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1 Beams WORKSHEET 8 to answer just click on the button or image related to the answer

2 floor joists are at 600mm centres and span 2.0m between bearers, draw the configuration Question 1 12 m 2 a 1.2 m 2 b 1,200,000 mm 2 c what is the tributary area for one joist?

3 given a floor 18 m x 18 m with columns on a 6m x 6m grid, draw the configuration Question 2 36 m 2 a 18 m 2 b 324 m 2 c what is the tributary area for an internal column

4 Question 3 36 m 2 a 18 m 2 b 324 m 2 c given a floor 18 m x 18 m with columns on a 6m x 6m grid, draw the configuration what is the tributary area for an edge column

5 Question 4 36 m 2 a 18 m 2 b 9 m 2 c given a floor 18 m x 18 m with columns on a 6m x 6m grid, draw the configuration what is the tributary area for a corner column

6 Question 5 what is the first thing we need to do? determine the load per metre on a truss a determine the tributary area for a truss b determine the bending moment on the truss c a roof weighing 0.4 kPa spans between roof trusses which are at 2.5 m centres and span 10m. We want to determine the total load on a truss.

7 Question 6 what is the tributary area for a truss? 2.5 m 2 a 10 m 2 b 25 m 2 c a roof weighing 0.4 kPa spans between roof trusses which are at 2.5 m centres and span 10m. We want to determine the total load on a truss.

8 Question 7 what is the total load on a truss? (neglecting the self-weight) 10 kPa a 10 kN b 100 kN c a roof weighing 0.4 kPa spans between roof trusses which are at 2.5 m centres and span 10m. We want to determine the total load on a truss.

9 Question 8 is this a UDL or a point load? UDL a point load b a roof weighing 0.4 kPa spans between roof trusses which are at 2.5 m centres and span 10m. We want to determine the total load on a truss.

10 Question 9 what is the load per metre on a truss? (neglecting the self-weight) 1 kPa a 1 kNm b 1 kN/m c a roof weighing 0.4 kPa spans between roof trusses which are at 2.5 m centres and span 10m. We want to determine the total load on a truss.

11 Question 10 what are the two main types of stress involved in beam action? buckling and shear a tension and compression b bending and shear c bending and buckling d

12 Question 11 in buildings which is more important? shear a bending b

13 Question 12 why? spans are large a we design for bending and check for shear b loads are light c spans are large relative to loads d in buildings bending is more important than shear

14 Question 13 what is the sign convention for BMD for sagging? positive a negative b

15 Question 14 what is the sign convention for BMD for hogging? positive a negative b

16 Question 15 what does a Shear Force Diagram tell you? where the maximum shear force occurs a where the maximum shear stress occurs b the values of the shear force along the beam c a and c d b and c e

17 Question 16 what does a Bending Moment Diagram tell you? where the maximum bending moment occurs a where the maximum load occurs b the values of the bending moment along the beam c a and c d a, b and c e

18 Question 17a what’s the first thing we do? calculate the maximum bending moment a calculate the maximum shear force b calculate the reactions c 16 kN 2m 4m given the beam loaded as shown

19 Question 17b what are the reactions? R L = 10 kN, R R = 6 kN a R L = 16 kN, R R = 16 kN b R L = 8 kN, R R = 8 kN c 16 kN 2m 4m given the beam loaded as shown RLRLR

20 Question 17c does the beam sag or hog? sag a hog b 16 kN 2m 4m given the beam loaded as shown draw the deflected shape R L = 8 kNR R = 8 kN

21 Question 17d is the Bending Moment? negative a positive b 16 kN 2m 4m given the beam loaded as shown draw the deflected shape R L = 8 kNR R = 8 kN positive and negative c

22 Question 17e is the SFD? block shaped a triangular shaped b 16 kN 2m 4m given the beam loaded as shown draw the Shear Force Diagram (SFD) R L = 8 kNR R = 8 kN

23 Question 17f what is the maximum Shear Force? 16 kNm a 16 kN b 2m 4m given the beam loaded as shown draw the Shear Force Diagram (SFD) R L = 8 kNR R = 8 kN 8 kN c 8 kNm d

24 Question 17g where does the maximum Shear Force occur? at the centre of the beam a at the ends of the beam b 16 kN 2m 4m given the beam loaded as shown draw the Shear Force Diagram (SFD) R L = 8 kNR R = 8 kN all along the beam c

25 Question 17h is the BMD? trapezoidal / triangular a parabolic b 16 kN 2m 4m given the beam loaded as shown draw the Bending Moment Diagram (BMD) R L = 8 kNR R = 8 kN

26 Question 17i what is the maximum Bending Moment? 16 kNm a 32 kNm b 16 kN 2m 4m given the beam loaded as shown draw the Bending Moment Diagram (BMD) R L = 8 kNR R = 8 kN 8 kNm c 16 kN/m d

27 Question 17j where does the maximum Bending Moment occur? 16 kN 2m 4m given the beam loaded as shown draw the Bending Moment Diagram (BMD) R L = 8 kNR R = 8 kN at the centre of the beam a at the ends of the beam b all along the beam c

28 Question 18a what’s the first thing we do? calculate the maximum bending moment a calculate the maximum shear force b calculate the reactions c given the beam loaded as shown UDL 5kN/m 2m

29 Question 18b what is the vertical reaction? 5 kN a 10 kN b 20 kN c given the beam loaded as shown RVRV UDL 5kN/m 2m

30 Question 18c does the beam sag or hog? sag a hog b given the beam loaded as shown draw the deflected shape R V = 10 kN UDL 5kN/m 2m

31 Question 18d is the Bending Moment? negative a positive b given the beam loaded as shown draw the deflected shape positive and negative c R V = 10 kN UDL 5kN/m 2m

32 Question 18e is the SFD? block shaped a triangular shaped b given the beam loaded as shown draw the Shear Force Diagram (SFD) R V = 10 kN UDL 5kN/m 2m

33 Question 18f what is the maximum Shear Force? 5 kNm a 10 kN b given the beam loaded as shown draw the Shear Force Diagram (SFD) 10 kNm c 5 kN d R V = 10 kN UDL 5kN/m 2m

34 Question 18g where does the maximum Shear Force occur? at the centre of the beam a at the left end of the beam / at the support b given the beam loaded as shown draw the Shear Force Diagram (SFD) at the right end of the beam c R V = 10 kN UDL 5kN/m 2m same all along the beam d

35 Question 18h is the BMD? trapezoidal / triangular a parabolic b given the beam loaded as shown draw the Bending Moment Diagram (BMD) R V = 10 kN UDL 5kN/m 2m

36 Question 18i what is the maximum Bending Moment? - 20 kNm a - 10 kNm b given the beam loaded as shown draw the Bending Moment Diagram (BMD) - 5 kNm c - 2.5 kNm d R V = 10 kN UDL 5kN/m 2m

37 Question 18j where does the maximum Bending Moment occur? given the beam loaded as shown draw the Bending Moment Diagram (BMD) at the centre of the beam a at the right end of the beam b at the support c R V = 10 kN UDL 5kN/m 2m

38 next question enough ! tributary area = 2 m x 0.6 m = 1.2 m 2 tributary area 600mm 2m

39 let me try again let me out of here how did you get that? The length of a joist is 2 m and the joists are at 0.6 m centres

40 let me try again let me out of here the number is right. But these are stupid units for an area this size

41 next question enough ! 6 m x 6 m = 36 m 2 tributary area 6m

42 let me try again let me out of here How did you get that? The columns are on a 6 m x 6 m grid. We are talking about an internal column

43 next question enough ! 6 m x 3 m = 18 m 2 tributary area 6m

44 let me try again let me out of here How did you get that? The columns are on a 6 m x 6 m grid. We are talking about a column on the edge. What are the distances to the neighbouring columns?

45 next question enough ! 3 m x 3 m = 9 m 2 tributary area 6m

46 let me try again let me out of here How did you get that? The columns are on a 6 m x 6 m grid. We are talking about a column at a corner. What are the distances to the neighbouring columns?

47 next question enough ! That’s right. Before we can work out the load on a member, We have to work out what that member is carrying

48 let me try again let me out of here We are after the TOTAL LOAD We need to find out what the truss is carrying

49 next question enough ! 10 m x 2.5 m = 25 m 2

50 let me try again let me out of here The length of the truss is 10 m The trusses are at 2.5 m centres

51 next question enough ! The tributary area is 25 m 2 the load is 0.4kPa 25m 2 x 0.4 kPa = 10 kN (remember 1 kPa = 1 kN/m 2 )

52 let me try again let me out of here We are talking about the TOTAL LOAD What are the units of a load (force)?

53 let me try again let me out of here How did you get that? What’s the tributary area? What’s the load per sq m?

54 next question enough ! If we look at the truss we see that the load is distributed over its length

55 let me try again let me out of here How did you arrive at the conclusion that all the load that the truss carries is concentrated at one point?

56 next question enough ! you’ve got it !! 10 kN / 10 m = 1 kN /m

57 let me try again let me out of here we are talking about a force over a distance

58 let me try again let me out of here we are talking about a force over a distance

59 next question enough ! you’ve got it !! Beam action means that a beam may fail in either bending or in shear

60 let me try again let me out of here Where did buckling come into it? We’re talking about a beam

61 let me try again let me out of here Yes but that’s what happens internally but it’s not the main actions that we ascribe to beams

62 next question enough ! brilliant

63 let me try again let me out of here Don’t guess

64 next question enough ! It’s not that loads are necessarily light or that spans are so large. It is relative

65 let me try again let me out of here Spans are not necessarily so large

66 let me try again let me out of here Yes, but that’s the consequence not the reason

67 let me try again let me out of here Loads in buildings are not necessarily light

68 next question enough ! + Yes, remember the happy smile

69 let me try again let me out of here Is it happy or sad?

70 next question enough ! Yes, remember the sad face -

71 let me try again let me out of here Is it happy or sad?

72 next question enough ! You can see the values of the shear force at any point along the beam and you can see where the maximum shear force occurs +12.5 kN -7.5 kN -5 kN

73 let me try again let me out of here It’s not the whole story

74 let me try again let me out of here Shear Force Diagrams show forces. Stresses depend on the section of the beam

75 next question enough ! You can see the values of the bending moment at any point along the beam and you can see where the maximum bending moment occurs ~+5.6 kNm -10 kNm

76 let me try again let me out of here It’s not the whole story

77 let me try again let me out of here Bending Moment Diagrams don’t show loads

78 next question enough ! have to determine ALL the forces and that means determining the reactions 16 kN 2m 4m RLRLR

79 let me try again let me out of here We could but that’s not the first thing we should do We really should determine all the forces acting

80 let me try again let me out of here How can we do that? What do we need to know first?

81 next question enough ! The system is symmetrical and so the reactions are also symmetrical. Since ΣV = 0 and the total downward load is 16 kN The total upward load equals 16 kN 16 kN 2m 4m R L = 8 kNR R = 8 kN

82 let me try again let me out of here have a look at the system work smarter

83 let me try again let me out of here What’s the downward load? Think of the equations of static equilibrium

84 next question enough ! Yes, as you might expect 16 kN

85 let me try again let me out of here and hogs may fly

86 next question enough ! + Yes, remember the happy smile

87 let me try again let me out of here Is it happy or sad?

88 next question enough ! Yes, point loads produce block-shaped Shear Force Diagrams

89 let me try again let me out of here try again Remember how you draw the SFD by following the forces

90 next question enough ! you just ‘follow’ the forces +8 kN - 8 kN up 8, across, down 16, across, up 8.

91 let me try again let me out of here we are talking about a shear force what are the units of force?

92 let me try again let me out of here how did you get that? Look at your answer for Question 17 b

93 The signing of positive and negative shear forces is just conventional. There is no difference in the effect. So the shear force of 8 kN is constant all along the beam next question enough ! +8 kN - 8 kN

94 let me try again let me out of here THINK !! Look again at the Shear Force Diagram

95 next question let me out of here triangular just like a string loaded similarly 16 kN

96 let me try again let me out of here remember the string

97 that’s it exactly! next question enough ! +16 kNm WL/4 = 16 x 4 / 4 =

98 let me try again let me out of here Not right !! Look up the formula for a single point load at the centre

99 let me try again let me out of here THINK !! This is not load per metre We are talking about moments – force x distance

100 that’s it exactly! next question enough ! 16 kN

101 let me try again let me out of here THINK !! GO BACK TO THE BMD In a simply supported beam the ends cannot produce moment reactions

102 let me try again let me out of here THINK !! GO BACK TO THE BMD

103 next question enough ! have to determine ALL the forces and that means determining the reactions UDL 5kN/m 2m

104 let me try again let me out of here We could but that’s not the first thing we should do We really should determine all the forces acting

105 let me try again let me out of here How can we do that? What do we need to know first?

106 next question enough ! Since ΣV = 0 and the total downward load is 5 x 2 = 10 kN The total upward load equals 10 kN UDL 5kN/m 2m V = 10 kN

107 let me try again let me out of here What is the TOTAL downward load?

108 next question enough ! Yes, as you would expect ! 10 kN

109 let me try again let me out of here not correct

110 next question enough ! Yes, remember the sad smile -

111 let me try again let me out of here Is it happy or sad? It’s not both

112 next question enough ! Yes, UDLS* produce triangular-shaped Shear Force Diagrams *UDL = Uniformly Distributed Load

113 let me try again let me out of here try again Remember how you draw the SFD by following the forces. Think of a UDL as a series of little point forces

114 next question enough ! you just ‘follow’ the forces up 10, across, down (a little), across, down (a little), …. +10 kN

115 let me try again let me out of here we are talking about a shear force what are the units of force?

116 let me try again let me out of here how did you get that? Look at your answer for Question 18 b

117 As shown in the SFD, the maximum shear force of 10kN occurs at the support next question enough ! +10 kN

118 let me try again let me out of here THINK !! Look again at the Shear Force Diagram

119 next question let me out of here parabolic UDLs produce parabolic-shaped BMDs

120 let me try again let me out of here Think again

121 that’s it exactly! next question enough ! -10 kNm -wL 2 /2 = -5 x 2 x 2 / 2 =

122 let me try again let me out of here Not right !! Look up the formula for a UDL load on a cantilever

123 You’ve graduated with honours! FINISH -10 kNm -wL 2 /2 = -5 x 2 x 2 / 2 =

124 let me try again let me out of here THINK !! GO BACK TO THE BMD

125 let me try again let me out of here THINK !! GO BACK TO THE BMD How can a free end of a cantilever produce a moment?

126


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