Unformatted text preview: ©Prep101 Civ100 Exam Solutions Dec 10 2010 May
Q1 STRESS AND SHEAR
INSTRUCTOR NOTE
First we need the simplified FBD of the object.
There are horizontal and veritacl reactions at B, a vertical reaction at D
The equivalent loads are
F = wL
= 6(0.6)
= 3.6 kN
at 0.3 m to the left of B, and
F = 3(1.8)
= 5.4 kN
at 0.9 m to the right of C. Take the moments around B
ΣMB = 0
+3.6 (0.3) – 2(0.3)  5.4(0.9) + Dy(1.2) = 0
Dy = 3.65 kN
ΣFx = 0
Bx = 0
ΣFy = 0
3.6 + By – 2 – 4.8 + 3.65 = 0
By = 7.35 kN Apr 2010 Q1 ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q1 Now redraw (or draw these values on the existing distributed load diagram) Now we go through the shear. Remember: the force is the rate of change of the shear.
This means that at any point
Shear = Σ(Distributed Loads + Point Forces)
There is no load at A so it starts at zero. Then there is a 6 kN/m force pointing down for
0.6 m, so
vB = 6(0.6) = 3.6
(the distributed load is pointing down, so that’s a negative area)
At B the shear is 3.6, then it leaps up by adding 7.35 to (3.6+7.35) = 3.75
Then there is no load, so it remains constant to C.
There it leaps down by 2 kN to 1.75, and starts sloping down at 3 kN/m.
We can find where it crosses zero. The negative area added at any point is (3)(distance)
v=0
v = 1.75 – 3x0 = 0
x0 = 0.58
It then continues on to
1.75 – 3(1.2) = 1.85
At D it leaps up by 3.65 to (1.85 + 3.65) = 1.8
Finally, it drops at 3 kN/m for 0.6 m
1.8 – 3(0.6) = 0 ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q1 Excellent – we know there is no load at E, so the shear must be zero, and that is what
we have. Next we find the moment, knowing that the shear is the rate of change of moment.
At any point we have
Moment = Σ(Areashear  Point moments)
There is no external moment at A so we start at 0 kNm.
From A to B the shear is sloped, so the moment is curved.
At B, the total shear area is
MomentB = ½ (3.6) (0.6) = 1.08
(At B there is a point force, so it changes direction but not value)
Between B and C, the shear is constant so the moment will be a sloped line
The total area is
3.75(0.3) = 1.125
so we have
MC = 1.08 + 1.125 = 0.045
This means we’ve crossed the xaxis. We must mark where, and can find it from
M = 0 = 1.08 + 3.75x0
x0 = 0.288
Between C and D the shear is sloped, so the moment will be curved.
The area between C and the point where the shear becomes zero is
Area = ½ (1.75)(0.58) = 0.507
So the moment at that point is ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q1 M = 0.045 + 0.507
= 0.553
The area between this point at D is
A = ½ (1.85)(1.20.58)
= 0.573
So
MD = 0.553 – 0.573
=  0.02
This means we crossed zero moment again, and must find where.
M=0
0 = 0.553 – ½ (1.85)x = 0
x = 0.60
Then the shear between D and E is positive and sloping, so the moment will curve back
up to 0 at E ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q1 BEAM DESIGN
Failure stress σF = 30 x 106 Pa
w = 2h
multiples of 20 mm only
Load factor = 2
So maximum allowed stress is
σA = 30 x 106/2 = 15 x 106
This is a shear/moment question, so compare to bending moment
σB = Mxy / I
The maximum moment from the problem is
Mx = 0.553 kNm
the value y is half the height
y = h/2
The moment of inertia of a rectangular member is
Ix = wh3/12
= 2hh3/12
= h4/6 .
σB = = . Comparing the bending and allowed moment, we have
15 x 106 = 1.659 x 103 / h3
h = 0.048 m
h must be at least 48 mm. It is available in integers of n x 20, so we must have 60 mm.
The width is twice this, so we have a beam of 120 x 60 mm
Is the beam used efficiently? ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q1 There aren’t any unloaded sections, nothing is wasted, there isn’t a huge load or
moment in only one region compared to tiny values everywhere else, so yes. It is used
efficiently. ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q2 Q2 Machine
EB is NOT a twoforce member – it is subjected to a moment M
The key is that the connection B is a SLOT, not a PIN. This means it can only exert
force at right angles to the direction of EB (Think of normal forces)
Draw the free body diagram of the entire assembly, finding any required angles.
tan1(0.15/0.36) = 22.6° ΣME = 0
M – 80(0.3) + Dy(0.45) – Dx(0.36) = 0
M = 24 – 0.45 Dy + 0.36 Dx
ΣFx= 0
Ex + Dx = 0
Dx =  Ex
ΣFy = 0
Ey – 40 – 80 + Dy = 0
Dy = 120  Ey ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q2 For the rod AD ΣMB = 0
40(0.15) 80(0.15) + Dy(0.3) = 0
Dy = 20 N
ΣFx = 0
Bx + Dx = 0
Dx = Bcos22.61
= 0.92 B
ΣFy = 0
40 + By 80 + Dy = 0
Bsin22.61 = 120 – 20
0.38 B = 100
B = 263.16 N
So
Dx = 0.92B
= 242.08
Now we can immediately use
M = 24 – 0.45 Dy + 0.36 Dx
= 24 – 0.45 (20) +0.36(242.08)
= 102.14 N m
Or if you hadn’t spotted that, you’d continue to the rod EB (The same method still
provides many routes to the answer, and they ALL work!) ©Prep101 Civ100 Exam Solutions Dec 10 ΣME = 0
M – BLEB = 0
M = BLEB
LEB = √[(0.36)2 + (0.15)2]
= 0.39 m
M = 263.16 (0.39)
= 102.63
or you could use the component method to find
ΣME = 0
M – Bx0.36 – By0.15 = 0
M – 0.36Bcos22.61 – 0.15Bsin22.61 = 0
M = 102.63
(the slight difference in M values is from rounding errors) Apr 2010 Q2 ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q3 Q3 TRUSS
First draw the diagram and work out any angles or distances. EFD is a zero force
member and can be ignored.
AB is a two force member and has a single force along its length.
We can work out the angles and distances
tan1 (4/5.2) = 37.6°
tan1 (4/6.4) = 32.0°
tan1 (2.4/4) = 73.3°
12tan73.3 = 3.6 m So we have (assuming AB is in compression)
Ax = ABx = ABcos55
AyABy = ABsin55
Now we look at the entire structure to find the reactions. Take pivot around H
ΣMH = 12(8) – 12(12) + Ay10.4 = 0
Ay = 23.08 kN
AB =  Ay / sin55
=  28.17 kN ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q3 Yes, it is in compression.
So
Ax = 28.17cos55 = 16.16
Ay = 28.17sin55 = 23.08
Now we can find the other support reactions
ΣFy = Hy + Ay = 0
Hy =  23.08 N
Hy is downwards.
ΣFx = Hx + 12 +12 – Ax = 0
Hx = 24 + 28.17cos55
Hx = 7.82 kN
Then we must proceed to make cuts to find CG and CF. We can use the method of
sections moving up from the base, or the method of joints startang at D, then F, then a
section between FC and GB. We will move up from the base. Taking the moments around G to eliminate the most unknowns
ΣMG = 0
7.82 (4) + 23.08(1.2) + BC(6.4) +23.08(9.2) 16.16(4) = 0
BC = 22.52
So BC is actualli in compression, 22.52
ΣFx = 0
7.82 +GFcos73.3 + CGcos32 – 16.16 = 0
0.29GF + 0.85CG = 23.98
GF = 82.69 – 2.93CG
ΣFy = 0 ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q3 23.08 + GFsin73.3 + CGsin32 +(22.52) + 23.08 = 0
0.96GF+ 0.53 CG = 22.52
0.96(82.69 – 2.93CG) + 0.53CG = 22.52
79.38 – 2.81 CG + 0.53CG = 22.52
2.28CG = 56.86
CG = 24.94 kN
and we can find
GF = 82.69 – 2.93(24.94)
= 9.61
Take another section between GF and DC Take moments around B to make things easier (get rid of the unwanted unknown CD,
but leave the desired value CF)
ΣMB = 0
7.82 (4) + 23.08(7.6) – 9.61(6.4)sin73.3 + CF(4) – 12(4) +23.08(2.8) – 16.16(4) = 0
37.20 + 4CF = 0
CF = 9.3 kN
CF is under tension (pulling) ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q3 BEAM DESIGN
The load in member CG = 24.94 kN
Square bar, so b = h
Increments of 2 mm
E = 200,000 MPa = 2 x1011 Pa
σy = 300 MPa
Load Factor = 3
So max allowable stress is
σa = σy/LF
= 300/3 = 100 MPa
The stress in CG is given by
σCG = CG/A
The area is
A = bh
= b2
So we have
σCG = σA
CG/A = 100 MPa
24.94 x 103 / b2 = 100 x 107
b = 4.99 x 103
The minimum breadth b is 4.99 mm. The increments are 2 mm, so the bar will measure
6 x 6 mm
The elongation of CG is given by
Δ=
We can find the length L from pythagoras
L = √(42 + 6.42)
= 7.55 m
The area is
A = b2
= (6 x 103)2
= 3.6 x 105
So
Δ= . .
. = 2.62 x102 m
CG will elongate by 2.62 cm ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q4 Q4 FLUID
The problem is 3 m wide, with one strut every 0.6 m. We will work out the values for a
0.6 m wide section this will give us the compression in one strut.
(You could also work out any problem for a 1 m wide section, then multiply it by the
desired width – here 0.6 m – at the end)
We find the equivalent force for the distributed load – the pressure at the surface is 0
Pa. The bottom is at a depth of 1.2sin60 = 1.04 m, so the distributed load is w = P(width)
= ρgh(width)
= 1000(9.81)1.04(0.6)
= 6.12 kN/m2
The equivalent force is thus
F= ½ WL
= ½ 6.12 (1.2)
= 3.67 kN
applie 1/3 of the way from the maximum load end, 1.2/3 = 0.4 m from A
The reactions at A are in the x and y direction, Ax and Ay
The force due to the compression of BC is in the direction of BC, pushing the barrier.
The free body diagram is thus ©Prep101 Civ100 Exam Solutions Dec 10 We take the moment around A to eliminate the reactions there.
ΣMA = 0
0 = 0.4(3.67) + 0.6BCsin60
1.47 = 0.52BC
BC = 2.82 kN
Each strut endures a compression of 2.82 kN Apr 2010 Q4 ©Prep101 Civ100 Exam Solutions Dec 10 Apr 2010 Q5 Q5 3D EQUILIBRIUM
First, work out the unit vectors
uBD = ((xD – xB)i +(yD– yB)j +(zD – zB)k)
((xD – xB)2 +(yD – yB) 2 +(zD – zB) 2) ½
= ((0 – 0)i +(0 – 9)j +(6 – 0)k)
((0)2 +(9) 2 +(6) 2) ½
= 0.85 j + 0.55 k
The tension in the cable is 24 kN, so we have
TBD = TuBD
= 24(0.85 j + 0.55 k) = 20.4 i + 13.2 j
Likewise, we find
uBC= 0.71 i – 0.71 j
and the tension in BC is
TBC = 0.71TBCi 0.71TBCj
The next part of setup is finding the relative position vectors, r. Since the problem has a
ball and socket joint at A we’ll take our pivot there.
rAB = 3i + 9j – 3k
The rod is uniform so the weight w will act through the halfway point
rAW = 1.5 i + 4.5 j – 1.5 k
The weight is unknown, but will act downwards, so the vector is
wj
Next, we use the equation – and like any 3D equilibrium problem, Σ=0 and we start with
the moment
ΣMA = Σ r x F = 0
0 = rA x FA + rAW x FW + rABx TBD + rAB x TBC ©Prep101 Civ100 Exam Solutions Dec 10 0 =0+ 1.5
0 4.5 1.5 + 3
0
0 9
3 +
20.4 13.2 3
0.71 Apr 2010 Q5 9
0.71 3
0 Taking each component one at a time
i component
0 = (4.5(0) – (w)(1.5)) + (9(13.2) – (20.4)(3)) + (9(0) – (0.71TBC)(3))
0 = 1.5 w + 57.6 – 2.13 TBC
That’s an equation with two unknowns, so we need more equations. But first we rewrite
it in terms of one of the unknowns.
w = 38.4 – 1.42 TBC
j component
0 = (1.5(0) – 0(1.5)) + (3(13.2) – 0(3)) + (3(0) – (0.71TBC)(3))
0 = 39.6 – 2.13 TBC
TBC = 18.6 kN
and then
w = 38.4 – 1.42(18.6)
= 12.0 kN ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2009 Q1 2009 Fall
Q1 3D EQUILIBRIUM
First we find the unit vector of the cable
uCF = ((xF – xC)i +(yF– yC)j +(zF – zC)k)
√[(xF – xC)2 +(yF – yC) 2 +(zF – zC) 2]
= ((0 – 6)i +(12– 0)j +(8 – 20)k)
√[(6)2 +(12) 2 +(10) 2]
(bottom line = length = 18)
= 0.33i + 0.67j – 0.67k
and for the tension we have
TCF = TuCF
= 0.33 T i + 0.67 T j – 0.67T k
Next we need the center of gravity. The plate is symmetric so it will be at z = 10, and in
the xz plane so it will be at y = 0. We find the x coordinate by
xcg = ΣAx / ΣA
Since the plate is symmetric we need only look at one half, a rectangle and a triangle
xcg = [10(3)1.5 + ½ 10 (3) (3 + (1/3)3)] / [10(3) + ½ 10(3)]
= 105/45
= 2.33 m
We want to know about the cable and B, so we take the moment around A to get rid of
the unknowns there.
so we have rAW = 2.33i + 0j + 10k
and W = 1500 j
The reactions at B are Bxi + Byj, as stated
ΣMA = Σ r x F
0 = rAB x FB + rAC x TCF + rAW x W ©Prep101 0= Civ100 Exam Solutions Dec 10 0 0 20 +
0 6
0.33 0
0.67 20 + 2.33
0.67
0 Dec 2009 Q1 0
10
1500 0 0 = i(0 – By20) –j (0Bx20) + k(0 – 0)
+ i(0 – 0.67T(20)) – j (6(0.67T) – (0.33T)20) + k(6(0.67T) – (0.33T)20)
+ i(0 – (1500)10)  j (0 – 0) + k(2.3(1500) – 0)
0 = (  20By – 13.4T + 15000)i
+ (20Bx + 4.0 T – 6.6 T) j
+ (4.0 T + 6.6T – 3450) k
Looking at each component on their own, from the kcomponent
2.6 T = 3450
T = 1326.9 kN
and then from the i component
20By – 13.4(1326.9) + 15000 = 0
By = 139.0 kN
j component
20Bx  2.6(1326.9) = 0
Bx = 172.5 kN
We have found the tension and reaction components at B. ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2009 Q2 Q2 Machine
There are no two force members here.
Draw the free body diagram without the pulley and ropes, working out any angles.
tan1(5/12) = 22.62° Looking at the entire assembly, take the moment around F
ΣMF = 0
TG(13) – 800 (2) + 800 (5) + 700 + Cy(2) – Cx (1.75) = 0
13TG = 3100 + 2Cy – 1.75 Cx
ΣFx = 0
800 – 800 + Fx + Cx= 0
Cx = Fx
ΣFy = 0
TG – 800 + Fy + Cy = 0
TG = 800 – Fy – Cy ©Prep101 Civ100 Exam Solutions Dec 10 Move on to beam ABC Take moments around C
ΣMC = 0
T(15) – By(12) – Bx(5) + 700 = 0
15T = 700 – 5Bx – 12By
ΣFx = 0
Bx + Cx = 0
Cx =  Bx
ΣFy = 0
T + By + Cy= 0
Cy =  T  By
Need to keep going through the system, look at BD Dec 2009 Q2 ©Prep101 Civ100 Exam Solutions Dec 10 Take moments around D
ΣMD = 0
Bx(5.75) + 2(800) = 0
Bx = 278.3
ΣFx = 0
Bx + Dx + 800 = 0
Dx =  521.7
ΣFy = 0
By + Dy = 0
B y = Dy
One last piece, look at DF Dec 2009 Q2 ©Prep101 Civ100 Exam Solutions Dec 10 ΣMF = 0
+Dy(10) + 800(5) = 0
Dy = 400 N
Now we can start feeding back through the equations!
B y = Dy
= 400 N
for the tension
15T = 700 – 5Bx – 12By
15T = 700 – 5(278.3) – 12(400)
T = 273.9N
And we have
Cx =  Bx
= 278.3 N
Cy = T  By
= 273.9 – (400)
= 126.1 N
We now have all the components. Dec 2009 Q2 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2009 Q3 Q3 TRUSS
The problem is not symmetric.
EK is a zero force member, so the force in EF is the same as the force in DE
Identify the reactions, and work out relevant angles and distances by trigonometry
J is halfway between H and K on a straight line, so it is half the vertical distance – 1.5/2
= 0.75 m below H.
tan1 4.5/4 = 48.37°
tan1 0.75/4 = 10.61° Find reactions over the whole structure. Take moment around A
ΣMA = 36(4) 120(8) – 120(16) + By(24) = 0
By = 126 kN
ΣFx = 0
Ax = 0
ΣFy = 0
Ay – 36 – 120 – 120 + 126 = 0
Ay = 150 kN
Now we want DK, JK and EF, which is the same as DK, JK and DE, which we can find
with one section. ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2009 Q3 Choose the pivot point to remove the maximum number of unknowns, but still leave
something we want – we can take it around K! (Taking it around D or another point
would of course still work, but this is faster)
ΣMK = 0
DE (4.5) + 120 (4) + 36 (8) 150(12) = 0
DE = 229.3 kN
De is actually in compression, pushing against D.
ΣFx = 0
DE + DFcos48.39 + JKcos10.61 = 0
229.3 + 0.66DF + 0.98 JK = 0
DF = 347.42 – 1.48 JK
ΣFy = 0
DFsin48.37 – JK sin10.61 = 0
(347.42 – 1.47JK)0.75 – 0.18JK = 0
260.57 + 1.10JK – 0.18 JK = 0
0.92JK = 260.57
JK = 283.2 kN
JK is in tension
DF = 347.42 – 1.48(283.2)
= 71.7 kN
DF is in compression ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2009 Q3 BEAM DESIGN
KL = 175 kN = 1.75 x 105 N
σy = 300 MPa = 3 x 108 Pa
It’s a truss question, so we must compare the stress in the beam to the maximum
allowed stress
σKL vs σa
For the minimum allowed value of anything, that means writing
σKL = σa
There is no load factor so σa = σy
The stress in the beam is given by
σKL= KL/A
The cross sectional area of the beam is given by
A = Aouter Ainner
= b2 – (0.2)2
= b2 – 4 x 102
So we have
σKL = σy
KL/A = 3 x 108
175 x 103/(b2 – 4 x 102) = 3 x 108
b2 = (175 x 103 / 3 x 108) + 0.04
b = 2.01 x 102 m
b = 201 mm.
c)
E = 200,000 MPa = 2 x 1011 Pa
Original length of KL is given by pythagoras
L = √(42 + 0.752)
= 4.07 m
The area is
A = (2.01 x 102)2 – (0.2)2
= 4 x 104 m
And the elongation is
Δ= ©Prep101 Civ100 Exam Solutions Dec 10 = . .
= 8.9 x 103 m The elongation is 8.9 mm Dec 2009 Q3 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2009 Q4 Q4 STRESS AND SHEAR
First we find the FBD for the whole object.
A is fixed, so there will be Ax, Ay and MA
An internal pin at B means Bx and By, and we must use machines.
A roller at D means Dy only.
The distributed load is
F = 30(6) = 180
Applied 6/3 = 3 m from A.
Draw FBD for the problem. Since it’s a machine, we must dismantle it and draw an
FBD for each. ΣMA = 0
MA – 180(3) + By(6) = 0
Two unknows, so rewrite in terms of one and move on
MA = 540 – 6 By
ΣFx = 0
Ax + Bx = 0
Ax = Bx
ΣFy = 0
Ay 180 + By =0
Ay = 180  By ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2009 Q4 On to the next component. Remember machines – the B components must point in the
opposite direction now! ΣMB = 0
80(3) + Dy(6) – 80(9) = 0
Dy = 160 kN
ΣFx = 0
Bx = 0
Bx = 0
ΣFy = 0
By – 80 + 160 – 80 = 0
By = 0
and we have
Ax =  Bx
Ax= 0
Ay = 180 – By
Ay = 180
M = 540 – 6(By)
M = 540
Now we can draw our total FBD and start working out shear ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2009 Q4 At A there is a jump up of 180 N
Between A and B the load is constant, so the shear is constantly decreasing
vB = 180 – 30(6)
=0
This is perfect – we know the shear here must be zero because we worked out the
vertical force at this hinge, By = 0. NOTE: the components at a hinge are not always
zero – they just happen to be in this case
Between B and C there is no load, so no change.
At C a load jumps down 80 kN to 80
Then no change until D where it jumps up 160 (going through zero, an important point)
Then no change until E when it jumps down 80 to 0. This is a good final check, as we
know we must finish at zero. ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2009 Q4 Now we work on the moments
At A there is an external moment of 540 in the plus direction, so the bar must bend back
at  540.
Remember
Moment = ΣAreashear  Point Moments
From A to B the shear is positive and sloping down, so the moment will be increasing
but curving to increasing less.
At B
MomentB =  540 + ½ (180)6
=0
At B the shear is zero so we will have a maximum moment.
This also makes sense – we know there is no reaction moment at a hinge, and here we
have zero moment.
Between B and C the shear is zero, so there is no change in moment.
Between C and D, there is a constant negative shear so the moment will be a straight
downward line. At D
MomentD = 0 – 80(3)
=  240
Between D and E there is a constant positive shear, so the moment will slope up. E is
an end with no external moment so it must return to zero, which we can check.
MomentE = 240 + 80(3)
=0 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2009 Q4 BEAM DESIGN
σmax = 300 MPa
It’s a shear/moment question, so compare the max allowed stress to the bending stress
σB = My/I
The maximum moment magnitude from the previous part is 540 kNm
y is the distance from the axis to the outermost part, y = (0.16+0.4)/2 = 0.1 m
We will have to calculate the moment of inertia ourselves. Time saving tip: The beam is symmetric about the axis of rotation, so we can calculate
I1/2 for half of it, sections 1 and 2, and double that to get the total moment.
I1/2 x = I1 x + I2 x
By the parallel axis theorem
I1 x = I1 + Ay2
= bh3/12 + bh(h/2)2
= bh3/3
= 0.02 (0.08)3/3
= 3.41 x106 kgm2
I2 x = I2 + Ay2
= bh3/12 + bh(0.08 + 0.01)2
= 0.2(0.02)3/12 + 0.2(0.02)(0.09)2
= 3.25 x 105 kgm2 ©Prep101 Civ100 Exam Solutions Dec 10
I1/2 x = 3.51 x 106 + 3.25 x 105
= 3.6 x 105 So
I = 7.2 x105
The bending stress is then
σB = Mxy/I
= 540 (0.1) / 7.2 x105
= 7.5 x105 Pa
= 0.75 MPa
300 >> 0.75
Yes, the beam is more than adequate. Dec 2009 Q4 ©Prep101 Civ100 Exam Solutions Dec 10 Q5 FLUID
The gate is 6m wide.
There is a horizontal reaction at D, Dx
There are horizontal and vertical reactions at the pin A, Ax and Ay
We will call the surface of the water on the section BC point E
We must find the equivalent forces due to the water
The length of BA, found from pythagoras √[32 + 42] = 5 m
The angle of BA is given by tan θ = 4/3, θ = 53.1° for EB, the pressure and load at the surface is 0 kN
The pressure at B is given by
PB =ρgh
= 1000(9.81)5
=49 kN
The width is 6m, so the load is
wB = P6
= 294 kN/m
and the equivalent force is thus
F = ½ wBL
= ½ 294 (5)
= 735 kN
applied 1/3 of the way from B, at 5/3 = 1.66 m above B
For section BA there are two equivalent forces
The load at A is given by
wA = PA(width)
= ρgh6
= 1000 (9.81)9(6)
= 529.74 kN
The first equivalent force is
F1 = wBL
= 294(5)
= 1470 kN
applied at the halfway point, 2.5 m from A
The second is
F2 = ½ (wA wB)L
= ½ (529.74 – 294)5 Dec 2009 Q5 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2009 Q5 = 589.35 kN
applied 1/3 of the way from A, 5/3 = 1.66 m from A.
The free body diagram is thus We take the moments around A to eliminate those reactions
ΣMA = 0
0 = + 589.35(1.66) + 1470(2.5) + 735 (4+1.66) – Dx(4+7)
11Dx = 8813.42
Dx = 801.22 kN
The value is positive, so Dx is to the right, as expected.
Now we apply ΣF = 0 to find the reactions at A. BA is at 53.1° to the horizontal, so the
forces acting at right angles to it are acting at 9053.1 = 36.9° to the horizontal
ΣFx = 0
0 = 801.22 – 735 – 1470cos36.9 – 589.35cos36.9 + Ax ©Prep101 Civ100 Exam Solutions Dec 10
Ax = 1580.61 kN
ΣFy = 0
0 =  1470sin36.9 – 589.35sin36.9 + Ay
Ay = 1236.47 kN Dec 2009 Q5 ©Prep101 Civ100 Exam Solutions Dec 10 2009 Winter
Q1 STRESS AND SHEAR
A pin at A means Ax and Ay
A roller at B means By
There are no other horizontal components, so we can say ΣFx = Ax = 0 kN
First we need the simplified FBD to find the reactions.
The distributed load is
F = 8(3) = 24 kN
applied at the midpoint of the region, (2 + 1.5) = 3.5 m from A.
Draw our FBD, labelling all interesting points. ΣMA = 0
24(3.5) 15(7) + By9 = 0
By = 21 kN
ΣFy = 0
Ay – 24 – 15 + 21 = 0
Ay = 18 kN
Draw our complete loading diagram May 2009 Q1 ©Prep101 Civ100 Exam Solutions Dec 10 At A the shear jumps up to 18
Between A and C there is no load, so no change.
From C to D the shear decreases at 8 kN per meter to
18 – 8(3) = 6
This means it crosses zero, and we must find where
v = 0 = 18 – 8x
x = 2,25
Mark this point on the diagram
Between D and E there is no load, so the shear remains constant.
At E it jumps down 15 to 615 = 21
Then no change until E, when it jumps up by 21 to 0. Excellent. Now we go through our moment diagram
Remember
Moment = ΣAreashear – Point moments
There is no external moment at A so we start at 0 Nm
Constant shear between A and C, so the moment rises linearly to
MC = 18(2) = 36 May 2009 Q1 ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q1 The moment rises to a maximum where the shear is zero. The shear is positive
decreasing to zero, so the moment is curving increasing to a maximum
Momentmax = 36 + ½ 18(2.25)
= 56.25
From there to D it continues to curve, now bending down because the shear is negative,
to
MD = 56.25 – ½ (6)(32.25)
= 54
Between D and E the shear is constant negative, so M slopes down linearly to
MD = 54 – 6(2)
= 42
Between E and B the shear is constant negative (21), so M slopes down linearly to
ME = 42 – 21(2)
=0
Perfect – E is an end with no external moment, so the moment is zero. ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q1 BEAM DESIGN
From the first part, Mmax = 56.25 kNm
σy = 30 MPa = 3 x 107 Pa
Load factor = 2
So max allowed stress σA = σy/2 = 1.5 x 107 Pa
w = 3h
Area = wh = 3h2
Find bending stress so we can compare it to max allowed stress
σB = My/I
The distance y is half the height
y = h/2
Moment of inertia of a rectangular beam
I = wh3/12
= 3hh3/12
= h4/4
So the bending stress is .
σB = = . Compare this to max allowed stress
σB = σA
1.13 x 105/h3 = 1.5 x 107
h = 1.96 x101 m
h is 196 mm and w = 3h = 588 mm
The increments are 5 mm, so the dimensions are 200 x 590 mm
NOTE: don’t round off to increments before finding the width, or you’ll get the wrong
answer of 200 x 600 mm ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q2 Q2 MACHINE
Draw the free body diagram without the pulley and cable.
(tan1 (2/6) = 18.43°) Take the moment around A
ΣMA =  700 (2) + 700 (1.5) – 700 (4) + Ey(7) = 0
Ey = 450 N ΣFx = 0
Ax + 700 – 700 + Ex = 0
Ax = Ex ΣFy = 0
Ay – 700 + 450 = 0
Ay = 250 N
We need another equation to find the xcomponents. Draw the FBD of ABC ©Prep101 Civ100 Exam Solutions Dec 10 Take the moment around C
ΣMC = 0
+Ax (3) – 250(1) + 700(1) = 0
Ax = 150 N
and we have
Ex = Ax
= 150 N May 2009 Q2 ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q3 Q3 TRUSS
The structure is symmetric
Note that BK and HF are zero force members (crooked T junctions), so AK = KJ = AJ as
AJ is all one beam. Likewise IH = HG = IG
Find the reactions and work out any angles or distances needed.
tan1 (16/14) = 41.19°
10tan41.19 = 8.75
tan1 ((148.75)/6) = 41.19°
tan1(4/6) = 33.69°
tan1(3/10) = 26.57° ΣFx = 0
Gx= 0
KEY – this means the loads and so the entire problem is symmetric. Ay = Gy
It also means if we’re looking for CD, JK, EI, FI, we can just find CD, JK, CJ, BJ. We
can find everything on one side of the structure, taking a section between BK and CG,
then a joint at C. ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q3 ΣFy = 0
Ay 50 – 150 – 200 – 150 – 50 + Ay = 0
Ay = Gy = 300 kN
This means that the load member EI = CJ, IF = BJ, and we can deal with everything on
one side of the structure
Take a method of sections cut across BC, BJ and KJ Take the moment around B
ΣMB = 0
300 (4) + JKx(5.25) + JKy(6) = 0
1200 + 5.25 JK cos41.19 + 6 JK sin41.19 = 0
7.90 JK = 1200
JK = 151.90 kN
JK is in tension
ΣFx = 0
BCx + BJ + JKx = 0
BCcos33.69 + BJ +151.9cos41.19 = 0
BJ = 0.83BC – 114.31 ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q3 ΣFy = 0
300 50 + BCy + JKy = 0
250 + BCsin33.69 + 151.90sin41.19 = 0
0.55BC = 350.03
BC = 636.43 kN
BC is under compression
BJ = 0.83(636.43) – 114.31
= 413.92 kN
BJ is under tension.
Now we can move on to the joint at C. We now know BC is under compression, so we
draw it that way. ΣFx = 0
+636.43 cos 33.69 + CD cos 26.57 + CJ sin 33.69 = 0
529.54 + 0.89 CD + 0.83 CJ = 0
CD =  594.99 – 0.93 CJ
ΣFy = 0
150 +636.43 sin33.69 + CD sin26.57 – CJ sin 33.69 = 0
203.03 + 0.45 (594.99 – 0.93 CJ) – 0.55 CJ = 0
54.72 – 0.97CJ = 0
CJ = 66.72 kN
CJ is under compression
CD = 594.99 – 0.93(87.92)
= 513.22 ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q3 CD is under compression.
(This all matches what we’d expect – the upper members are under compression, the
lower members under tension, and the things in between could be either..) BEAM DESIGN
INSTRUCTOR NOTE
The load in EI = CJ = 66.72 kN = 6.67 x103 N
Apply
Δ = CJ x LCJ / (E x A) ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q4 Q4 FLUID
First, work out any numbers, angles or equivalent forces.
The gate AB is 0.5 x 0.8 m
In the diagram the length of AB, from Pythagoras, is √(0.642 + 0.482) = 0.8 m
So the problem is 0.5 wide.
The angle of BA is given by tan θ = 0.48/0.64 so θ = 36.87°
The angle of BC is given by tanθ = (0.48 + 0.45 + 0.27)/0.64 and θ = 61.93°
The angle between BA and BC is thus 61.93 – 36.87 = 25.06
The tension T acts in the direction of BC
There are reactions in the x and y direction at A, Ax and Ay
When T is just sufficient to start opening the gate, the total moment around A is zero
and there is no normal force from the stop at B – the gate is no longer leaning on it, but
is being held up by the cable.
Equivalent forces:
The distributed load at A is given by
wA = PA(width)
= ρgh(0.5)
= 1000 (9.81)0.45(0.5)
= 2.21 kN/m
The distributed load at B is given by
wB = PB(width)
= ρgh(0.5)
= 1000(9.81)(0.45 + 0.48)(0.5)
= 4.56 kN/m
There are two equivalent forces over AB. The first is
F1 = wAL ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q4 = 2.21(0.8)
= 1.77 kN
applied at the midway point, 0.4 m from A
The second is
F2 = ½ (wB – wA)L
= ½ (4.56 – 2.21)(0.8)
= 0.94 kN
applied 1/3 of the way from B, 0.8/3 = 0.27 m from B, or 0.53 m from A
The free body diagram for gate AB is thus Taking the moments around A to eliminate the reaction components, we have
ΣMA = 0
0 = T(0.8)sin25.06 + 0.94(0.53) + 1.77(0.4)
0.34T = 1.21
T = 3.56 kN ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q5 Q5 3D EQUILIBRUM
This is a 3D equilibrium problem, but with the trick of two ballandsocket junctions –
where do we take the moments around? Remember why we take the moment around
ballandsocket joints – to make things simple for ourselves, to eliminate as many
variables as possible. We must do that, and there is a great way to make things simpler
– take the moment around the AE axis!
We must keep to SI units, so use m
First, find the unit vector for DF
uDF = (0 – 0.16)i + (0.110)j + (0.16 – 0.24)k
√((0.16)2 + (0.11)2+ (0.08)2)
(length DF = 0.21 m)
=  0.76 i + 0.52 j – 0.38 k
So the tension in DF
TDF = TuDF
The axis uAE
uAE = (0 – 0.07)i + (0 – 0)j + (0.24 – 0)k
√[(0.07)2 + (0.24)2]
(length AE = 0.25 mm)
= 0.28i + 0.96 k
A is a point on the axis AE, so we take distances relative to there (you could also use E)
rAC = 0.09i + 0.10 k
Fc = 600 j
rAD = 0.09i + 0.24 k
And we have
ΣMAE = uAE .Σ(r x F)
= (0.28 i + 0.96 k) . [ rAC x FC + rAD x TDF]
We can disregard the reactions at A and E because we know they are on the axis
= (0.28 i + 0.96 k) . [ 0.09
0 0
0.1 +
600 0 0.09
0.76 0
0.52 0.24
0.38 ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q5 = (0.28 i + 0.96k) . [ i (0 – (600)(0.1) – j (0 – 0) + k (0.09(600) – 0)
+ i (0 – (0.52T)(0.24)) –j (0.09(0.38T) – (0.76T)(0.24)) + k (0.09 (0.52T) – 0)]
0 = (0.28 i + 0.96 k) . [(60  0.12T)i
0.15 Tj
+ (54 + 0.05N)k] 0 = 0.28(60 – 0.12T) + 0.96(54 + 0.05T)
0 = 16.8 + 0.03T – 51.84 + 0.05 T
68.64 = 0.08 T
T = 858 N ©Prep101 Civ100 Exam Solutions Dec 10 May 2009 Q5 Alternate method
You can still work this out by taking the moment around points, E and then A
(here I use the component method to demonstrate that method, but could still
also be done with the vector method)
First, find the unit vector for DF
uDF = ( 0.16 i + 0.11 j – 0.08 k) / square root ((0.16)2 + (0.11)2+ (0.08)2)
=  0.76 i + 0.52 j – 0.38 k
DF = 0.76 DF i + 0.52 DF j  0.38DF k
Use ΣF = 0 in the three axes
ΣFx= Ay + Ey – 0.76 DF – 600 = 0
ΣFy = Ay + Ey + 0.52 DF = 0
ΣFz = Az + Ez – 0.38 DF = 0
There are still far too many unknowns, so we start taking ΣM = 0. We choose points to
eliminate as many unwanted variables as possible – we want DF.
What can we work out easily? Since E and D are in a line, by taking ΣMEx = 0
 600 (0.14) + Ay (0.24) = 0
Ay= 350 N
The next simplest case is taking ΣMDZ = 0
Ey(0.16) + Ay(0.09) = 0
Ey = 197 N
Now we want something using what we have in y and DF, so we take ΣMAx
600 (0.1) – Ey (0.24) – 0.52 DF (0.24) = 0
60 – (197)0.24 – 0.125 DF = 0
DF = 858 N ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2008 Q1 2008 Fall
Q1 FLUID The problem has a width of 3 m
It’s an odd, angled shape so the most convenient way to do the problem is in x y
components – we will break things into x and y
We add names to the points, making the rod ACDB The length of AC is given by pythagoras, √[1.22 + 0.92] = 1.5 m
The angle of AC is given by tan θ = 1.2/0.9 so θ = 53.13°
The reaction at the roller A will be at right angles to this, at 9053.13 = 36.87°
We can break this up into x and y components
Ax = Acos36.87 = 0.80 A
Ay = Asin36.87 = 0.60 A
The reactions at B will be in the x and y directions, Bx and By
Next we find the equivalent forces due to the fluid pressures
The distributed load at C is given by
wC = PC (width)
= ρgh(width)
= 1000(9.81)1.2(3)
= 35.3 kN/m
So the equivalent force over AC is given by ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2008 Q1 FAC = ½ wCL
= ½ 35.3(1.5)
= 106.0 kN
This is at right angles to the section AC, so the components are
FACx= FACcos36.87
= 84.8 kN
FACy = FACsin36.87
= 63.6 kN
This is located at 1/3 of the distance from C, 1.5/3 = 0.5 m from C.
You can easily find the horizontal and vertical distances the same way
1.2/3 = 0.4 m above C, 0.9/3 = 0.3 m to the left of C
Over the section CD, the equivalent force is
FCD = wCL
= 35.3(1.2)
= 42.4 kN
This is vertically downwards, at right angles to CD, at the midway point, 0.6 m to the
right of D
Finally, there are two equivalent forces on DB
wB = PB(width)
= ρgh(width)
= 1000(9.81)2.7(3)
= 79.5 kN/m
FDB1 = wDL
D is at the same depth as C, so wD = wC
= 35.3(1.5)
= 53.0 kN
located at the halfway point, 0.75 m above B
FDB2 = ½ (wB – wD)L
= ½ (79.5 – 35.3)1.5
= 33.2 kN
located 1/3 of the distance from B, 1.5/3 = 0.5 above B ©Prep101 Civ100 Exam Solutions Dec 10 So our free body diagram is Taking the moment around B to eliminate the reactions,
ΣMB = 0
0 = Ay(2.1) –Ax(2.7)
+ 84.8(1.5+0.4) + 63.6(1.2+0.3)
+ 42.4(0.6)
+53(0.75) + 33.2(0.5)
0 =  0.8A(2.1) – 0.6A(2.7) + 338.3
3.3A = 338.3
A = 102.5 kN
Then, to find the reactions at B we use the ΣF=0
ΣFx = 0
0 = Ax – 84.8 – 53 – 33.2 + Bx
0 = 0.6(102.5) – 171 + Bx
Bx = 109.5 kN
ΣFy = 0
0 = Ay – 63.6 – 42.4 + By
0 = 0.8(102.5) – 106 + By
By = 24 kN Dec 2008 Q1 ©Prep101 Q2 Civ100 Exam Solutions Dec 10 Dec 2008 Q3 STRESS AND SHEAR INSTRUCTOR NOTE
First we must draw the simplied FBD of the bar
Pin at A means Ax and Ay
A quick look tells us there are no other horizontal components, so ΣFx = Ax = 0
Roller at B means By
Pin at C means Cx and Cy, dismantle object as machine
Angled piece at D means reactions at D:
A force of Dy = 7.5 kN (pushing down on D)
A moment of MD =  7.5(1.5) = 11.25 kN m applied at D
Roller at E means Ey
Distributed load equivalent
F = 10(4) = 40
applied at midway point, 2 m from A
Draw simplified FBD of first piece ΣMA = 0
40(2) + By(4) + Cy(5) = 0
By = 80 – 5Cy
ΣFx = 0
Cx = 0
ΣFy = 0
Ay – 40 + By + Cy = 0
Ay = 40 –By  Cy
Move on to the next piece ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2008 Q3 Remember machines – the components at C must now point in the opposite directions ΣMC = 0
7.5(1.5) – 11.25 + Ey(4.5) = 0
Ey = 5 kN
ΣFx = 0
[If you had not already noticed Ax = 0, here you would find
Cx = 0
and then from the previous component
Ax = Cx = 0]
ΣFy = 0
Cy – 7.5 + 5 = 0
Cy = 2.5 kN
So we have
By = 80 – 5Cy
= 80 – 5(2.5)
= 92.5 kN
Ay = 40 – By  Cy
= 40 – 92.5 – (2.5)
= 50 kN
Now we have all the reactions, we can draw our full FBD for the entire bar. NOTE: the
pin forces at C are internal to the entire bar and do not appear on its FBD
At A the shear jumps down by 30 to 30 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2008 Q3 Between A and B there is a constant load, so the shear drops by 10 kN/m to
vB = 50 – 10(4) = 90
There it jumps up by 92.5 to 2.5
Between B and D there is no load, so the shear stays constant at 2.5 kN
Nice! We’ve already worked out Cy = 2.5, so that matches up.
At D it jumps down 7.5 to (57.5) = 5 kN
Moment at D has no effect on shear.
No load between D and E so it stays constant.
At E it jumps up by 5 to 0, which is what we expect. Perfect. Now on to the moments.
At any point
M = ΣAreashear – Point moments
There is no external moment at A so we start at 0 kNm
The shear is negative decreasing, so the moment start with a negative slope, since the
shear starts negative, and curve downards with increasing slope to
MB = Area of trapezoid
=  Area of rectangle – Area of triangle
=  50(4) – ½ (9050)4
= 280 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2008 Q3 Between B and D we have constant positive shear, so the moment will linearly slope
upwards to
MD = 280 + 2.5(2.5)
= 273.75
At D there is a point moment of (11.25), so we have
MD = 273.75 – (11.25)
= 262.5
Then, since there is a constant shear until the end point at E, the moment moves
linearly to zero. ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2008 Q3 BEAM DESIGN
The maximum moment magnitude from the problem is Mmax = 280 kNm
σy = 8 MPa = 8 x 106 Pa
Safety factor = 1.9
Max allowed stress
σa = σy/1.9 = 4.21 x 106 Pa
Area = 0.15d
This is a shear question, so compare the bending stress to the max allowed stress
σB = My/I
y=½d
I = 0.15d3/12 = 1.25 x102 d3
σB = 280 x 103 ½ d / [1.25 x 102 d3]
= 1.12 x 107 / d2
Compare this to max allowed stress
σB = σA
1.12 x 107/d2 = 4.21 x 106
d = 1.63 m
INSTRUCTOR NOTE ©Prep101 Q3 Civ100 Exam Solutions Dec 10 Dec 2008 Q3 3D EQUILIBRUM We want to know the tension and reactions at D, A is a ball and socket, so we take the
pivot around A
First, find the unit vector of the cable
uBE = (xE – xB)i + (yE – yB)j + (zE zB)k
√[(xE – xB)2 + (yE – yB)2 + (zE – zB)2]
= (3 – 2)i + (2 – 0)j + (0 – 1)k
√[12 + 22 + (1)2]
(length BE = 2.45 m)
= 0.41i + 0.82j 0.41k
so the tension is
TBE = TuBE
= 0.41T i + 0.82T j – 0.41 T k
The relative position of B to A is rAB= 2i
The weight acts through the center of AC, at a position rAW = 1.5 i
The weight is 2(3) = 6 N, with a vector FW = 6j
The reactions at D are FD = Dx i + Dy j
The relative position of D is rAD = 3i – 1 k
The moment around A is thus
0 = rAW
0 = 1.5
0 0
6 ΣMA = Σ(r x F)
x FW + rAB x TBF + rAD x FD 0 + 2
0 0.41 0
0.82 0
0.41 + 3 0 0 = i(0 – 0) – j (0 – 0) + k (1.5(6) – 0)
+ i (0 – 0) –j (2(0.41T) 0) + k (2(0.82T) – 0)
+ i (0 – Dy (1)) – j (0 – Dx(1)) + k (3Dy – 0)
0 = (Dy)i
+ (0.82T – Dx) j
+ (9 + 1.64T + 3Dy) k
Looking at each component: i component
Dy = 0
j component 1
0 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2008 Q3 Dx = 0.82 T
k component
9 + 1.64 T + 3(0) = 0
T = 5.49 N
then
Dx = 0.82(5.49)
= 4.50 N
ALTERNATE METHOD You can also find the moments around the axis AD, then use
ΣF = 0 to find the reactions at D ©Prep101 Q4 Civ100 Exam Solutions Dec 10 MACHINE No 2 force members. Draw the FBD with the pulleys and ropes removed. Take the moment around F
ΣMF = 0
Ay(3) + Ax(0.5) + 4(1.5) 4(2) + 4(1) – 2(2.5) = 0
3Ay + 0.5 Ax 3 = 0
Ax = 6 + 6Ay
ΣFx = 0
Ax – 2 + 4 – 4 + Fx = 0
Ax = 2  Fx
ΣFy = 0
Ay – 4 + Fy = 0
Ay = 4  Fy Dec 2008 Q4 ©Prep101 Civ100 Exam Solutions Dec 10 Draw the FBD of ABC Take the moment around A
ΣMA = 0
+ By(1) – 4(1.5) – 4(2) = 0
By = 14 kN
ΣFx = 0
Ax + Bx – 4 = 0
Ax = 4  Bx
ΣFy = 0
A y + By – 4 = 0
Ay + 14 – 4 = 0
Ay = 10 kN
Then we have
Ax = 6 + 6Ay
= 6 + 6(10)
= 66 kN
and then we have
Ax = 4  Bx Dec 2008 Q4 ©Prep101 Civ100 Exam Solutions Dec 10
66 = 4  Bx
Bx = 62 kN Then we have
Ax = 2  Fx
66 = 2  Fx
Fx =  64
Ay = 4  Fy
10 = 4  Fy
Fy = 6
[CHK] Dec 2008 Q4 ©Prep101 Civ100 Exam Solutions Dec 10 [OPTIONAL If you looked at the other piece FD] Dec 2008 Q4 ©Prep101 Q5 Civ100 Exam Solutions Dec 10 Dec 2008 Q5 TRUSS The structure is not symmetric.
By looking at joints C and I, members CL and IF are zero force members. IF = 0 kN.
We identify the supports and work out any necessary angles (there’s only one repeating
through the whole problem, tan1 (4/3) = 53.1°) Take the moment around A
ΣMA=0
60 (3) + 18 (4) 40 (12) – 40 (15) + Hy18 = 0
Hy = 66 kN
ΣFx = 0
Ax 18 = 0
Ax = 18
ΣFy = 0
Ay – 60 – 40 – 40 + 66 =0
Ay = 94
To find LK, we take a section from A cutting between CL and DK ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2008 Q5 We would find the values with almost any pivot, but the best is at point D – removing the
two unknowns we don’t want and only leaving LK
ΣMD = 0
94(18) + 18(4) +60(6) +LK(4) = 0
LK = 315 kN
LK is under tension
To find EJ, we take a section from H between FJ and E ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2008 Q5 Taking the moment around J, we eliminate the unwanted unknowns JK and EF, leaving
only EJ
ΣMJ = 0
+66(9) 40(6) + EJ (3) = 0
EJ = 118 kN
EJ is under compression. ©Prep101 Civ100 Exam Solutions Dec 10 BEAM DESIGN
From previous part of question, LK = 315 kN = 3.15 x 105 N
σy = 100 MPa = 1 x108 Pa
Load factor = 1.8
Max allowed stress σa = σy/1.8 = 5.56 x 107
Square cross section, so for width w
A = w2
Compare stress in LK to max allowed stress
σLK = σA
LK/A = 5.56 x 107
3.15 x 105 / w2 = 5.56 x 107
w = 7.53 x 102 m
E = 200,000 MPa = 2 x1011 Pa
A = w2
= (7.53 x102)2
= 5.67 x103
Elongation is given by
Δ= = .
. = 8.33 x 104 m Dec 2008 Q5 ©Prep101 Civ100 Exam Solutions Dec 10 2007
Q1 TRUSS The structure is not symmetric.
CG and HJE are zero force members.
Find the reactions and any necessary angles (by using tan)
tan1 (3/4) = 36.87°
tan1 (5/4) = 30.96° Take the moments around B
ΣMB = 0
+Fx(6) + Fy(15) – Ay(4) = 0
70cos60(6) + 70sin60(15) = 4Ay
Ay = 279.83 kN
ΣFx = 0
Fx + Bx = 0
Bx = 70cos60
= 35 kN
ΣFy = 0
Fy + Ay + By = 0
By = 70sin60 – 279.83
=  219.21 kN Dec 2007 Q1 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2007 Q1 We can find both CD and HE by taking a section, from B, by cutting between D and CH Take the moment around E to eliminate the most unknowns (we don’t need HD, but
finding it will make finding the others possible).
ΣME = 0
HD(4) 279.83(4) + 35(3) = 0
HD = 253.58 kN
HD is under compression.
ΣFx = 0
CD – EHx + 35 = 0
CD – EHcos36.87 + 35 = 0
CD = 35 – 0.80EH
ΣFy = 0
HD + 279.82 + EHy – 219.21 = 0
253.58 + 60.61 + EHsin36.87 = 0
0.60 EH = 192.97
EH = 321.62 kN
EH is under tension
CD = 35 – 0.8(321.62)
CD = 222.30 kN
CD is under compression ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2007 Q3 a) Q3 a)
Note that we are asked for the moment due to the fluid about A – nothing to do with the
weight of the dam (in fact, we’re not even given the dam’s density)
The width of the problem is 5 m
We are asked for the moment of the silt and water around A
We must find the equivalent forces
The distributed load at the bottom of the water is given by
ww = Pw(width)
= ρwghw (width)
= 1000(9.81)4(5)
= 196.2 kN
The equivalent force is
Fw = ½ wwL
= ½ 196.2(4)
= 392.4 kN
applied 1/3 of the distance from the bottom of the water, 4/3 = 1.33 m above the silt.
The distributed load at the bottom of the silt is given by
ws = Ps(width)
The pressure at the base of the silt is due to BOTH the silt and the woter pressing down
ws = (ρwghw + ρsghs)(width)
= (1000(9.81)4 + 1760(9.81)2)(5)
= 368.9 kN
There are two equivalent forces
Fs1 = wwL
= 196.2(2)
= 392.4 kN
applied at the midpoint, 1m from the surface of the silt.
Fs2 = ½ (ws – ww)L
= ½ (368.9 – 196.2)2
= 172.7 kN
applied 1/3 of the way from the bottom of the silt, 2/3 = 0.67 m from the base of the silt. ©Prep101 Civ100 Exam Solutions Dec 10 Our free body diagram is thus Taking the moment around A
ΣMA= +392.4(3.33) + 392.4 (1) + 172.7(0.67)
= 1814.8 kNm Dec 2007 Q3 a) ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2007 Q4 Q4 STRESS AND BEAM DESIGN
Pin at A, reactions Ax and Ay
Roller at B, reaction By
Angled piece BC – find reaction and moment at B
The angle of BC is tan1(5/4) = 51.34°
The height is thus h = 2tan51.34 = 2.5
Bxc = 37 kN
MBC = + 37(2.5) = 92.5 kNm
Distributed load equivalent force is given by
F = 25(7)
= 175 kN
located at midpoint, (2+3.5) = 5.5 m from A
Draw simplifed FBD to find reactions. Label interesting points to make the math clearer. ΣMA = 0
60(2) 175(5.5) + By(9) +92.5 = 0
By = 110 kN
ΣFx = 0
Ax – 37 = 0
Ax = 37
ΣFy = 0
Ay 60 – 175 + 110 = 0
Ay = 125 kN
Draw the complete FBD for AB to find the shear. ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2007 Q4 At A the shear jumps up by 125 kN
Between A and D there is no load, so it remains constant.
At D it jumps down by 60 to 12560 = 65
Between D and B it decreases at 25 kN/m to
vB = 65 – 25(7)
= 110
This means it crosses zero. We must find where
v = 0 = 65 – 25x
x = 2.6 m
Mark this on the diagram
At B the shear jumps up 110 to 0. Perfect. There is no external moment at A, so it starts at 0 kN m
Between A and D the shear is constant positive, so the moment slopes up to
MD = 125(2)
= 250 kNm
Between D and the zero shear point, the shear is positive and decreasing, so the
moment will curve up with decreasing slope. At the zero shear point it will reach a
maximum of
Mmax = 250 + ½ 65 (2.6)
= 334.5 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2007 Q4 Between the maximum and B the shear is negative and dropping, so the moment will
curve down with increasing slope to
MB = 334.5 – ½ 110 (72.6)
= 92.5
At point E there is a point moment of +92.5, so the moment jumps to
MB = 92.5 – 92.5
= 0 kNm
Perfect. ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2007 Q5 Q5 Truss
First we do our prep work.
The unit vector for AD
uAD = (xD – xA)i + (yD – yA)j + (zD – zA)k
√[(xD – xA)2 + (yD – yA)2 + (zD – zA)2]
use distances in meters
uAD= (0 – 0.9)i + (0.3 – 0) j + (0.60)k
√[(0.9)2 + (0.3)2 + (0.6)2]
(length AD = bottom line = 1.12 m)
uAD = 0.80i – 0.27j + 0.53k
and so the Tension is
TAD = TuAD
= 0.8T i – 0.27T j + 0.53T k
We will take moments around B because that’s the ball and socket, to get rid of the
reactions in 3 dimensions there.
Pin connected bars act only in their direction:
CE acts only in the x direction
CF acts only in the z direction
To find the center of gravity, we find the centroid (the plate is uniform) We see that the large triangle AOC is equilateral – AO and OC are the same length.
The area of a triangle is ½ base (height) ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2007 Q5 The centroid of a triangle is 1/3 of the distance from the base
xcg = ΣAx/ΣA
= [ ½ 0.9 (0.9) (0.9/3) – ½ 0.2(0.9) (0.9/3)] / [ ½ 0.9 (0.9) – ½ 0.2 (0.9)]
= 0.3
ycg = [ ½ 0.9 (0.9) (0.9/3) – ½ 0.2 (0.9) (0.2/3)] / [ ½ 0.9 (0.9) – ½ 0.2 (0.9)]
= 0.115 / 0.315
= 0.365
The center of gravity is located at rcg = 0.3 i + 0.365 j
The weight vector is Fw =  20 k
The force at Q is FQ =  40 k
We must also find the location of Q.
There are several ways to find Q, all involving trigonometry. The simplest is to look at
the big triangle OAC We can find the angle at C from
tan θ = OA/OC = 0.9/0.9 = 1
θ = 45°
Then we can find the x component of Q by
tan 45 = Qx /0.3
Qx = 0.3 tan45 = 0.3 m
rQ = 0.3 i + 0.6 j ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2007 Q5 We want to take the moment around B, so we need the relative positions relative to B.
rB = 0.2 j
rBcg= rcg – rB = 0.3 i + 0.165 j
rBC= 0.7 j
rBA= 0.9 i – 0.2 j
rBQ= 0.3 i + 0.4 j
Now we take the moments around B
ΣMB = 0
0 = Σ(r x F)
0 = rBcg x Fw + rBQx FQ + rC x CE + rBC x CF + rBC CE + rBA x TAD
0=
0.3 0.165
0
0 0 + 0.3 0.4
0
0
20
0.9
0.8 0 + 0
40 0
0.2
0.27 0.7
0 0
0.53 0 + 0 0.7
0 0 +
0 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2007 Q5 0 = i (0.165(20) – 0) –j (0.3(20) – 0) + k (0 – 0)
+ i (0.4 (40) – 0) – j (0.3 (40) – 0) + k (0 – 0)
+ i (0.7CF – 0)  0 + 0
+ 0  0 + k(0 – CE(0.7)
+ i (0.2(0.53T) – 0) – j(0.9(0.53T) – 0) + k (0.9 (0.27T) – (0.8T)(0.2)) 0 = 3.3i + 6 j
16i + 12j
+ 0.7CFi
0.7CEk
 0.11Ti – 0.48Tj 0.40 Tk 0 = (19.3 +0.7 CF – 0.11T) i
+ (18 – 0.48 T)j
+ (0.7CE – 0.4 T)k
Each component must equal zero
j component
18  0.48T = 0
T = 37.5 kN
i component
19.3 + 0.7 CF 0.11(37.5) = 0
CF = 33.5 kN
k component
0.7CE – 0.4(37.5) = 0
CE = 21.4 kN ©Prep101 2006 Dec
Q1 Civ100 Exam Solutions Dec 10 Dec 2006 Q1 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2006 Q2 Q2 Machine
Draw the diagram. We can easily find the vertical and horizontal distances from the
angle of θ = 30 °. There is no reaction at O because this is raised airplane landing gear
– it is off the ground! Also, there are no dimensions leading to O, another hint it is
irrelevant. The key is that the member DC is a twoforce member (there is a hint is when the
question tells you BC and DC are massless – so they have no force along their
midpoint. BC is not a twoforce member because there is a moment applied at one end.)
This means DC will have a single force along its direction, DC
Drawing the FBD of BC. We must guess if DC is in compression or tension – it doesn’t
matter if we get it wrong, the math will tell us. Since it is being used to raise the heavy
landing gear, we will say tension (it is pulling the gear up) ©Prep101 Civ100 Exam Solutions Dec 10 ΣMB = 0
M – DC (0.5) sin60 = 0
(using M = Frsinθ)
M = 0.25DC
Draw the FBD of AG Take moments around A
ΣMA = + DC(0.8)sin60 – 490 (1) sin30
DC = 353.6 N
M = 0.25 DC = 153.1 Nm Dec 2006 Q2 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2006 Q3 Q3 Truss
This is two problems and you just have to do both.
First, with the 120 kN loading at C. Work out the reactions and any necessary angles.
In this configuration GD and EH are zero force members.
tan1 (3/4) = 36.87° Take the moments around A
ΣMA = 120 (4) +By(16) = 0
By = 30 kN
ΣFx = 0
Ax = 0
ΣFy = 0
Ay 120 + 30 = 0
Ay = 90
Now we can find FD by making a section from A by cutting between FC and GD ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2006 Q3 We notice FD is the only unknown with a ycomponent, so we can skip straight to ΣFy =
0 (Even if you didn’t, you’d be applying it in a couple of lines anyway, so there’s no trick)
ΣFy = 0
90 – 120 – FDsin36.87 = 0
0.60FD = 30
FD = 50 kN
FD is under compression.
For the second case, with the load of 120 kN applied to D, the problem becomes
symmetric. Now GD, HE and FC are zero force members. By symmetry (or by appling ΣMA as normal), Ay = By = 120/2 = 60 kN
ΣFx = Ax = 0
We find FD by cutting a section between FC and GD ©Prep101 Civ100 Exam Solutions Dec 10 Again, use
ΣFy = 0
60 – FDsin36.87 = 0
0.60 FD = 60
FD = 100 kN
FD is under tension. Dec 2006 Q3 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2006 Q5 Q5
This is a two part problem – first we find the relation between T and the force exerted on
the gate. When the T is the minimum required, the gate is barely closed and therefore
not pressing against B – there is no reaction at B.
Looking at the pin connected assembly, we see that with T pulling up, DC and DE are in
compression. We also see from the joint D that they’re symmetric, so CD = ED
In they ydirection
ΣFy= 0
0 = + T – CDsin20 – EDsin20
T = 2EDsin 20
= 0.68 ED
ED = 1.47 T
Now we can look at the gate and find the T required.
ED is in compression so it is pushing the gate at E. The components are
EDx = EDcos20
= 1.38 T
EDy = EDsin20
= 0.50 T
We must find the equivalent force due to the fluid pressing on the gate.
The gate is 3.2 m wide.
The distributed load at B is
wB = PB(width)
= ρgh(width)
= 1040(9.81)2.6(3.2)
= 84.9 kN/m
The equivalent force is thus
F = ½ wBL
= ½ 84.9(2.6)
= 110.4 kN
located 1/3 of the distance from B, 2.6/3 = 0.87 m from B ©Prep101 Civ100 Exam Solutions Dec 10 The free body diagram is thus Taking the moment around A to eliminate the reactions
ΣMA = 0
0 = + 1.38T(1.1) – 110.4(2.13)
1.52 T = 235.2
T = 154.7 N Dec 2006 Q5 ©Prep101 Civ100 Exam Solutions Dec 10 2005 DECEMBER
Q1
First we find the equivalent force due the water
The problem is 6 m wide.
The distributed load at C is
wC = PC(width)
= ρgh(width)
= 1000(9.81)3(6)
= 176.6 kN/m
The equivalent force is then
F = ½ wCL
= ½ 176.6(3)
= 264.9 kN
applied 1/3 of the distance from C, 1 m above C
Next we must find the weight of the gate and its center of gravity.
The weight is given by
weight = mg
= ρ Volume g
= ρ Area (width) g
The area can be seen as a triangel minus another triangle.
The area of a triangle is ½ base (height) Dec 2005 Q3 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2005 Q3 Area = ½ 5 (5) – ½ 2(5)
= 7.5 m2
So the weight of the gate is
weight = 4000(7.5)(6)9.81
= 1765.8 kN
Now, to find the center of gravity / centroid (we are told the gate is uniform), we use
xcg = ΣAx / ΣA
knowing that the center of gravity of a triangle is 1/3 of the way from the base in that
axis
= [ ½ 5 (5) (5/3) – ½ 2 (5) (5/3)] / [ ½ 5(5) – ½ 2(5)]
= 5/3
= 1.57 from B
[We don’t need to know the vertical location, since the weight is a vertical force, but
here’s how to get it anyway
ycg = ΣAy/ΣA
= [ ½ 5(5) (5/3) – ½ 2 (5) (2/3)] / [ ½ 5(5) – ½ 2(5)]
= 17.5 / 7.5
= 2.33 m below A ] ©Prep101 Civ100 Exam Solutions Dec 10 The hinge A will have reactions in x and y, Ax and Ay
The contact point C will have a reaction at right angles to the ground, Cy
Drawing our free body diagram Taking the moment around A to eliminate the reactions there
ΣMA = 0
0 = +1765.8(3.43) + 264.9(4)  Cy(5)
Cy = 1423.3 kN
Use ΣF = 0
ΣFx = 0 Dec 2005 Q3 ©Prep101 Civ100 Exam Solutions Dec 10
0 = 264.9 + Ax
Ax = 264.9 kN Ax is 264.9 kN to the left
ΣFy = 0
0 = Ay – 1765.8 + 1423.3
Ay = 342.5 kN Dec 2005 Q3 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2005 Q3 Q2 TRUSS
You just have to solve both cases as different problems.
With the load at C, work out angles and reactions. There are no zero force members. Take moments around A
ΣMA = 0
117(4) + By12 = 0
By = 39 kN
ΣFx = 0
Ax = 0
ΣFy = 0
Ay – 117 + 39 = 0
Ay = 78
Take a section between EC and FD to find ED ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2005 Q3 The usual procedure will always work, but we notice that ED is the only unknown with a
ycomponent so we skip ahead to
ΣFy = 0
78 – 117 – EDsin36.87 = 0
39 – 0.60 ED = 0
ED =  65 kN
ED is under compression
With the load at D, EC is a zero force member. Find the reactions ΣMA = 0
117(8) + By12 = 0 ©Prep101 Civ100 Exam Solutions Dec 10
By = 78 kN
ΣFx = 0
Ax = 0
ΣFy =0
Ay 117 + 78 = 0
Ay = 39 Take a section cut between EC and FD to find ED Again,
ΣFy = 0
39 – EDsin36.87 = 0
0.60 ED = 39
ED = 65 kN
ED is under tension. Dec 2005 Q3 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2005 Q3 Q3
The 2.2m uniform bar AB is supported by a ballandsocket at A and two smooth walls
at B, one in the xz plane and one in the yz plane. The bar weighs 400N/m and the
tension in the vertical cable CD is 2kN. Determine the reaction components at A and B. REDO HANDILY, MAKE SURE 2D IS MENTIONED IN BOOKLET
Draw the free body diagram of the bar AB
By
Bx 2000
mg
Ax Ay
Az ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2005 Q3 Because in the xy plane, both ends act as pin connections, we know that the total of
the XY components must act in the direction of the member because it is a twoforce
member.
Bx
tan = 1.4 / 1.2 = 49.4º By FB 1.2
Ax Ay
1.4 FAXY Look at the bar from the side
B 0.6
1.84 1.206 O 0.5 1.6 1.84
1.34 1.4
A 1.2 ©Prep101 Civ100 Exam Solutions Dec 10 Dec 2005 Q3 Using similar triangles
1.84/2.2 = x / 1.6
x = 1.34
2 B 2 0.5 OB = (2.2 – 1.84 )
OB =1.206 1.206 1.1 Determine the weight of the bar.
F = (400N/m)(2.2m)
F = 880N 1.84
O 0.92 A
0.92 This is applied halfway along the bar.
The free body diagram looking at the bar in the OA plane is:
FB 2000
880
FAXY
FAZ Take the moment about A
MA = 0 (CCW +)
2000(1.34) + 880(0.92) – FB (1.206) = 0
FB = 2893.5N
Remember that this is a combination of Bx and By components.
49.4
By Bx ©Prep101 Civ100 Exam Solutions Dec 10 By = FB sin49.4
By = 2197N
Bx = FB cos49.4
Bx = 1883N
Now we can just sum the forces in all three global directions.
Fx = 0
Bx + Ax = 0
Ax = Bx
Ax = 1883 Fy = 0
By + Ay = 0
Ay = By
Ay = 2197 Fz = 0
Az – 880 – 2000
Az = 2880N Dec 2005 Q3 ©Prep101 Civ100 Exam Solutions Dec 10 2001 Fluid 2001
Fluid Question
The wall that separates liquid mud (density = 1760 kg/m3) and fresh water is supported
by a continuous hinge at the base A and anchor bolts BC which are embedded in the
bedrock at C. The attachments of the anchor bolts at the wall and the bedrock are
equivalent to pin connections. What is the maximum safe uniform spacing of the anchor
bolts in metres if the failure load for each bolt is 120kN and the load factor is 3. mud
2.25m B C
1.25m water 2m A
We don’t know the spacing of the bolts, ie the width of the problem, so we will work out
everything for 1 m and then work out the spacing at the end.
Width = 1 m
Determine the distribution of water pressure.
At a depth of 2m below the surface the load is:
Pw = gz
= 1000 (9.81) 2
= 19.62kN/m 19.62 kN/m This results in a total equivalent force of:
Fw = 19.62 (2) / 2
= 19.62 kN applied at a height of 2/3 m from the hinge 19.62 kN/m ©Prep101 Civ100 Exam Solutions Dec 10 2001 Fluid Determine the distribution of mud pressure.
At a depth of 3.5m below the surface the load is:
Pmud = gz
= 1760 (9.81) 3.5
= 60.43kN/m 60.43 kN/m This results in a total equivalent force of:
Fmud = 60.43 (3.5) / 2
= 105.75 kN 105.75 kN/m applied at a height of (1/3)(3.5) m = 1.167m from the hinge Draw the free body diagram of the gate Fbolt
Fmud Fw Since the gate doesn’t move, take a the moment about the hinge. Counterclockwise is
positive. M =0
= Fmud (1.167) – Fw (2/3) – Fbolt (1.25)
= 105.75(1.167) – 19.62(2/3)  Fbolt (1.25) Fbolt = 88.26 kN
88.26 kN needs to be provided by the bolts per meter.
If one bolt can only take (120/3) = 40kN ©Prep101 Civ100 Exam Solutions Dec 10 Then to take 88.26 per meter, we must have 88.26/40 = 2.21 bolts per meter the spacing is 1/2.21 = 0.453 m 2001 Fluid ©Prep101 Civ100 Exam Solutions Dec 10 2001 Truss and Beam Truss
The problem is not symmetric.
BC and DE are zero force members, so BD and DF are one beam BF and have the
same force.
We must find the force in AB, BF and FH
Find the angles, distances and reactions
h = 3tan30 = 1.73 m Take moments around A
ΣMA = 0
32(6) – 20 (9) + Hy (12) = 0
Hy = 31 kN
ΣFx = Ax = 0
ΣFy =0
Ay 32 – 20 + 31 = 0
Ay = 21 kN
To find AB we take a joint at A ©Prep101 Civ100 Exam Solutions Dec 10 ΣFx = 0
ABcos30 + AC = 0
AC = 0.87 AB
ΣFy = 0
21 + ABsin30 = 0
AB =  42 kN
AB is under compression.
AC = 0.87 (42)
= 36.5 kN
AC is under tension.
To find BD = DF = BF we take a section between BC and DE
Note that CE = AC because BC is a zero force member 2001 Truss and Beam ©Prep101 Civ100 Exam Solutions Dec 10 2001 Truss and Beam To eliminate the most unknowns while leaving the one we want, we take moments
around E (though taking moments around almost any point will give you the answer
quite soon anyway)
ΣME = 0
21 (6) – BD(1.73) = 0
BD = 72.8 kN
BD is under compression
To find FH we take a joint at H ΣFx = 0
FHcos30 – GH = 0
GH = FHcos30 ©Prep101 Civ100 Exam Solutions Dec 10
ΣFy = 0
FHsin30 + 31 = 0
FH =  62 kN FH is under compression of 62 kN 2001 Truss and Beam ©Prep101 Civ100 Exam Solutions Dec 10 2001 Truss and Beam BEAM DESIGN
Since all four members are in compression they must have sections big enough for axial
stress as well as buckling.
For axial stress the critical sections will be DF and BD as they have the largest load.
The safety factor (SF) = 2.8
The load Pmax = 72.748 = Pmax SF /A = 72.748 (2.8) /b2
The maximum stress can be 40000kPa for the material to just yield.
b2 = 72.748(2.8) / 40000
b = 71.36 mm
For buckling we can’t tell which will be the critical member. It will either be DF (or BD)
or HF.
For DF, L = 3m
P (SF) = 2EI/L2
72.748 (2.8) = 212000000(I)/32
I = 15.478*106 m4
For HF, L = 3.464m
P(SF) = 2EI/L2
62 (2.8) = 212000000(I)/3.4642
I = 17.588*106 m4
The moment of inertia for a square section is: I = 1/12(b)(b3) = 17.588*106 m4
b = 120.53mm
Choose a member with a dimension of 120.6 mm ©Prep101 Civ100 Exam Solutions Dec 10 2001 Shear Shear and Stress Question
a) Sketch the shear and moment diagrams for the horizontal beam AD.
b) Select the most economical wideflange steel section for the beam AD. Assume
that the yield stress for the steel is 360 Mpa and the load factor is 2.0. Draw the free body diagram of the whole structure to determine the external reactions.
200 50 70
By
Sum the moments about point B
MB = 0 (CCW +)
200(2) + 70 + Dy(9) – 50(6) = 0
Dy = 70kN
Sum the forces in the Y direction.
FY = 0
50 + By – 200 + 70 = 0
By = 180kN Dy ©Prep101 Civ100 Exam Solutions Dec 10 2001 Shear From A to B the shear starts at zero and decreases linearly to point B where the shear
is:
V = 25(2) = 50kN
50 130 At B, the shear is increased by a positive 180kN.
50
130
It then decreases linearly again until point C where the shear is:
25(8) +180  V = 0
V = 20 50 20 The point where the shear is zero is at x from B:
25(2) + 180 – 25(x) = 0
x = 5.2m
At C the shear decreases by 50kN and then stays constant to the end, where it
increases by 70kN.
130 50 5.2m 20
70 The moment starts out a zero at A and ends at zero at D. ©Prep101 Civ100 Exam Solutions Dec 10 2001 Shear From A to B the slope is parabolic and negative, starting out small and increasing. The
moment at B is:
25(2)(1) + M = 0
M = 50 50 Between B and C the slope is increasing parabolically until it reaches a maximum at
5.2m from B. At this point the moment is:
288
25(2)(6.2) + 25(5.2)(5.2/2) – 180(5.2) + M = 0 0.4m
M = 288 kNm 50 5.2m The location where the moment is zero is at a distance of x from B:
50(x+1) 180x + 25(x)(x/2) = 0
12.5x2 130x + 50 = 0
x = 0.4
From the maximum to point C the curve decreases parabolically and the moment at
point C is:
50(7) – 180(6) + 150(3) + M = 0
M = 280kNm
At C the moment decreases by 70kNm to 210 kNm
Between C and D the moment decreases linearly back to zero at D.
288
280 0.4m 50 210 5.2m ©Prep101 Civ100 Exam Solutions Dec 10 Beam Design
The maximum moment is 288kNm
With a load factor of 2, the moment is 576kNm.
The bending stress is:
y
= My/I
360 = 576 y / I
I/y= section modulus = 1600000mm3
Therefore choose W45789. 2001 Shear ©Prep10
01 Civ100 Exam Solutions Dec 10
C
m
s 2000 Q5
0 3D Equ
uilibrium Question
n
The ass
sembly sho
own consis of two lengths of pipe AB a
sts
and BC wh
hich have b
been
welded t
together at B. The as
ssembly is s
supported b special b and so
by
ball
ocket at A w
which
permits movement in the x direction, a regular ball and sock at C and by cable DE.
t
ket
e
e
2
etermine th tension i the cable due to th weight o the
he
in
e
he
of
The pipe weighs 20 N/m. De
pipe. VECTORS AROUND A
A
AC
REDO V
Determine the com
mponents of force DE
f
FDE = 1.5i 3 j 2k
1.52 32 22 = 0.384i + 0.768 j – 0.512 k FDEX = 0
0.384FDE
FDEY = 0
0.768FDE
FDEZ = 0
0.512FDE
(We drop the – sign on the z c
n
component, but we dra it in the negative z direction)
aw
The 4m portion of the pipe we
t
eighs 4*20N = 80N
N/m
The diag
gonal portio of the pip weighs 2
on
pe
2.5*20N/m = 50N ©Prep101 Civ100 Exam Solutions Dec 10 Draw the free body diagram
Ay 2000 Q5 Cy FDEY
FDEZ Az FDEX Cx Cz 50 80 Use the force equilibrium equations in all three directions.
Fx = 0
FDEX + Cx = 0
0.384FDE + Cx = 0 (i) Fy = 0
Ay – 80 – 50 + Cy + FDEY = 0
Ay – 130 + Cy + 0.768FDE = 0 (ii) Fz = 0
Az  FDEZ + Cz = 0
Az  0.512FDE + Cz = 0 (iii) Sum the moments about the X axis. Drawing a diagram looking along the x axis.
Y FDEY
Ay
Z Az Cy
FDEZ 80 Cz
50 1m 1m + ©Prep101 Civ100 Exam Solutions Dec 10 Mx = 0
50(1) + 80(2) – (Ay)(2) – (FDEY)(2) = 0
210 – 2Ay – (0.768FDE)(2) = 0
210 – 2Ay – 1.536FDE = 0 (iv) 2000 Q5 ©Prep101 Civ100 Exam Solutions Dec 10 2000 Q5 Sum the moments about the Y axis. Drawing a diagram looking along the y axis.
Cx
Cz
3m
FDEZ FDEX X Z
Az My = 0
(FDEX)(2) + (FDEZ)(2.5) – Cz(5.5)= 0
(0.384FDE)(2) + (0.512FD)(2.5) – Cz(5.5)= 0
0.768FDE + 1.28FDE – 5.5Cz= 0
2.048FDE – 5.5Cz= 0 +
2.5m (v) Sum the moments about the Z axis. Drawing a diagram looking along the z axis. Y
Cy FDEY
FDEX Ay Cx Ax
80 2m 50 0.5m 2.25m 0.75m X + ©Prep101 Civ100 Exam Solutions Dec 10 Mz = 0
80(2) – 50(4.75) + (FDEY)(2.5) + Cy(5.5)= 0
397.5 + (0.768FDE)(2.5) + 5.5Cy = 0
397.5 + 1.92FDE + 5.5Cy = 0 (vi) Using equations (vi), (ii) and (iv) we can solve for Cy, Ay and FDE.
From (ii)
Ay – 130 + Cy + 0.768FDE = 0
Ay = 130 – Cy  0.768FDE
Sub into (iv)
210 – 2(130 – Cy  0.768FDE ) – 1.536FDE = 0
50 + 2Cy + 1.536 FDE – 1.536FDE = 0
Cy = 25N
Sub into (vi)
397.5 + 1.92FDE + 5.5(25) = 0
FDE = 135.4N 2000 Q5 ...
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