III DISCUSIÓN, CONCLUSIONES y FUTURAS
3.1 Discusión y conclusiones de la investigación
3.1.2 Conclusiones del estudio
EXAMPLE 5.2
A steel [E 30,000 ksi] bar of rectangular cross section consists of a uniform-width segment (1) and a tapered segment (2), as shown. The width of the tapered segment varies linearly from 2 in. at the bottom to 5 in. at the top. The bar has a constant thick-ness of 0.50 in. Determine the elongation of the bar resulting from application of the 30-kip load. Neglect the weight of the bar.
Plan the Solution
The elongation of uniform-width segment (1) may be determined from Equation (5.2).
The tapered segment (2) requires the use of Equation (5.5). An expression for the varying cross-sectional area of segment (2) must be derived and used in the integral for the 75-in. length of the tapered segment.
SOLUTION
For the uniform-width segment (1), the deformation from Equation (5.2) is
1 1 1
in. in. 30,0000 in.
ksi) 0 0250.
␦ ⫽ ⫽ ⫽
For tapered section (2), the width w of the bar varies linearly with position y. The cross-sectional area in the tapered section can be expressed as
A y2 wt 2 3 y
75 0 5 1
( ) in. in. ( ) ( . )
in. in. in. 0 02. y in.2
⫽ ⫽ ⫹ ⫽ ⫹
Since the weight of the bar is neglected, the force in the tapered segment is constant and simply equal to the 30-kip applied load. Integrate Equation (5.5) to obtain
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The total elongation of the bar is the sum of the segment elongations:
uA ⫽␦1⫹␦2 ⫽0 0250. in.⫹0 0458. in.⫽0 0708. in. Ans.
Note: If the weight of the bar had not been neglected, the internal force F in both uniform-width segment (1) and tapered segment (2) would not have been constant, and Equation (5.5) would be required for both segments. To include the weight of the bar in the analysis, a function should be derived for each segment, expressing the change in internal force as a function of the vertical position y. The internal force F at any position y is the sum of a constant force equal to P and a varying force equal to the self-weight of the axial member below position y. The force due to self-weight will be a function that expresses the volume of the bar below any position y, multiplied by the specifi c weight of the material that the bar is made of. Since the internal force F varies with y, it must be included inside the integral in Equation (5.5).
M5.1 Use the axial deformation equation for three introductory problems.
MecMovies M M Exercises
FIGURE M5.1
FIGURE M5.2
M5.2 Apply the axial deformation concept to compound axial members.
P5.1 A steel [E 200 GPa] rod with a circular cross section is 7.5-m long. Determine the minimum diameter required if the rod must transmit a tensile force of 50 kN without exceeding an allow-able stress of 180 MPa or stretching more than 5 mm.
P5.2 An aluminum [E 10,000 ksi] control rod with a circular cross section must not stretch more than 0.25 in. when the tension
PROBLEMS PROBLEMS
in the rod is 2,200 lb. If the maximum allowable normal stress in the rod is 12 ksi, determine
(a) the smallest diameter that can be used for the rod.
(b) the corresponding maximum length of the rod.
P5.3 A 12-mm-diameter steel [E 200 GPa] rod (2) is con-nected to a 30-mm-wide by 8-mm-thick rectangular aluminum
98
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[E ⫽ 70 GPa] bar (1), as shown in Figure P5.3. Determine the force P required to stretch the assembly 10.0 mm.
P5.7 Aluminum [E ⫽ 70 GPa] member ABC supports a load of 28 kN, as shown in Figure P5.7. Determine
(a) the value of load P such that the defl ection of joint C is zero.
(b) the corresponding defl ection of joint B.
P5.4 A rectangular bar of length L has a slot in the central half of its length, as shown in Figure P5.4. The bar has width b, thick-ness t, and elastic modulus E. The slot has width b/3. If L ⫽ 400 mm, b ⫽ 45 mm, t ⫽ 8 mm, and E ⫽ 72 GPa, determine the overall elongation of the bar for an axial force of P ⫽ 18 kN.
P5.5 An axial member consisting of two polymer bars is sup-ported at C, as shown in Figure P5.5. Bar (1) has a cross-sectional area of 540 mm2 and an elastic modulus of 28 GPa. Bar (2) has a cross-sectional area of 880 mm2 and an elastic modulus of 16.5 GPa. Determine the defl ection of point A relative to support C.
P5.6 The roof and second fl oor of a building are supported by the column shown in Figure P5.6. The column is a structural steel W10 ⫻ 60 wide-fl ange section [E ⫽ 29,000 ksi; A ⫽ 17.6 in.2].
The roof and fl oor subject the column to the axial forces shown.
Determine
(a) the amount that the fi rst fl oor will defl ect.
(b) the amount that the roof will defl ect.
P5.8 A solid brass [E ⫽ 100 GPa] axial member is loaded and supported as shown in Figure P5.8. Segments (1) and (2) each have a diameter of 25 mm, and segment (3) has a diameter of 14 mm.
Determine
(a) the deformation of segment (2).
(b) the defl ection of joint D with respect to the fi xed support at A.
(c) the maximum normal stress in the entire axial member.
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P5.9 A hollow steel [E 30,000 ksi] tube (1) with an outside diameter of 2.75 in. and a wall thickness of 0.25 in. is fastened to a solid aluminum [E 10,000 ksi] rod (2) that has a 2-in.-diameter and a solid 1.375-in.-diameter aluminum rod (3). The bar is loaded as shown in Figure P5.9. Determine
(a) the change in length of steel tube (1).
(b) the defl ection of joint D with respect to the fi xed support at A.
(c) the maximum normal stress in the entire axial assembly.
P5.10 A solid 5/8-in. steel [E 29,000 ksi] rod (1) supports beam AB as shown in Figure P5.10. If the stress in the rod must not exceed 30 ksi and the maximum deformation in the rod must not exceed 0.25 in., determine the maximum load P that may be supported.
P5.11 A 1-in.-diameter by 16-ft-long cold-rolled bronze [E 15,000 ksi and 0.320 lb/in.3] bar hangs vertically while suspended from one end. Determine the change in length of the bar due to its own weight.
P5.12 A homogeneous rod of length L and elastic modulus E is a truncated cone with a diameter that varies linearly from d0 at one end to 2d0 at the other end. A concentrated axial load P is applied to the ends of the rod as shown in Figure P5.12. Assume that the
taper of the cone is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid.
(a) Determine an expression for the stress distribution on an arbitrary cross section at x.
(b) Determine an expression for the elongation of the rod.
P5.13 Determine the extension, due to its own weight, of the conical bar shown in Figure P5.13. The bar is made of aluminum alloy [E 10,600 ksi and 0.100 lb/in.3]. The bar has a 2-in.
radius at its upper end and a length of L 20 ft. Assume that the taper of the bar is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid.
P5.14 The wooden pile shown in Figure P5.14 has a diame-ter of 100 mm and is subjected to a load of P 75 kN. Along the length of the pile and around its perimeter, soil supplies a constant frictional resistance of w 3.70 kN/m. The length of the pile is L 5.0 m and its elastic modulus is E 8.3 GPa.
Calculate
(a) the force FB needed at base of the pile for equilibrium.
(b) the magnitude of the downward displacement at A relative to B.
FIGURE P5.9
(1) (2) (3)
A B C D
34 kips 18 kips
25 kips
34 kips 18 kips
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101 EXAMPLE 5.3
The assembly shown consists of rigid bar ABC, two fi ber-reinforced plastic (FRP) rods (1) and (3), and FRP post (2). The modulus of elasticity for the FRP is E ⫽ 18 GPa. Determine the ver-tical defl ection of joint D relative to its initial position after the 30-kN load is applied.
Plan the Solution
The defl ection of joint D relative to its initial position must be computed. The defl ection of D relative to joint C is simply the elongation in member (3).
The challenge in this problem, how-ever, lies in computing the defl ection at C. The rigid bar will defl ect and rotate due to the elongation and con-traction in members (1) and (2). To
de-termine the fi nal position of the rigid bar, we must fi rst compute the forces in the three axial members, using equilibrium equations. Then, Equation (5.2) can be used to compute the deformation in each member. A deformation diagram can be drawn to defi ne the re-lationships between the rigid bar defl ections at A, B, and C. Then, the member deforma-tions will be related to the rigid bar defl ecdeforma-tions. Finally, the defl ection of joint D can be computed from the sum of the rigid bar defl ection at C and the elongation in member (3).
SOLUTION Equilibrium
Draw a free-body diagram (FBD) of the rigid bar and write two equilibrium equations:
F F F F re-sult, we can simultaneously solve the two equa-tions to give F1⫽ 22.5 kN and F2⫽ –52.5 kN.
The three axial members are connected to the rigid beam by pins. Assume that member (1) is pinned to the foundation at F and member (2) is fi xed in the foundation at E.
Many structures consist of more than one axially loaded member, and for these structures, axial deformations and stresses for a system of pin-connected deformable bars must be determined. The problem is approached through a study of the geometry of the deformed system, from which the axial deformations of the various bars in the system are obtained.
In this section, the analysis of statically determinate structures consisting of homoge-neous, prismatic axial members will be considered. In analyzing these types of structures, begin with a free-body diagram showing all forces acting on the key elements of the struc-ture. Then, investigate how the structure as a whole defl ects in response to the deformations that occur in the axial members.
5.4 Deformations in a System of Axially Loaded Bars
A homogeneous, prismatic member (a) is straight, (b) has a constant cross-sectional area, and (c) consists of a single material (i.e., one value of E).
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Force–Deformation Relationships
Compute the deformations in each of the three members.
1 1
The negative value of ␦2 indicates that member (2) contracts.
3 3 Sketch the fi nal defl ected shape of the rigid bar. Member (1) elon-gates, so A will defl ect upward.
Member (2) contracts, so B will defl ect downward. The defl ection of C must be determined.
(Note: Joint defl ections trans-verse to the rigid bar are denoted by v.)
The rigid bar defl ections at joints A, B, and C can be related
How are the rigid bar defl ections vA and vB shown on the sketch related to the member deformations ␦1 and ␦2? By defi nition, deformation is the difference between the initial and fi nal lengths of an axial member. Using the defl ected rigid bar sketch, we can defi ne the deformation in member (1) in terms of its initial and fi nal lengths:
1 ⫽ Lfinal⫺Linitial ⫽(L1⫹vA)⫺L1 ⫽vA ⬖vA ⫽ 1 ⫽ 9 00. mm
␦ ␦
Similarly, for member (2),
2 ⫽ Lfinal⫺Linitial ⫽(L2⫺vB)⫺L2 ⫽⫺vB ⬖vB ⫽⫺ 2 ⫽⫺ ⫺( 7 0. 00mm)⫽7 00. mm
␦ ␦
With these results, the magnitude of the rigid bar defl ection at C can now be computed:
vC ⫽ 0 75. (vA ⫹vB)⫹vB ⫽ 0 75 9 00. ( . mm⫹7 00. mm)⫹7 00. mm⫽19 00. mm The direction of the defl ection is shown on the deformation diagram; that is, joint C defl ects 19.00 mm downward.
Defl ection of D
The downward defl ection of joint D is the sum of the rigid bar defl ection at C and the elongation in member (3):
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The preceding examples considered structures consisting of parallel axial bars, making the geometry of deformation for the structure relatively straightforward to analyze.
Suppose, for example, that one is interested in a structure in which the axial members are not parallel. The structure shown in Figure 5.9 consists of three axial members (AB, BC, and BD), which are connected to a common joint at B. In the fi gure, the solid lines represent the unstrained (i.e., unloaded) confi guration of the system and the dashed lines represent the confi gurations due to a force applied at joint B. From the Pythagorean theorem, the actual deformation in bar AB is
AB ⫽ (L⫹y)2⫹x2 ⫺L
␦
Transposing the last term and squaring both sides gives
AB L AB L L Ly y x
2 2 2 2 2 2 2
␦ ⫹ ␦ ⫹ ⫽ ⫹ ⫹ ⫹
If the displacements are small (the usual case for stiff materials and elastic action), the terms involving the squares of the displacements may be neglected; hence, the deformation in bar AB is
AB y
␦
In a similar manner, the deformation in bar BD is
BD x
␦ The axial deformation of bar BC is
BC ⫽ ( cosR ⫹x)2 ⫹( sinR ⫹y)2 ⫺R
␦
Transposing the last term and squaring both sides gives
BC2 ⫹2R BC ⫹R2
␦ ␦
R2cos2 ⫹2Rxcos ⫹x2⫹R2sin2 ⫹2Rysin ⫹y2
⫽
An assembly consists of three rods attached to rigid bar AB. Rod (1) is steel, and rods (2) and (3) are aluminum. The area and elastic modulus of each rod is noted on the sketch. A force of 80 kN is applied at D. Determine the vertical defl ections of points A, B, C, and D.