# Mechanics - Bar Of Uniform Strength

6 de Jan de 2023
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### Mechanics - Bar Of Uniform Strength

• 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