Stress-calculation-pdf

STRESS CALCULATION EXAMPLE For the simply supported beam shown, determine the maximum stresses at midspan section due to its own weight and the following cases of loading and prestressing. 1.) A uniform live load of 13.2KN/m 2.) A uniform live load of 13.2KN/m and an axial centroidal compressive force of P= 1129.5 kN 3.) A uniform live load of 30.63 kN/m and an eccentric longitudinal compressive force of P=1129.5 KN acting at an eccentricity e = 100mm 4.) a uniform live load of 39.35 kN/m and an eccentric longitudinal compressive force of P=1129.5 KN acting at the maximum practical eccentricity for this section equal to 150mm.

STRESS CALCULATION EXAMPLE 1.)

Stress due to live load: 𝑤𝐿² 13.2(7.2)² 𝑀= = = 85.536 𝐾𝑁. 𝑚 8 8

Unit weight of concrete = 23.5 KN/m³ Stress due to self weight: Beam weight = 23.5(0.3)(0.6) = 4.23 KN/m ²

𝑀=

=

𝑓= =

(

.

= ³

.

( . )²

= )

(

)²

= 27.41 𝐾𝑁. 𝑚

²

= ± 1.523MPa

𝑓=

6(85.536𝑥10 ) = ± 4.752MPa 300(600)²

Stress at Midspan: 𝐹 = −1.523 − 4.752 = −6.275 𝑀𝑝𝑎(𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛) 𝐹 = 1.523 + 4.752 = +6.275 𝑀𝑝𝑎(𝑇𝑒𝑛𝑠𝑖𝑜𝑛)

STRESS CALCULATION EXAMPLE Stress Diagram

The tensile stress is +6.275 which is higher than the modulus of rupture of concrete (𝑓 = 0.7 𝑓′ ), hence the concrete will collapse.

STRESS CALCULATION EXAMPLE 2.)

Stress due to P: 𝑓=

=

. (

³ )

= -6.275 Mpa

Stresses at Midspan: 𝑓 𝑓

= −1.523 − 6.275 − 4.725 = −12.55𝑀𝑝𝑎 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 = 1.523 − 6.275 + 4.725 = 0 𝑀𝑝𝑎

STRESS CALCULATION EXAMPLE Stress Diagram:

Final stress due to live load and beam weight with the prestressing force in the top and bottom fiber are -12.55 Mpa and 0 Mpa. In case the prestressing force has doubled the compressive stress at the top fiber reduced the tensile stress at the bottom to ). Comparing the maximum compression stress =12.55Mpa< allowable compressive stress=14 Mpa, therefore the section is safe.

STRESS CALCULATION EXAMPLE 3.)

Stress due to prestress: Moment due to eccentricity M = Pe = 1129.5(0.1) = 112.95 kN.m 𝑓=− ±

=-6.275±

(

²

. (

) )

Stress due to live load: 𝑤𝐿² 30.62(7.2)² 𝑀= = = 198.48 𝑘𝑁. 𝑚 8 8 6(198.48𝑥10 ) 𝑓= = ±11.027 𝑀𝑃𝑎 300(600 ) Stress at Midspan: 𝑓

𝑓

=-6.275±6.275 Mpa 𝑓 =0 𝑓 = −12.55 𝑀𝑃𝑎

= −1.523 + 0 − 11.027 = −12.55 𝑀𝑃𝑎(𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛)

= 1.523 − 12.55 + 11.027 = 0 𝑀𝑃𝑎

STRESS CALCULATION EXAMPLE Stress Diagram:

Final Stress due to live load and dead load in addition to the prestressing force is exactly the same as the previous case. The stresses at the top and bottom fiber are 12.55 Mpa and 0 Mpa respectively. Therefore the same beam can support greater live load by adding eccentricity to the system.

STRESS CALCULATION EXAMPLE 4.)

Stress due to prestress: Moment due to eccentricity M = Pe = 1129.5(0.15 = 169.425 kN.m 𝑓=− ±

=-6.275±

(

²

=-6.275±9.413 Mpa 𝑓 == 3.138 𝑀𝑃𝑎 𝑓 = −15.688 𝑀𝑃𝑎

. (

) )

Stress due to live load: 𝑤𝐿² 39.35(7.2)² 𝑀= = = 254.988 𝑘𝑁. 𝑚 8 8 6(254.988𝑥10 ) 𝑓= = ±14.116 𝑀𝑃𝑎 300(600 ) Stress at Midspan: 𝑓

𝑓

= −1.523 + 3.138 − 14.116 = −12.55 𝑀𝑃𝑎(𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛)

= 1.523 − 12.55 + 14.116 = 0 𝑀𝑃𝑎

STRESS CALCULATION EXAMPLE Stress Diagram:

Final stresses due to live load and dead load plus eccentric force on the top and bottom fibers are 12.55 Mpa and 0 Mpa; which are exactly the same of the previous cases yet the live load has been increased to 39.35 kN/m. a tensile stress of +1.615 Mpa is developed when prestresses force is applies; this stress is less than modulus of rupture of concrete hence crack will not develop in the beam.