Steel Design Test 1

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Steel Design Test 1

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Question 1
1 / -0

If p and d are pitch and gross diameter of rivets, the efficiency (η) of the riveted joint is given by

A
η = p/(p - d)

B
η = p/(p + d)

C
η = (p - d)/p

D
η = (p + d)/p

Solution

Concept:
Efficiency of Riveted Joints: It is the ratio of the strength of the riveted joint to the strength of the connected plates.
Efficiency of riveted Joint is given by \(= \frac{{p - d}}{p} \times 100\)
Where, p = pitch of rivets
d = Gross diameter of rivets
Question 2
1 / -0

A steel plate subjected to a factored tensile force of 500 kN is connected to another plate by lap connection using 5 number of bolts in a row. The strength of each bolt in shear is 100 kN. What would be the possible value of bearing strength (P) of each bolt?

A
P = 100 kN only

B
P ≥ 100 KN

C
P < 100 KN

D
P can take any value.

Solution

Concept:
Bolt Value, BV = Min (S, P)
Where ,
S is the strength of single bolt in shear.
P is the strength of single bolt in bearing.
Strength of joint or bolt = n × BV
Where n is the number of bolts in a row
For joint to be safe,
Applied load ≤ Strength of joint
Calculation:
Given
Applied load, T = 500 kN
Strength of single bolt in shear, S = 100 kN
Strength of single bolt in bearing = P kN
Case 1:
Let S ≤ P ⇒ BV = S = 100 kN
Now, for safe joint
Applied load ≤ strength of joint
500 ≤ 5 × 100
⇒ 500 ≤ 500 ⇒ hence, ok
This means that P ≥ S or P ≥ 100 kN
Case 2:
Let S > P ⇒ BV = P kN
Now, for safe joint
Applied load ≤ strength of joint
500 ≤ 5 × P ⇒ P ≥ 100 kN
In both cases, we get the same answer. Hence, P ≥ 100 kN is the correct answer.
Question 3
1 / -0

For a lap connection with axial pull, shear strength is 80 KN, bearing strength is 97 KN and tearing strength of bolt is 50 KN. The gross area of the flat plate is 700 mm². The yield and ultimate strength of plate is 250 MPa & 410 MPa respectively. Partial safety factor γ_mo = 1.1 & γ_m1 = 1.25 respectively. Find the efficiency of the connection in percentage ?

A
30.5-33.5

B
30.54-33.51

C
30.57-33.56

D
30.59-33.52

Solution

The connection is lap connection with axial pull so there will not be tearing failure of bolt. So the strength of connection is = 50 KN
Efficiency, \({\rm{\eta \;}} = {\rm{\;}}\frac{{{\rm{strength\;of\;connection}}}}{{{\rm{Strength\;of\;soild\;plate}}}} \times 100\% \)
Strength of solid plate \(= \;\frac{{{A_g}.{f_y}}}{{{\gamma _{mo}}}}\)
\(\begin{array}{l} = \;\frac{{700 \times 250}}{{1.1}}{\rm{\;}} = {\rm{\;}}159.09{\rm{\;KN}}\\ \therefore \eta \; = \;\frac{{50}}{{159.09}} \times 100\% \; = \;31.42\% \end{array}\)