This set of Design of Steel Structures test focuses on “Behavior and Design Strength of Tension Members”.

1. Which of the following statement is correct?

a) stress and strain calculated using initial cross section area and initial gauge length are referred to as true stress and true strain

b) stress and strain calculated using current cross section area and initial gauge length are referred to as true stress and engineering strain

c) stress and strain calculated using initial cross section area and initial gauge length are referred to as engineering stress and engineering strain

d) stress and strain calculated using current cross section area and gauge length are referred to as engineering stress and engineering strain

View Answer

Explanation: Stress and strain calculated using initial cross section area and initial gauge length are referred to as engineering stress and engineering strain. Stress and strain calculated using current cross section area and gauge length are referred to as true stress and true strain.

2. Arrange the regions of engineering stress-strain curve in order from right to left as in graph

a) strain softening region, strain hardening region, yield plateau, linear elastic region

b) strain hardening region, strain softening region, linear elastic region, yield plateau

c) strain softening region, yield plateau, linear elastic region, strain hardening region

d) strain hardening region, linear elastic region, yield plateau, strain softening region

View Answer

Explanation: The engineering stress-strain curve is typically represented by four regions : linear elastic region, yield plateau, strain hardening region, strain softening (unloading)region.

3. Which of the following is true regarding engineering stress-strain curve?

a) it gives true indication of deformation characteristics of metal because it is entirely based on true dimensions of specimen

b) it does not gives true indication of deformation characteristics of metal because it is entirely based on true dimensions of specimen

c) it gives true indication of deformation characteristics of metal because it is not entirely based on true dimensions of specimen

d) it does not gives true indication of deformation characteristics of metal because it is not entirely based on true dimensions of specimen

View Answer

Explanation: The engineering stress-strain curve does not provide true indication of deformation characteristics of metal. It is entirely based on true dimensions of specimen and these dimensions change continuously as the load increases.

4. Choose the correct option

a) post ultimate strain softening in engineering stress strain curve is present in true stress strain curve

b) post ultimate strain softening in true stress strain curve is absent in engineering stress strain curve

c) post ultimate strain softening in true stress strain curve is present

d) post ultimate strain softening in engineering stress strain curve is absent in true stress strain curve

View Answer

Explanation: The post ultimate strain softening in engineering stress strain curve caused by necking of cross section is absent in true stress strain curve as engineering stress strain curve are based on true dimensions of specimen and true stress strain curve are based on actual cross sectional area of specimen.

5. What is the yield point for high strength steel?

a) 0.5% of offset load

b) 0.2% of offset load

c) 0.1% of offset load

d) 1.5% of offset load

View Answer

Explanation: High-strength steel tension members do not exhibit well defined yield point and yield plateau. Hence, 0.2% of offset load is usually taken as yield point for such high strength steel.

6. True stress strain curve is also known as

a) flow curve

b) un-flow curve

c) elastic curve

d) parabolic curve

View Answer

Explanation: True stress strain curve is also known as flow curve since it represents basic plastic flow characteristics if the material. Any point on the flow curve can be considered as local stress for metal strained in tension by magnitude shown on the curve.

7. The design strength of tension member corresponding to gross section yielding is given by :

a) γ_{m0} f_{y}A_{g}

b) γ_{m0}f_{y}/A_{g}

c) f_{y}/A_{g} γ_{m0}

d) f_{y}A_{g}/ γ_{m0}

View Answer

Explanation: The design strength of tension member corresponding to gross section yielding is given by T

_{dg}= f

_{y}A

_{g}/ γ

_{m0}, where f

_{y}= yield strength of material in MPa, A

_{g}= gross cross-sectional area in mm

^{2}, γ

_{m0}= partial safety factor for failure in tension by yielding = 1.10.

8. Which of the following relation is correct?

a) Net area = Gross area x deductions

b) Net area = Gross area + deductions

c) Net area = Gross area – deductions

d) Net area = Gross area / deductions

View Answer

Explanation: Net area = Gross area – deductions, that is net area of tensile members is calculated by deducting areal of holes from the gross area.

9. The design strength of tension member corresponding to net section rupture is given by :

a) A_{n}f_{y}γ_{m1}

b) 0.9A_{n}f_{y}γ_{m1}

c) 0.9A_{n}/f_{y}γ_{m1}

d) 0.9A_{n}f_{y}/γ_{m1}

View Answer

Explanation: The design strength of tension member corresponding to net section rupture is given by T

_{dn}= 0.9A

_{n}f

_{y}/γ

_{m1}, where An = net effective area of cross section in mm

^{2}, f

_{y}= ultimate strength of material in MPa, γ

_{m1 }= partial safety factor for failure due to rupture of cross section = 1.25.

10. The block shear strength at an end connection for shear yield and tension fracture is given by :

a) (A_{vg}f_{y}/√3 γ_{m0})+(0.9A_{tn}f_{u}/γ_{m1})

b) (A_{tg}f_{y}/√3 γ_{m0})+(0.9A_{vn}f_{u}/γ_{m1})

c) (0.9A_{tg}f_{y}/√3 γ_{m0})+( A_{vn}f_{u}/γ_{m1})

d) (0.9A_{vg}f_{y}/√3 γ_{m0})+(A_{tn}f_{u}/γ_{m1})

View Answer

Explanation: The block shear strength at an end connection for shear yield and tension fracture is given by T

_{db1}= (A

_{vg}f

_{y}/√3 γ

_{m0})+(0.9A

_{tn}f

_{u}/γ

_{m1}), where A

_{vg}= minimum gross area in shear along line of action of force, A

_{tn}= minimum net area of cross section in tension from hole to toe of angle or last row of bolts in plates perpendicular to line of force, f

_{y}and f

_{u}are yield and ultimate stress of material respectively, γ

_{m1}= 1.25, γ

_{m0}= 1.10.

11. The block shear strength at an end connection for shear fracture and tension yield is given by :

a) (A_{vg}f_{y}/√3 γ_{m0})+(0.9A_{tn}f_{u}/γ_{m1})

b) (A_{tg}f_{y}/ γ_{m0})+(0.9A_{vn}f_{u}/√3 γ_{m1})

c) (0.9A_{vg}f_{y}/√3 γ_{m0})+(A_{tn}f_{u}/γ_{m1})

d) (0.9A_{tg}f_{y}/√3 γ_{m0})+( A_{vn}f_{u}/γ_{m1})

View Answer

Explanation: The block shear strength at an end connection for shear fracture and tension yield is given by T

_{db2}= (A

_{tg}f

_{y}/ γ

_{m0})+(0.9A

_{vn}f

_{u}/√3 γ

_{m1}), where A

_{vn}= minimum net area in shear along line of action of force, A

_{tg}= minimum gross area in tension from hole to toe of angle or last row of bolts in plates perpendicular to line of force, f

_{y}and f

_{u}are yield and ultimate stress of material respectively, γ

_{m1}= 1.25, γ

_{m0 }= 1.10.

12. The block shear strength of connection is ________

a) block shear strength at an end connection for shear fracture and tension yield

b) block shear strength at an end connection for shear yield and tension fracture

c) larger of block shear strength at an end connection for (shear fracture, tension yield) and (shear yield, tension fracture)

d) smaller of block shear strength at an end connection for (shear fracture, tension yield) and (shear yield, tension fracture)

View Answer

Explanation: The block shear strength of connection is smaller of block shear strength at an end connection for shear yield, tension fracture T

_{db1}= (A

_{vg}f

_{y}/√3 γ

_{m0})+(0.9A

_{tn}f

_{u}/γ

_{m1}) and block shear strength at an end connection for shear fracture, tension yield T

_{db2}= (A

_{tg}f

_{y}/ γ

_{m0})+(0.9A

_{vn}f

_{u}/√3 γ

_{vm1}).

13. The design tensile strength of tensile member is

a) minimum of strength due to gross yielding, net section rupture, block shear

b) maximum of strength due to gross yielding, net section rupture, block shear

c) strength due to gross yielding

d) strength due to block shear

View Answer

Explanation: The design tensile strength of tensile member is taken as the minimum of strength due to gross yielding (T

_{dg}=f

_{y}A

_{g}/1.1), net section rupture(T

_{dn}=0.9A

_{n}f

_{}y/γ

_{m1}), block shear (T

_{db1}=(A

_{vg}f

_{y}/√3 γ

_{m0})+(0.9A

_{tn}f

_{u}/γ

_{m1}), T

_{db2}=(A

_{tg}f

_{y}/ γ

_{m0})+(0.9A

_{vn}f

_{u}/√3 γ

_{m1})).

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