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RENSSELAER POLYTECHNIC INTITUTE, TROY, NY

EXAM NO. 3 - INTRODUCTION TO ENGINEERING ANALYSIS

ENGR 1100 April 11, 2007 8:00-9:50 AM

NAME________________________ SECTION_________ RIN______________________ INSTRUCTOR________________

Problem Points Score

1 25

2 30

3 25

4 20

Total 100

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Problem 1 The beam AB and a cable CB lie in a common plane and support a weight of 3,600 lb. The beam AB has a uniform cross section, a length of 24 ft, and a weight of 600 lb. The beam AB is attached to a pinned support at point A and the cable is attached to a wall at point C. (a) Draw a free body diagram of the beam AB. This FBD must be a figure that is separate from the figure given above. (5 pts) (b) Determine the tension in cable CB. (10 pts) (c) Determine the reaction at the pinned support at A. Express your result in terms of the magnitude of the reaction and its orientation with respect to a horizontal axis. (10 pts) ________________________________________________________________________

C

A

10º

B

30º

3,600 lb

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Solution: (a) FBD 5 pts total (1 pt for each force) ___________________________________________________________________________ (b) Moment equilibrium about point A:

o o o o o oA CB CBM 600(12cos30 ) 3600(24cos30 ) T sin10 (24cos30 ) T cos10 (24sin30 ) 0=− − − + =∑

8 pts total (2 pts for each term in equation) TCB = 9875 lb = 9.88 kips 2 pts ___________________________________________________________________________ (c) Equilibrium in horizontal direction:

x x CBF A -T cos10 0= °=∑ 2 pts total (1 pt for each term) Ax = 9725 lb = 9.73 kips 1 pt Equilibrium in vertical direction:

y y CBF A T sin10 600 3,600 0= − °− − =∑ 4 pts total (1 pt for each term) Ay = 5915 lb = 5.92 kips 1 pt Resultant force and orientation:

( ) ( )2 22 2x yA A A 9.73 5.92 11.4 kips= + = + = 1 pt

y1 1 o

xx

A 5.92tan tan 31.3A 9.73

− −⎛ ⎞ ⎛ ⎞θ = = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

1 pt

3,600 lb 600 lb

x

y TCB

Ay Ax

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Problem 2 A mast AB is subjected to a horizontal force F of 20 kN at point M (midway between A and B). The mast is supported by a ball-and-socket joint at B and 2 cables AC and AD.

a) Draw a free-body diagram of the mast. This FBD must be a figure that is separate

from the figure given above. (6 pts) b) Express all the forces shown on your FBD in Cartesian vector form. (7 pts) c) Using equilibrium, determine the magnitude of the tension in cables AC and AD.

(17 pts)

________________________________________________________________________

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Solution:

(a) FBD 6 pts total (2 pts for each force)

___________________________________________________________________________

(b)

F = 20 (sin 20o i + cos 20o j) = 6.84 i + 18.79 j kN (2 pts) B = Bx i + By j + Bz k (1 pt) TAC = TAC eAC eAC = (4 i - 5 j - 10 k)/11.87 TAC = (4 i - 5 j - 10 k) TAC / 11.87 (2 pts) TAD = TAD eAD eAD = (- 4 i - 5 j - 10 k)/11.87 TAD = (- 4 i - 5 j - 10 k) TAD / 11.87 (2 pts)

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(c) ΣM

B = r

BC × T

AC+ r

BD × T

AD + r

BM × F = 0 (3 pts)

r

BC = 4 i - 2j m r

BD = -4 i - 2j m r

BM = 1.5 j + 5 k m (3 pts)

ΣM

B = ( 4 i - 2j ) × [(4 i - 5 j - 10 k) TAC / 11.87] +

( -4 i - 2j ) × [(- 4 i - 5 j - 10 k) TAD / 11.87] + (1.5 j + 5 k) × (6.84 i + 18.79 j)

AC ADB

T T4 2 0 4 2 0 0 1.5 511.87 11.87

4 5 10 4 5 10 6.84 18.79 0

i j k i j k i j kΣM = − + − − +

− − − − −

ΣM

B = ( 20 i + 40 j - 12 k) TAC / 11.87 +

( 20 i - 40 j + 12 k) TAD / 11.87 + ( - 93.95 i + 34.2 j - 10.26 k) (6 pts) ΣM

B = (20 TAC / 11.87 + 20 TAD / 11.87 - 93.95) i +

(40 TAC / 11.87 - 40 TAD / 11.87 + 34.2) j + (-12 TAC / 11.87 + 12 TAD / 11.87 - 10.26) k = 0 20 TAC / 11.87 + 20 TAD / 11.87 - 93.95 = 0 40 TAC / 11.87 - 40 TAD / 11.87 + 34.2 = 0 (3 pts) -12 TAC / 11.87 + 12 TAD / 11.87 - 10.26 = 0 Multiplying first equation by 2 and adding to second equation gives: 80TAC / 11.87 - 153.7 = 0 TAC = 22.81 kN (1 pt) Substituting TAC back into first equation: 38.43 + 20 TAD / 11.87 - 93.95 = 0 TAD = 32.96 kN (1 pt)

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Problem 3 The truss shown in the figure below is subjected to applied forces at joints F, H, and I and is supported by a pinned support at A and a roller support at G. The support reactions are given in the figure. The horizontal distance between the truss joints is 5 m for a total horizontal length of 25 m. The height of the truss is 5 m.

a) Using the Method of Sections, determine the force in member IH. Clearly indicate the

magnitude of the force and whether the member is in tension or compression. (11 points) b) Using the Method of Sections, determine the force in member BI. Clearly indicate the

magnitude of the force and whether the member is in tension or compression. (11 points) c) By inspection (i.e., there is no need for calculations), identify all members in the truss

that are zero-force members.(3 points) NOTE: Any free-body diagrams that are needed to solve this problem MUST be drawn separately from the figure given below (i.e., do NOT draw any free-body diagrams on top of the figure shown below). NOTE: No credit will be given for this problem if the Method of Joints is used.

Gy = 52.5 kNAy = 7.5 kN

Ax = 10 kN

Gy = 52.5 kNAy = 7.5 kN

Ax = 10 kN

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Solution: PART a)

FCD

FIH

FID

7.5 kN

10 kN

DFCD

FIH

FID

7.5 kN

10 kN

D

D IHM F (5) 20(5) 10(5) 7.5(15) 0= + − − =∑

IH20(5) 10(5) 7.5(15)F 12.5 kN

5− + +

= = FIH = 12.5 kN (T)

5 pts (free-body diagram showing all applied forces and reactions) 5 pts (moment equilibrium equation and magnitude of FIH) 1 pt (indication that FIH is tensile) Alternate Solution:

FCD

FIH

FID

52.5 kN

FCD

FIH

FID

52.5 kN

FCD

FIH

FID

52.5 kN

D IHM F (5) 52.5(5) 20(10) 0= − + =∑

y

IH

G (5) 20(10) 52.5(5) 20(10)F 12.5 kN5 5− −

= = = FIH = 12.5 kN (T)

5 pts (free-body diagram showing all applied forces and reactions) 5 pts (moment equilibrium equation and magnitude of FIH) 1 pt (indication that FIH is tensile)

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PART b)

FBC

FIJ

FBI

7.5 kN

10 kN

FBC

FIJ

FBI

7.5 kN

10 kN

o

y BIF F (cos 45 ) 7.5 0= − + =∑

BI o

7.5F 10.6 kNcos 45

= = FBI = 10.6 kN (T)

5 pts (free-body diagram showing all applied forces and reactions) 5 pts (force equilibrium equation and magnitude of FBI) 1 pt (indication that FBI is tensile) Alternate Solution:

FBC

FIJ

FBI

52.5 kN

FBC

FIJ

FBI

52.5 kN

oy BIF F (cos 45 ) 20 20 52.5 20 0= − − + − =∑

BI o

20 20 52.5 20F 10.6 kNcos 45

+ − += = FBI = 10.6 kN (T)

5 pts (free-body diagram showing all applied forces and reactions) 5 pts (force equilibrium equation and magnitude of FBI) 1 pt (indication that FBI is tensile) PART c)

3 pts (1 point for each zero-force member correctly identified; take off one point for a member that is identified as a zero-force member but that is actually not a zero-force member; maximum number of points that can be earned or taken off is 3)

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Problem 4 The simple three-joint truss shown below is subjected to two forces at the top joint. The unknown support reactions, R1, R2, and R3, can be determined by writing equilibrium equations for the complete truss. The resulting equilibrium equations represent a system of non-homogeneous linear algebraic equations in the three unknown forces R1, R2, and R3 and can therefore be written in matrix form (AX = B) as follows:

1

2

3

1 0 0 R 5000 1 1 R 3001 1 1 R 0

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

A X B

a) Determine the inverse of A (i.e., determine A-1). (8 points) b) Solve for the truss reactions using X = A-1B. (6 points) c) Compute the determinant of A. (6 points)

NOTE: No credit will be given to any numerical results unless ALL intermediate work is shown. ___________________________________________________________________________

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Solution: Part a)

1 0 0 1 0 00 1 1 0 1 01 1 1 0 0 1

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

R1*(-1) + R3

1 0 0 1 0 00 1 1 0 1 00 1 1 1 0 1

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

R2 + R3

1 0 0 1 0 00 1 1 0 1 00 0 2 1 1 1

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

R3 / 2

1 0 0 1 0 00 1 1 0 1 00 0 1 0.5 0.5 0.5

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

R3*(-1) + R2

1 0 0 1 0 00 1 0 0.5 0.5 0.50 0 1 0.5 0.5 0.5

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

1

1 0 00.5 0.5 0.50.5 0.5 0.5

−

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

A

8 pts (2 pts for each step) NOTE: The row operations shown above are not unique. Other row operations could be used to obtain the same result.

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Part b)

1

1 0 0 500 1( 500) 0(300) 0(0) 5000.5 0.5 0.5 300 0.5( 500) 0.5(300) 0.5(0) 1000.5 0.5 0.5 0 0.5( 500) 0.5(300) 0.5(0) 400

X=A B−

− − + + −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = − + − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − + +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

1

2

3

R 500R 100R 400

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

X=

R1 = -500 N R2 = -100 N R3 = 400 N 6 pts (2 pts for each result) ___________________________________________________________________________

Part c)

( ) ( ){ }1 0 0

1 1det( ) 0 1 1 1 1 1 1 1 1 2

1 11 1 1

= = = − − =−

−A

6 pts (2 pts for 1 1

11 1− , 2 pts for first term in ( ) ( ){ }1 1 1 1 1− − , 2 pts for second term in

( ) ( ){ }1 1 1 1 1− − , 1 pt for final result. NOTE: The determinant above was obtained via cofactor expansion along the first row. Cofactor expansion along other rows or columns can also be used to obtain the same result. Also, the duplicate column method could be used to evaluate the determinate.