The document discusses stresses in bolts and nuts when they are tightened. It provides two examples: (1) A steel bolt connecting a copper tube that is tightened by 1/4 turn, (2) A steel bolt and tubular piece tightened by 1/4 turn and then subjected to an external tensile load. It calculates the stresses induced in the bolt and tube for each example, accounting for both the tightening and any external loads based on principles of static equilibrium and compatibility. The final stresses are provided for the bolt and tube in each example.

1. Stresses in
Bolts & Nuts
Shikha Singh

2. Stresses in Bolts & Nuts
• The bolt will be subjected to tension while the washer & tube will be
subjected to compression.
• Following criteria will apply for the above arrangement:
a. From static equilibrium, tensile load in the bolt will be equal to compressive
load in the tube.
b. From compatibility point of view , axial advancement of the nut is equal to
be sum of extension of the bolt and the contraction of the tube.

3. Examples: Stresses in Bolts & Nuts
(1) A Steel bolt 650mm2 cross-sectional area passes centrally through a copper tube
to 1200mm2cross-sectional area. The tube is 500mm long and is closed by rigid
washers, which are fastened by the threads on the steel bolt.
The nut is tightened by 1/4th of a turn. Find the stress in the bolt and the tube if the
pitch of the thread is 3 mm. Take Es=2.05 x 105 N/mm2 and Ec=1.1 x 105 N/mm2.
Solution:
Let us use suffix s for steel bolt and c for copper tube,
ps x As = pc x Ac
ps (650) = pc (1200), which gives ps = 1.846 pc .............................(i)
Since the final length of the bolt and tube is the same, we have
∆s + ∆c = movement of the nut

4. Examples: Stresses in Bolts & Nuts
...............................(ii)
Substituting the values of ps from (i) we get
(1.846 pc) 243.9 + pc x 454.5= 0.75 x 105
Which gives, pc = 82.89 N/mm2 and ps =1.846 x 82.89 =153 N/mm2

5. Examples: Stresses in Bolts & Nuts
(2) Two steel plugs fit freely into the ends of a steel tubular distance piece 400mm long
and are drawn together by a steel bolt (500mm long) and nut, the nut being tight fit
in the beginning. The nut is further tightened by 1/4 turn to draw the pieces together,
the pitch of the bolt thread being 2 mm. The pieces are then subjected to forces of
50 kN tending to pull them apart. Calculate the stresses in the bolt and the tube. The
area of cross-section of the bolt is 700 sq.mm and that of the tube is 500 sq.mm.
Take E for steel as 200 GN/m2

6. Examples: Stresses in Bolts & Nuts
Solution:
Let us consider both the effects separately.
(a) Effect of tightening the nut:
Due to tightening of the nut, tensile stress will be induced in the bolt and compressive
stress will be induced in the tube.
Let us use suffix 1 for the bolt and suffix 2 for the tube.
From static equilibrium, p1 A1 = p2 A2 ................................................(i)
Also, ∆1+ ∆2 =movement of nut
...............................(ii)

7. Examples: Stresses in Bolts & Nuts
Substitute this value of p1 in (i) , we get
Substituting the numerical values and noting that
E= 205 GN/m2 = 2.05 x 105 N/mm2, we get
From which, p2 = 130.45 N/mm2 (Compressive)
Hence,
= 93.19 N/mm2 (Tensile)

8. Examples: Stresses in Bolts & Nuts
(b) Effect of external load:
Let p1’ and p2’ be the additional stresses, both tensile, due to external tensile load of
50 kN
p1’ A1 + p2’ A2 =50000 ...................................(iii)
Also from compatibility,
............................. (iv)
Hence from (iii) , 1.25 p2’ (700) + p2’(500)= 50000
From which p2’= 36.36 N/mm2 (tensile) and p1’= 1.25 x 36.36 =45.45 N/mm2 (tensile)
(c) Final Stresses:
Total stresses in bolt =93.19 + 45.45 = 138.64 N/mm2 (tensile)
Total stress in tube = 130.45 - 36.36 = 94.09 N/mm2 (Compressive)

9. Examples: Stresses in Bolts & Nuts
Alternative Solution:
In previous solution, both the effects have considered separately and the final
solution obtained by adding the effects together. This is a long and cumbersome
procedure. However, the solution can be obtained directly, and in short, by
considering both the effects together.
Let p1 (tensile) and p2 (compressive) be the final stresses in the bolt and the tube
respectively.
From static equilibrium,
we have:
p1 A1-p2 A2 =50000
700 p1- 500 p2 =50000
p1- 0.714 p2 =71.43 ...............................(1)

10. Examples: Stresses in Bolts & Nuts
Also from compatibility,
(Total extension of bolt) + (Total compression of tube) = Movement of nut
p1 x 400 +p2 x 500 =0.5 x 2.05 x 105
p1+1.25 p2 = 256.25 ....................................(2)
Subtracting (1) from (2), we get 1.964 p2=184.82
From which, p2 = 94.1 N/mm2 (Compressive)
Hence,p1= 71.43+0.714x 94.1 =138.62 N/mm2 (tensile)

11. Thank You