3.
11 Solutions Problem Set # 6
Problem #1 Determine the maximum shear stress and rate of twist of the given shaft if a
10 kNm torque is applied to it. If the length of the shaft is 15 m, how much would it
rotate by? Let G = 81 GPa, D = 75 mm
Which equates to :
If the shaft is 15 m long, the angle of rotation at the free end is 34.157o degrees.
Problem #2 GERE Problem 3.3-3
Aluminum Bar in Torsion
a) kT=GIp/L = Gd4/32L = (3.8e6psi)( )(1.0in)^4/(32*48in) = 7770lb-in
b) = 5 = 5*/180 rad = 0.087rad
= TL/GIp T = GIp/L
max = Tr/Ip = Td/2Ip plug in T, so
max = Gd /2L = 3450psi
Max Shear Strain
Hookes law: so max = max/G = 909e-6 rad
Problem #3 GERE Problem 3.4-2
Polar Moments of Inertia for AB, BC and CD
Calc I = /32 * d^4
AB: I = 4.021e6mm^4
BC: I = 1.272e6
CD: I = 0.2513e6
AB
a) Shear Stresses
AB: = TAB rAB/IpAB = 57.7 MPa
BC: = TBC rBC/IpBC = 66.0 MPa
CD: = TCD rCD/IpCD = 63.7 MPa
max = 66.0Mpa
b) angle of twist
AB: AB = TABLAB/G(Ip)AB = 0.00902rad
BC: BC = TBCLBC/G(Ip)BC = 0.01376rad
CD: CD = TCDLCD/G(Ip)CD = 0.01990rad
D = AB + BC + CD = 0.04268 rad = 2.45degrees
Stresses and Strains within a Beam
Problem #4 GERE Problem 5.5-3
a) Maximum Bending Stress
* = L so = L/
max = Ey/ = E(t/2) /L = 50,600psi
b) Change in stress
If the angle increases, the stress max increases.
Problem #5 GERE Problem 5.5-6
Mmax = Pb
S = /32 * d^3
max = Mmax/S = 32Pb/ (d^3) = 185.0MPa
Problem #6 Horizontal Shear Stresses
STATICS & STRENGTH OF MATERIALS
A loaded, simply supported W 10 x 45 beam is shown below. For this beam:
A. Determine the maximum bending stress 6 feet from the left end of the
beam.
B. Determine the horizontal shear stress at a point 4 inches above the bottom
of the beam cross section and 6 feet from the left end of the beam.
Unless otherwise indicated, all joints and support points are assumed to be pinned or
hinged joints.
Solution:
Part A:
STEP 1: Determine the external support reactions:
1.) FBD of structure (See Diagram)
2.) Resolve all forces into x/y components
3.) Apply equilibrium conditions:
Sum Fx = 0 none
Sum Fy = By + Dy - 2,000 lbs/ft (4 ft) - 5,000 lbs = 0
Sum TB = 5,000 lbs (4 ft) - 2,000 lbs/ft (4 ft) (6 ft) + Dy(8 ft) = 0
Solving: By = 9,500 lbs; Dy = 3,500 lbs
STEP 2: Determine the shear force and bending moment at x=6 ft.
1.) Cut beam at 6 ft. Draw the FBD of left end of beam, showing and labeling
all external forces.
2.) Resolve all forces into x/y directions.
3.) Apply equilibrium conditions:
Sum Fx = 0 none
Sum Fy = -5,000 lbs + 9,500 lbs - V6 = 0
Sum TA = 9,500 lbs (4 ft) - 4,500 lbs (6 ft) + M6 = 0
Solving: V6 = 4,500 lbs; M6 = -11,000 ft-lbs
STEP 3: Apply the Flexure Formula to determine the Maximum Bending Stress
(MBS) at 6'.
MBS = M6'/S (Where M6' is the bending moment at 6 ft, and S is the section
modulus for the beam. The section modulus is available from the Beam Tables.
The W 10 x 45 beam has a section modulus for the beam from the beam tables
is 49.1 in3.)
MBS = -11,000 ft-lbs(12 in/ft)/49.1 in3 = -2,688 lbs/in2
Part B:
STEP 4: To determine the Horizontal Shear Stress (HSS) at 6 ft from the end of
the beam and 4 inches above the bottom of the beam, apply the horizontal
shear stress formula.
The form we will use is: HSS = Vay'/Ib
Where:
V = Shear force 6 ft from the end of the beam
a = cross sectional area from 4 in above the bottom of the beam to bottom of
beam
y' = distance from neutral axis to the centroid of area a
I = moment of inertia of the beam (249 in4 for W 10 x 45 beam)
b = width of beam a 4 in above the bottom of the beam
HSS = [(4,500 lbs)(6.153 in2)(4.37 in)]/[(249 in4)(.35 in)] = 1,388 psi
