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**Q.1 A 16 mm thick gusset plate is connected to the 12 mm thick flange plate of an l-section using fillet welds on both sides as shown in figure (not drawn to scale). The gusset plate is subjected to point of 350 kN acting at a distance of 100 mm from the flange plate. Size of fillet weld is 10 mm.The maximum resultant stress (in MPa, round off to 1 decimal place) on the fillet weld along the vertical plane would be_______. [2019 : 2 Marks, Set-I]Solution:**

P = 150 kN, e = 100 mm

t

t

S = Size of weld = 10 mm

âˆ´ t

Shear stress due to direct force,

Normal stress due to bending,

Resultant stress,

The force in bolt P is [2018 : 2 Marks, Set-II]

(a) 32.50 kN

(b) 69.32 kN

(c) 82.50 kN

(d) 119.32 kN

Ans.

Solution:

Factored normal stress, f

Factored shear stress, q = 50 MPa

According to IS-800 : 2007, clause 10.5.10.1.1 The equivalent stress,

â‡’

So, equivalent stress 147.99 MPa.

Solution:

Direct bending tensile stress,

f

Direct shear stress,

q = 50 MPa

According to IS 800: 2007, clause 10.5.10.1.1

The equivalent stress,

â‡’

So, equivalent stress of 173.21 MPa.

The permissible stresses in the plate and the weld are 150 MPa and 110 MPa, respectively. Assuming the length of the weld shown in the figure to be the effective length, the permissible load P (in kN) is ____________ . [2017 : 2 Marks, Set-II]

Solution:

= 110 (100 + 100 + 50) x 0.7 x 6

= 115.5 kN

The resultant force (in kN, up to one decimal place) in the bolt 1 is _________ . [2017 : 2 Marks, Set-I]

Solution:

Which one of the following conditions must be ensured to have higher net tensile capacity of configuration shown in figure 2 than that shown in figure 1? [2016 : 2 Marks, Set-Il]

(a) p

(b) p

(c) p

(d) p > 4gd

Ans.

Tensile strength of plate in arrangement (2) will be greater than in arrangement (1),

* As per IS code 800:2007 close 6.3

As per the Limit State Method of IS 800 : 2007, the minimum length (rounded off to the nearest higher multiple of 5 mm) of each weld to transmit a force P equal to 270 kN (factored) is [2016 : 2 Marks, Set-T]

(a) 90 mm

(b) 105 mm

(c) 110 mm

(d) 115 mm

Ans.

Maximum force carried by plates,

Strength of weld = Load carried by weld

= 105 mm (rounded off to nearest higher multiple of 5 mm)

â‡’

(a) 30.33 and 20.00

(b) 30.33 and 25.00

(c) 33.33 and 20.00

(d) 33.33 and 25.00

Ans.

Shear in each bolt =

(a) 100 mm

(b) 105 mm

(c) 110 mm

(d) 115 mm

Ans.

Maximum force transmitted by plate = P

Force resisted by weld,

P= 272.72 kN ...(i)

Let length of weld on either side = l

Size of weld (S) = 10 mm

force resisted by weld =

Size= 10 mm

Throat thickness = 0.7 x 10 = 7 mm

...(ii)

From equations (i) and (ii),

(a) tension

(b) compression

(c) flexure

(d) shear

Ans.

(a) 0.61s

(b) 0.65s

(c) 0.70s

(d) 0.75s

Ans.

Effective throat thickness is the shortest distance from the root of fillet weld to the face of the line joining the toes. The effective throat thickness should not be less than 3 mm.

Effective throat thickness = Ks.

where s is the size of weld in mm and K is a constant. The value of K depends upon the angle between the fusion faces.

(a) 245. 3 mm

(b) 229.2 mm

(c) 205.5 mm

(d) 194.8 mm

Ans.

The maximum size of a fillet weld is obtained by subtracting 1.5 mm from the thickness of the thinner member to be jointed.

âˆ´ S = 10 - 1.5 = 8.5 mm

Strength of the thinner plate

= 100 x 10 x 150 x 10

Strength of the weld

(a) 1083.6 kN

(b) 871.32 kN

(c) 541.8 kN

(d) 433.7 kN

Ans.

Strength of one rivet in double shear

Strength of the riveted joint in double shear

= 12 x 72.61 = 871.32 kN

Strength of one rivet in bearing

= (20 + 1.5) x 14 x 300 x 10

Strength of the riveted joint in bearing

= 12 x 90.3 = 1083.6 kN

Thus the strength of joint will be governed by shearing and it will be equal to 871.32 kN

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