PSAD COMPREHENSIVE Flashcards
A square post is made of two parts, glued at plane wxyz with an angle of θ = 25° with the vertical. The compressive stress at the glue is 6.3 MPa and the allowable shearing stress is 4.4 MPa. The axial load P = 40 kN.
4. What is the minimum post dimension B (mm) without exceeding the allowable compressive stress at the glue?
a. 33.68
b. 49.31
c. 72.22
d. 36.11
5. What is the minimum post dimension B (mm) without exceeding the allowable shearing stress at the glue?
a. 59.01
b. 40.30
c. 86.41
d. 80.60
6. What angle θ will give the maximum shearing stress?
a. 30 deg
b. 45 deg
c. 90 deg
d. 60 deg
A, A, B
The connection shown in FIG. SD-003 is composed of four 25-mm bolts and carries an eccentric, horizontal load P. Given that a = 75 mm and b = 50 mm.
FIG. SD-003
- If P150 kN, determine the maximum shear stress in the bolts, in MPa.
a. 76.4
b. 97.8
c. 137.5
d. 125.0 - If P=150 kn, determine the minimum shear stress in the bolts.
a. 96.7
b. 64.5
c. 56
d. 79.1 - If the allowable shear stress in the bolts is 150 Mpa, determine the safe load P that the
connection can carry.
A. 230
B. 180
C. 160
D. 290
B, D, A
- Find the maximum shear (KN)
A.60
B.40
C.30
D.20 - Find the maximum bending moment (kN.m) at C.
A.7.30
B.27.2
C.110.8
D.64 - Find the maximum twisting moment (kN.m) at C.
A.17.2
B.64
C.73.9
D.110.8
B, D, A
A steel beam is simply supported on a span of 12 m.
Section: W410mm x 10kg/m
A = 12,710 mm2 tf = 17mm Zy = 0.6 x 106 mm3 d = 410mm Ix = 400 x 106 mm4 Zx = 2.2 x 106 mm3 tw=10 mm Iy = 50 x 106 mm4 E = 200 GPa bf = 260mm
Loads causing bending about the major x-axis:
DL (with beam weight) = 12 kN/m
Moving LL = P kN
13. Determine the load P (kN) based on the design flexural strength of the beam, Mu. Resistance factor for flexure, Φ = 0.9
U = 1.2D + 1.6L
A. 134.8
B. 116.8
C. 83.8
D. 99.1
14. Determine the load P (kN) based on the design shear strength of the beam, Vu.
Factored shear stress, Vu = 0.6Fy
Resistance factor for shear = 1.0
A. 1165
B. 762
C. 953
D. 582
15. At service load, the allowable midspan deflection due to LL = 1/360 of the span. Determine P (kN) based on the allowable deflection.
A. 46.3
B. 74.1
C. 88.2
D. 80.8
B, D, B
Light gage cold-formed steel channels used as purlins are simply supported on roof trusses 6m apart. Sagrods are provided at midspan.
Given:
Roof Slope: 1V:4H
Dead Load (purlin weight included) = 720 Pa
Live Load = 1000 Pa
Wind Load = 1440 Pa
Properties of C-Purlin: C200x76
Sx = 6.19x10^4 mm^3
Sy = 1.38x10^4 mm^3
Fbx = Fby = 207 Mpa
Wind Pressure Coefficient:
@ Windward side, pressure = 0.18
@ Leeward side, suction = 0.60
- What is the safe purlin spacing (m) due to D+L?
A. 0.8
B. 1.0
C. 1.3
D. 1.5 - What is the maximum spacing of purlins (m) due to loads, D,L and W at the leeward side?
A. 1.4
B. 1.8
C. 2.0
D. 3.0 - What is the maximum spacing of purlins (m) due to loads, D +L + W at the winward side?
A. 1.3
B. 1.6
C. 1.9
D. 2.2
C, D, B
The truss shown in FIG. PSAD-004 is to be subjected to the following loadings:
Moving concentrated live load = 40 kN
Uniform highway live load = 12.7 kN/m
Uniform dead load = 10 kN/m
The ordinates for the influence line of the force acting on member CH are also given as follows: 0,-0.417, 0.833, 0.417, 0.
- Determine the length of span where the uniform highway live load be placed to obtain maximum tensile force in member CH.
A. 8.67 m
B. 5.33 m
C. 7.33 m
D. 10.67 m - Determine the maximum tensile force in member CH
A. 134.22 kN
B. 89.78 kN
C. 123.11 kN
D. 109.00 kN - If the cross-sectional area of member Ch is 1200 mm2, determine the maximum tensile stress in member CH.
A. 111.85 MPa
B. 102.59 MPa
C. 990.83 MPa
D. 74.85 MPa
D, C, B
A bar is subjected to uniaxial stress and the corresponding mohr circle is drawn.
Find the area (cm2) of the bar if it is acted on by a tensile force of 120 kN.
A. 30.0
B. 7.5
C.10.0
D. 15.0
23. Find the normal stress (MPa) on a plane 25 degrees from the horizontal.
A. 3.8
B. 25.7
C. 14.3
D. 9.4
24. Find the shear stress (MPa) on a plane inclined at 25 degrees from the direction of loading
A. 16.9
B. 25.7
C. 14.3
D. 30.6
D, C, D
.Refer to FIG RC-004. The design of a simple beam yields the following.
As’ = 3 of 25 mm dia bar
h1 = 125 mm
As = 5 of 25 mm dia bar
h2 = 525 mm
a (measured center to center = 60 mm b = 400 mm
Concrete compressive strength, f’c = 28 MPa
Shear Strength reduction factor = 0.75
Longitudinal steel yields strength, fy = 415 MPa
Concrete cover to 12 mm dia ties = 40 mm
Stirrups yield strength, fyv = 275 MPa
- Calculate the shear strength (kN) provided by 12 mm diameter ties spaced at 125 mm on center
A. 437
B. 392
C. 419
D. 415 - Calculate the shear strength provided by the concrete.
A. 211
B. 189
C. 199
D. 202 - If the allowable concrete shear stress is 0.80MPa, determine the required spacing of 12mm diameter stirrups for a factored shear force of 500kN.
a. 107
b. 300
c. 280
d. 140
C, D, A
SITUATION.
The slab shown in the figure carries the following loads.
Floor Finish = 1.1 kPa F’c = 21 Mpa
Ceiling load = 0.30 kPa Fy = 275 Mpa
Live Load = 4.5 kPa L1 = 6.5m L2 = 7.0m
Beam Dimension, 350 by 450 mm S1 = 2.8m S2 = 3.2 m
Unit weight of concrete, 24 kN/m^3
0
- Calculate the minimum thickness (mm) of the slab bounded by GHKL
a. 120
b. 140
c. 100
d.110 - Using a uniform thickness of 100mm, calculate the negative factored moment (kN-m) bounded by the slab CDGH.
a. 5.88
b. 6.88
c. 7.96
d. 8.82 - Calculate the positive factored moment (kN-m) bounded by slab GHKL.
a. 6.82
b. 5.90
c. 5.04
d. 8.60
C, B, A
- A tied column 350x500mm is reinforced with 8-25mm bars equally distributed on its shorter sides, with the eccentricity measured parallel to the longer side. Neglect slenderness effects, fc’=20.7 MPa and Fy=415 MPa, use 40mm clear cover with tie diameter 10mm. Es=200 GPa.
Which of the following gives the nominal load (kN) that the column could carry?
A. 3712
B. 3016
C. 2413
D. 4640 - Which of the following gives the balance load in kN?
A. 1613
B. 815
C. 1354
D. 925 - Determine the balance moment in kN-m.
A. 876
B. 495
C. 236
D. 480
A, C, B
Refer to the FIG. PSAD-005. A uniform load of 112kN/m is acting downward and supported by and upward pressure of 48kN/m.
- Determine the maximum shear.
A. 48
B. 144
C. 96
D. 84 - Determine the maximum moment.
A. 24
B. 48
C. 42
D. 72 - Determine the distance from the left to which the flexural stress is zero.
A. 1.75
B. 3
C. 3.5
D. 1.5
A, C, C
A hollow pipe 1.8m long has an inner diameter of 350mm and an outer diameter of 450mm. The pipe is subjected to a compressive force of 1200kN. Given E = 110 GPa.
37. Determine the compressive stress (MPa) in the pipe.
A. 19.1
B. 12.6
C. 11.9
D. 24.5
38. Compute the deformation of the pipe in mm.
A. 0.331
B. 0.113
C. 0.313
D. 0.425
39. If the strain of the pipe resides 0.0002 mm/mm what axial load must be applied?
A. 1328.3 kN
B. 145.28 kN
C. 1204.1 kN
D. 1382.3 kN
A, C, D
Refer to the figure shown.
Given: a = 0.5 m P = 300 kN
b = 0.3 m
C = 0.7 m
Allowable weld shear = 193 MPa
- If the force P is acting on the centroid of the weld group, determine the shear force(N) per millimeter of the weld
A. 231
B. 273
C. 375
D. 333 - Determine the maximum force(N) carried by the welds per millimeter.
A. 1543
B. 1051
C. 963
D. 1247 - IF R= 100 N/mm, determine the required weld thickness (mm).
A. 6
B. 8
C. 10
D. 12
B, D, B
The flooring of a warehouse is made up of double tee joists (DT) in FIG. PCD-OOI. The joists are simply supported on a span of 7.5m and are pre tensioned with one tendon in each stem with in initial force of 745 KN each, located at 75mm above the bottom fiber, loss of stress at service load is 20%.
Load imposed on the joists are:
Dead load = 2.3 KPa
Live load = 6.0 KPa
Properties of DT:
A = 200,000mm^2
I = 1880 x I0^6 mm^4
43.Calculate the resulting stress at the neutral axis (MPa) at the supports.
a. 0
b. 7.45
c. 5.96
d. 4.75
44.Compute the resulting stress at the top fibers (MPa) of the DT at the support due to service loads and prestressing force.
a. 5.94 (T)
b. 48.08 (C)
c. 4.75 (T)
d. 38.46 (C)
45. Compute the resulting stress at the top fibers (MPa) of the DT at midspan due to service loads and prestressing force?
A. 1.81
B. 31.91
C. 18.58
D. 15.14
C, C, A
- When a material is to repeated cycle of stress or strain, what causes its structure to breakdown ultimately leading to fracture?
A. Creep
B. Plasticity
C. Fatigue
D. Yielding
C