P5.2. Write the equations for shear and moment B between ... HW 4 F18.pdfP5.47. For the frame in Figure P5.47, draw the shear and moment curves for all members. Next sketch the deflected

5-3

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6(8) 13(3 0

19 kips

3 0

11 kips

8

Σ 10) 18

Σ

0

8 kip

1

s

0

Σ

A y

y

y y y

y

x x

x

E

A

A

E

M

F

F

E

E

E

=- + =

= = - =

= = - =

=

⋅ -

⋅ +

2

Σ

Σ

3(10 ) 19

3 11 kips

2

319 kip

2

3 19( ) 0

ft

y

A x

x

x

x

F

M

V x

V xxM x x

M x x

= - - +

= +æ ö÷ç+ ÷ç ÷ç ÷è ø

=

-

⋅

=

= -

⋅

P5.2. Write the equations for shear and moment

between points D and E. Select the origin at D.

10ʹ3ʹ5ʹ

6ʹ

8 kips

A ED

C

B

w = 3 kips/ft

x

P5.2

3 kips/ft

8 kips

Ay Ey

D

B

C

3 kips/ft

EyD x 10-x

VxMx

5-13


Σ 0; 8 4 48 8 14 0

22 kips

Σ 0; 8 48 22 0

34 kips

B C

C

y B

B

M R

RF R

R

+

= ´ - ´ + =

= = - - + + =

=

(a) A–B Origin at “A;” 0 4x£ £

N k

Σ 0; 8 0

8

Σ 0; 8 0

8

Σ 0; 8 34 3( 4) 0

3 38 (EQ. 1)

( 4)Σ 0; 8 34( 4) 3( 4) 0

2

y

k

z

y

z

F V

VM M x

M x

F x VV x

xM x x x M

+

+⋅

+

+

= - - =

=-= + =

=-

= - + - - - ==- +

-= - - + - + =

2 1.5 38 160 (EQ. 2)M x x=- + -

(b) Moment at (1) set 9x =

2

11.5(9) 38 9 160

121.5 340 160

60.5 kip ft

M =- + ´ -=- + -

= ⋅

P5.12. Consider the beam shown in

Figure P5.12.

(a) Write the equations for shear and moment

using an origin at end A.

(b) Using the equations, evaluate the moment at

section 1.

(c) Locate the point of zero shear between

B and C.

(d) Evaluate the maximum moment between

points B and C.

(e) Write the equations for shear and moment

using an origin at C.

(f ) Evaluate the moment at section 1.

(g) Locate the section of maximum moment and

evaluate Mmax

.

(h) Write the equations for shear and moment

between B and C using an origin at B.

(i) Evaluate the moment at section 1.

AB

C

P = 8 kips1

5ʹ

16ʹ4ʹ

w = 3 kips/ft

P5.12

(c) Locate Point 0V B C= -

Use EQ. 1 0 3 38

12.67

x

x

=- +

=

max

2

max

max

( ) : Set 12.67 in (EQ. 2)

1.5(12.67) 38 12.67 160

240.79 481.46 160

80.67 kip ft

d M xM

M

==- + ´ -=- + -

= ⋅

3 kips/ft

5’

A C

16’

Cy = 22 kips

8 kips

By = 34 kips

1

4’

8 kips

x

M(x)V(x)

z

8 kips

4’

M(x)

V(x)

z

3(x – 4)

x – 4

By = 34 kips

x

A

5-14


P5.12. Continued

(e) B–C

2

Σ 0; 3 22 0

22 3

Σ 0; 3 22 02

122

2

y

z

F V x

V x

xM M x x

M x x

+

+

= - + =

=- +

= + ⋅ - =

= -

A–B Σ 0; 8 0

8

Σ 0; 8(20 ) 0

8 160

ky

k

Z

F V

V

M x M

M x

+

+ = - - =

=-

= - - - =

= -

(f ) M at section (1) 211AR x+ =

2322(11) (11) 60.5 kip ft

2M = - = ⋅

M V x

V

M

M

gmax

2

2

max

max

( ) , 0; 22 3 0

227.33

33

22 7.33 (7.33)2

161.2

Set

6 80.59

80.67 kip ft

= - + =

= =

= ´ -

= -

= ⋅

2

( ) Σ 0; 8 34 3 0

26 3

3Σ 0; 8(4 ) 34 0

2

332 26

2

y

z

h F x V

V xx xM K x M

M x x

+

+

= - + - - =

= -⋅= - + + - - =

=- + -

(i) Moment at section (1)

Let 5x ¢=

2332 26(5) 5

2

60.5 kip ft

M

M

æ ö÷ç=- + - ÷ç ÷ç ÷è ø= ⋅

8 kips

4’

M(x)

V(x)

z

3x

xBy = 34 kips

x

A

x

M(x)

V(x)

z

Cy = 22 kips

3x

8 kips

20 – x

M(x)V(x)

zA

5-19


P5.17. For each beam, draw the shear and

moment curves label the maximum values of

shear and moment, locate points of inflection,

and sketch the deflected shape.

P = 20 kips

w = 12 kips/ft

CB DA

10ʹ 15ʹ5ʹ

hinge

P5.17

5-42


Segment ABC: Origin CD; Range 0 ≤ x2 = 2.4

2 2

2 2

Σ 0;

24.8 2.4 0

24.8 2.4

yk

Fx V

V x

+ =- - =

= -

2

2 2 2

2

2 2 2

Σ 0;

24.8 4 (15 ) 2.4 02

60 24.8 1.2

O

k

Mx

x x M

M x x

+

=æ ö÷ç¢ ÷+ - - =ç ÷ç ÷çè ø

= + -

P5.40. (a) Draw the shear and moment curves

for the frame in Figure P5.40. Sketch the

deflected shape. (b) Write the equations for

shear and moment in column AB. Take the

origin at A. (c) Write the shear and moment

equations for girder BC. Take the origin at

joint B. x1

x2

4 kips12 kips

3ʹ24ʹ

10ʹ

5ʹ

w = 2.4 kips/ft

A

B C

D

E

F15ʹ

P5.40

Segment AB, Orgin @ A, Range 0 ≤ x, ‒ 15′

1 1 1 1Σ 0; 4 0 4kOM x M M+

= + - = =

Segment BC

Segment CE

5-49


P5.47. For the frame in Figure P5.47, draw the

shear and moment curves for all members. Next

sketch the deflected shape of the frame. Show

all forces acting on a free-body diagram of

joint C.

A

B

D

C E

20 kips

w = 5 kips/ft

4ʹ

hinge

4ʹ

4ʹ

6ʹ

6ʹ

P5.47

M Diagrams Deflected Share

Joint C

1 VukazichCE160ReviewProblemSolution[5]

CE 160 Review Problem: Shear and Moment Diagram

The beam shown is pin supported at point A; roller supported at points B, D, and F (note that roller supports resist movement both up and down); and has internal hinges at points C and E. Neglecting the weight of the beam, for the point loads applied at points C and G as shown:

1. Show that the beam is statically determinate and find the support reactions at points A, B, D, and F.

Show that beam is statically determinate

To evaluate determinacy, we must cut beam at supports and all places where we know an internal force (i.e. M = 0 at hinges at C and E.

F.B.D.s

X = total number of unknowns = 9 n = number of F.B.D. 3n = total number of equations of equilibrium = 3(3) = 9

beam is statically determinate

12 ft 15 ft 15 ft 12 ft

A

B

C

D

E

15 ft 12 ft F

G

2 k 3 k

12 ft 15 ft

Ay By Dy Fy

15 ft 12 ft

2 k 3 k

VC

Ax FC

15 ft 12 ft

VC

FC

VE FE

VE

FE


Find support reactions

F.B.D. of piece EFG

Apply equations of equilibrium to find unknown forces

Moment Equilibrium

𝑀" = 0

Counterclockwise moments about point E positive

− 3𝑘)(27𝑓𝑡 +𝐹1 15𝑓𝑡 = 0 𝑭𝒚 = 𝟓. 𝟒𝒌 (Fy is positive so Fy acts upward as assumed)

Force Equilibrium

+↑ 𝐹1 = 0

Upward forces positive

𝑉" − 3𝑘 +𝐹1 = 0 𝑉" − 3𝑘 + 5.4𝑘 = 0

𝑽𝑬 = −𝟐. 𝟒𝒌 (VE is negative so VE acts downward)

Force Equilibrium

+→ 𝐹A = 0

forces positive to the right

Fy

15 ft 12 ft

3 k

VE

FE

+


−𝐹" = 0

𝑭𝑬 = 𝟎 (no axial force in beam segment)

F.B.D. of piece CDE with known forces shown


Moment Equilibrium

𝑀C = 0

Counterclockwise moments about point C positive

2.4𝑘)(27𝑓𝑡 +𝐷1 15𝑓𝑡 = 0 𝑫𝒚 = −𝟒. 𝟑𝟐𝒌 (Dy is negative so Dy acts downward)

Force Equilibrium

+↑ 𝐹1 = 0


𝑉C + 2.4𝑘 +𝐷1 = 0

𝑉C + 2.4𝑘 + −4.32𝑘 = 0 𝑽𝑪 = 𝟏. 𝟗𝟐𝒌 (VC is positive so VC acts upward as assumed)

Force Equilibrium

+→ 𝐹A = 0


Dy

15 ft 12 ft

VC

FC

2.4 k

+


−𝐹C = 0

𝑭𝑪 = 𝟎 (no axial force in beam segment)

F.B.D. of piece ABC with known forces shown


Moment Equilibrium

𝑀J = 0

Counterclockwise moments about point A positive

− 2𝑘)(27𝑓𝑡 − 1.92𝑘)(27𝑓𝑡 + 𝐵1 12𝑓𝑡 = 0 𝑩𝒚 = 𝟖. 𝟖𝟐𝒌 (By is positive so By acts upward as assumed)

Force Equilibrium

+↑ 𝐹1 = 0


𝐴1 +𝐵1 − 2𝑘 − 1.92𝑘 = 0

𝐴1 + 8.82𝑘 − 2𝑘 − 1.92𝑘 = 0 𝑨𝒚 = −𝟒. 𝟗𝒌 (Ay is negative so Ay acts downward)

12 ft 15 ft

Ay By

2 k

1.92 k

Ax

+


Force Equilibrium

+→ 𝐹A = 0


𝐴A = 0

𝑨𝒙 = 𝟎 (no axial force in beam segment)

F.B.D. of entire beam showing support reactions

12 ft 15 ft 15 ft 12 ft

A B C D E

15 ft 12 ft

F G

2 k 3 k

5.4 k 4.32 k 8.82 k 4.9 k


2. Construct the internal shear and bending moment diagrams for the beam. Label all local maximum and minimum values and their locations. Use the “usual” Civil Engineering sign convention for each of your diagrams (i.e. the sign convention we used in Lab #2). Show your results on the axes provided on the next page.

Find internal shear and moment from A to B Make a cut at an arbitrary point between points A and B F.B.D of beam to left of cut, show unknown internal forces V and M in their positive sense

Moment Equilibrium of segment

𝑀S = 0

Counterclockwise moments about point a are positive

4.9 𝑥 + 𝑀 = 0 𝑴 = −𝟒. 𝟗𝒙 (units in k and ft)

Force Equilibrium

+↑ 𝐹1 = 0


−4.9 − 𝑉 = 0 𝑽 = −𝟒. 𝟗 (units in k)

x

F V 4.9 k

M

+


Force Equilibrium

+→ 𝐹A = 0


𝐹 = 0

𝑭 = 𝟎 (no axial force in beam segment)

In general, F = 0 for all of the beam segments Plot V and M between points A and B

At point A Evaluate at x = 0

𝑉J = −𝟒. 𝟗𝒌 𝑀J = − 4.9)(0 = 𝟎𝒌 − 𝒇𝒕

At point B- just the left of point B (x = 11.999999… ft) Evaluate at x = 12 ft

𝑉XY = −𝟒. 𝟗𝒌 𝑀XY = − 4.9)(12 = −𝟓𝟖. 𝟖𝒌 − 𝒇𝒕

Note that M is linear in x between A and B

Shear between B and C, C and D, D and F, and F and G will be constant since there is no distributed load in these regions. Find internal shear and moment from B to C

Make a cut at an arbitrary point between points B and C F.B.D of beam to left of cut, show unknown internal forces V and M in their positive sense


Moment Equilibrium of segment

𝑀S = 0

Counterclockwise moments about point a are positive

4.9 𝑥 − 8.82 𝑥 − 12 + 𝑀 = 0 𝑴 = 𝟑. 𝟗𝟐𝒙 − 𝟏𝟎𝟓. 𝟖𝟒 (units in k and ft)

Force Equilibrium

+↑ 𝐹1 = 0


−4.9 + 8.82 − 𝑉 = 0

𝑽 = 𝟑. 𝟗𝟐 (units in k)

Plot V and M between points B and C

At point B+ just the right of point B (x = 12.000…001 ft) Evaluate at x = 12

𝑉XZ = 𝟑. 𝟗𝟐𝒌 𝑀XZ = 3.92 12 − 105.84 = −𝟓𝟖. 𝟖𝒌 − 𝒇𝒕

At point C- just the left of point C (x = 26.999999… ft) Evaluate at x = 27 ft

𝑉CY = 𝟑. 𝟗𝟐𝒌 𝑀CY = 3.92 27 − 105.84 = 𝟎𝒌 − 𝒇𝒕

12 ft

F V

M

x

4.9 k 8.82 k

+


Note that M is linear in x between B and C, the bending moment should be zero at the internal hinge at C, and there is continuity in the moment diagram across point B.


Can also use the area under shear diagram to find moment at C

MC – MB = Area of V diagram between points B and C

Area of shear diagram between B and C = (3.92 k) (15 ft) = 58.8 k-ft

𝑀C −𝑀X = 58.8𝑘 − 𝑓𝑡 𝑀C = 𝑀X + 58.8𝑘 − 𝑓𝑡 𝑀C = −58.8 + 58.8 = 𝟎𝒌 − 𝒇𝒕

Shear between C and D, D and F, and F and G will be constant since there is no distributed load in these regions. Make cuts in these regions and fill in the rest of the shear diagram. Find internal shear from C to D

F.B.D of beam to left of cut between C and D, showing unknown internal forces V and M in their positive sense

Force Equilibrium

+↑ 𝐹1 = 0


−4.9 + 8.82 − 2 − 𝑉 = 0

𝑽 = 𝟏. 𝟗𝟐 (units in k)

12 ft

F

V

M

4.9 k 8.82 k

2 k

A B C

15 ft


Find internal shear from D to F F.B.D of beam to left of cut between D and F, showing unknown internal forces V and M in their positive sense

Force Equilibrium

+↑ 𝐹1 = 0


−4.9 + 8.82 − 2 − 4.32 − 𝑉 = 0

𝑽 = −𝟐. 𝟒 (units in k)

Find internal shear from F to G F.B.D of beam to left of cut between F and G, showing unknown internal forces V and M in their positive sense

Force Equilibrium

+↑ 𝐹1 = 0


−4.9 + 8.82 − 2 − 4.32 + 5.4 − 𝑉 = 0

12 ft

F

V

M

4.9 k 8.82 k

2 k

A B C

15 ft 15 ft 12 ft

4.32 k

D E

12 ft 15 ft 15 ft 12 ft

F

V

M

4.9 k 8.82 k

2 k

A B C 4.32 k

D E 5.4 k

F

15 ft


𝑽 = 𝟑(units in k)

Note that we get the same result looking at the F.B.D. to the right of the cut

Force Equilibrium

+↑ 𝐹1 = 0


𝑉 − 3 = 0

𝑽 = 𝟑(units in k, checks with previous result)

Now use areas under shear diagram to complete the moment diagram Moment at point D

Area of shear diagram between C and D = (1.92 k) (15 ft) = 28.8 k-ft 𝑀[ −𝑀C = 28.8𝑘 − 𝑓𝑡 𝑀[ = 𝑀C + 28.8𝑘 − 𝑓𝑡 𝑀[ = 0 + 28.8𝑘 − 𝑓𝑡

𝑴𝑫 = 𝟐𝟖. 𝟖𝒌 − 𝒇𝒕

M is linear in between C and D

Moment at point E

Area of shear diagram between D and E = (-2.4 k) (12 ft) = -28.8 k-ft 𝑀" −𝑀[ = −28.8𝑘 − 𝑓𝑡 𝑀" = 𝑀[ − 28.8𝑘 − 𝑓𝑡 𝑀" = 28.8 − 28.8𝑘 − 𝑓𝑡

𝑴𝑬 = 𝟎𝒌 − 𝒇𝒕

M is linear in between D and E and the bending moment should be zero at the internal hinge at E

3 k

G F

V

M


Moment at point F

Area of shear diagram between E and F = (-2.4 k) (15 ft) = -36 k-ft 𝑀\ −𝑀" = −36𝑘 − 𝑓𝑡 𝑀\ = 𝑀" − 36𝑘 − 𝑓𝑡 𝑀\ = 0 − 36𝑘 − 𝑓𝑡

𝑴𝑭 = −𝟑𝟔𝒌 − 𝒇𝒕

M is linear in between E and F

Moment at point G

Area of shear diagram between F and G = (3k) (12 ft) = 36 k-ft 𝑀_ −𝑀\ = 36𝑘 − 𝑓𝑡 𝑀_ = 𝑀\ + 36𝑘 − 𝑓𝑡 𝑀_ = −36 + 36𝑘 − 𝑓𝑡

𝑴𝑮 = 𝟎𝒌 − 𝒇𝒕

M is linear in between F and G and the bending moment should be zero at the free end at point G.


Plot V and M diagrams under the F.B.D of the beam

Note that a shear discontinuity will be expected at B, C, D, and F due to the point loads.

V

A B C D E F G

M

A B C D E F G

12 ft 15 ft 15 ft 12 ft

A B C D E

15 ft 12 ft

F G

2 k 3 k

5.4 k 4.32 k 8.82 k 4.9 k

- 4.9 k

3.92 k 1.92 k

-2.4 k

3 k

28.8 k-ft

-58.8 k-ft -36 k-ft

0 0


Note the differential and integral relationships between V, M, and w:

𝑑𝑉𝑑𝑥

= 𝑤 slopeoftangenttoVdiagramatapoint

= distributedloadintensityatthatpoint

𝑑𝑀𝑑𝑥

= 𝑉 slopeoftangenttoMdiagramatapoint = valueofVatthatpoint

VB – VA = Area of distributed load between points A and B

MB – MA = Area of V diagram between points A and B

Documents

P5.2. Write the equations for shear and moment B between ... HW 4 F18.pdfP5.47. For the frame in Figure P5.47, draw the shear and moment curves for all members. Next sketch the deflected