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CE 371 - Section 02
Homework No. 6 - Solutions
8-3
Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates.
Specify the slope at A and the maximum deflection. EI is constant
Sol:
M (x) = Px1
x
1
P
1
P
a x - a2
M (x) = Pa2
P
P
2
2
dx
vdEI = M(x)
For M1(x) = Px1
EI d2v1/dx1
2 = Px1
EI dv1/dx1 = Px12/2 + C1 (1)
EIv1 = Px12/6 + C1x1 + C2 (2)
For M2(x) = Pa
EId2v2/dx2
2 = Pa
EIdv2/dx2 = Pax2 + C3 (3)
EIv2 = Pax22/2 + C3x2 + C4 (4)
Boundary Conditions
v1 = 0 at x = 0
From eq. (2)
C2 = 0
Due to symmetry:
dv2/dx2 = 0 at x2 = L/2
From eq.(3)
0 = PaL/2 + C3
C3 = -PaL/2
Continuity conditions:
v1 = v2 at x1 = x2 = a
Pa3/6 + C1a = Pa
3/2 Pa2L/2 + C4
C1a - C4 = Pa3/3 Pa2L/2 (5)
dv1/dx1 = dv2/dx2 at x1 = x2 = a
Pa2/2 + C1 = Pa
2 PaL/2
C1 = Pa2/2 PaL/2
Subtitute C1 into eq. (5)
C4 = Pa3/6
dv1/dx1 = P(x12 + a
2 aL) /2EI
A = dv1/dx1 x1=0 = Pa(a L)/2EI Ans (Points 3)
v1 = Px1{(x12 + 3a(a L)}/6EI Ans (Points 3)
v2 = Pa{(3x(x L) + a2}/6EI Ans (Points 3)
vmax = v2 x= L/2 = Pa(4a2 3L2)/24EI Ans (Point 1)
8-4
Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates.
Specify beams maximum deflection. EI is constant
Sol:
P
P/2 3P/2
L L/2
M(x ) = -P/2x1 1
V(x )1
P/2
x1
P
V(x )2
M(x ) = -Px2 2
x2
Slope and Elastic Curve
EId2v/dx
2 = M(x)
For M(x1) = -Px1/2,
EId2v1/dx
21 = - Px1/2
EIdv1/dx1 = -Px12/4 + C1 (1)
EIv1 = -Px13/12 + C1x1 +C2 (2)
For M(x2) = -Px2,
EId2v2/dx2
2 = -Px2
EIdv2/dx2 = -Px22/2 + C3 (3)
EIv2 = -Px23/6 = C3x2 + C4 (4)
Boundary Conditions:
v1= 0 at x1 = 0
From eq.(2), C2 = 0
v1 = 0 at x1 = L
From eq. (2)
0 = -PL3/12 + C1L C1 = PL
2/12
v2 = 0 at x2 = L/2
From eq. (4),
0 = -PL3/48 + LC3/2 + C4 (5)
Continuity Conditions:
At x1 = L and x2 = L/2, dv1/dx1 = -dv2/dx2
From eqs (1) and (3),
-PL2/4 + PL
2/12 = -(PL
2/8 + C3)
C3 = 7PL2/24
From eq. (5)
C4 = -PL3/8
The Slope:
Substitute the value of C1 into eq. (1)
dv1/dx1 = P(L2 3x1
2)/12EI
dv1/dx1 = 0 = P(L2 3x1
2)/12EI
x1 = L/3
The Elastic Curve:
Substitute the values of C1, C2, C3 and C4 into eqs (2) and (4), respectively,
v1 = Px1(-x12 + L
2) /12EI Ans (Points 4)
vD = v1 x1 = L/3 = P(L/3)(-L2/3 + L
2)/12EI = 0.0321 PL
3/EI
v2 = P(-4x23 + 7L
2x2 3L
3)/24EI Ans (Points 4)
vc = v2 x2 = 0 = -PL3/8EI
Hence,
vmax = vc = PL3/8EI Ans (Points 2)
8-12
Use the moment- area theorems to determine the slope and deflection at C. EI is constant
Sol:
15 kip
CA B
7.5 kip 22.5 kip
30' 15'C
tan A
tC/A
tan B
tan C
B/At
C/A
-225/EI
M/EI
A = tB/A /30
tB/A = (-225/EI)(30)(10) = -33750/EI
A = 1125/EI
C/A = (-225/EI)(30) + (-225/EI)(15) = -5062.5/EI = 5062.5/EI
C = C/A + A
C = 5062.5/EI 1125/EI = 3937.5/EI Ans (Points 5)
C = tC/A - 45/30 tB/A
tC/A = (-225/EI)(30(25) + (-225/EI)(15)(10) = -101250/EI
C = 101250/EI 45/30(33750/EI) = 50625/EI Ans (Points 5)
8-20
Use the moment- area theorems and determine the deflection at C and the slope of the beam
at A, B and C. EI is constant.
Sol:
CA B
1.333kN
6m 3m
tan AtC/A
tan B
tan C
B/At
-8/EI
M/EI
1.333kN
8kN.m
tB/A = (-8/EI)(6)(2) = -48/EI
tC/A = (-8/EI)(6)(3+2) + (-8/EI)(3)(1.5) = -156/EI
C = tC/A - 9/6 tB/A = 156/EI 9(48)/6EI = 84/EI Ans (Points 3)
A = tB/A /6 = 8/EI Ans (Points 3)
B/A = (-8/EI)(6) = -24/EI = 24/EI
B = B/A + A
B = 24/EI 8/EI = 16/EI Ans (Points 2)
C/A = (-8/EI)(6) + (-8/EI)(3) = -48/EI = 48/EI
C = C/A + A
C = 48/EI 8/EI = 40/EI Ans (Points 2)
A
C/A B/A
C
8-24
Use the moment- area theorems and determine the slope at B and the displacement at C. The
member is an A-36 steel structure Tee for which I = 76.8in4.
Sol:
CA B
B/C
tan B
tan C
750kip.ft/EIM/EI
3' 3'
5kip
1.5(6) = 9.0kip
7.0kip 7.0kip
6.75kip.ft/EIM/EI
3 6
3 6
t
Moment Area Theorems:
Due to symmetry, the slope at midspan C is zero. Hence the slope at B is
B = B/C = (7.50/EI) (3) + 2/3(6.75/EI) (3)
= 24.75kip.ft2/EI
= 24.75(144)/ {29(103) (76.8)}
= 0.00160 rad Ans (Points 5)
C
The displacement at C is
C = tA/C = (7.50/EI)(3)(2/3)(3) + 2/3(6.75/EI)(3)(5/8)(3)
= 47.8125 kip.ft3/EI
= 47.8125(1728)/ {29(103)(76.8)}
= 0.0371 in Ans (Points 5)