ort beams, and positive at the centre of each span. The location and amount of the reinforce- thence varies at Sections A and B. Determine the magnitude of uniform applied load (w, in kN/m) required to crack the slab at on A Assuming that the slab is cracked but still linear, calculate the stress in the reinforcing steel he maximum stress in the concrete at Section B under a uniform applied load of w=12 kN/m Determine the maximum uniform load that the slab can support. Where does failure occur (i.e.,

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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The bending moment diagram for a continuous concrete slab is shown below as a function of the
applied load (w, in kN/m) and the span length (L = 2800 mm). The moment is negative at the
support beams, and positive at the centre of each span. The location and amount of the reinforce-
ment hence varies at Sections A and B.
4a) Determine the magnitude of uniform applied load (w, in kN/m) required to crack the slab at
Section A
4b) Assuming that the slab is cracked but still linear, calculate the stress in the reinforcing steel
and the maximum stress in the concrete at Section B under a uniform applied load of w=12 kN/m
4e) Determine the maximum uniform load that the slab can support. Where does failure occur (i.e.,
at Section A or Section B)? Verify any assumptions
Mag-(1/11)wL²
120
k
Continuous Concrete
Slab (see Sections
I
M=
pos= (1/16)wL²
Meg=(1/11)wz²
2800
15M @ 175
175
Section A
Section A
Mpos= (1/16)wL²
Bending Moment Diagram
*
0 0 O
20
Mneg = (1/11)wL²
Section B
2800
120
O
Mpos=(1/16)wL²
Support Beams
10M @ 225
225
2800
0
Maeg = (1/11)wL²
Section B
T
25
Important Notes:
• The steel and concrete strengths are 400 MPa and 35 MPa, respectively, with Young's
moduli of Ec=26000 MPa and E, = 200,000 MPa, respectively.
The area of a 10M reinforcing bar is 100 mm².
• The area of a 15M reinforcing bar is 200 mm².
Transcribed Image Text:The bending moment diagram for a continuous concrete slab is shown below as a function of the applied load (w, in kN/m) and the span length (L = 2800 mm). The moment is negative at the support beams, and positive at the centre of each span. The location and amount of the reinforce- ment hence varies at Sections A and B. 4a) Determine the magnitude of uniform applied load (w, in kN/m) required to crack the slab at Section A 4b) Assuming that the slab is cracked but still linear, calculate the stress in the reinforcing steel and the maximum stress in the concrete at Section B under a uniform applied load of w=12 kN/m 4e) Determine the maximum uniform load that the slab can support. Where does failure occur (i.e., at Section A or Section B)? Verify any assumptions Mag-(1/11)wL² 120 k Continuous Concrete Slab (see Sections I M= pos= (1/16)wL² Meg=(1/11)wz² 2800 15M @ 175 175 Section A Section A Mpos= (1/16)wL² Bending Moment Diagram * 0 0 O 20 Mneg = (1/11)wL² Section B 2800 120 O Mpos=(1/16)wL² Support Beams 10M @ 225 225 2800 0 Maeg = (1/11)wL² Section B T 25 Important Notes: • The steel and concrete strengths are 400 MPa and 35 MPa, respectively, with Young's moduli of Ec=26000 MPa and E, = 200,000 MPa, respectively. The area of a 10M reinforcing bar is 100 mm². • The area of a 15M reinforcing bar is 200 mm².
An engineer has decided to strengthen a timber post against buckling by adding one 1 mm thick
x 350 mm wide strips of glass fibre reinforced polymer (GFRP) laminate to each side of the column
to create a stiffer, composite section. The modulus of elasticity of the timber is 9 GPa and the
modulus of elasticity of the GFRP is 45 GPa.
5a) Determine the buckling capacity of the unreinforced timber column. The post is 5 m tall and
can be considered to be pinned at the bottom and fixed at the top for buckling about both
principal axes.
5b) Determine the percent increase in buckling capacity of the post by adding the GFRP.
5c) If the timber has an allowable stress of 9 GPa, and the GFRP has an allowable stress of 1200
GPa, will the post fail by buckling or yielding?
241
394
Timber Post before
Strengthening
GFRP laminate 1 mm thick
x 350 mm wide on each side
Timber Post after
Strengthening
Hint: Buckling of a composite section can be calculated by using the "transformed moment of
inertia" (IT) in the appropriate equation. Yielding can be checked by using a "transformed area".
Transcribed Image Text:An engineer has decided to strengthen a timber post against buckling by adding one 1 mm thick x 350 mm wide strips of glass fibre reinforced polymer (GFRP) laminate to each side of the column to create a stiffer, composite section. The modulus of elasticity of the timber is 9 GPa and the modulus of elasticity of the GFRP is 45 GPa. 5a) Determine the buckling capacity of the unreinforced timber column. The post is 5 m tall and can be considered to be pinned at the bottom and fixed at the top for buckling about both principal axes. 5b) Determine the percent increase in buckling capacity of the post by adding the GFRP. 5c) If the timber has an allowable stress of 9 GPa, and the GFRP has an allowable stress of 1200 GPa, will the post fail by buckling or yielding? 241 394 Timber Post before Strengthening GFRP laminate 1 mm thick x 350 mm wide on each side Timber Post after Strengthening Hint: Buckling of a composite section can be calculated by using the "transformed moment of inertia" (IT) in the appropriate equation. Yielding can be checked by using a "transformed area".
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