Fundamentals of Structural Analysis
Fundamentals of Structural Analysis
5th Edition
ISBN: 9780073398006
Author: Kenneth M. Leet Emeritus, Chia-Ming Uang, Joel Lanning
Publisher: McGraw-Hill Education
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Chapter 2, Problem 22P
To determine

Calculate the hydrodynamic load applied at floor level on the wall IJKL due to outflow for load cases 2 and 3.

Calculate the hydrostatic loading on the adjacent outside walls due to water retained by the floor and the debris impact load applied to the free standing column CD.

Expert Solution & Answer
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Answer to Problem 22P

The hydrodynamic load applied at floor level on the wall IJKL due to outflow for load case 2 is FdK=8,624kips_.

The hydrodynamic load applied at floor level on the wall IJKL at J due to outflow for load case 3 is FdJ=616.6kips_.

The hydrodynamic load applied at floor level on the wall IJKL at K due to outflow for load case 3 is FdK=1,096.2kips_.

The hydrostatic loading on the adjacent outside walls due to water retained by the floor is Fh=11.1kips_.

Explanation of Solution

Given information:

The width of the building (B) is 35 ft.

The maximum inundation height (hmax) is 33 ft.

The flow velocity (umax) is 20ft/sec.

Assume Itsu=1.0 and Cd=1.25.

Calculation:

For load case 2:

Calculate the height of the water applicable to the structure (hdes) as shown below.

hdes=23hmax=23×33=22ft

Calculate the velocity of flow applicable to the structure (udes) as shown below.

udes=umax=20ft/sec

Sketch the load applied at floor level on the wall IJKL due to outflow for load case 2 as shown in Figure 1.

Fundamentals of Structural Analysis, Chapter 2, Problem 22P , additional homework tip  1

Refer to Figure 1.

Calculate the height of the water applicable to the structure for K as shown below.

hdes,K=Tributoryheight of point K=8+6=14ft

Consider the specific weight density of sea water (γs) is 70.4lb/ft3.

Calculate the hydrodynamic load applied at floor level on the wall IJKL (FdK) due to outflow for load case 2 as shown below.

FdK=12γsItsuCdCcxBhdes,Kudes2=12×(70.4lb/ft3)×1×1.25×1×(35ft)×(14ft)×(20ft/sec)2×1kip1,000lb=8,624kips

Hence, the hydrodynamic load applied at floor level on the wall IJKL due to outflow for load case 2 is FdK=8,624kips_.

For load case 3:

Calculate the height of the water applicable to the structure (hdes) as shown below.

hdes=hmax=33ft

Calculate the velocity of flow applicable to the structure (udes) as shown below.

udes=13umax=203=6.67ft/sec

Sketch the load applied at floor level on the wall IJKL due to outflow for load case 3 as shown in Figure 2.

Fundamentals of Structural Analysis, Chapter 2, Problem 22P , additional homework tip  2

Refer to Figure 2.

Calculate the height of the water applicable to the structure for J as shown below.

hdes,J=Tributoryheight of point J=8+1=9ft

Calculate the hydrodynamic load applied at floor level on the wall IJKL at J (FdJ) due to outflow for load case 3 as shown below.

FdJ=12γsItsuCdCcxBhdes,Judes2=12×(70.4lb/ft3)×1×1.25×1×(35ft)×(9ft)×(6.67ft/sec)2×1kip1,000lb=616.6kips

Hence, the hydrodynamic load applied at floor level on the wall IJKL at J due to outflow for load case 3 is FdJ=616.6kips_.

Calculate the height of the water applicable to the structure for K as shown below.

hdes,K=Tributoryheight of point K=8+8=16ft

Calculate the hydrodynamic load applied at floor level on the wall IJKL at K (FdK) due to outflow for load case 3 as shown below.

FdK=12γsItsuCdCcxBhdes,Kudes2=12×(70.4lb/ft3)×1×1.25×1×(35ft)×(16ft)×(6.67ft/sec)2×1kip1,000lb=1,096.2kips

Hence, the hydrodynamic load applied at floor level on the wall IJKL at K due to outflow for load case 3 is FdK=1,096.2kips_.

Sketch the hydrostatic loading on the adjacent outside walls due to water retained by the floor as shown in Figure 3.

Fundamentals of Structural Analysis, Chapter 2, Problem 22P , additional homework tip  3

Refer to Figure 3.

Calculate the height of the water applicable to the structure (hdes) as shown below.

hdes=3ft

Calculate the hydrostatic loading on the adjacent outside walls due to water retained by the floor as shown below.

Fh=12γsBhdes2=12×(70.4lb/ft3)×(35ft)×(3ft)2×1kip1,000lb=11.1kips

Hence, the hydrostatic loading on the adjacent outside walls due to water retained by the floor is Fh=11.1kips_.

Consider the orientation coefficient (Co) as 0.65.

Calculate the debris impact load applied to the free standing column CD as shown below.

Fi=330CoItsu=330×0.65×1=214.5kips

Therefore, the debris impact load applied to the free standing column CD is Fi=214.5kips_.

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