This article presents a worked example on evaluating the bending capacity of a prestressed concrete beam.
After verifying the stresses within a prestressed beam is within limit, it is also important to design the beam for ultimate limit state such as bending and shear. The beam to be evaluated for flexural capacity in this article has already been designed here for serviceability using magnel diagram. The beam was subjected to prestress force of 2217.6KN at an eccentricity of 1335mm at mid span. The cross-section of the beam showing the position of the tendons at mid-span is provided in the figure below. We shall calculate the bending capacity of the beam as shown in the below worked example.
Material properties
Properties of tendon
Fpk = 1860MPa
Fpk0.1 = 0.85fpk = 1581MPa
Fpd = Fpk0.1/1.15 = 1374MPa
Capacity of tendon = fpd x As = 1374 x 112 x 10-³ = 154KN
Properties of concrete
Concrete grade (fck) = 40MPa
Material safety factor = 1.5
Design strength of concrete (fcd) = 40/1.5 = 26.7MPa
Loads on the Beam
Permanent load = 37.5KN
Impose load = 80KN
Ultimate Load = 1.35 x 37.5 + 1.5 x 80 = 170.6KN
Ultimate moment on the beam = (170.6 x 20000²)/ (8 x 1000) = 8531KNm
To evaluate the beam bending capacity, all the tendons would be assumed to have yielded. Subsequently neutral axis depth will be assumed for the concrete at the top fiber to get the depth of compression block so that equilibrium is sought between the tensile force from tendons and the compressive force from the concrete block. If equilibrium is achieved, then the ultimate moment of resistance of the beam is determined by taking the moment of all forces about the soffit of the beam.
STEP 1: Evaluate the total tensile force produced by the tendons.
Here we will assume all the tendons have yielded, hence the tension in the tendons equals;
Tensile force = fpd x As = 16 x 1374 x 112 x 10-³ = 2464KN
STEP 2: Assume the depth of stress block
Assume the depth of neutral axis and calculate the depth of the compressive stress block
Let assume a depth of neutral axis (x) = 37.5
Depth of compressive stress block (s) = λx = 0.8 x 37.5 = 30mm
STEP 3: Calculate the compressive force.
Comp force = fcd x area of rectangular stress block
= (26.7 x 30 x 3000)/1000 = 2400KN
The difference between the compressive and tensile force = 2464-2400 = 64KN.
This is small and can be ignored, so the compressive and tensile force can be assumed to be in equilibrium.
STEP 4: Calculate the Ultimate Moment of the beam.
Calculate the ultimate moment by taking moment of forces about the soffit of the beam. Take tensile force as positive.
Moment due to compressive force
Compressive force = 2400
Lever arm of compressive force to the soffit (z) = 2500 – 30/2 = 2485mm
Compressive moment = (2400 x 2485)/1000 = 5964KNm
Moment due to tensile force (starting from bottommost tendons to top)
First Layer
Tensile fore = fpd x As x no of tendons
Tensile fore = 1374 x 112 x 10-³ x 4 = 615.9KN
Lever arm (z) = 100mm
Moment = (615.9 x 100)/1000 = 61.6KNm
Second Layer
Tensile fore = fpd x As x no of tendons
Tensile fore = 1374 x 112 x 10-³ x 4 = 615.9KN
Lever arm (z) = 150mm
Moment = (615.9 x 150)/1000 = 92.4KNm
Third Layer
Tensile fore = fpd x As x no of tendons
Tensile fore = 1374 x 112 x 10-³ x 4 = 615.9KN
Lever arm (z) = 200mm
Moment = (615.9 x 200)/1000 = 123.2KNm
Fourth Layer
Tensile fore = fpd x As x no of tendons
Tensile fore = 1374 x 112 x 10-³ x 4 = 615.9KN
Lever arm (z) = 250mm
Moment = (615.9 x 250)/1000 = 154KNm
Ultimate Bending Capacity of the beam = 61.6 + 92.4 + 123.2 + 154 – 5964 = -5532.9KNm
The ultimate bending capacity of the beam (5532.9KNm) is lesser than the ultimate bending moment (8531KNm), then it is required to provided conventional bonded reinforcement bars to complement the tendons and improve its flexural capacity.
Determine the Bending Capacity of the beam with both tendons and conventional reinforcement
Provide conventional steel reinforcement bars
Let’s provide 2 layers of 4 numbers of 25mm steel bars at each layer
Properties of rebar
Fyk = 500MPa
Fpd = 500/1.15 = 434.8MPa
Diameter = 25mm
Area = (πd²)/4 = 490.9mm2
Force of rebar = fpd x As = 434.8 x 490.9 x 10-³ = 213.5KN
STEP 1: Evaluate the total tensile force produced by the tendons and rebars.
Tensile force of tendons = 2464
Tensile force of rebars = fpd x As = 8 x 434.8 x 490.9 x 10-³ = 1708KN
Total tensile force = 2464 + 1708 = 4171KN
STEP 2: Assume the depth of stress block
Assume the depth of neutral axis and calculate the depth of the compressive stress block
Let assume a depth of neutral axis (x) = 65
Depth of compressive stress block (s) = λx = 0.8 x 65 = 52mm
STEP 3: Calculate the compressive force.
Comp force = fcd x area of rectangular stress block
= (26.7 x 52 x 3000)/1000 = 4160KN
The difference between the compressive and tensile force = 4171-4160 = 11KN.
This is small and can be ignored, so the compressive and tensile force can be assumed to be in equilibrium.
STEP 4: Calculate the Ultimate Moment of the beam.
Calculate the ultimate moment by taking moment of forces about the soffit of the beam. Take tensile force as positive.
Moment due to compressive force
Compressive force = 4160
Lever arm of compressive force to the soffit (z) = 2500 – 52/2 = 2474mm
Compressive moment = (4160 x 2474)/1000 = 10291.8KNm
Moment due to tensile force (starting from bottommost tendons to top)
Since the moment due to the tendons in the four layers have been evaluated, we are left with evaluating the moment due to the unstressed reinforcement bars
Fifth layer
Tensile fore = fpd x As x no of tendons
Tensile fore = 434.8 x 490.9 x 10-³ x 4 = 853.8KN
Lever arm (z) = 300mm
Moment = (853.8 x 300)/1000 = 256KNm
Sixth layer
Tensile fore = fpd x As x no of tendons
Tensile fore = 434.8 x 490.9 x 10-³ x 4 = 853.8KN
Lever arm (z) = 350mm
Moment = (853.8 x 350)/1000 = 298.8KNm
Ultimate Bending Capacity = 61.6 + 92.4 + 123.2 + 154 + 256 + 298.8 – 10291.8 = -9305.7KNm
The ultimate bending capacity of the beam is 9305.7KNm which is greater than the ultimate moment in the beam (8531KNm). The beam now has adequate flexural capacity
Click here to Study a Worked Example where the beam is designed for Shear
