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)