1. BAR OF UNIFORM STRENGTH
ASSIGNMENT MECHANICS
Central University Of Kashmir
PREPARED BY: SAMEEM MAKHDOOMI
ENROLLMENT NO. : 2124CUKmr09
SUBMITTED TO: AMAN SHEHERYAR

2. S.NO. CONTENTS PAGE NO.
01 CASE OF HORIZONTAL BAR 3
02 CASE OF TAPERED BAR 4
03 DEFORMATION DUE TO SELF WEIGHT 6
04 Proof of : Log A/A2 = (W/P)x 8
05 Numerical 10

3. BAR OF UNIFORM STRENGTH:
UNIFORM STRENGTH:
• Uniform strength is defined when stress is constant along the length of a
material or structure.
• In case of a horizontal bar, the cross sectional area is constant
• On the application of load on both ends of the bar, the stress value at any cross
section will be same i.e. stress value will be constant throughout the length.
p = W/A
Where
p represents stress
W represents load and
A represents area of cross section
W
W

4. TAPERED BAR:
• In case of tapered bar, the cross sectional area is different at different positions.
• On the application of load W on both ends of tapered bar, the stress value will be different
at different positions.
W W

5. • Thus we can say that stress is a function of area.
• For the first case, the bar is of uniform strength and for the second case, the bar is of
non –uniform strength.

6. DEFORMATION DUE TO SELF WEIGHT:
• Let the bar with length ‘L’ be suspended with no external force beign applied to it.
• This is the case where we are having the deformation due to self weight.
• In case of deformation due to self weight even though the cross section of the bar is
constant , the stress will not be constant.
F= ma
F= 0
L

7. • Force is zero at the free end and the force is max at the fixed end.
• Therefore, the stress at the lower end will be zero and at the fixed
end, stress will be maximum.
• This indicates that the bar is not a uniform strength bar as the
stress is varying when we are moving from free to the fixed end,

8. Prove that:
Log A/A2 = (W/P)x
Proof:
• Suppose that area of the upper and lower ends be A1 & A2 respectively.
• Let the area of section be ‘A’ be at a distance 'x' from the lower end.
• Let the area be (A+dA) at a distance (x+dx) from the lower end.
• Let weight/unit volume of the member be W.
• Consider equilibrium at EFGH.

9. • Total force acting upwards = Total force downwards
• P(A+dA) = PA + Wadx
• PA + PdA = PA + Wadx
• PdA = Wadx
• Integrating both sides,
• dA/A = (W/P)x
• Log A = (W/P)x +c
• Let c = Log A2
• Log A = (W/A)x + Log A2
• Log A - Log A2 = (W/P)x
• Log (A/A2) = (W/P)x

10. Q. A vertical tile of uniform strength is 18m long . If the area of the bar at the lower end is
500m². Find the area at the upper end when the tile is to carry a load of 700kN. The material
of the tile weighs 8×10-⁵ N/mm³
Ans
A2=500mm^2
p=F/A
=700000/500
=1400N/mm^2
weight /unit volume = 8*10^-5 N/mm^2
Using 2.3 log10 A1/A2 = W/p x
2.3 log10 A1/A2=(8*10^-5/1400) *18000

11. 2.3 log10 A1/A2 = 0.10*10^-5*10^3
2.3 log10A1/A2=0.001
log10A1/A2=0.001/2.3
A1/A2 =0.004
A1 =0.004*A2
A1=0.004*500
A1=0.2mm^2

12. REFRENCES:
Notes
https://youtu.be/bKJF8GdOTd8