi and

Sys = 30 ksi

Solution:: The area properties of the shaft are

#3)? s2 z@% A= q =T, 069 in.*, J = 3 = 7 7.952 in 3

CHAPTER 4 e DEFLECTION AND IMPACT

The atigular velocity equals

2z Qn

o=n (5) == 150 (5) = Sx rad/s

(a): The kinetic energy of the flywheel must be absorbed by the shaft So, substituting

Bq (4.43) into Eq (4.41), we have

Wor?

28 1205x302 2686)

Ey = (a)

= 3835 in-Ib From Eq (4.39a);

*4.10 BENDING OF THIN PLATES

A plate is an jnitiaily flat structural member with thickness small compared with remaining dimensions it is usual to divide the plate thickness f into equal halves by a plane parallel

to the faces This plane is called the midsurface of the plate The plate thickness is mea-

sured in a direction normal to the midsurface at each point under consideration Plates of technical importance are usually defined as thin when the ratio of the thickness to the smaller span length is less than 1/20 Here, we discuss briefly the bending of thin plates

For a detailed treatment of the subject, see [13~15]

Basic ASSUMPTIONS Consider a plate before deformation, depicted in Figure 4.19a, where the xy plane coin-

cides with the midsurface and hence the z deflection is 0 When deformation occurs due to external loading, the midsurface at any point x4, ya undergoes a deflection w Referring to

2 Straight lines (such as mn) initially normal to the midsurface remain straight and nor-

mal to that plane after bending:

3 No midsurface straining occurs due to bending This is equivalent to stating that Strains 1⁄s¿, Yee, and €, are negligible

4 The component of stress normal to the midsurface, o,, may be neglected

These presuppositions are analogous to those associated with the simple bending theory of beams

tures at the midsurface in planes parailel to the zx (Figure 4.19b), yz, and xy planes are,

Examining the preceding relationships, we are left to conclude that a circle of curva-

ture can be constructed similarly to Mohr’s circle of strain The curvatures hence trans-

form in the same manner as strains It can be verified by using Mohr’s circle that

(1/7) + (1/ry) = VẦw, The sum of the curvatures in perpendicular directions, called the

average curvature, is invariant with respect to rotation of the coordinate axis This asser- tion is valid at any location on the midsurface

PLATE STRESS, CURVATURE, AND MOMENT RELATIONS

For a thin plate, substituting Eqs (4.44) into Hooke’s law, we obtain

1/2 1/2 / oy dy dz = ay f 20, dz = M, dy

Here, t is the thickness of the plate Substituting into this expression the stress o, given

by Eqs (4.46), we obtain M, in terms of curvatures Expressions involving M, and

181

{a)

Figure 4.20 (a) Plate segment in pure bending

{b) Positive stresses on an element in the bottom half of a plate

Myy = My, are derived in a like manner In so doing, we have

a factor 1/(1 — v) or about 10% for v = 0.3

According to the sign convention, a positive moment is one that results in positive stresses in the positive (bottom) half of the plate (see Section 1.13), as depicted in Fig- ure 4.20b The maximum stresses occurring on the surface of the plate are obtained by sub- stituting z = 1/2 into Eqs (4.46), together with the use of Eqs (4.48), as

Since there is a direct correspondence between the moments and stresses, the equation for transforming the stresses should be identical with that used for the moments Mohr’s circle

4.11 DEFLECTION OF PLATES BY INTEGRATION

Consider an element dx dy of the plate subject to a uniformly distributed load per unit area p

(Figure 4.21) In addition to the moments M,, My, and M,y previously discussed, we now find vertical shear forces Q, and Q, (force per unit length) acting on the sides of the element With the change of location, for example, by dx, one of the moment components, say, M,, acting on the negative x face, varies in value relative to the positive x face as

M, + (0M,,/0x) dx Treating all components similarly, the state of stress resultants shown

Variations in the moment and force resultants are governed by the conditions of equi- librium Application of the equations of statics to Figure 4.21 leads to a single differential equation in terms of the moments This, when combined with Eqs (4.48), results in [13]

Oxt (4.514)

or, in concise form,

WỶw= (4.51b)

The preceding expression, first derived by Lagrange in 1811, is the governing differential

equation for deflection of thin plates

Other typical conditions at the boundaries may be expressed similarly

To determine the deflection w, we must integrate Eq (4.51) with the constants of inte- gration dependent on the appropriate boundary conditions Having the deflection available, the stress (as well as strain and curvature) components are obtained using the formulas derived in the preceding section This is illustrated in the solution of the following problem

EXAMPLE 4.13 | Determination of Deflection and Stress in'a Plate

A long, narrow plate of width b and thickness ¢, a so-called plate strip, is simply supported at edges y= O-and y= b, as depicted in Figure 4.23 The plate carries the loading of the form

Solution: Due'to symmetry in the loading and end restraints about the x axis, the plate deforms

into: a cylindtical’ surface with its generating line parallel to the x’ axis Since, for this situation, 8w/8x = Ú and 82w/2xôy =s 0, Eqs (4.48) reduce to

Mỹ= T19: My == gã (4.54) Also Bg (4:51) becomes

CHAPTER 4 8 DEFLECTION AND IMPACT

(a): Introducing Eq (a) into Eq (4.55), integrating, and satisfying the boundary conditions

Pig w= 0, Se = 0, (y = 0, y= b)

we obtain the deflection

b pe (ay

‘The largest deflection of the plate occurs at y = b/2 Therefore,

B\t Winax = (=) = (4.57) The moments are now readily determined carrying Eq (4.56) into (4.54) Then, the maxi- mum stresses, occurring at y = b/2, are found applying Eqs (4.50) as

2

Øy max =2 0.6 Po 7}" Os trax = VOy.max = 0.2 Po () > Try = 0 (4.58)

(}- Substituting the given data into Eq (4.49), we have

Ởy max = 344) =: 4.8 MPa

Comment: The result, Wma, /t = 0.02, shows that the deflection surface is extremely flat, as is

often the case for small deflections

3 Ugural, A C Mechanics of Materials New York: McGraw-Hill, 1991

4 West, H H Fundamentals of Structural Analysis, 2nd ed New York: Wiley, 2002

Harris, C M Shock and Vibration Handbook New York: McGraw-Hill, 1988

9 Burr, A H., and J B Cheatham, Mechanical Analysis and Design, 2nd ed Upper Saddle River, NJ: Prentice Hall, 1995

10 Goldsmith, W Impact London: Edward Arnold Ltd., 1960

11, Koisky, H Stress Waves in Solids New York: Dover, 1965

12 Varley, E., ed “The Propagation of Shock Waves in Solids.” AMD-17 Symposium New York:

ASME, 1976

13 Ugural, A C Stresses in Plates and Shells, 2nd ed New York: McGraw-Hill, 1999

14 Szilard, R Theory and Analysis of Plates Upper Saddle River, NJ: Prentice Hall, 1974

15 Timoshenko, S P., and S Woinowsky-Krieger Theory of Plates and Shells, Ind ed New York:

(a) What is the required diameter of the rod?

(b) Calculate the spring rate for the rod

Given: P = 10 KN, E = 200 GPa, S$, = 250 MPa, L = 6m, 6 == S mm Design Decision: The rod will be made of ASTM-A242 steel Take » = 1.2

4.2 Before loading there is a gap A between the wall and the right end of the copper rod of diameter

d (Figure P4.2), Calculate the reactions at A and B, after the rod is subjected to an axial load of P

Given: A = 0.0)4in., d = 4 in, P = 8 kips, E = 17 x 10° psi

E——8in.—— [10 fo, ke A pat,

Figure P4.2 4.3 Atroom temperature (20°C), a gap A exits between the wall and the right end of the bars shown

in Figure P4.3 Determine (a) The compressive axial force in the bars after temperature reaches 140°C

Aq = 1000 mm?, Eq =70GPa, a = 23 x 10°6/°C, A= tmm

A, = 500 mm?, E, = 210 GPa, oy = 12 x 107S/°C, Redo Problem 4.3 for the case in which A = 0

Two steel shafts are connected by gears and subjected to a torque 7, as shown in Figure P4.5

Calculate

(a) The angle of rotation in degrees at D

(b) The maximum shear stress in shaft AB

Given: G = 79 GPa, T = 500N-m, d; = 45 mm, @ = 35 mm

Figure P4.5 Determine the diameter d, of shaft AB shown in Figure P4.5, for the case in which the maximum shear stress in each shaft is limited to 150 MPa

Design Decisions: d; = 65 mm The factor of safety against shear is n = 1.2

A disk is attached to a 40-mm diameter, 0.5-m long steel shaft (G = 79 GPa) as depicted in

Figure P4.7

40 mm

Figure P4.7

401

412

Design Requirement: To achieve the desired natural frequency of torsional vibrations, the

stiffness of the system is specified such that the disk will rotate ].5° under a torque of 1 KN-m

How deep (A) must a 22-mmi diameter hole be drilled to satisfy this requirement?

A solid round shaft with fixed ends is under a distributed torque of intensity T(x) = 7), Ib- in,/in., as shown in Figure P4.8 Determine the reactions at the walls

Figure P4.8 Resolve Problem 4.8, for the case in which T(x) = (x/L)T)

A simply supported beam AB carries a triangularly distributed load of maximum intensity wy (Figure P4.10)

(a) Employ the fourth-order differential equation of the deflection to derive the expression for the elastic curve

(6) Determine the maximum deflection vax

Figure P4.10

Asimply supported beam is loaded with a concentrated moment M,, as shown in Figure P4.i1

Derive the equation of the elastic curve for the segment AC of the beam

Figure P4.12

The overhanging beam ABC supports a concentrated load P at the free end, as shown in

Figure P4.13 For the segment BC of the beam,

(a) Derive the equation of the elastic curve

(b) What is the maximum deflection Đmax?

(ec) Calculate the value of the Umax for the following data:

I1=5.12x 10 mm, E=200GPa, - P=25kN L=2m, a=0.5m

Figure P4.13

Two cantilever beams AB and CD are supported and loaded as shown in Figure P4.14 What

is the interaction force R transmitted through the roller that fits snugly between the two

beams at point C? Use the method of superposition and the deflection formulas of the beams

from Table A.9

Figure P4.15 shows a compound beam with a hinge at point B It is composed of two parts: a

beam BC simply supported at C and a cantilevered beam AB fixed at A Apply the superposi- tion method using Table A.9 to determine the deflection vg at the hinge

Figure P4.15