Exam 2020, questions PDF

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TEST 2 SEMESTER 1, 2019/2020 STATICS (SETK 1213/SKTK1213)

QUESTION 1 (20 Marks) CLO

Marks

CLO2 (i)

(ii)

8

12

A 500 N force is applied to a bent plate as shown. Determine: (i)

An equivalent force-couple system at A

(8 marks)

(ii)

An equivalent system formed by parallel forces at A and B

(12 marks)

QUESTION 2 (20 Marks) CLO

Marks

CLO2 (i)

(ii)

5

15

A light rod AD is supported by frictionless pegs at B and C and rests against a frictionless wall at A. A vertical 120 N is applied at D.

(i)

Draw the Free Body Diagram

(5 marks)

(ii)

Determine the reactions at A, B and C and also their directions. (15 marks)

120 N

Question 3 (20 marks) CLO CLO2 a Marks 4

CLO2 b 16

The truss shown in Figure 2 carries a load of 20 N. It is connected by a pin at both A and C. a.

From your observation (without calculation), is there any zero-force member within the truss? Give your reasoning.

b.

(4 marks)

Using method of joint, determine the force of all members and identify if they are tensile force or compression force or zero-force. Summarize the forces in Table 1 below.

(16 marks)

Figure 2: Truss Table 1: Summary of Truss Member Member Magnitude (N)

State

SOLUTION - Question 1

Question 2

Question 3 a.

Member BD is zero-force member. The joint has 3 members. Members DE and DC are collinear. there is no external load at Joint D and the joint isn’t a support reaction.

Joint E FEB FED W 1.4

∑ 𝐹𝑦 = 0 ;

∑ 𝐹𝑥 = 0 ; 𝐹𝐸𝐷 (

𝐹𝐸𝐷 (1.66) − 20 = 0

𝐹𝐸𝐵 = 0

0.9

1.66

𝐹𝐸𝐷 = 23.71 𝑁 (𝐶)

) −

𝐹𝐸𝐵 = 12.85 𝑁 (𝑇)

Joint D FDB

FDE

FDC ∑ 𝐹𝑥 = 0 ;

∑ 𝐹𝑦 = 0 ;

𝐹𝐷𝐶 − 𝐹𝐷𝐸 = 0

𝐹𝐷𝐵 = 0

𝐹𝐷𝐶 = 𝐹𝐷𝐸 = 23.71 𝑁 (𝑇) Joint B FBE FBA FBC

∑ 𝐹𝑥 = 0 ;

0.9

𝐹𝐵𝐸 − 𝐹𝐴𝐵 (1.66) = 0 𝐹𝐴𝐵 = 23.71 𝑁 (𝑇) MEMBER AB BC BD BE CD DE

∑ 𝐹𝑦 = 0 ;

MAGNITUDE (N) 23.71 19.99 0 12.85 23.71 23.71

𝐹𝐵𝐶 − 𝐹𝐴𝐵 (

𝐹𝐵𝐶

1.4

)=0 = 19.99 𝑁 (𝐶) 1.66

State Tensile Compression Zero Tensile Compression Compression...