MTE119 - Solutions Hw7 - University of Waterloomte119/s/MTE119 - Solutions Hw7.pdf · HOMEWORK 7...
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Mechatronics Engineering NAME & ID DATE MTE 119 – STATICS HOMEWORK 7 SOLUTIONS PAGE PROBLEM 1 For the ladder rung of Problem 2: a. Draw the shear force and bending moment diagram b. What is the maximum bending moment in the ladder rung and where does it occur? Ans. 54.2 M = N-m, 0.233 x m = SOLUTION : a. The free-body for the rung is: 0 0.2 x m < < 250 V N = , 250 M x = N-m 0.2 0.3 m x m < < ( ) 250 7500 0.2 V x N = - - ( ) 2 1 250 7500 0.2 2 M x x = - - N-m 1 15
Transcript of MTE119 - Solutions Hw7 - University of Waterloomte119/s/MTE119 - Solutions Hw7.pdf · HOMEWORK 7...
Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
PAGE
PROBLEM 1
For the ladder rung of Problem 2:
a. Draw the shear force and bending
moment diagram
b. What is the maximum bending
moment in the ladder rung and
where does it occur?
Ans. 54.2M = N-m, 0.233x m=
SOLUTION:
a. The free-body for the rung is:
0 0.2x m< <
250V N= , 250M x= N-m
0.2 0.3m x m< <
( )250 7500 0.2V x N= − −
( )21
250 7500 0.22
M x x= − − N-m
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
PAGE
0.3 0.375m x m< <
500V N= − ,
( )500 0.375M x= − N-m
Shear force and bending moment diagrams:
b. The maximum moment occurs in the interval 0.2 0.3m x m< < in which:
( )2
250 3750 0.2M x x= − − N-m
Setting 0dM
dx= � ( )250 7500 0.2 0x− − =
We find that the maximum moment occurs at:
0.233x m= Ans
Substituting this value into the expression for M :
54.2M = N-m Ans
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
PAGE
Problem 2
1. Problem 7.47 (Text book –Page 347)
The shaft is supported by a thrust bearing at A and Journal bearing at B. If L=10 ft the
shaft will fail when the maximum moment is max kip.ftM k= . Determine the largest
uniform distributed load w the shaft will support.
Solution:
2
wL
For 0 x L≤ ≤
( )
02
2
22
wLwx V
wLV wx
wV L x
− − =
= − +
= −
0;M =∑
( )
2
2
02 2
2 2
2
wL xx wx M
wL wxM x
wM Lx x
− + + =
= −
= −
3
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
PAGE
2
wL−
2
wL / 2L
2
8
wL
From the moment diagram:
2
max
2
8
(10)5000
8
400 lb/ft
wLM
w
w
=
=
=
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
PAGE
2. Problem 7.49 (Text book page -347)
Draw the shear and bending –moment diagrams for the beam.
Solution:Support Reaction:
0;
1000(10 200 (20) 0
490 lb
B
y
y
M
A
A
=
− − =
=
∑
M
50x
V490
M
V
490
V (lb)
9.80x
-510
M (lb.ft) 2401
-200
(Fig-A)
(Fig-B)
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
PAGE
Shear and Moment Functions: For 0 20x≤ < ft (Fig A)
0;
490 50 0
{490 50 } lb
yF
x V
V x
=
− − =
= −
∑
2
0;
50 490 02
{490 25 } lb.ft
M
xM x x
M x x
=
+ − =
= −
∑
For 20 ft 30 ft [Fig B]x< ≤
0; 0
0;
200 0
200 lb.ft
yF V
M
M
M
= =
=
− − =
= −
∑∑
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
PAGE
Problem 3
1. Problem 7.67 (Text book – page: 357)
Draw the shear and moment diagrams for the beam ABCDE. All pulleys have a radius of
1 ft. Neglect the weight of the beam and pulley arrangement. The load weighs 500 lb.
Solution:
Support reactions: From FBD (A)
0; (15) 500(7) 500(3) 0 333.33 lb
0; 333.33 500 0 166.67 lb
A y y
y y y
M E E
F A A
= − − = =
= + − = =
∑∑
Shear and Moment Diagrams:
The load on the pulley at D can be replaced by equivalent force and couple moment at D
as shown on FBD (B)
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
PAGE
Fig (A)
8 ft 2 ft 2 ft 3 ft
Ay=166.67lb
500 lb
1000 lb 500 lb
1000 lb Ey=333.33 lb
1000 lb.ft500 lb
500 lb
500 lb
Ay=166.67lb Ey=333.33 lb
V (lb)
167
-833
-333
x (ft)
x (ft)
M (lb.ft)1333
1000
-333
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
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Problem 3
2. Problem 7.79 (Text book – page: 359)
The beam consists of two segments pin connected at B. Draw the shear and moment
diagrams for the beam.
Solution:
700 lb
8 ft 4 ft 6 ft
9400 lb.ft 150 lb/ft
V (lb)
x (ft)
x (ft)
M (lb.ft)
1017
0
317
-583
8 14.118
8
-9400
-1267
16.2
-800
334
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
PAGE
Problem 3
3. Problem 7.82 (Text book – page: 359)
Draw the shear and moment diagrams for the beam.
Solution:
Support reactions:
0; (6) 3.00(1) 3.00(5) 0 3.00 kN
0; 3.00 3.00 3.00 0 3.00 kN
A y y
y y y
M C C
F A A
= − − = =
= + − − = =
∑∑
Shear and Moment Diagrams:
The peak value of the moment diagram can be evaluated using the method of sections.
The maximum moment occurs at the midspan (x=3 m) where V=0. From FBD (B),
0; 3.00(1) 0 3.00 kN.mM M M= − = =∑
( )( )1
2 3 3.0 kN2
= ( ) ( )1
2 3 3.0 kN2
=
( ) ( )1
2 3 3.0 kN2
=
3.0 kN
3.0 kNy
A = 3.0 kNyC =
1 m 2 m
1 m 1 m4 m
0V =
M
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
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V (kN)3.0
3.0−
( )x m
( )x m
M (kN.m)
3
30
6
3.0
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
PAGE
Extra Practice Problems: Problem 7.53
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
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Problem 7.59:
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
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Problem 7.77
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS
HOMEWORK 7
SOLUTIONS
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Problem 7.88:
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