A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the stress is limited to 120 MN/m2.Solution 104

where:

thus,

 answer

Strength of Materials 4th Edition by Pytel and SingerProblem 105 page 12Given:Weight of bar = 800 kgMaximum allowable stress for bronze = 90 MPaMaximum allowable stress for steel = 120 MPaRequired: Smallest area of bronze and steel cablesSolution 105

By symmetry:

For bronze cable:

 answerFor steel cable:

 answer

106. Given:Diameter of cable = 0.6 inchWeight of bar = 6000 lbRequired: Stress in the cableSolution 106

 answer

Strength of Materials 4th Edition by Pytel and SingerProblem 107

Given:Axial load P = 3000 lbCross-sectional area of the rod = 0.5 in2

Required: Stress in steel, aluminum, and bronze sections

Solution 107

For steel:

 answer

For aluminum:

 answer

For bronze:

 answer

Strength of Materials 4th Edition by Pytel and SingerProblem 108 page 12

Given:Maximum allowable stress for steel = 140 MPaMaximum allowable stress for aluminum = 90 MPaMaximum allowable stress for bronze = 100 MPa

Required: Maximum safe value of axial load P

Solution 108

For bronze:

For aluminum:

For Steel:

For safe  , use   answer

Strength of Materials 4th Edition by Pytel and SingerProblem 109 page 13Given:Maximum allowable stress of the wire = 30 ksiCross-sectional area of wire AB = 0.4 in2

Cross-sectional area of wire AC = 0.5 in2

Required: Largest weight W Solution 109

 

For wire AB: By sine law (from the force polygon):

 For wire AC:

 

Safe load   answer

Strength of Materials 4th Edition by Pytel and SingerProblem 110 page 13

Given:Size of steel bearing plate = 12-inches squareSize of concrete footing = 12-inches squareSize of wooden post = 8-inches diameterMaximum allowable stress for wood = 1800 psiMaximum allowable stress for concrete = 650 psi

Required: Maximum safe value of load P

Solution 110For wood:

From FBD of Wood:

For concrete:

From FBD of Concrete:

Safe load   answer

Strength of Materials 4th Edition by Pytel and SingerProblem 111 page 14

Given:Cross-sectional area of each member = 1.8 in2

Required: Stresses in members CE, DE, and DF

Solution 111From the FBD of the truss:

At joint F:

At joint D: (by symmetry)

At joint E:

Stresses:Stress = Force/Area

 answer

 answer

 answer

Strength of Materials 4th Edition by Pytel and SingerProblem 112 page 14Given:Maximum allowable stress in tension = 20 ksiMaximum allowable stress in compression = 14 ksiRequired: Cross-sectional areas of members AG, BC, and CE Solution 112

 

 

 Check:

 (OK!) For member AG (At joint A):

 

 

 

 answer For member BC (At section through MN):

 

 Compression 

 answer For member CE (At joint D):

 

 

 At joint E:

 

 

 Compression 

 answer Strength of Materials 4th Edition by Pytel and SingerProblem 113 page 15Given:Cross sectional area of each member = 1600 mm2.Required: Stresses in members BC, BD, and CFSolution 113

For member BD: (See FBD 01)

 Tension

 answerFor member CF: (See FBD 01)

 Compression

 answerFor member BC: (See FBD 02)

 Compression

 answerStrength of Materials 4th Edition by Pytel and SingerProblem 114 page 15Given:Maximum allowable stress in each cable = 100 MPaArea of cable AB = 250 mm2

Area of cable at C = 300 mm2

Required: Mass of the heaviest bar that can be supportedSolution 114

Based on cable AB:

Based on cable at C:

Safe weight 

 answer

SHEAR STRESS

Strength of Materials 4th Edition by Pytel and SingerProblem 115 page 16

Given:Required diameter of hole = 20 mmThickness of plate = 25 mmShear strength of plate = 350 MN/m2

Required: Force required to punch a 20-mm-diameter hole

Solution 115

The resisting area is the shaded area along the perimeter and the shear force   is equal to the punching force  .

 answer

Strength of Materials 4th Edition by Pytel and SingerProblem 116 page 16Given:Shear strength of plate = 40 ksiAllowable compressive stress of punch = 50 ksiThe figure below:

Required:a. Maximum thickness of plate to punch a 2.5

inches diameter holeb. Diameter of smallest hole if the plate is 0.25 inch

thickSolution 116

a. Maximum thickness of plate:Based on puncher strength:

 Equivalent shear force of the plateBased on shear strength of plate:

 answerb. Diameter of smallest hole:

Based on compression of puncher:

 Equivalent shear force for plateBased on shearing of plate:

 answer

Strength of Materials 4th Edition by Pytel and SingerProblem 117 page 17

Given:Force P = 400 kNShear strength of the bolt = 300 MPa

The figure below:

Required: Diameter of the smallest bolt

Solution 117The bolt is subject to double shear.

 answer

Strength of Materials 4th Edition by Pytel and SingerProblem 118 page 17

Given:Diameter of pulley = 200 mmDiameter of shaft = 60 mmLength of key = 70 mmApplied torque to the shaft = 2.5 kN·mAllowable shearing stress in the key = 60 MPa

Required: Width b of the key

Solution 118

Where:

 answer

Strength of Materials 4th Edition by Pytel and SingerProblem 119 page 17

Given:Diameter of pin at B = 20 mm

Required: Shearing stress of the pin at B

Solution 119

From the FBD:

 shear force of pin at B

 double shear

 answer

Strength of Materials 4th Edition by Pytel and SingerProblem 120 page 17Given:Unit weight of each member = 200 lb/ftMaximum shearing stress for pin at A = 5 000 psiRequired: The smallest diameter pin that can be used at A Solution 120For member AB: 

 Length, Weight,  

 Equation (1)

 For member BC:

 

Length, Weight,  

 Equation (2)

 Add equations (1) and (2)

 From equation (1):

 From the FBD of member AB

 

 

 shear force of pin at A 

 answerStrength of Materials 4th Edition by Pytel and SingerProblem 121 page 18Given:Allowable shearing stress in the pin at B = 4000 psiAllowable axial stress in the control rod at C = 5000 psiDiameter of the pin = 0.25 inchDiameter of control rod = 0.5 inchPin at B is at single shearRequired: The maximum force P that can be applied by the operatorSolution 121

 Equation (1)

From Equation (1), 

From Equation (1), 

 Equation (2)Based on tension of rod (equation 1):

Based on shear of rivet (equation 2):

Safe load   answerStrength of Materials 4th Edition by Pytel and SingerProblem 122 page 18Given:Width of wood = Thickness of wood = Angle of Inclination of glued joint = Cross sectional area =  Required: Show that shearing stress on glued joint  Solution 122 

 Shear area, Shear area, Shear area, Shear force,  

 (ok!)

BEARING STRESSProblem 125In Fig. 1-12, assume that a 20-mm-diameter rivet joins the plates that are each 110 mm wide. The allowable stresses are 120 MPa for bearing in the plate material and 60 MPa for shearing of rivet. Determine (a) the minimum thickness of each plate; and (b) the largest average tensile stress in the plates. 

 Solution 125Part (a): From shearing of rivet:

 From bearing of plate material:

 answer Part (b): Largest average tensile stress in the plate:

 answer

Strength of Materials 4th Edition by Pytel and SingerProblem 126 page 21

Given:Diameter of each rivet = 3/4 inchMaximum allowable shear stress of rivet = 14 ksiMaximum allowable bearing stress of plate = 18 ksi

The figure below:

Required: The maximum safe value of P that can be applied

Solution 126

Based on shearing of rivets:

Based on bearing of plates:

Safe load   answer

Strength of Materials 4th Edition by Pytel and SingerProblem 127 page 21

Given:Load P = 14 kipsMaximum shearing stress = 12 ksiMaximum bearing stress = 20 ksiThe figure below:

Required: Minimum bolt diameter and minimum thickness of each yoke

Solution 127

For shearing of rivets (double shear)

 diameter of bolt answer

For bearing of yoke:

 thickness of yoke answer

Strength of Materials 4th Edition by Pytel and SingerProblem 128 page 21

Given:Shape of beam = W18 × 86Shape of girder = W24 × 117Shape of angles = 4 × 3-½ × 3/8Diameter of rivets = 7/8 inchAllowable shear stress = 15 ksiAllowable bearing stress = 32 ksi

Required: Allowable load on the connection

Solution 128Relevant data from the table (Appendix B of textbook): Properties of Wide-Flange Sections (W shapes): U.S. Customary Units

Designation Web thicknessW18 × 86 0.480 inW24 × 117 0.550 in

Shearing strength of rivets:There are 8 single-shear rivets in the girder and 4 double-shear (equivalent to 8 single-shear) in the beam, thus, the shear strength of rivets in girder and beam are equal.

Bearing strength on the girder:The thickness of girder W24 × 117 is 0.550 inch while

that of the angle clip   is   or 0.375 inch, thus, the critical in bearing is the clip.

Bearing strength on the beam:The thickness of beam W18 × 86 is 0.480 inch while that of the clip angle is 2 × 0.375 = 0.75 inch (clip angles are on both sides of the beam), thus, the critical in bearing is the beam.

The allowable load on the connection is   answer

Strength of Materials 4th Edition by Pytel and SingerProblem 129 page 21

Given:Diameter of bolt = 7/8 inchDiameter at the root of the thread (bolt) = 0.731 inchInside diameter of washer = 9/8 inchTensile stress in the nut = 18 ksiBearing stress = 800 psi

Required:Shearing stress in the head of the boltShearing stress in threads of the boltOutside diameter of the washer

Solution 129

Tensile force on the bolt:

Shearing stress in the head of the bolt:

 answer

Shearing stress in the threads:

 answer

Outside diameter of washer:

 answer

Strength of Materials 4th Edition by Pytel and SingerProblem 130 page 22

Given:Allowable shear stress = 70 MPaAllowable bearing stress = 140 MPaDiameter of rivets = 19 mm

The truss below:

Required:Number of rivets to fasten member BC to the gusset plateNumber of rivets to fasten member BE to the gusset plateLargest average tensile or compressive stress in members BC and BE

Solution 130At Joint C:

 (Tension)

Consider the section through member BD, BE, and CE:

 (Compression)

For Member BC:Based on shearing of rivets:

 Where A = area of 1 rivet × number of rivets, n

 say 5 rivets

Based on bearing of member:

Where Ab = diameter of rivet × thickness of BC × number of rivets, n

 say 7 rivets

use 7 rivets for member BC answer

For member BE:Based on shearing of rivets:

Where A = area of 1 rivet × number of rivets, n

 say 5 rivets

Based on bearing of member:

Where Ab = diameter of rivet × thickness of BE × number of rivets, n

 say 3 rivets

use 5 rivets for member BE answer

Relevant data from the table (Appendix B of textbook): Properties of Equal Angle Sections: SI Units

Designation AreaL75 × 75 × 6 864 mm2

L75 × 75 × 13 1780 mm2

Tensile stress of member BC (L75 × 75 × 6):

 answer

Compressive stress of member BE (L75 × 75 × 13):

 answer