Review in Strength of Materials
September 3, 2022 | Author: Anonymous | Category: N/A
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STRENGTH OF MATERIALS
= :: = ,, = , , =
at the positions indicated. indicated. 2780 psi, comp. ) ) Al
Determine the stress in e each ach section.
( Ans.
Stress – unit strength of materials. The stress n any member under loading is
Normal stress – either tensile or compressive stress produced by force acting perpendicular to the area Shearing stress – is produced whenever the applied load causing sliding to the sections. It is either a single shear or double shear. Bearing stress – is the contact pressure between separate bodies
Solution: Axial Diagram, (AD) Assume positive (tension)
Problems
1. 1. The homogeneous bar shown in the figure is supported by a pin at A and a cable that runs from B to C around the smooth peg at D. Find the stress in the cable if its diameter is 0.6 in and the bar weighs 6 kips.
= ⊥ =
Solution: F.B.D.
= .. = .. = .. = . = .. = .
3. An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown in the figure. Find the maximu maximum m value of P that will not exceed a stress in
steel of 140 MPa, in aluminum of 90 MPa or in bronze of 100 MPa. ( Ans. P 10.0 kN . )
∑ =
√ = .=. = = = . ⊥ = .
R
Solution:
2. 2. A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in the figure. Axial loads are applied
→ ∑ ∑ = = =
Axial Diagram, (AD) Assume positive (tension)
a. maximum thickness of the plate from which a hole 2.5 in. in diameter can be punched.
= ⊥ = = ∥ = =... = = =
= ⊥
70kN
== =
= = = = = = =
36kN
10kN
Therefore use the smallest value of P P = 10Kn
b. b. the diameter of the smallest hole that can be punched
4. 4. A hole is to be punched out of a plate having a shearing shearing strength of 40 ksi. ksi. The compressive stress in the punch is limited to 50 ksi. (a) Compute the maximum thickness of the plate from which a hole 2.5 in. in diameter can be punched. (b) If the plate is 0.25 in. thick, determine the diameter of the smallest hole that can be punched. ( Ans.t = 0.781 in, d = 0.800 in )
= =..
5. 5. From the figure shown
6. 6. A square post is made up of t wo parts glued at points a, b, c, and d. Plane a-b-c-d is at an angle of 30° from the vertical axis. Axial load P = 36 kN. Given: Allowable stresses in the glued joint. Compressive stress = 5.2 MPa Shear stress = 3.5 MPa What is the minimum width w (mm) required to prevent failure in compressi compression? on? Determine w (mm) based on the allowable shear stress. At what angle (degrees) will the maximum shear stress occur?
Calculate the tensile stress in the body of the bolt(70.5)
= = = = . = . = =
Calculate the tensile stress at the root of the threads(99. threads(99.5) 5)
Find the compressive stress at the head as the bolt bears on the surface to resist the tensile load (40.3) Solution:
= .^ = . ^ = = = . = . °
°
a. a. minimum width w (mm) required to prevent failure in compression
=
°
= . = = . . = . = = . . °
°
°
b. minimum width w (mm) based on the allowable shear stress ° °
°
c. c. angle (degrees) which the maximum shear stress occur
= = = = = = ==
°
7. 7. The strut shown in the figure carries an axial load of P = 148 kN.
c. c. shearing stress in the bolts bolts
= = , = = .
°
8. 8. A member AB 3 m long is supported by a pin at A and a tensile rod BC at B (See Fig. shown). The pin at A is under double shear while the pin at B is under single shear. The allowable shear stress for the pins is τallow = 90 MPa and the allowable tensile stress for the rod is ( σt)allow = 115 MPa.
Determine the bearing stress between the pin and the strut Determine the shearing stress in the pin. Determine the shearing stress in the bolts Solution: a. a. bearing stress between the pin and the strut strut
, , = ( () = = ^ , , = = = . , = = = = .
b. b. shearing stress in the pin pin
shearing stress in pin: (double shear)
The smallest diameter of the pin at A is nearest to: The smallest diameter of the pin at B is nearest to: The smallest diameter of the tension rod BC is nearest to: SOLUTION:
a. a. smallest diameter diameter of the pin at A A
∑ = = = ∑ = = = ∑ = = = = = . .
= ⊥
= . = . . = = ..
= ⊥
Smallest diameter of the pin at B
smallest diameter of the tension rod BC
= ⊥
= = ..
9. 9. The inclined member in the figure is subjected to a compressive force of 600 lb. Determine the average compressive stress along the smooth areas of contact defined by AB and BC, and the average shear stress along the horizontal plane 2400 psi, BC 16 1600 psi, DB 80 ps psii ) defined by DB. ( AB 24
10. 10. If the allowable bearing stress for the material under the supports at A and B is b allow 1.5 MPa, determine the size of square bearing plates required to support the load. Dimension the plates to the nearest mm. The reactions at the supports are vertical. (A=130 mm, B=300 mm)
Solution:
Solution:
= . =
= . = = . =
∑ = ∑ =
....// . = = .. = =
Bearing stress,
=
= = . = = =. = .
The 20-kg chandelier is suspended from the wall and ceiling using rods AB and BC, which have diameters of 3 mm and 4 mm respectively. Determine the angle θ so that t he average normal stress in both rods is the same. If θ = 450, determine the largest mass of the chandelier that can be supported if the average normal stress in both rods is not allowed to exceed 150 MPa. a. 148 kg c. 168 kg
Solution : a. a. Angle θ so that the average normal stress in both rods is the same
b. 158 kg d. 178 kg
= = = = = = = .
By sine law:
°
°
Solution:
°
°
the largest mass of the chandelier that can be supported if the average normal stress in both rods is not allowed to exceed 150 MPa MPa
= = = = . = = . = = = = .. °
Thin Walled Vessels A spherical gas tank has an inner radius radius of 1.5 m. If it is subjected to an internal pressure of 300 kPa, determine its required thickness if the maximum normal stress is not to exceed 12 MPa. (18.8 mm) Solution :
°
°
Therefore the largest mass, m = 147.64kg
∑= = = = = = . .
A vertical cylindrical gasoline is 30ism0.74. in4.diameter and is filledoftothe a depth of 15 m with gasoline whosestorage specifictank gravity 0.7 If the yield point shell
plating is 250 MPa and a safety factor of 2.5 is adequate, calculate the required wall thickness at the bottom of the tank. (16.3 mm)
Consider section a-a
∑= =
= = A vertical cylindrical gasoline storage tank is 30 m in diameter and is filled to a depth of 15 m with gasoline whose specific gravity is 0.74. 0.74. If the yield point of the shell plating is 250 MPa and a safety factor of 2.5 is adequate, calculate the required wall thickness at the bottom of the tank. (16.3 mm)
F.B.D.
= = = .. = = . = . = .
Wall thickness,t :
Consider section b-b
∑= = = =
The tank shown in the figure is fabricated from 1/8” steel plate. Calculate the maximum longitudinal and circumferential stress caused by an internal pressure of 125 psi. ( σL = 6566 psi)
=
Deformation of Members Under Axial Loading Where
= = , , = ,, , = ,
AXIAL DEFORMATION
∑ =
=
/" = . = ∑ = = .. . . = = . .
A steel rod having a cross-sectional area of 300 mm2 and a length of 150 m is suspended vertically from one end. end. It supports a tensile tensile load of 20 kN at the lower end end.. If the unit mass of steel is 7,850 kg/m 3 and E = 200 GPa, find the total elongation of the rod. (54.3 mm) mm)
= == ∫ = = = == ∫ == ∫ = = = = / . = = ..
Alternate solution
= = . = . = ..
. = = .
; = = = . = = . ; =
A bronze bar is fastened between a steel bar and an aluminum bar as shown in the following figure. Axial loads are applie applied d as indicated. Find the largest value of P that will not exceed an overall deformation of 3.0 mm, or the following stresses: 140 MPa in the steel, 120 MPa in the bronze, and 80 MPa in the aluminum. Assume that the assembly is suitably braced against against buckling. Use Est 200 200 GPa, Eal 70 GPa , and Ebr 83 GPa . (12.8 kN)
Therefore use the smallest value of P = 12.8Kn
Rigid beam AB beam AB rests rests on the two short posts shown in the figure. AC is AC is made of steel and has a diameter of 20 mm, and BD is BD is made of aluminum and has a diameter of 40 mm. Determine the displacement of point F if if a vertical load of 90 kN is applied over this point. Take Est = 200 GPa, Eal = 70 GPa. (0.225 mm)
Solution: Draw Axial diagram
= ∑
Based on allowable total deformation
. == [ . . ] .] = .
=
Based on allowable stress
; = = = .
∑ =
.. .. = = = =
∑ =
. = .. = . .
=
The rigid bars AB bars AB and CD are supported by pins at A and C and the two rods. Determine the maximum force P that can be applied as shown if its vertical movement is limited to 5 mm. Neglect weight of all mem members. bers. ( Ans. P = 76.3 kN )
= . = = = = = . =
Solution:
∑ =
= =
As shown in the figure, two aluminum rods AB AB and BC, BC, hinged to rigid supports are pinned together at B to carry a vertical load P = 6000 lb. If each rod has a cross sec sectional tional area of 0.6 in2 and E 10 106 psi , compute the elongation of each rod and the horizontal and vertical displacements of point B. ( h 0.0023 ft, v 0.016 ft )
∑ =
= =
Deformation relationship
Solution:
= = +
≤
= / = = .
By sine law
= = = = = °
°
°
Compatibility Equation
==
=
= . = . = . = . .
F.B.D.-Deformation
= .° =
= . = = . = . = .
A reinforced concrete column 200 mm in diameter is designed to carry an axial compressive load of 300 kN. Determine the required area of the reinforcing steel if the allowable streEsses stresses areGPa. 6 MPa and mm 120 2MPa respecti vely. Use Ec = 14 GPa and (1,400 ) ) for the concrete and steel, respectively. s = 200
. = = . / . = = ./ . = . = .° =.="
= = =
= = = , =? = . . > , = = , =? = = . < ,, = ., = = = . = .
The composite rod in the figure is stress-free before the axial loads are applied. Assuming that the walls are rigid, calculate the stress in each material if P1 = 150 kN and P2 = 90 kN. kN. ( Ans. Al = 86.2 MPa )
If
If
Solution:
From
P1 = 150 P2 = 90kN Assume R >240 Assume
→
Since rigid support:
= ∑ = = = =
= . = = . = = . = . . = = . . . ... ==. . = . = .. . = . = .
Therefore:
∑ =
∑ =
= =
= =
Deformation:
The horizontal beam is assumed to be rigid and supports the distributed load shown. Determine the angle of tilt of the beam after the load is applied. Each support consists of a wooden post having a diameter of 120 mm and an unloaded (original) length of 1.40 m. Take Ew = 12 GPa. (1.14x10-3°) °)
Compatibility Eq.
= = = = , = , = .. = = =.°
By similar triangle
Same material, length and area
Solution: Angle of tilt:
Three rods, each of area 250 mm 2, jointly supports a 7.5 kN load as shown in the figure. Assuming that there is no slack or stress in the rods before the load was applied, find the stress in each rod. Use Est 200 GPa and Ebr 83 GPa . (σs = 18.53 MPa)
Deformation :
= = ° ° . = . = ° =
Solution:
∑ =
° ° . = . ° = . ° ° . = . ° . =
° . = °
From
= . . = . = .
Solution:
Thermal Stress:
A steel rod is stretched between two rigid walls and carries a tensile load of 5000 N at 20°C. If the allowable stres stresss is not to exceed 13 130 0 MPa at –20°C, what is the mm) minimum diameter diameter of the rod? Assume 11.7m /m C and Est 200 GPa. (13.2 mm)
= = ∆ = ∆ ∆ − = . =. =
=
= = = = =
The rigid horizontal bar of negligible mass is connected connected to two rods as shown. If the system is initially stress free, calculate the temperature change that will cause a tensile stress of 90 MPa in the brass rod. Assume that both rods are subjected to the change in temperature. temperatur e. Coefficients of linear expansion are 18.7m / m C for brass and 16.8m / m C for copper. (56.01°C)
∑ =
Compatibility Eq. Relax support for brass and assume temperature increase
= ∆∆ ∆∆ = .−∆ = .− ∆∆ . ∆ ∆ = ..∆ ∆ . ∆ =∆.=. ° °
Calculate the increase in stress for each segment of the compound bar shown if the temperature increases by 100°F. Assume that the s upports are unyielding and that the bar is suitably braced against buckling.
Solution: Due to pull
C C (decrease)
Due to temperature: Assume temperature increase
∑ = =
. . − . − = . = . = . = .. , = ... = .
The composite bar is firmly attached to unyielding supports. The bar is stress free at 60 0F. compute the stress in each material after the 50kip force is applied and the t emperature is increased to 1200F. Use = 6.5x10^-6/0F for steel and = 12.8x10^-6/0F for aluminum.
Solution: Assume R > 50
→
CompatibilityEquation: Temperature increase
∑ = 0
= 0 .− .− = = . . = = . = .
TORSION:
For solid shaft,
, , =
= = = = = = = , = = = = = = = = , = = =
= . . = =. = . = . = . = = . =
For hollow shaft:
Angle of twist, :
241.201000 = 3452.85 = 3101. 5 = 5.424 32 32
Two solid shafts of different materials are rigidly fastened together and attached to rigid supports as shown. shown. The torque, T = 10 kip-in kip-in,, is applied at the junc junction tion of the two segments. Compute the maximum shearing stress developed in the assembly. ( Ans. St 3450 psi )
A shaft composed of bronze, aluminum and steel segments is fastened to rigid supports and loaded as shown the figure. For bronze, G = 35 GPa; for aluminum, G = 28 GPa, and for steel, G = 83 GPa. Determine the maximum shearing stress developed in each segment.
Solution: Solution: Assume T>1000
For rigid connection:
∑ = 0
= . = . , , = 1.5 4.576761.51000 = 332 = 332 = 863.2
For rigid connection:
∑ = 0
. . . = = . . , = = =
Shearing stress:
= . = . = . = .
=
= = . . = = . . . = = = < = = .. = .
= . = .
A composite shaft is manufactured by shrink-fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters diameters of the two parts are db = 40 mm for the brass core and d s = 50 mm for the steel sleeve. The shear moduli of elasticity are G b = 35 GPa for the brass and G s = 80 GPa for the steel. Allowable shear stresses in the brass and steel are 48 MPa and 80 MPa, respectively. Determine the maximum permissible permissible torque in N-m that may be applied to the shaft. (1,521) (1,521)
= = = = = = = . . = = .. = . = . > ,,
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