p5.2. write the equations for shear and moment b between ... hw 4 f18.pdfp5.47. for the frame in...
TRANSCRIPT
5-3
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6(8) 13(3 0
19 kips
3 0
11 kips
8
ฮฃ 10) 18
ฮฃ
0
8 kip
1
s
0
ฮฃ
A y
y
y y y
y
x x
x
E
A
A
E
M
F
F
E
E
E
=- + =
= = - =
= = - =
=
โ -
โ +
2
ฮฃ
ฮฃ
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
= - - +
= +รฆ รถรทรง+ รทรง รทรง รทรจ รธ
=
-
โ
=
= -
โ
P5.2. Write the equations for shear and moment
between points D and E. Select the origin at D.
10สน3สน5สน
6สน
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
Copyright ยฉ 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
ฮฃ 0; 8 4 48 8 14 0
22 kips
ฮฃ 0; 8 48 22 0
34 kips
B C
C
y B
B
M R
RF R
R
+
= ยด - ยด + =
= = - - + + =
=
(a) AโB Origin at โA;โ 0 4xยฃ ยฃ
N k
ฮฃ 0; 8 0
8
ฮฃ 0; 8 0
8
ฮฃ 0; 8 34 3( 4) 0
3 38 (EQ. 1)
( 4)ฮฃ 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
+
+โ
+
+
= - - =
=-= + =
=-
= - + - - - ==- +
-= - - + - + =
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 =- + ยด -=- + -
= โ
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
5สน
16สน4สน
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
==- + ยด -=- + -
= โ
3 kips/ft
5โ
A C
16โ
Cy = 22 kips
8 kips
By = 34 kips
1
4โ
8 kips
x
M(x)V(x)
z
8 kips
4โ
M(x)
V(x)
z
3(x โ 4)
x โ 4
By = 34 kips
x
A
5-14
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P5.12. Continued
(e) BโC
2
ฮฃ 0; 3 22 0
22 3
ฮฃ 0; 3 22 02
122
2
y
z
F V x
V x
xM M x x
M x x
+
+
= - + =
=- +
= + โ - =
= -
AโB ฮฃ 0; 8 0
8
ฮฃ 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 = - = โ
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
= - + =
= =
= ยด -
= -
= โ
2
( ) ฮฃ 0; 8 34 3 0
26 3
3ฮฃ 0; 8(4 ) 34 0
2
332 26
2
y
z
h F x V
V xx xM K x M
M x x
+
+
= - + - - =
= -โ = - + + - - =
=- + -
(i) Moment at section (1)
Let 5x ยข=
2332 26(5) 5
2
60.5 kip ft
M
M
รฆ รถรทรง=- + - รทรง รทรง รทรจ รธ= โ
8 kips
4โ
M(x)
V(x)
z
3x
xBy = 34 kips
x
A
x
M(x)
V(x)
z
Cy = 22 kips
3x
8 kips
20 โ x
M(x)V(x)
zA
5-19
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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
10สน 15สน5สน
hinge
P5.17
5-42
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Segment ABC: Origin CD; Range 0 โค x2 = 2.4
2 2
2 2
ฮฃ 0;
24.8 2.4 0
24.8 2.4
yk
Fx V
V x
+ =- - =
= -
2
2 2 2
2
2 2 2
ฮฃ 0;
24.8 4 (15 ) 2.4 02
60 24.8 1.2
O
k
Mx
x x M
M x x
+
=รฆ รถรทรงยข รท+ - - =รง รทรง รทรงรจ รธ
= + -
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
3สน24สน
10สน
5สน
w = 2.4 kips/ft
A
B C
D
E
F15สน
P5.40
Segment AB, Orgin @ A, Range 0 โค x, โ 15โฒ
1 1 1 1ฮฃ 0; 4 0 4kOM x M M+
= + - = =
Segment BC
Segment CE
5-49
Copyright ยฉ 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
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
4สน
hinge
4สน
4สน
6สน
6สน
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
2 VukazichCE160ReviewProblemSolution[5]
Find support reactions
F.B.D. of piece EFG
Apply equations of equilibrium to find unknown forces
Moment Equilibrium
๐" = 0
Counterclockwise moments about point E positive
โ 3๐)(27๐๐ก +๐น1 15๐๐ก = 0 ๐ญ๐ = ๐. ๐๐ (Fy is positive so Fy acts upward as assumed)
Force Equilibrium
+โ ๐น1 = 0
Upward forces positive
๐" โ 3๐ +๐น1 = 0 ๐" โ 3๐ + 5.4๐ = 0
๐ฝ๐ฌ = โ๐. ๐๐ (VE is negative so VE acts downward)
Force Equilibrium
+โ ๐นA = 0
forces positive to the right
Fy
15 ft 12 ft
3 k
VE
FE
+
3 VukazichCE160ReviewProblemSolution[5]
โ๐น" = 0
๐ญ๐ฌ = ๐ (no axial force in beam segment)
F.B.D. of piece CDE with known forces shown
Apply equations of equilibrium to find unknown forces
Moment Equilibrium
๐C = 0
Counterclockwise moments about point C positive
2.4๐)(27๐๐ก +๐ท1 15๐๐ก = 0 ๐ซ๐ = โ๐. ๐๐๐ (Dy is negative so Dy acts downward)
Force Equilibrium
+โ ๐น1 = 0
Upward forces positive
๐C + 2.4๐ +๐ท1 = 0
๐C + 2.4๐ + โ4.32๐ = 0 ๐ฝ๐ช = ๐. ๐๐๐ (VC is positive so VC acts upward as assumed)
Force Equilibrium
+โ ๐นA = 0
forces positive to the right
Dy
15 ft 12 ft
VC
FC
2.4 k
+
4 VukazichCE160ReviewProblemSolution[5]
โ๐นC = 0
๐ญ๐ช = ๐ (no axial force in beam segment)
F.B.D. of piece ABC with known forces shown
Apply equations of equilibrium to find unknown forces
Moment Equilibrium
๐J = 0
Counterclockwise moments about point A positive
โ 2๐)(27๐๐ก โ 1.92๐)(27๐๐ก + ๐ต1 12๐๐ก = 0 ๐ฉ๐ = ๐. ๐๐๐ (By is positive so By acts upward as assumed)
Force Equilibrium
+โ ๐น1 = 0
Upward forces positive
๐ด1 +๐ต1 โ 2๐ โ 1.92๐ = 0
๐ด1 + 8.82๐ โ 2๐ โ 1.92๐ = 0 ๐จ๐ = โ๐. ๐๐ (Ay is negative so Ay acts downward)
12 ft 15 ft
Ay By
2 k
1.92 k
Ax
+
5 VukazichCE160ReviewProblemSolution[5]
Force Equilibrium
+โ ๐นA = 0
forces positive to the right
๐ดA = 0
๐จ๐ = ๐ (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
6 VukazichCE160ReviewProblemSolution[5]
2. Construct the internal shear and bending moment diagrams for the beam. Label all local maximum and minimum values and their locations. Use the โusualโ 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
๐S = 0
Counterclockwise moments about point a are positive
4.9 ๐ฅ + ๐ = 0 ๐ด = โ๐. ๐๐ (units in k and ft)
Force Equilibrium
+โ ๐น1 = 0
Upward forces positive
โ4.9 โ ๐ = 0 ๐ฝ = โ๐. ๐ (units in k)
x
F V 4.9 k
M
+
7 VukazichCE160ReviewProblemSolution[5]
Force Equilibrium
+โ ๐นA = 0
forces positive to the right
๐น = 0
๐ญ = ๐ (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
๐J = โ๐. ๐๐ ๐J = โ 4.9)(0 = ๐๐ โ ๐๐
At point B- just the left of point B (x = 11.999999โฆ ft) Evaluate at x = 12 ft
๐XY = โ๐. ๐๐ ๐XY = โ 4.9)(12 = โ๐๐. ๐๐ โ ๐๐
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
8 VukazichCE160ReviewProblemSolution[5]
Moment Equilibrium of segment
๐S = 0
Counterclockwise moments about point a are positive
4.9 ๐ฅ โ 8.82 ๐ฅ โ 12 + ๐ = 0 ๐ด = ๐. ๐๐๐ โ ๐๐๐. ๐๐ (units in k and ft)
Force Equilibrium
+โ ๐น1 = 0
Upward forces positive
โ4.9 + 8.82 โ ๐ = 0
๐ฝ = ๐. ๐๐ (units in k)
Plot V and M between points B and C
At point B+ just the right of point B (x = 12.000โฆ001 ft) Evaluate at x = 12
๐XZ = ๐. ๐๐๐ ๐XZ = 3.92 12 โ 105.84 = โ๐๐. ๐๐ โ ๐๐
At point C- just the left of point C (x = 26.999999โฆ ft) Evaluate at x = 27 ft
๐CY = ๐. ๐๐๐ ๐CY = 3.92 27 โ 105.84 = ๐๐ โ ๐๐
12 ft
F V
M
x
4.9 k 8.82 k
+
9 VukazichCE160ReviewProblemSolution[5]
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.
10 VukazichCE160ReviewProblemSolution[5]
Can also use the area under shear diagram to find moment at C
MC โ 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
๐C โ๐X = 58.8๐ โ ๐๐ก ๐C = ๐X + 58.8๐ โ ๐๐ก ๐C = โ58.8 + 58.8 = ๐๐ โ ๐๐
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
+โ ๐น1 = 0
Upward forces positive
โ4.9 + 8.82 โ 2 โ ๐ = 0
๐ฝ = ๐. ๐๐ (units in k)
12 ft
F
V
M
4.9 k 8.82 k
2 k
A B C
15 ft
11 VukazichCE160ReviewProblemSolution[5]
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
+โ ๐น1 = 0
Upward forces positive
โ4.9 + 8.82 โ 2 โ 4.32 โ ๐ = 0
๐ฝ = โ๐. ๐ (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
+โ ๐น1 = 0
Upward forces positive
โ4.9 + 8.82 โ 2 โ 4.32 + 5.4 โ ๐ = 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
12 VukazichCE160ReviewProblemSolution[5]
๐ฝ = ๐(units in k)
Note that we get the same result looking at the F.B.D. to the right of the cut
Force Equilibrium
+โ ๐น1 = 0
Upward forces positive
๐ โ 3 = 0
๐ฝ = ๐(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 ๐[ โ๐C = 28.8๐ โ ๐๐ก ๐[ = ๐C + 28.8๐ โ ๐๐ก ๐[ = 0 + 28.8๐ โ ๐๐ก
๐ด๐ซ = ๐๐. ๐๐ โ ๐๐
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 ๐" โ๐[ = โ28.8๐ โ ๐๐ก ๐" = ๐[ โ 28.8๐ โ ๐๐ก ๐" = 28.8 โ 28.8๐ โ ๐๐ก
๐ด๐ฌ = ๐๐ โ ๐๐
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
14 VukazichCE160ReviewProblemSolution[5]
Moment at point F
Area of shear diagram between E and F = (-2.4 k) (15 ft) = -36 k-ft ๐\ โ๐" = โ36๐ โ ๐๐ก ๐\ = ๐" โ 36๐ โ ๐๐ก ๐\ = 0 โ 36๐ โ ๐๐ก
๐ด๐ญ = โ๐๐๐ โ ๐๐
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 ๐_ โ๐\ = 36๐ โ ๐๐ก ๐_ = ๐\ + 36๐ โ ๐๐ก ๐_ = โ36 + 36๐ โ ๐๐ก
๐ด๐ฎ = ๐๐ โ ๐๐
M is linear in between F and G and the bending moment should be zero at the free end at point G.
15 VukazichCE160ReviewProblemSolution[5]
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
16 VukazichCE160ReviewProblemSolution[5]
Note the differential and integral relationships between V, M, and w:
๐๐๐๐ฅ
= ๐ค slopeoftangenttoVdiagramatapoint
= distributedloadintensityatthatpoint
๐๐๐๐ฅ
= ๐ slopeoftangenttoMdiagramatapoint = valueofVatthatpoint
VB โ VA = Area of distributed load between points A and B
MB โ MA = Area of V diagram between points A and B