Bachelor of Design in Architecture
DESIGN STUDIES 1B
Year 1, July Semester
DOMAIN:
Constructing the Built
Environment
SECTION:
TOPIC:
Structures
DATE
Lecturer
Sections: Extra Exercises
Mike Rosenman
Name…………………………………………………………SID: …………………..
Where appropriate, include all units. Use appropriate units
Q1. Find the Moment of Inertia, I, and the Section Modulus, Z, of the following
timber sections: ( include units).
a) 50mm width x 250mm deep
(i) I = ………………………
(ii) Z = ………………………….
b) 75mm x 175mm
(i) I = ………………………
(ii) Z = ………………………….
c) 100mm x 150mm
(i) I = ………………………
Q2.
(ii) Z = ………………………….
Referring to 1(a), 1(b) and 1(c),
a) Which is the heaviest? ………………………………………………..
b) Which is stiffest, and by what proportion?…………………………….
c) Which is strongest, and by what proportion? …………………………………
Q3.
a) You want a beam as stiff as a 300 x 100, but you only have
space for 200mm depth. How wide would it have to be?
…………………………………………………………………………………..
b) You want a beam as strong as a 300 x 100, but you only have
space for 200mm depth. How wide would it have to be?
…………………………………………………………………………………..
Q4.
We are thinking of using a Universal (steel) Beam UB410UB59.7 to span 12 m
carrying a UDL of 15 kN/m.
Assume a maximum allowable stress of 200MPa and a Modulus of Elasticity of
200,000MPa. Use the Table of Properties for Universal Beams attached.
(a)
is the given beam strong enough? ………………………………………….
(b)
if the beam is not strong enough, which beam do we need? ……………..
(c)
is this beam stiff enough? (assume a maximum allowable deflection of
3
L/500. The deflection formula is 5/384 x WL /EI ……………………….
…………………………………………………………………………………
(d)
if the new section is not stiff enough, which beam do we need? ………
…………………………………………………………………………………
Q5.
We want a timber beam, grade F8, to span 5.5 m carrying a point load at the
centre of 6kN. Assume an allowable compressive stress of 8 MPa and a
Modulus of Elasticity of 9000 MPa.
(a)
find a beam section that would be strong enough.
…………………………………………………………………………………..
(b)
is this section stiff enough? Assume a maximum allowable deflection of
3
L/200. The deflection formula for a point load is WL / 48EI
…………………………………………………………………………………..
………………………………………………………………………………….
(c)
if it’s not stiff enough, which section is? ……………………………………
(d)
what span-to-depth ratio do you end up with, for the beam? ……………
………………………………………………………………………………….
ANSWERS & WORKINGS OUT
Q1. Find the Moment of Inertia, I, and the Section Modulus, Z, of the following
timber sections: ( include units).
a) 50mm width x 250mm deep
(i) I = bd3/12 = 50 x 2503 / 12 = 65.1 x 106 mm4
(ii) Z = bd2/6 = 50 x 2502 / 6 = 520.8 x 103 mm3
b) 75mm x 175mm
(i) I = bd3/12 = 75 x 1753 / 12 = 33.5 x 106 mm4
(ii) Z = bd2/6 = 75 x 1752 / 6 = 382.8 x 103 mm3
c) 100mm x 150mm
(i) I = bd3/12 = 100 x 1503 / 12 = 28.1 x 106 mm4
(ii) Z = bd2/6 = 100x 1502 / 6 = 375 x 103 mm3
Q2.
Referring to 1(a), 1(b) and 1(c),
a) Which is the heaviest?
area of 50 x250 = 12500 mm2
area of 75 x 175 = 13125 mm2 1.05 heavier than section 1(a)
area of 100 x150 = 15000 mm2
1.20 heavier than section 1(a)
1.14 heavier than section 1(b)
section 1(c) is the heaviest, i.e. uses most material
b) Which is stiffest, and by what proportion?
Section 1(a) is the stiffest – I = 65.1 x 106 mm4
with respect to Section 1(b) I(a) / I(b) = 65.1 / 33.5 = 1.94
with respect to Section 1(c) I(a) / I(c) = 65.1 / 28.1 = 2.32
section 1(a) is 1.9 times stiffer than section 1(b) and 2.3 times stiffer than section
1(c). This means that section 1(a) will have a deflection approximately half that of
sections 1(b) or 1(c)
c) Which is strongest, and by what proportion?
Section 1(a) is the strongest – Z = 520.8 x 103 mm3
with respect to Section 1(b), Z(a) / Z(b) = 520.8 / 382.8 = 1.36
with respect to Section 1(c), Z(a) / Z(c) = 520.8 / 375 = 1.39
section 1(a) is 1.3.6 times stronger than section 1(b) and 1.39 times stronger than
section 1(c). This means that section 1(a) can carry approximately 40% more load
than sections 1(b) or 1(c) for the same span.
Note that while section 1(a) uses less material than the others, it is stronger and
stiffer – all because of its greater depth.
Q3.
a) You want a beam as stiff as a 300 x 100, but you only have
space for 200mm depth. How wide would it have to be?
First find the Moment of Inertia, I, of the beam
I = bd3/12 = 100 x 3003 / 12 = 225 x 106 mm4
So we need a beam 200 mm deep with the same value of I. Let the width = B
Bd3/12 = 225 x 106
B = 225 x 106 x 12 / 2003
= 225 x 106 x 12 / 23 x 1003
= 2700 /8 = 337.5 mm
So the 200mm deep beam would have to be 350 mm wide, a huge increase.
b) You want a beam as strong as a 300 x 100, but you only have
space for 200mm depth. How wide would it have to be?
First find the Section Modulus, Z, of the beam
Z = bd2/6 = 100 x 3002 / 6 = 1500 x 103 mm3
So we need a beam 200 mm deep with the same value of Z. Let the width = B
Bd2/6 = 1500 x 103
B = 1500 x 103 x 6 / 22 x 1002
= 9000 x 103 / 4 x 1002
= 900 /8 = 225 mm
So the 200mm deep beam would have to be 225 mm wide, a huge increase.
Q4.
We are thinking of using a Universal (steel) Beam UB410UB59.7 to span 12 m
carrying a UDL of 15 kN/m.
Assume a maximum allowable stress of 200MPa and a Modulus of Elasticity of
200,000MPa. Use the Table of Properties for Universal Beams attached.
(a)
is the given beam strong enough?
First, work out the maximum Bending Moment
Using max BM = wL2/8 (for a simply supported beam with a UDL)
Max BM = 15 x 122 / 8 = 270 kNm
Next, get the Section Modulus of the beam from the Table
Zx = 1060 x 103 mm3
Using the formula for stress, f = M / Z, we get
f = 270 x 103 x 103 / 1060 x 103 (bring everything to Newtons and mm)
= 254.7 MPa (N/mm2)
Since the allowable stress is 200 MPa, the beam is not strong enough
(b)
if the beam is not strong enough, which beam do we need?
We need a beam which has a Section Modulus 254.7 / 200 as big as the old one
New Required Zx = 254.7 / 200 x 1060 x 103
= 1.274 x 1060 x 103
= 1350 x 103 mm3
So look up the Table for a beam which has a Zx of at least 1350 x 103 mm3
We find the beam 460UB74.6 has a Zx of 1460 103 mm3 and a depth of 457 mm
(c)
is this beam stiff enough? (assume a maximum allowable deflection of
3
L/500. The deflection formula is 5/384 x WL /EI
From the Table we find that the Ix value for the 460UB74.6 beam is 334 x 106
mm4
3
The Total load on the beam, W = 15 x 12 = 180 KN = 180 x 10 N
3
3 3
5
Deflection = 5 x 180 x 10 x (12 x 10 ) / 384 x 2x 10 x 334 x 106
(bringing everything to Newtons and mm)
3
12
11
= 5 x 180 x 12 x 10 / 384 x 2 x 334 x 10
= 1555200 x 10 / 25651
= 60.6 mm
Maximum allowable deflection = 12 x 1000 / 500 = 24 mm
The new beam deflects 60.6 / 24 = 2.53 times as much as allowed. The new
beam is not stiff enough and a newer beam needs to be found with a greater
Moment of Inertia
(d)
if the new section is not stiff enough, which beam do we need?
New I required = Old I x 2.53 = 334 x 2.53 = 845.0 x 106 mm4
We look up the Table for a beam which has an Ix of at least 845.0 x 106 mm4
We find the beam 610UB113 which has an Ix of 874 x 106 mm4
and a depth of 607mm
Q5.
We want a timber beam, grade F8, to span 5.5 m carrying a point load at the
centre of 6kN. Assume an allowable compressive stress of 8 MPa and a
Modulus of Elasticity of 9000 MPa.
(c)
find a beam section that would be strong enough.
First, work out the maximum Bending Moment
Using max BM = WL/4 (for a simply supported beam with a central point load)
Max BM = 6 x 5.5 / 4 = 8.25 kNm
Next, using the formula for stress and assuming the maximum allowable stress as
the stress required, get the required Section Modulus of the beam
f=M/Z
Zreqd = M / f
= 8.25 x 103 x 103 / 8
= 1031 x103 mm3
Using the formula, z = bd2/6, we can try different widths of beams
(i) with width, b, = 50 mm
d2 = 1031 x103 x6 / 50 = 123,720
d = 351
so a beam that would be strong enough would be a 355 x 50 mm F8 beam
(ii) with width, b, = 75 mm
d2 = 1031 x103 x6 / 75 = 82,480
d = 287.2
so a beam that would be strong enough would be a 300 x 75 mm F8 beam
(iii) with width, b, = 100 mm
d2 = 1031 x103 x6 / 100 = 61,860
d = 248.7
so a beam that would be strong enough would be a 250 x 100 mm F8 beam
(d)
is this section stiff enough? Assume a maximum allowable deflection of
3
L/200. The deflection formula for a point load is WL / 48EI
Let us assume that a 355 x 50 mm beam is too slender and might buckle. Let us
try the 300 x 75 mm as being more likely to be stiffer than the 250 x 100 mm
beam because of its greater depth.
First find the Moment of Inertia, I
I = bd3 / 12 = 75 x 3003 /12
= 168.8 x 106 mm4
deflection = 6 x 103 x (5.5 x 103) 3 / 48 x 9000 x 168.8 x 106
(bring everything to Newtons and mm)
= 998 x 1012 / 729 x 1011
= 13.7mm
maximum allowed deflection = 5.5 x1000 / 200 = 27.5 mm
(d)
if it’s not stiff enough, which section is?
The section is stiff enough. No need to upsize it.
(e)
what span-to-depth ratio do you end up with, for the beam?
span / depth = 5.5 x 1000 / 300
= 18.5
this compares well with span/depth ratios of 18 – 20 given for timber beams