BendingStressMomOfInertiaC

B

MECHANICS OF MATERIALS REVIEW with static loading conditions

DESIGN PROCESS

! Must identify and find worst case

Disclaimer from Prof. Caldwell – some of these are “as they appeared” on tests. They may have additional calculations on the page and may not be documented very clearly.

Calculate the Stress

!

Find the loading

Condition

s!

Free-bodydiagram(s)

Statics equations

Find cross-section

property

!

Calculate the design

Stress

Find the Guideline

!

Find the material property

Compare

Mechanics of Materials 1Equation Sheet

Loading Conditions P - Axial (Tension, Compression) Ps - Direct Shear no bending T - Torsion M - Bending Moment with V - Transverse Shear

Cross-section Properties A - Area As - Shear Area J - Polar Moment of Inertia Z - Polar Section Modulus I - Moment of Inertia S - Section Modulus Q - First moment of area

Link AB is designed to be 5 inches long and 0.25 inches in diameter and made of 1020 hot rolled steel. a) Find the elongation of AB after the load is applied. b) Is AB safe for static load?c) Is pin C (diameter 0.25 in) safe if it is also made of 1020 hot rolled steel?

CyCs

AB

3K

a)

b)

Guideline

SAFE

c)

Guideline (note Sys = Sy/2) Mott

NOT SAFE

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All links and pins in the assembly show are made of AISI 1040 Hot Rolled and are carrying a static load. Knowing that a 0.6 in diameter pin is used at each connection, a Is link AB safe at the hole?b Is pin C safe in shear?

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An electric motor must be purchased to deliver 6 kip∙in of torque to the applications when rotating at 100 rpm. a) Find the motor size needed in horsepower. b) The shaft is aluminum (G=3.7x106 psi). Find the angle of twist from B to D. c) Find the torsional shear stress in shaft CD.

C

B

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Ix Ix Ady244 384 428 in4

84kip f t12inf t 5in

428 in4

11.78kipin2

Ain2Ixin4dyinAdy2in41 26224

2 62312

8

4 2442384

2 26312

36

0 0 44 384

Sketch the shear and moment diagram for the beam and find the worst case of bending moment and the resulting stress. The cross section is symmetric with all members 2 inches x 6 inches.

A=12 kip B=12 kip

2 kip/ft

V kip

x ft

12

-12

M kip ft

x ft

48

84

6 in

6 in1

2

1

124 48

48 12612 84

2 in

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SIMPLY SUPPORTED STRUCTURE (Mott)

WELDED STRUCTURE (Mott)

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Design Side Rails for Distributed Load of 2000 lb in bed

2000 lb

48 in 48 in72 in

Rtire Rhitch

2000 lb

72 in + 48 inRtire Rhitch

12 in

DESIGN SIDE RAILS

-800

Bending StressUse 2 cross-sections

Guideline for impact

Find needed section modulus

Factor of Safety

FS = 7.7

https://www.metalsupermarkets.com/MSC-MetalGuide.aspx?CategoryID=STEEL-HOT_ROLLED&ProductID=TUBE_RECT

Shear and Moment Diagram

ASTM A500 Grade BMatweb.com

For more advanced analysis use the Mechanical Design text

7 cross members, 2000 lb distributed load

Check cross members – use Welded Structure Diagram

72

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Short-Column BucklingProblem: Determine the critical load on a steel column having a square cross section 12 mm on a side with a length of 300 mm. The column is to be made of AISI 1040, hot-rolled. It will be rigidly welded to a firm support at one end and connected by a pin joint at the other. Also compute the allowable load on the column for a design factor of N = 3.

Solution: Given: L = 300 mm, cross section is square, each side is b = 12 mm.One pinned end; one fixed end. Column is Steel; AISI 1040 hot-rolled.From Steel properties table: sy = 414 MPa; E = 207 GPa = 207 x 109 N/m2

Step 1: Determine the end-fixity factor. For the fixed-pinned-end column, K = 0.8 is a practical valueStep 2: Compute the effective length.

Le=KL=0.8 (300mm )=240mmStep 3: Compute the smallest value of the radius of gyration. From properties of Areas table, for a square cross section,r=b/√12. Then,

r= b√12

=12mm√12

=3.46 mm

Step 4: Compute the slenderness ratio, SR = Le/r.

SR=Le

r= 240mm

3.46mm=69.4

Step 5: Compute the column constant, Cc

C c=√ 2π 2 Es y

=√ 2π2(207 x 109 Nm2 )

414 x106 Nm2

=99.3

Step 6: Compare Cc with SR and decide if column is long or short. Then use the appropriate column formula to compute the critical buckling load. Since SR is less than Cc, the Johnson formula applies.

Pcr=A sy [1− s y (SR)2

4 π2 E ]The area of the square cross section is

A=b2=(12mm)2=144mm2

Then,

Pcr=(144mm2)(414 Nmm2 ) [1−

(414 x 106 Nm2 )(69.4 )2

4π 2(207 x109 Nm2 ) ]

Pcr=45.1kNStep 7: A design factor of N = 3 is specified.Step 8: The allowable load, Pa , is

Pa=Pcr

N=45.1 kN

3=15.0kN

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Long-Column BucklingProblem: A round compression member with both ends pinned and made of AISI 1020 cold-drawn steel is to be used in a machine. Its diameter is 25mm, and its length is 950mm. What maximum load can the member take before buckling would be expected? Also compute the allowable load on the column for a design factor of N=3.

Solution: Given: L = 950 mm, cross section is circular, D = 25mm, Pinned endsColumn is Steel; AISI 1020 cold-drawnFrom Steel properties table: sy = 441 MPa; E = 207 GPa = 207 x 109 N/m2

Step 1: Determine the end-fixity factor. For pinned-end column, K=1.0Step 2: Compute the effective length.

Le=KL=1.0 (950mm )=950mmStep 3: Compute the smallest value of the radius of gyration. From properties of Areas table, for any axis of a circular cross section, r = D/4. Then,

r=D4

=25mm4

=6.25mm

Step 4: Compute the slenderness ratio, SR = Le/r.

SR=Le

r= 950mm

6.25mm=152

Step 5: Compute the column constant, Cc

C c=√ 2π 2 Es y

=√ 2π2(207 x 109 Nm2 )

441 x 106 Nm2

=96.2

Step 6: Compare Cc with SR and decide if column is long or short. Then use the appropriate column formula to compute the critical buckling load. Since SR is greater than Cc, Euler formula applies.

Pcr=π 2 EA(SR )2

The area is

A=π D2

4=

π (25mm )2

4=491mm2

Then,

Pcr=

π 2(207 x109 Nm2 ) ( 491mm2 )

(152 )2x 1m2

(103 mm )2=43.4 kN

Step 7: A design factor of N = 3 is specified.Step 8: The allowable load, Pa , is

Pa=Pcr

N=43.4 kN

3=14.5 kN

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Combined Torque and Bending Problem: Specify a suitable material for the shaft shown in the figure below. The shaft has a uniform diameter of 55 mm and rotates at 120 rpm while transmitting 3.75 kW of power. The chain sprockets at B and C are keyed to the shaft with sled-runner keyseats. Sprocket C receives the power, and sprocket B delivers it to another shaft. The bearings at A and D provide simple support for the shaft.

Solution: Given: Shaft and loading if figure above

Power = P = 3.75 kW. Rotational speed n = 120 rpmShaft diameter = D =55 mmSled-runner keyseats at B and C.Simple supports at A and D.

Step 1: The desirable unit for torque is N*m. Then it is most convenient to observe that the power unit of kilowatts is equivalent to the units of kN*m/s. Also, rotational speed must be expressed in rad/s.

n=

120 revmin

∗2π rad

rev∗1min

60 s=12.57 rad / s

We can now compute the torque.

T=Pn

=

3.75 kN∗ms

∗1

12.57 rad /s=0.298 kN∗m

Step 2: The tensions in the chains are indicated in the figure by the forces F1 and F2. For the shaft to be in equilibrium, the torque on both sprockets must be the same in magnitude but opposite in direction. On either sprocket the torque is the product of the chain force times the radius of the pulley that is,

T=F1 R1=F2 R2

The forces can now be computed.

F1=TR1

=

0.298 kN∗m75mm

∗103mm

m=3.97 kN

F2=TR2

=

0.298 kN∗m50mm

∗103mm

m=5.96 kN

Step 3: The figure below shows the complete shearing force and bending moment diagrams. The maximum bending moment is 1.06 kN*m at section B, where one of the sprockets is located

Step 4: At section B, the applied bending stress can be calculated below, using the bending stress for circular cross

section equation. Where S= π32

D3.

σ= MS

σ x=(1060∗103 N∗mm )

π32

(55mm)3

σ x=64.90 MPaStep 5: Find the polar section modulus

Zp=π D3

16=π (55mm )3

16=3.267 ¿104 mm3

Step 6: Find the applied torsional shear stress can be calculated below.

τ xy=TZp

=298∗103 N∗mm3.267 ¿104 mm3 =9.12MPa

Step 7: Find maximum stress

τ max=√( σx

2 )2

+ τxy2=√(64.90 MPa

2 )2

+(9.12 MPa)2

τ max=33.71 MPaStep 8: Find effective maximum stress considering Kt (same for both bending and torsion)

τ max=(33.71 MPa )∗1.6τ max=53.94

Step 9: Let τ max=τd=Sy

2 N

Then sy=2N τmax=(2 ) (4 ) (53.94 MPa )=431.5MPa

Step 10: Refer to Typical Properties of Alloys table there are many different types of alloys we could use. For example AISI 1040, cold-drawn, has a yield strength of 490 MPa. Alloy AISI 1141 OQT 1300 has a yield strength of 469 MPa and also a very good ductility, as indicated by the 28% elongation. Either of these would be a reasonable choice.

σ= k PA

1ksi=1 K¿2 1K=1000 lb

1 MPa=1 Nmm2=106 Pa=106 N

m2

1GPa=1 kNmm2 =103 MPa=109 Pa

σ d=su

N∨

s y

Nuse guideline

E=∆σ∆ε

=slope of linear region

ε= δL0

δ=Lf −L0

A¿̊=π d2

4 C ¿̊=πd ¿ ¿

δ= PLAE

δ temp=αL(∆T )

σ B=PB

ABσ B pin∧hole=

PB

t ∙ d

τ=Ps

A sG=

τ xy

γ xy

ν= E2G

−1

τ max=kTcJ

= kTZ

J¿̊= πd 4

32 Z¿̊=π d3

16¿¿

θ=TLGJ

θ ∙ r=γ ∙ L

2π radians=360o=1revPower=T ∙ω

1hp=550 ft lbs

1watt=1 N ms

746watts=1hp

Y=Σ(A ∙ y)

ΣA

I total=Σ I+Σ ( A d2 ) I x=rx2 A

I rectangle=bh3

12I

¿̊= πd 4

64¿

σ= k MCI

= k MS

Srect=bh2

6

∫V 0

V

dV=∫0

x

−wdxw=−dVdx

∫M 0

M

dM=∫0

x

V dxV=dMdx

τ=VQ¿ Q=Σ(Ap ∙ y )

(alpha)Coefficient of thermal expansion. /oC, /oF. (theta) Angle of twist, radians (delta or del) Linear deformation = Lf – L0, in, mm.ε (epsilon)Normal strain, in/in, mm/mm (gamma)Shear strain, rad (sigma)Normal stress, psi, ksi, MPa, GPa. (nu)Poission’s ratio, no units. (omega)Rotational speed, rad/s (theta)Shear stress, psi, ksi, MPa, GPa.B Bearing Stress , psi, ksi, MPa, GPa.d Design or safe stress , psi, ksi, MPa, GPa.A Area, in2, mm2

AB Bearing area, in2, mm2

Ap Used in (Ap y), area of that part of the cross-section outside the plane of interest, in2, mm2

As Shear area, in2, mm2

c distance from center of rotation or neutral axis to outer most fiber, in, mmC Circumference, in, mm.d or D Diameter, in, mm.d Used in (Ad2), distance from major centroid (Y-bar) to section centroid (y), in, mm.E Modulus of elasticity, psi, ksi, MPa, GPa.G Modulus of rigidity or modulus of elasticity in shear, psi, ksi, MPa, GPa.GPa Giga-Pascals, unit of stress = kN/mm2

hp Horsepower.

I Moment of inertia, in4 mm4

J Polar moment of inertia, in4, mm4.k Stress concentration factor, no units.K or kip = 1000 poundsksi K per square inch, K/in2 or kip/in2

L Length (sometimes gage length), in, mm.M Moment (bending) in lb, N mm.MPa Mega-Pascals, unit of stress = N/mm2

N Design factor, no unitsP Axial force, lb, N.Pa Pascals, unit of stress, N/m2.PB Bearing force, lb, N.PS Shear force no bending, lb, N.psi Pounds per square inch, lb/in2

Q First moment of the area, in3, mm3

r Radius (always use d/2)rx Radius of gyration about the x axis, in, mm.S Section modulus, in3, mm3

Su Tensile or ultimate strength, psi, ksi, MPa, GPa.Sus Ultimate shear strength , psi, ksi, MPa, GPa.Sy Yield strength, psi, ksi, MPa, GPa.Sys Yield shear strength, psi, ksi, MPa, GPa.t Thickness, in mm.T Torque or Torsion, in lb, N m.V Shear force (transverse) occurs with bending, lb, N.w Distributed load, lb/in or N/mm.

Mechanics of Materials I EQUATIONS

y Used in (Ay2), distance from self selected datum (y=0) to centroid of A, in mm.Y Y-bar, Centroid in the y direction, in, mm.y Used in (Ay ),distance from major centroid (Y-bar) to the centroid of Ap, in, mm.Z Polar section modulus , in3, mm3