RENSSELAER POLYTECHNIC INSTITUTE TROY, NY

FINAL EXAM INTRODUCTION TO ENGINEERING ANALYSIS (ENGR-1100)

NAME: Solution ` Section: ___________

RIN: _______________________________ Wednesday, December 15, 2010

Problem Points Score

1 20

2 20

3 20

4 20

5 20

6 20

Total 100 N.B.: You will be graded on 5 problems, 20 points per problem. Problems 1, 2, and 3 are mandatory and will be graded. Before turning in your exam, please make sure you have circled the two problems you want to be graded out of problems 4, 5 and 6.

Problem 2 (

(20 points)

Problem 3 ( A uniform bN/m. The lo

Determine:

(1) The (2) The (3) The (4) The

Solution:  

(1) The

The resultanThe resultanThe resultan

2

3

2

3 A

A

R

R

unit) The total res

21 RRR

(2) The Shap

1

2

3

total

(20 points)

beam AB weoad distribut

resultant forline of actiofree-body direaction sup

distributed l

nt force of trnt force of rent force of th

22

33 R

A

AR

sultant force105032 R

calculation vpe R

10

21

23

l 55

ights 100 N.tion in the ri

rce of the dison of the resuiagram of th

pport at B

load can be c

iangle is R1 ectangle is Rhe semi-circl

2157

252

2

e of the distri221000

values are sh[N]

050

100

356.2

506.2

. The load oght 10-m seg

stributed loaultant force ohe beam usin

considered a

= (300×7)/2R2 = 300×7 =le is:

2.2356100

ibuted load:2.356 =5506

hown in the x-centroid [m

4.67 (1)

10.5 (1)

19 (1)

on the beam gment of the

ad of the distrib

ng the concen

as a composi

2 = 1050 N. (= 2100 N. (1

2 N (1 for eq

6.2 N (1 for e

table below:m] My

490

22,

44,

71,

varies with xe beam is a s

buted load wntrated load

ite area:

(1 for eq; 0.5for eq; 0.5 f

q or other me

eq; 0.5 for v

: y = R × xc [N

03.5

050

764

717.5 (1 for

x with a maxsemi-circle.

with respect t

5 for value; 0for value; 0.5

ethods; 0.5 f

value; 0.5 for

N-m] (1)

r value and

ximum load

(8) to A. (7) (3) (2)

0.5 for unit)5 for unit)

for value; 0.5

r unit)

of 300

5 for

The d = 7

(3) FBD

(4) Sum

B

M

line of actio71717.5/550

D (1 for 5506

m the momen

24

5.78784

100

y

A

B

M

on of R w.r.t.06.2 = 13 m

6.2 N and loc

nts about A:

N7.30325

2.550612

. A is: (1 for eq; 1

cation; 1 for

N

24132 yB

uni

for value an

100 N and l

04 (1 for

it)

nd unit)

location; 1 f

r eq; 1 for va

for By and A)

alue and unit

)

t)

Problem 4 ((20 points)

Problem 5 ( The bent balength 0.5m

a) Exprb) Repl

Cartc) Calcd) Calc

Solution

1- M d)

1- R c)

105-

) (30

10

00

) (0.3

b)

) 75 (-

(-. 200

10- )a

1

1

1

2

1

i

i

ji

FxrC

kr

FR

iM

F

F

1

(20 points)

r, which conm and 0.4m re

ress both forlace this systesian vector

culate the maculate the ma

48105

10160

80 48 -

48- (-60

0100

3.0

pt) (1 m;

(-160

( pt) (2 N.m

N ) .6 - 8

pt) (1 N 00

2

2

2

2

kj

j i

kj

MFxr

iF

ki

j

2

nsists of a veespectively, rces and the tem by a res

r form agnitude of tagnitude of t

N 808

12000

pt) (2 N.m

75- ) 80

0160

5.0

nofor 1pt (-

120 - 100 -

pt .5subtract

12 - (-160

f .5subtract )

22

2

ik

ji

j

i

ertical part ofis subjected moment in Cultant force

the resulting the resulting

140.46 N.m

223.6 N

75

120

3.

thincludingt

pt) (2 N )

missingfor t

pt 2 ( N ) 0

u missingfor

2

i

i

k

k

k

f length 0.3mto two force

Cartesian veand a resulta

force at Ocouple at O

pt) 2 ( N.m6

( pt) 2 ( N 61

pt) (3

moment) he

(0.5

)

fopt .5 sign, g

for vept (1 )

unit

r2

m and two pes F1 and F2

ector formant couple a

O

for vpt (1.5 )

for vapt (1.5

pt) (2 )

m ) 0.3

coo for wrong

.5 form ector

kj

arts parallel 2, and a mom

at O and expr(1

fopt .5 value,

forpt .5 alue,

pt) (1 m

p .5 ,ordinate

correcfor pt

to the X-Y pment as show(5) ress your an11) (2) (2)

unit)or

unit)r

missinfor pt

pt .5 signs,ct

plane of wn.

swer in

unit) g

missingfor t unit)