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MECHANICS OF STRUCTURES (CI3233)

Semester 231 - Class P01

ASSIGNMENT

Lecturer: NGUYỄN HỒNG ÂN

Student: Phạm Thị Thanh Tâm ID 2014439

DOB : 25/10/2002

1.1 Problem 1.1 31.2 Problem 1.5 8

2.1 Problem 2.1 132.2 Problem 2.5 16

3.1 Problem 3.4 193.2 Problem 3.6 24

4.1 Problem 4.2 294.2 Problem 4.5 32

a Determine the normal forces of bars.

b Draw the normal force diagram

SolutionGive the nodes some indices

ˆ Calculation of Reaction Forces

As the structure is statically determinate, the reaction forces can be calculated with the

3 Equilibrium equations

In our case, we are calculating the support forces HA, VA, and VC

X

H = 0 : HA= −4PX

M/A= 0 : 2a × VC = a3P +

√3

3 P (compression)Horizontal Equilibrium:

X

H = 0 : N1+ N4× cos 60◦ = 4P

⇒ N1 = 4P − 6 − 10

√3

3 × P × cos 60◦

⇒ N1 = 4P − 9 + 5

√3

3 × P (tension)Node C

3 P (compression)

3 P × cos 60

◦

⇒ N2 = 3 + 4

√3

3 P (tension)Section 1-1 (On the left side)

4P 4P

√3 2a

X

H = 0 : N3+ N5× cos 60◦+ 4P − 4P + 9 + 5

√3

3 P = 0

⇒ N3 = −N5× cos 60◦− 9 + 5

√3

⇒ N3 = (−2 − 2√

3)P (compression)

Node D

2P D

3 P (tension)

a The normal force of bars

N1 = 4P − 9 + 5

√3

3 P (tension)

N2 = 3 + 4

√3

3 P (tension)

N7 = −6 + 8

√3

3 P (compression)

b Plot the normal force diagram

ˆ Normal force diagram

Data λ1 λ2 λ3 λ4 λ5

Requirements:

a Determine the normal forces of bars

b Draw the normal force diagram

SolutionGive the nodes and the bars some indices

ˆ Calculation of the internal forces

Node E

H = 0 : N3+ N4 × sin 45◦ = 0

⇒ N3 = −5√

2P × sin 45◦ = −5P (compression)

Section 1-1 (On the left side)

N2

N3

N645°

b Plot the normal force diagram

ˆ Normal force diagram

7√2)P5√2)P

5P

8P

P

10P

a Determine the reaction forces.

b Draw the shear force and bending moment diagrams

SolutionGive the nodes some indices

Free body diagram :

ˆ Analysis of complementary structure

X

H = 0 : HC = 0X

M/C = 0 : 2L × VD = qL2 ⇒ VD = 1

2qLX

27qL 4

a Determine the reaction forces.

b Draw the shear force and bending moment diagrams

SolutionGive the nodes some indices

Free body diagram :

ˆ Analysis of complementary structure

H = 0 : HA= 0X

V = 0 : VA+ 7

2qL = 5qL ⇒ VA =

3

2qLX

B

C 2L L

3

2qL

3

2qL7

M

A frame is subjected to loads as shown in above figure Using the force method:

a Determine the reaction forces of supports

b Draw the M, V, N of the frame

c Determine the vertical displacement of the point where the force P is applied

Solution

ˆ Degree of indetermeinacy: DI = 1

ˆ The normal equation is :

δ11+ X1+ ∆1P = 0

ˆ Select the released structure :

P=5qL

C

qEI

M/B = 0 : 2L × VA= L × 5qL + 3qL2+ L × 2qL ⇒ VA = 5qLX

V = 0 : VB = VA+ 5qL = 5qL + 5qL = 10qL

ˆ Calculation of primary beam subjected to X1 = 1 :

X

H = 0 : HB = 1X

M/B = 0 : 2L × VA= 2L × 1 ⇒ VA= 1X

V = 0 : VB = VA= 1

ˆ The bending moment diagram for primary beam due to external load and

X1 = 1 is shown in below:

q EI

10qL28qL2

3qL22qL2

50qL43EI = 0

→ X1 = −25

8 qL

a Calculation of Reaction Forces:

M/B = 0 : 2L × VA = 3ql2+ 5qL2+ 2qL × L −25

8 qL × 2L ⇒ VA=

15

8 qLX

q EI

EI

EI A

qL 55 8

4

25qL

8

25qL 8

qL 55 8

qL 55 8

c Calculation of the vertical displacement of the point where the force P isapplied:

Calculation of Reaction forces due to PK = 1 as follows:

X

H = 0 : HB = 0X

M/B = 0 : 2L × VA= L ⇒ VA= 1

2X

V = 0 : VB = VA+ 1 = 1

2+ 1 =

32Plot the bending moment due to PK = 1 and external load is shown in below:

qL558

B

D PK=11

2

32

"m"

L

Using the graph multiplication method:

A frame is subjected to loads as shown in above figure Using the force method:

a Determine the reaction forces of supports

b Draw the M, V, N of the frame

c Determine the vertical displacement of the point where the force P is applied

Solution

ˆ Degree of indetermeinacy: DI = 1

ˆ The normal equation is :

δ11+ X1+ ∆1P = 0

ˆ Select the released structure :

P=qL M=2qL2

M/A = 0 : L × HB = 2qL2+ 4L × qL − qL × 1

2L ⇒ HB =

11

2qLX

M/A= 0 : L × HB = 1 ⇒ HB = 1

LX

H = 0 : HA= HB = 1

L

ˆ The bending moment diagram for primary beam due to external load and

X1 = 1 is shown in below:

P=qL M=2qL2

VA= 0

HA= 1

L

HB=1L

2qL24qL2

2qL26qL2

8

qL2

1

1 1

H = 0 : HA= HB+ qL → HA= 269

56qL

b Plot the M, V, N of the frame below:

P=qL M=2qL2

95qL 2

56

2qL24qL2

8

qL2

V

M N

qL 213 56

269qL56

95qL 2

56 241qL2

56 17qL 2

56

269qL

56

269qL56

c Calculation of the vertical displacement of the point where the force P isapplied:

Calculation of Reaction forces due to PK = 1 as follows:

X

V = 0 : VA= 1X

M/A= 0 : L × HB = 4L × 1 ⇒ HB = 4X

H = 0 : HA= HB = 4Plot the bending moment due to PK = 1 and external load is shown in below:

P=qL M=2qL 2 C

95qL 2

56

2qL24qL2

ˆ Select the Kinematically Determinate Structure:

q

P=3qL M=4qL2

q

P=3qL M=4qL2

8EI 3L

8EI L

4EI L

ˆ Determination of Coefficients by using equilibrium equations:

4EI L

4EI L

2EI L

3EI 2L

ˆ Shear force and bending moment diagram of Kinematically IndeterminateStructure: With M = M1Z1+ Mo

11qL 54