Problem 1A column 3m high has a solid circular cross section and
carries an axial load of 10000KN. Direct stress in the column is
limited to 150MPa. Poissons ratio is 0.30 and modulus of elasticity
is 200GPa.1. Determine the minimum allowable diameter of the
column.2. Determine the shortening of the column.3. Determine the
increase in its diameter.Solution:1. S = P/A
150 =
D = 291.3mm
2.
3.
Problem 2A concrete column of height 5m has a square cross
section of side 300mm. It is designed to support an axial load of
200KN. At mid-height, a recess is out in one face of the column to
receive a floor beam. The cut is 300mm vertical and 75mm deep. E =
15000MPa.1. Determine the stress of the column before the recess is
cut.2. Determine the strain energy of the column produced by the
axial load before the recess is cut out.3. Determine the strain
energy of the column after the recess is cut.Solution:1. S =
P/A
S = 200000/300(300)
S = 2.22MPa
2. U = P2L/2AE
U = 2000002(5000)/2(300)(300)(15000)
U = 74.07N-m
3. U = P2L1/2AE + P2L2/2AE
U = 2000002(4700)/2(300)(300)(15000) +
2000002(300)/2(300)(225)(15000)
U = 75.55N-m
Problem 3An aluminum bar having a cross-sectional area of 160mm2
carries the axial loads as shown. Assume that the bar is suitably
braced to prevent buckling and E = 70000MPa.1. Determine the axial
load acting on bar B2.2. Determine the axial stress of bar B1.3.
Determine the total deformation of the bars B1,B2 and B3.
B1B2B335KN15KN30KN10KN
B1 = 0.8mB2 = 1.0mB3 = 0.6m
Solution:1. P2 = 35 15 = 20KN 2. S1 = P1/A1 = 35000/160 =
218.75MPa3. Y = PL/AEYT = Y1 + Y2 + Y3YT = 35000(800)/160(70000) +
20000(1000)/160(70000) 10000(600)/160(70000) = 3.75mm
(elongation)
Problem 4A compound shaft is attached to a rigid
supports.Properties:Bronze:Diameter = 75mmG = 35GPaAllowable stress
= 60MPaLength = aSteel:Diameter = 50mmG = 83GPaAllowable stress =
80MPaLength = b
1. Determine the ratio of the lengths b/a so that each material
will be stressed to its maximum permissible limit.2. Determine the
maximum torque carried by the bronze.3. Determine the maximum
torque that could be applied to the compound shaft.
Solution:1. (TL/JG)B = (TL/JG)SS = Td/2JT = 2SJ/d(2SJL/dJG)B =
(2SJL/dJG)S(SL/dG)B = (SL/dG)S60(a)/75(35) = 80(b)/50(83)b/a =
1.186
2. S = 16T/d360 = 16TB(1000)2/(75)3TB = 4.97KN-m
3. 80 = 16TS(1000)2/(50)3TS = 1.96KN-m
T = TS + TBT = 6.93KN-m
Problem 5A simple beam having a span of 20m carries a uniform
load w along its entire span and a clockwise moment of 100KN-m at
the right end of the beam. Length of the beam is equal to
20m.Cross-sectional data:Width (b) = 400mmDepth (d) = 600mm1.
Determine the value of w when the maximum moment occurs at a
distance of 9m from the left support of the beam.2. Give the value
of this maximum moment.3. Determine the flexural stress of the
beam.Solution:1. R1 w(9) = 0R1 = 9wMR2 = 0R1(20) + 100 =
w(20)(10)9w(20) + 100 = 200ww = 5KN/m
2. R1 = 9(5) = 45KNMmax = 45(9)/2 = 202.5KN-m
3. f = Mc/If = 6M/bd2f = 3.44MPa
