Floor Framing
In floor framing, the subfloor is supported by light beams
called floor joists or simply joists which in turn supported by
heavier beams called girders then girders pass the load to columns.
Typically, joist act as simply supported beam carrying a uniform
load of magnitude p over an area of sL, where
p = floor load per unit area
L = length (or span) of joist
s = center to center spacing of joists and
wo = sp = intensity of distributed load in joist.
Problem 538
Floor joists 50 mm wide by 200 mm high, simply supported on a 4-m
span, carry a floor loaded at 5 kN/m2. Compute the center-line
spacing between joists to develop a bending stress of 8 MPa. What
safe floor load could be carried on a center-line spacing of 0.40 m?
Solution 538
Part 1:
(fb)max=McI(fb)max=McI
where:
(fb)max=8MPa(fb)max=8MPa
M=18(5s)(42)M=18(5s)(42)
M=10skN⋅mM=10skN⋅m
c=h/2=200/2c=h/2=200/2
c=100mmc=100mm
I=bh312=50(2003)12I=bh312=50(200
3)12
I=33.33×106mm4I=33.33×106mm4
Thus,
8=10(100)(10002)33.33×1068=10(100)(10002)33.33×106
s=0.267ms=0.267m answer
Part 2:
(fb)max=McI(fb)max=McI
where:
M=18woL2=18(0.4p)(42)M=18woL2=1
8(0.4p)(42)
M=0.8pM=0.8p
Thus,
8=0.8p(100)(10002)33.33×1068=0.8p(100)(10002)33.33×106
p=3.33kN/m2p=3.33kN/m2 answer
Problem 539
Timbers 12 inches by 12 inches, spaced 3 feet apart on centers, are
driven into the ground and act as cantilever beams to back-up the
sheet piling of a coffer dam. What is the maximum safe height of
water behind the dam if water weighs = 62.5 lb/ft3 and ( fb)max =
1200 psi?
Solution 539
wo = 62.5hlb/ft2
F = ½ woh(3) = 3/2 (62.5h)h
F = 93.75h2lb
M = (1/3 h)F = 1/3h(93.75h2)
M = 31.25h3lb⋅ft
(fb)max = Mc
I
1200 = 31.25h3(12)(12/2)
12(123)
12
h = 9.73ft answer
Problem 540
Timbers 8 inches wide by 12 inches deep and 15 feet long,
supported at top and bottom, back up a dam restraining water 9
feet deep. Water weighs 62.5 lb/ft3. (a) Compute the center-line
spacing of the timbers to cause fb = 1000 psi. (b) Will this spacing
be safe if the maximum fb, (fb)max = 1600 psi, and the water reaches
its maximum depth of 15 ft?
Solution 540
Part (a)
wo = 62.5(9s) = 562.5slb/ft
Fw = ½ wo(9)
Fw = ½ (562.5s)(9)
Fw = 2531.25slb
ΣMR1 = 0
15R2 = 12Fw
15R2 = 12(2531.25s)
R2 = 2025s
ΣMR2 = 0
15R1 = 3Fw
15R1 = 3(2531.25s)
R1 = 506.25s
Location of Maximum Moment
y/x = 562.5s
9
y = 62.5s
506.25s – ½ xy = 0
506.25s – ½ x(62.5sx) = 0
x2 = 16.2
x = 4.02ft
Maximum Moment
M = (506.25s)(6) + 2/3 (x)(506.25s)
M = 3037.5s + 337.5(4.02s)
M = 4394.25s
Required Spacing
(fb)max = Mc
I
100 = 4394.25s(12)(12/2)
8(123)
12
s = 3.64ft answer
Part (b)
wo = 62.5(15)(3.64) = 3412.5lb/ft
Fw = ½ wo(15)
Fw = ½ (3412.5)(15)
Fw = 25593.75lb
ΣMR1 = 0
15R2 = 10Fw
15R_2 = 10(25\,593.75)
R2 = 17062.5lb
ΣMR2 = 0
15R1 = 5Fw
15R1 = 5(25593.75)
R1 = 8531.25lb
Location of Maximum Moment (Shear = 0)
y/x = 3412.5
15
y = 227.5x
8531.25 – ½ xy = 0
8531.25 – ½ x(227.5x) = 0
x2 = 75
x = 8.66ft
Maximum Moment
M = 2/3 x(8531.25)
M = 2/3 (8.66)(8531.25)
M = 49255.19lb⋅ft
Actual Stress
fb = Mc
I
fb = (49255.19)(12)(12/2)
8(123)
12
fb = 3078.36psi > 1600psi
Therefore, the 3.64 ft spacing of timbers is not safe when water reaches its
maximum depth of 15 ft.
Problem 541
The 18-ft long floor beams in a building are simply supported at
their ends and carry a floor load of 0.6 lb/in2. If the beams have
W10 × 30 sections, determine the center-line spacing using an
allowable flexural stress of 18 ksi.
Solution 541
wo = (0.6lb/in2)(12in/ft)2(sft)
wo = 86.4slb/ft
R1 = R2 = ½ (86.4s)(18)
R1 = R2 = 777.6slb
Mmax = 777.6s(9) − 86.4s(9)(4.5)
Mmax = 3499.2slb⋅ft
From Table B-7 in Appendix B of textbook: Properties of Wide-Flange
Sections (W-Shapes): US Customary Units.
Designation Section Modulus, S
W10 × 30 32.4 in3
fb = M/S
18000 = 3499.2s(12)
32.4
s=13.9ft answer
Problem 542
Select the lightest W shape sections that can be used for the beams
and girders in Illustrative Problem 537 of text book if the allowable
flexural stress is 120 MPa. Neglect the weights of the members.
Solution 542
For Beams (B - 1)
Total Load,
W = 5(2×;4)
W = 40kN
Distributed Load,
wo = W/L = 40/4
wo = 10kN/m
R1 = R2 = ½ W = ½ (40)
R1 = R2 = 20kN
Mmax = R1(L/2) − 10(L/2)(L/4)
Mmax = 20(4/2) − 10(4/2)(4/4)
Mmax = 20kN⋅m
Srequired = M = 20(10002)
fb 120
Srequired = 166666.67mm3
From Appendix B, Table B-2 Properties of Wide-Flange Sections (W
Shapes): SI Units, of text book:
Designation Section Modulus
W130 × 28 167 × 103 mm3
W150 × 24 168 × 103 mm3
W200 × 22 194 × 103 mm3
W250 × 18 179 × 103 mm3
Consider W250 × 18 with S = 179 × 103 mm3
Checking:
Sresisting ≥ Slive−load + Sdead−load
Sresisting = 179 × 103mm3
Slive−load = 166666.67mm3
Sdead−load = Mdead−load = [1/8 [0.17658)(4)]1000
fb 120
3
Sdead−load = 2.943mm
(179 × 103) ≥ 166666.67 + 2.943
(179 × 103) ≥ (166.67 × 103)
Use W250 × 18 for B-1. answer
For Girder (G - 1)
Slive−load = M = 40(10002)
fb 120
Slive−load = 333.33 × 103mm3
From Appendix B, Table B-2
Properties of Wide-Flange
Sections (W Shapes): SI Units,
of text book:
Section
Designation
Modulus
342 ×
W200 × 36
103 mm3
379 ×
W250 × 33
103 mm3
351 ×
W310 × 28
103 mm3
Consider W310 × 28 with S = 351 ×
103 mm3
Checking:
Ssupplied ≥ Srequired + Sown−weight
Ssupplied = 1790 ×10mm3
Srequired = 333.33 × 103mm3
Sown−weight = Mown−weight = [1/8 (274.68)(62)](1000)
Fb 120
Sown−weight = 10300.5mm3
(1790 × 103) ≥ (333.33 × 103) + 10300.5
(1790 × 103) ≥ (343.63 × 103)
Use W310 × 28 for G-1. answer
For Beams (B - 2)
ΣMR2 = 0
6R1 = 20(4) + 10(2)(5) + 15(4)(2)
R1 = 50kN
ΣMR1 = 0
6R2 = 20(2) + 10(2)(1) + 15(4)(4)
R2 = 50kN
Location of Maximum Moment
x = 4−x
50 10
10x = 200 – 50x
x = 10/3m
Mmax = ½ (10/3)(50)
Mmax = 250/3 kN⋅m
Srequired = M = 250/3 (10002
fb 120
Srequired = 695 × 103mm3
From Appendix B, Table B-2 Properties of Wide Flange Sections (W
Shapes): SI Units, of text book:
Section
Designation
Modulus
709 ×
W200 × 71
103 mm3
806 ×
W250 × 67
103 mm3
747 ×
W310 × 52
103 mm3
796 ×
W360 × 51
103 mm3
773 ×
W410 × 46
103 mm3
Consider W410 × 46 with S = 773 × 103 mm3
Checking:
Ssupplied ≥ Srequired + Sown−weight
Ssupplied = 773 × 103mm3
Srequired = 695 × 103mm3
Sown−weight = Mown−weight = 1/8 (451.26)(42)(1000)
fb 12
Sown−weight = 7521mm3
Srequired + Sown−weight = (695 × 103) + 7521
Srequired + Sown−weight = 702521mm3
(Ssupplied = 773 × 103mm3) > 702521mm3
Use W410 × 46 for Beams (B - 2). answer
For Beams (B - 3)
Mmax = 1/8 (20)(62)
Mmax = 90kN⋅m
Srequired = Mmax =
2
90(1000 )
fb 120
3 3
Srequired = 750 × 10 mm
From Appendix B, Table B-2 Properties of Wide Flange Sections (W
Shapes): SI Units, of text book:
Section
Designation
Modulus
853 ×
W200 × 86
103 mm3
806 ×
W250 × 67
103 mm3
849 ×
W310 × 60
103 mm3
796 ×
W360 × 51
103 mm3
773 ×
W410 × 46
103 mm3
Consider W410 × 46 with S = 773 × 103 mm3
From the Checking of B - 2
Sown−weight = 7521mm3
Srequired + Sown−weight = (750 × 103) + 7521
Srequired + Sown−weight = 757521mm3
(Ssupplied = 773 × 103mm3) > 757521mm3
Use W410 × 46 for B-3, this section is the same to B - 2. answer
For Girders (G - 2)
Srequired = M = 120(10002)
fb 120
Srequired = 1000 × 103mm3
From Appendix B, Table B-2 Properties of Wide-Flange Sections
(W Shapes): SI Units, of text book:
Section
Designation
Modulus
1100 ×
W250 × 89
103 mm3
1060 ×
W310 × 74
103mm3
1030 ×
W360 × 64
103mm3
1060 ×
W410 × 60
103mm3
1120 ×
W460 × 60
103 mm3
There are two options, both exceeds the required S of 1000 ×
103 mm3. One is W410 × 60 with theoretical mass of 59.5 kg/m and
the other is W460 × 60 with theoretical mass of 59.6 kg/m. For
economic reason,
we prefer W410 × 60.
Checking:
Ssupplied ≥ Srequired + Sown−weight
Ssupplied = 1060 × 103mm3
Srequired = 1000 × 103mm3
Sown−weight = Mown−weight = 1/8 (588.6)(62)(1000)
Fb 120
3
Sown−weight = 22072.5mm
Srequired + Sown−weight = (1000 × 103) + 22072.5
Srequired + Sown−weight = 1 022 072.5mm3
(Ssupplied = 1060 × 103mm3) > 1 022 072.5mm3
Use W410 × 60 for G-2. answer
Problem 543
A portion of the floor plan of a building is shown in Fig. P-543. The
total loading (including live and dead loads) in each bay is as
shown. Select the lightest suitable W-shape if the allowable flexural
stress is 120 MPa.
Solution 543
For Member B - 1
Srequired = M = 1/8 (22.5)(52)(10002)
Fb 120
3 3
Srequired = 586 × 10 mm
From Appendix B, Table B-2 Properties of
Wide-Flange Sections (W Shapes): SI Units, of text book:
Use W410 × 39 with S = 634 × 103 mm3
for member B - 1. answer
For Member G - 1
M = 2.5(28.125)
M = 70.3125kN⋅m
Srequired = M = 70.3125(10002)
Fb 120
Srequired = 586 × 103mm3
From Appendix B, Table B-2 Properties of Wide-Flange Sections (W
Shapes): SI Units, of text book:
Use W410 × 39 with S = 634 × 103 mm3 for member G - 1. answer
For Member B - 2:
ΣMR2 = 0
7R1 = 28.125(5) + 18.75(2)(6) + 30(5)(2.5)
R1 = 105.804kN
ΣMR1 = 0
7R2 = 28.125(2) + 18.75(2)(1) + 30(5)(4.5)
R2 = 109.821kN
Location of Maximum Moment:
x = 5−x
109.821 40.179
40.179x = 549.105 − 109.821x
x = 3.6607m
Maximum Moment
M = ½ (3.6607)(109.821)
M = 201.01kN⋅m
Srequired = M = 201.01(10002)
Fb 120
Srequired = 1675 × 10 mm3
3
From Appendix B, Table B-2 Properties of Wide-Flange Sections (W
Shapes): SI Units, of text book:
Use W610 × 82 with S = 1 870 × 103 mm3 for member B - 2. answer
For Member B - 3
Srequired = M = 1/8 (37.5)(72)(10002)
Fb 120
3 3
Srequired = 1914 × 10 mm
From Appendix B, Table B-2 Properties of
Wide-Flange Sections (W Shapes): SI Units,
of text book:
Use W610 × 92 with S = 2 140 × 103 mm3 for member B - 3. answer
Summary
