1. WORKSHEET 9c let's go !! Sections
to answer just click on the button or image related to the answer
let's go !!

2. what do we need to determine?
Question 1a
when we want to design a beam
what do we need to determine?
its shape and dimensions a
its material b
its Section Modulus c
a and c d
a and b e

3. what do we need to know to start?
Question 1b
when we want to design a beam
what do we need to know to start?
the span and support types a
the total load on the beam b
the shape and dimensions of the beam c
the maximum bending moment d
the moment of inertia and section modulus e
a and b f
a, b and e g

4. what else do we need to know?
Question 1c
when we want to design a beam
what else do we need to know?
the maximum allowable bending stress a
the Modulus of Elasticity b
the maximum allowable deflection c
whether it is elastic, plastic or brittle d
all the above e
a, b and c f

5. what is the tributary area?
Question 2a
In a two-story house, we are spanning across a double garage 7m wide with steel beams at
3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in Q 19 spans between these steel beams (200 x 50 mm softwood
600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and the dead load is 0.4 kPa
(including the self-weight of the joist)
what is the tributary area?
4.2 m2 a
49.0 m2 b
25.2 m2 c

6. what is the total load on a beam?
Question 2b
In a two-story house, we are spanning across a double garage 7m wide with steel beams at
3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
what is the total load on a beam?
47.9 kN a
100.8 kN b
47.9 kPa c

7. Question 2c uniformly distributed load (UDL) a point load
In a two-story house, we are spanning across a double garage 7m wide with steel beams at
3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
is this load a?
uniformly distributed load (UDL) a
a point load b

8. do we design for strength or for stiffness
Question 2d
In a two-story house, we are spanning across a double garage 7m wide with steel beams at
3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
do we design for strength or for stiffness
stiffness a
strength b

9. in order to get the dimensions of the beam
Question 2e
In a two-story house, we are spanning across a double garage 7m wide with steel beams at
3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
in order to get the dimensions of the beam
what do we need to find?
the stress in the beam a
the minimum required Section Modulus b
the minimum required Moment of Inertia c

10. Question 2f in order to find the minimum required
In a two-story house, we are spanning across a double garage 7m wide with steel beams at
3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
in order to find the minimum required
Section Modulus, what do we need to find?
the maximum Bending Moment a
the stress in the beam b
the maximum Shear Force c

11. what is the maximum Bending Moment?
Question 2g
In a two-story house, we are spanning across a double garage 7m wide with steel beams at
3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
what is the maximum Bending Moment?
293.3 kNm a
83.8 kNm b
41.9 kNm c

12. what is the minimum required Section Modulus?
Question 2h
In a two-story house, we are spanning across a double garage 7m wide with steel beams at
3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
we want to find a Universal Beam which is strong enough.
The maximum allowable stress of grade 300 steel is 200 MPa
what is the minimum required Section Modulus?
209.5 x 103 mm3 a
477.3 x 103 mm3 b
4.8 x 103 mm3 c

13. click here to see the Table of Universal Beams
Question 2i
we want to find a Universal Beam which is strong enough. Given the we need a Section
Modulus of at least x 103 mm3, we can look up a Table of Universal Beams.
click here to see the Table of Universal Beams
which beam would we select as being strong enough?
200 UB 29.8 a
200 UB 25.4 b
180 UB 22.2 c
250 UB 31.4 d

14. back to
Q2i
back to
Q2n
back to
Q2t

15. Question 2j what do we need to do now? a use it b
having found a beam section 200UB 25.4 (with a depth of 203 mm) as being strong enough
what do we need to do now?
use it a
check it for strength b
check it for stiffness c

16. Question 2k what must we do? a check its width b check its deflection
to check the stiffness of a beam
what must we do?
check its width a
check its deflection b
check its span-to-depth ratio c

17. Question 2l what must we do? a b c
to check the deflection of a beam
what must we do?
compare the actual deflection with the maximum allowable deflection a
check that it doesn’t deflect more than the
span-to-depth ratio b
make the beam twice as deep c

18. what else do we need to know?
Question 2m
to check the deflection of the 200 UB 25.4 beam
what else do we need to know?
the total load and the span a
the Moment of Inertia (I) and the Modulus of Elasticity, E b
the maximum allowable deflection c
b and c d
a, b and c e

19. Question 2n what is the Moment of Inertia of the beam? a b c
given that the Modulus of Elasticity of structural steel is MPa
and the maximum allowable deflection is span / 500, we need to find the Moment of Inertia of
the200 UB 25.4 beam.
We can do that by going back to the Table of Properties
click here to see the Table of Universal Beams
what is the Moment of Inertia of the beam?
28.9 x 106 mm4 a
44.4 x 106 mm4 b
23.6 x 106 mm4 c

20. what is the deflection of the beam?
Question 2o
Given that the Moment of Inertia of the beam is 23.6 x 106 mm4 and that the Modulus of
Elasticity of structural steel is MPa, we need to find the deflection
The deflection formula for a simply supported beam with a uniformly distributed load is
5 x WL3 / 384 E I
(remember the total load is kN, the span is 7 m)
what is the deflection of the beam?
45.3 mm a
453.0 mm b
35.5 mm c

21. is the section stiff enough?
Question 2p
Given that the deflection of the beam is 45.3 mm and
the maximum allowable deflection is span / 500
is the section stiff enough?
yes a no b

22. Question 2q so the section is not stiff enough what do we need to do?
upsize it by increasing its Section Modulus a
upsize it by increasing its Moment of Inertia b
increase its Modulus of Elasticity c

23. how much stiffer must the section be? By how much must we increase
Question 2r
so the section is not stiff enough.
A Moment of Inertia of 23.6 x 106 mm4 produced a deflection of 45.3 mm
The maximum allowable deflection is 14.0 mm
how much stiffer must the section be?
By how much must we increase
the Moment of Inertia ?
by a factor of 3.24 a
by a factor of 1.92 b
by a factor of 1.69 c

24. what value of the Moment of Inertia do we need?
Question 2s
so the section is not stiff enough.
A Moment of Inertia of 23.6 x 106 mm4 produced a deflection of 45.3 mm
So we need to increase the Moment of Inertia by a factor of 3.24
what value of the Moment of Inertia do we need?
330.4 x 106 mm4 a
45.3 x 106 mm4 b
76.4 x 106 mm4 c

25. Question 2t what section has a satisfactory Moment of Inertia
So we need a section that has a Moment of Inertia of 76.4 x 106 mm4
We need to go back to the Table of Properties to find a satisfactory section
click here to see the Table of Universal Beams
what section has a satisfactory Moment of Inertia
and, therefore is stiff enough?
250UB 37.3 a
310UB 46.2 b
310UB 40.4 c

26. a Great Start next question enough !
when we design anything what we are doing is determining its
material, shape and dimensions
next question
enough !

27. Partly correct let me try again let me out of here
that’s only part of it
let me try again
let me out of here

28. No! let me try again let me out of here
we will need to use the Section Modulus
but that’s not what we are aiming at
let me try again
let me out of here

29. You Bewdy ! next question enough !
yes, we always need to know what the
span, support type and loads are
next question
enough !

30. Partly correct let me try again let me out of here
that’s only part of it
let me try again
let me out of here

31. isn’t that what we are trying to find?
Oh my gosh!!
isn’t that what we are trying to find?
let me try again
let me out of here

32. No! let me try again let me out of here
we will need to calculate the maximum Bending Moment
but we can do that from the information we need to know
let me try again
let me out of here

33. No! let me try again let me out of here
we will need to find the minimum required I and Z
but we can calculate that from the information we will have
let me try again
let me out of here

34. you've got it it!! next question enough !
in order to calculate some of the things that we will have to
we need to be given the maximum allowable bending stress,
the Modulus of elasticity and the maximum allowable deflection.
This we will get from codes once we decide on a material
next question
enough !

35. Partly correct let me try again let me out of here
that’s only part of it
let me try again
let me out of here

36. Oh my gosh!! let me try again let me out of here
Why would we want to know whether the beam will be of
an elastic, plastic or brittle material at this stage?
What would that give us?
let me try again
let me out of here

37. you've got it it!! next question enough ! 3.6 m 3.6 m 7 m steel beams
timber joists
@ 600mm crs
tributary area
for beam = 7 x 3.6
= 25.2 m2
3.6 m
7 m
next question
enough !

38. The beams span 7m and they are 3.6m centres
Oh my gosh!!
How did you get that?
The beams span 7m and they are 3.6m centres
let me try again
let me out of here

39. Fantastic next question enough !
The load on the beam (neglecting the weight of the joists) is
the sum of the live load (1.5kPa) and the dead load (0.4kPa)
Tributary area = 25.2 m2
load per sq m = 1.9 kPa
TOTAL LOAD = 25.2 x 1.9 = 47.9 kN (kPa x m2 = kN)
next question
enough !

40. What is the total load per sq m? What is the tributary area?
Not right !!
How did you get that?
What is the total load per sq m?
What is the tributary area?
let me try again
let me out of here

41. Oh my Goodness!! let me try again let me out of here you should learn
We are talking about total load.
So what are the units for a load (force)?
let me try again
let me out of here

42. the load is distributed over the length of the beam
Yeaaahh!
that’s right
the load is distributed over the length of the beam
next question
enough !

43. it's enough to make one cry
is the load acting at just one point
or is it acting all along the length of the beam?
let me try again
let me out of here

44. we design for strength and check for stiffness
terriffic !!
we design for strength and check for stiffness
next question
enough !

45. Uhhhh ???
No, not really
let me try again
let me out of here

46. Fantastic next question enough !
once we have the Z value we can find the dimensions
since we are designing for strength we need to find the
minimum required Section Modulus since that is the property
of a beam’s section which determines its strength
i.e. a beam with the minimum required Z will be strong enough.
next question
enough !

47. we already have that given as the maximum allowable bending stress
Uhhhh ???
we already have that given as the maximum allowable bending stress
let me try again
let me out of here

48. The Moment of Inertia of a beam’s X-section determines its stiffness
it's enough to make one cry
The Moment of Inertia of a beam’s X-section determines its stiffness
NOT its strength
let me try again
let me out of here

49. Yipee !! next question enough !
the formula for bending stress is f = M / Z (stress = BM / Section Modulus)
so Z = M / f
since we know f (max allowable bending stress)
we need to first find the maximum BM
next question
enough !

50. we already have that given as the maximum allowable bending stress
Uhhhh ???
we already have that given as the maximum allowable bending stress
let me try again
let me out of here

51. what has the shear force to do with
No, No, No !!
what has the shear force to do with
the Section Modulus?
let me try again
let me out of here

52. Yipee !! next question enough !
since this is a simply supported beam with a UDL
Max BM = W L / 8 (where W = total load)
= x 7 / 8
= 41.9 kNm
next question
enough !

53. Not right!! let me try again let me out of here
How did you get that? Don’t guess!
what is the max BM formula for a simply supported beam with a UDL?
What is the total load? (see Q2b) What is the span?
let me try again
let me out of here

54. What is the max BM formula for a simply supported beam with a UDL?
No, No, No !!
What is the max BM formula for a
simply supported beam with a UDL?
let me try again
let me out of here

55. yes, yes, yes !! next question enough ! Z = M / f
= 41.9 / 200 (since M is in kNm bring to N and mm)
= 41.9 x 106 / 2 x 102
= x 103 mm3
next question
enough !

56. Sorry but, No !! let me try again let me out of here
How did you get that?
What did we say the formula for Z was?
What is the max BM?
What is the stress that we can allow the beam to take?
let me try again
let me out of here

57. Brilliant !! next question enough !
a universal beam 200UB 25.4 has a Zx value of 232 x 103 mm3
since this > the min required Zx of x 103 mm3 it is strong enough.
A smaller beam would not be strong enough and
a larger beam would be wasteful
next question
enough !

58. Sorry but, No !! let me try again let me out of here
The Zx value is certainly adequate and
the beam would be strong enough
But it is more than necessary and would be wasteful
let me try again
let me out of here

59. Sorry but, No !! let me try again let me out of here
The Zx value of a 180 UB 22.2 is only 171 x 103 mm3
such a beam would not be strong enough
since we need at least a Zx value of x 103 mm3
let me try again
let me out of here

60. Yes, the beam may be strong enough but maybe it deflects too much
Cheers !!
Yes, the beam may be strong enough
but maybe it deflects too much
next question
enough !

61. It is strong enough but what else could go wrong?
You devil !
It is strong enough but what else could go wrong?
let me try again
let me out of here

62. You devil ! let me try again let me out of here
but we’ve already designed it for strength
we know that it’s strong enough
let me try again
let me out of here

63. Cheers !! next question enough !
Exactly! We have to compare its actual deflection with
the maximum allowable deflection
next question
enough !

64. You devil ! let me try again let me out of here
what will that give us?
let me try again
let me out of here

65. Yes but .... let me try again let me out of here
We could do that. That will give us an indication
But we really need to be precise.
let me try again
let me out of here

66. this will tell us if the beam is stiff enough
Good on You !!
this will tell us if the beam is stiff enough
next question
enough !

67. A devil of an answer ! let me try again let me out of here
how do you compare a deflection with a span-to depth ratio?
One is a distance, the other is a ratio
let me try again
let me out of here

68. how do you know whether that’s enough or too much?
A devil of an answer !
how do you know whether that’s enough or too much?
let me try again
let me out of here

69. Good on You !! next question enough !
remembering the deflection formula d = k x WL3 / E I
we see that we need to also know E and I
and of course we need to know the max allowable deflection
next question
enough !

70. we already know the total load and the span
Don't cry !!
we already know the total load and the span
let me try again
let me out of here

71. It’s not the whole story
Yes but ....
It’s not the whole story
let me try again
let me out of here

72. from the Table we see that a 200UB 25.4
Good on You !!
from the Table we see that a 200UB 25.4
has a value of Ix of 23.6 x 106 mm4
next question
enough !

73. It’s a 200 UB 25.4 we are looking at
Not correct!
check again
It’s a 200 UB 25.4 we are looking at
not a 200 UB 29.8
let me try again
let me out of here

74. It’s a 200 UB 25.4 we are looking at
Not correct!
check again
It’s a 200 UB 25.4 we are looking at
not a 250 UB 31.4
let me try again
let me out of here

75. Good on You !! next question enough ! = 5 x W L3 / (384 x E I)
= 5 x x 73 / (384 x 2 x 105 x 23.6 x 106)
(the load is in kN and the span is in m, so bring everything to N & mm)
= 5 x x 103 x (7 x 103)3 / (384 x 2 x 105 x 23.6 x 106)
= 5 x x 343 x 1012 / (384 x 2 x 23.6 x 1011)
= x 10 /
= 45.3 mm
next question
enough !

76. don’t you think that that’s a rather large deflection
Not right !
don’t you think that that’s a rather large deflection
check your zeros
let me try again
let me out of here

77. you have all the necessary values
Don't guess !
work it out
you have all the necessary values
let me try again
let me out of here

78. so the section is not stiff enough
Cheers !!
the actual deflection is 45.3 mm
the max allowable deflection is span / 500 = 7000 / 500 = 14 mm
so the actual deflection is much greater than the max allowable deflection
so the section is not stiff enough
next question
enough !

79. You devil ! let me try again let me out of here
what is the deflection of this beam?
what is the maximum allowable deflection?
is the actual deflection less or greater than the max allowable deflection?
let me try again
let me out of here

80. Cheers !! next question enough ! that’s exactly it.
increasing the Moment of Inertia of a beam’s section
will increase its stiffness
next question
enough !

81. we are talking about stiffness – not strength
You devil !
we are talking about stiffness – not strength
let me try again
let me out of here

82. Yes but .... let me try again let me out of here
theoretically we could but where would we find a material
with such a high Modulus of Elasticity?
let me try again
let me out of here

83. Good on You !! next question enough ! the actual deflection is 45.3 mm
the deflection we want is mm
so we want the beam to be 45.3 / 14 times stiffer
so we must increase the Moment of Inertia by a factor of 3.24
next question
enough !

84. Don't guess !! let me try again let me out of here work it out
what is the actual deflection?
what is the maximum allowable deflection that we want?
so how much stiffer must the beam be?
let me try again
let me out of here

85. Good on You !! next question enough !
the Moment of Inertia of the 200UB25.4 is 23.6 x 106 mm4
we need to increase that by a factor of 3.24
this gives us a required I of 23.6 x 3.24 = 76.4 x 106 mm4
next question
enough !

86. Don't guess !! let me try again let me out of here work it out
what is the Moment of Inertia of the 200UB 25.4?
what is the factor by which we have to increase it?
let me try again
let me out of here

87. Well Done ! FINISH You’ve graduated with honours!
a 310UB 40.4 has an Ix value of 85.2 x 106 mm4
which is the lowest we can find that is
greater than the required min value of 76.4 x 106 mm4
FINISH

88. but we need an Ix of at least 76.4 x 106 mm4
Not Correct !
look again
the 250UB 37.3 has an Ix of 55.6 x 106 mm4
but we need an Ix of at least 76.4 x 106 mm4
let me try again
let me out of here

89. Not Correct ! let me try again let me out of here look again
the 310UB 46.2 has an Ix of 99.5 x 106 mm4
that is greater than 76.4 x 106 mm4
but somewhat wasteful as we could use a smaller beam.
let me try again
let me out of here