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:
P
P
M1 (x) = Px1
M2 (x) = Pa
x1
a
x2 - a
P
P
EI
d 2v
= M(x)
dx2
For M1(x) = Px1
EI d2v1/dx12 = Px1
EI dv1/dx1 = Px12/2 + C1
2
EIv1 = Px1 /6 + C1x1 + C2
(1)
(2)
For M2(x) = Pa
EId2v2/dx22 = 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 = Pa3/2 – Pa2L/2 + C4
C1a - C4 = Pa3/3 – Pa2L/2
(5)
dv1/dx1 = dv2/dx2 at x1 = x2 = a
Pa2/2 + C1 = Pa2 – PaL/2
C1 = Pa2/2 – PaL/2
Subtitute C1 into eq. (5)
C4 = Pa3/6
dv1/dx1 = P(x12 + a2 –aL) /2EI
θA = dv1/dx1
x1=0
= Pa(a – L)/2EI
v1 = Px1{(x12 + 3a(a –L)}/6EI
v2 = Pa{(3x(x – L) + a2}/6EI
vmax = v2
x= L/2
2
2
= Pa(4a – 3L )/24EI
Ans
(Points 3)
Ans
(Points 3)
Ans
(Points 3)
Ans
(Point 1)
8-4
Determine the equations of the elastic curve for the beam using the x1 and x2 coordinates.
Specify beam’s maximum deflection. EI is constant
Sol:
P
L/2
L
P
3P/2
P/2
V(x1)
V(x2)
M(x2) = -Px2
M(x1) = -P/2x1
x1
x2
P/2
Slope and Elastic Curve
EId2v/dx2 = M(x)
For M(x1) = -Px1/2,
EId2v1/dx21 = - Px1/2
EIdv1/dx1 = -Px12/4 + C1
3
EIv1 = -Px1 /12 + C1x1 +C2
(1)
(2)
For M(x2) = -Px2,
EId2v2/dx22 = -Px2
EIdv2/dx2 = -Px22/2 + C3
3
EIv2 = -Px2 /6 = C3x2 + C4
Boundary Conditions:
v1= 0 at x1 = 0
(3)
(4)
From eq.(2),
C2 = 0
v1 = 0 at x1 = L
From eq. (2)
0 = -PL3/12 + C1L
C1 = PL2/12
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 + PL2/12 = -(PL2/8 + C3)
C3 = 7PL2/24
From eq. (5)
C4 = -PL3/8
The Slope:
Substitute the value of C1 into eq. (1)
dv1/dx1 = P(L2 – 3x12)/12EI
dv1/dx1 = 0 = P(L2 – 3x12)/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 + L2)/12EI
vD = v1
x1 = L/√3
Ans
(Points 4)
Ans
(Points 4)
Ans
(Points 2)
= P(L/√3)(-L2/3 + L2)/12EI = 0.0321 PL3/EI
v2 = P(-4x23 + 7L2x2 – 3L3)/24EI
3
vc = v2 x2 = 0 = -PL /8EI
Hence,
vmax = vc = PL3/8EI
8-12
Use the moment- area theorems to determine the slope and deflection at C. EI is constant
Sol:
15 kip
tan A
θ
θC/A
Δ’
t B/A
t C/A
A
ΔC
tan B
tan C
30'
7.5 kip
C
B
15'
22.5 kip
M/EI
-225/EI
θA = tB/A /30
tB/A = ½ (-225/EI)(30)(10) = -33750/EI
θA = 1125/EI
θC/A = ½ (-225/EI)(30) + ½ (-225/EI)(15) = -5062.5/EI = 5062.5/EI
θC = θC/A + θA
θC = 5062.5/EI – 1125/EI = 3937.5/EI
Ans
(Points 5)
∆C = tC/A - 45/30 tB/A
tC/A = ½ (-225/EI)(30(25) + ½ (-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:
θB/A
θC/A
tan A
t B/A
θA
t C/A
A
tan B
6m
ΔC
tan C
1.333kN
C
B
8kN.m
3m
1.333kN
M/EI
-8/EI
tB/A = ½ (-8/EI)(6)(2) = -48/EI
tC/A = ½(-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)
θA = tB/A /6 = 8/EI
Ans
(Points 3)
Ans
(Points 2)
Ans
(Points 2)
θB/A = ½ (-8/EI)(6) = -24/EI = 24/EI
θB = θB/A + θA
θB = 24/EI – 8/EI = 16/EI
θC/A = ½ (-8/EI)(6) + (-8/EI)(3) = -48/EI = 48/EI
θC = θC/A + θA
θC = 48/EI – 8/EI = 40/EI
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:
5kip
1.5(6) = 9.0kip
tan B
ΔC
A
t B/C
tan C
C
3'
B
3'
7.0kip
M/EI
7.0kip
750kip.ft/EI
3
M/EI
6
6.75kip.ft/EI
3
6
Moment Area Theorems:
Due to symmetry, the slope at midspan C is zero. Hence the slope at B is
θB = θB/C = ½ (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)
The displacement at C is
∆C = tA/C
= ½ (7.50/EI)(3)(2/3)(3) + 2/3(6.75/EI)(3)(5/8)(3)
= 47.8125 kip.ft 3/EI
= 47.8125(1728)/ {29(103)(76.8)}
= 0.0371 in
Ans
(Points 5)