Problem #2 (25 Points)

a) Composite approach: Table worth 4pts

The force is the area: 247.5 kN

The vertical distance from point A to the y-centroid: �̅� =Σ𝐴�̃�

Σ𝐴= 4.49 m

b)

FBD (4pts) = Each arrow (0.5pts x 5 = 2.5pts) + Label and information about positions (1.5 pt)

Note that x-coordinate at point C can be any.

Forces in Cartesian form:

FA = Axi + Ayj, FB = FB cos45o I + FB sin45o j and FC = -247.5i kN

Moments about A (can be any point):

𝐌𝐴𝐵 = |

𝐢 𝐣 𝐤8 −8 0

√2

2𝐹𝐵

√2

2𝐹𝐵 0

| = 8√2𝐹𝐵𝐤 and 𝐌𝐴𝐶 = |𝐢 𝐣 𝐤

4.49 −4.49 0−247.5 0 0

| = −1112.5𝐤

Eqs of equilibrium:

{

𝐴𝑥 +

√2

2𝐹𝐵 − 247.5 = 0

𝐴𝑦 +√2

2𝐹𝐵 = 0

𝑀𝐴 + 8√2𝐹𝐵 − 1112.5 = 0

So:

{

𝐴𝑥 = −

√2

2𝐹𝐵 + 247.5

𝐴𝑦 = −√2

2𝐹𝐵 kN

𝑀𝐴 = 1112.5 − 8√2𝐹𝐵

c) |𝐴𝑥| is minimum when |Ax|=0. Thun 𝐴𝑥 = −√2

2𝐹𝐵 + 247.5 = 0 (2pts) FB = 350 kN.

d) |MA| is minimum when MA = 0. Thus 𝑀𝐴 = 1112.5 − 8√2𝐹𝐵 = 0 (2pts) FB = 98.3 kN.

A

B

4 m

45 kN/m

FB

x

y

30 kN/m

30 kN/m

1 m

3 m45o

45o

0 kN/m

i A (m2) �̃� from A (m) 𝐴�̃� (m3)

1 (30 × 1)/2 2/3 10

2 30 × 7 1 + 7/2 945

3 (45 − 30) × 3/2 8-1 157.5

247.5 1112.5

Full mark for correct answers with

written processes. If not, give the

partial points shown below.

Take 1pt off if the answer does not

have unit.

(1pt)

(1pt)

(2pts)

(1pt)

(1pt)

(2pts)

(1pt)

(2pts)

(1)

(2)

(3)

Ax

Ay

MA

F

247.5 kN

A(0,0)

C(4.49,-4.49)

B(8,-8)

x

y

45o

Problem #3 (25 Points)

a) (6pts)

FBD (1pt).

EofE:

{

𝐴𝑥 = 0𝐴𝑦 + 𝐽𝑦 − 80 = 0

𝑀𝐴 = −60 − 120 − 120 − 180 + 12𝐽𝑦 = 0

{

𝐴𝑥 = 0 kN𝐴𝑦 = 40 kN

𝐽𝑦 = 40 kN (1pt x3 = 3pts):

b) FBD (1pt). c) FBD (1pt).

EofE (1pt x 2 = 2pts): EofE (1pt x 2 = 2pts):

{𝐴𝑥 +

1

√5𝐴𝐵 + 𝐴𝐶 = 0

2

√5𝐴𝐵 + 𝐴𝑦 = 0

{−

1

√5𝐴𝐵 +

1

√2𝐵𝐷 +

1

√2𝐵𝐶 = 0

−2

√5𝐴𝐵 +

1

√2𝐵𝐷 −

1

√2𝐵𝐶 = 0

{𝐴𝐵 = −44.7 kN𝐴𝐶 = 20 kN

{𝐵𝐶 = 14.1 kN𝐵𝐷 = −42.4 kN

d) FBD (1pt).

EofE:

{

𝐴𝑥 +3

√10𝐷𝐹 +

3

5𝐷𝐸 + 𝐶𝐸 = 0

𝐴𝑦 +1

√10𝐷𝐹 −

4

5𝐷𝐸 − 20 = 0

𝑀𝐷 = −3𝐴𝑦 + 4𝐴𝑥 + 4𝐶𝐸 = 0

→ {𝐶𝐸 = 30 kN𝐷𝐸 = 25.4 kN𝐷𝐹 = −47.7 kN

𝑀𝐷𝐴 = |

𝐢 𝐣 𝐤−3 −4 0𝐴𝑥 𝐴𝑦 0

| , 𝑀𝐷𝐶 = |𝐢 𝐣 𝐤0 −4 0𝐶𝐸 −20 0

|

A

B

C

D

E

F

G

H

3 m 3 m 3 m 3 m

2 m

20 kN 20 kN 20 kN

J

I

2 m

1 m

20 kN

1 m2 m

2 m

1 m

1 m

Full mark for correct answers

with written processes. If not,

give partial points shown below.

Ok if positive + “compression”

instead of negative.

-1pt if no FBD.

-0.5pt if no unit.

AAx

Ay

AB

AC1

2

B

BD

BCAB

1

2

1

1

1

1

(5pts) (5pts)

(9pts)

(1.5pts)

(1.5pts)

(2pts)

(0.5pts)

(0.5pts)

(1pt)

(1pt)

(1pt)

(1pt)

(1pt)

(1pt)

(1pt)

(1pt)

C(3,0)

D

E(6,0)

G

H(9,0)

20 kN 20 kN 20 kN

J(12,0)

I

20 kN

F(6,5)

B

A(0,0)Ax

Ay

Jy

A(-3,-4)

B

C(0,-4)

D(0,0)

20 kN

Ax

Ay

DF

DE

CE

3

4

13