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ARC341 – Structural Analysis II Lecture 7: Sections SA1.3, SA2.1, SA4.4– SA4.5

SA1.3) Loads SA2.1) Idealized Structures SA2.2) Principle of Superposition SA2.3) Equations of Equilibrium SA2.4) Determinacy & Stability SA4.4) Shear and Moment Diagrams for a Frame SA4.5) Moment Diagrams by Superposition Method

ARC 341 – Structural Analysis II Dr. Ammar T. Al-Sayegh

SA1.3) Loads The designing structural engineer is responsible for determining

the loads applied on the structure using information obtained from design codes and other sources. Loads are classified as: → Dead Loads: weights of structural members and other objects

permanently attached to the building (Table 1.2 & Table 1.3) → Live Loads: loads that can vary in magnitude and location, such

as objects temporarily located in buildings, moving vehicles, or natural forces (Table 1.4)

Types of live loads include: → Building loads: loads assumed to be applied to flooring of

buildings. Some codes allow for reducing the building loads for large floors according to Equation 1-1.

→ Highway bridge loads, Railway bridge loads, Impact loads, Wind loads, Snow loads, Earthquake loads.

ARC 341 – Structural Analysis II Dr. Ammar T. Al-Sayegh 1

Minimum Design Dead Loads


Minimum Design Live Loads


Load and Resistance Factor Design (LRFD) When designing a structure, we must account for both material

and load uncertainties. This is done in Load and Resistance Factor Design (LRFD) by multiplying the loads by a factor of safety, then combining different load types into a single factored load that is assumed to be the maximum applied load as follows: → 1.4 (dead loads) … [ I ] → 1.2 (dead load) + 1.6 (live load) + 0.5 (snow load) … [ II ] → 1.2 (dead load) + 1.5 (earthquake) + 0.5 (live load) … [ III ]


Problem SA1.6 The precast T-beam has the cross-section shown. Determine its weight per meter of length if it is made from reinforced stone concrete and eight 20-mm cold-formed steel reinforcing rods.


SA2.1) Idealized Structures Exact structural analysis cannot be performed because:

→ Dimensions and geometries are inexact → Material properties are estimated. → Loads and their points of application are estimated. → Connection conditions are approximated

The structure must be idealized to make the analysis practical.

Element-to-element connections (joints) and element-to-support connections (supports) are commonly idealized to one of 3 most common types of connections: pin, roller, and fixed.


Pin Connections


Fixed Connections


In real life, supports lie between these two. Thus, a more accurate representation would include a torsional spring.


In this case, when k = 0 the joint is a pin. When k = ∞ the joint is fixed.

Now, idealize this structure:

Idealized Floors A Girder is the main

load-carrying element of a floor, whereas the smaller elements having shorter spans and connected to the girders are called Beams. When idealizing a framing plan, fixed connections are shown as connected joints, whereas pinned connections are shown as disconnected joints.


Tributary Loadings Tributary loadings are loadings coming from flat surfaces, such as

floors and roofs. These loadings are assumed to be transmitted to beams in girders in one of two systems.

One-way slab system is assumed when the span ratio 𝐿2𝐿1≥ 2,

because as L1 becomes smaller, the closer beams provides greater stiffness to carry the load. Otherwise, the system is considered a two-way slab.


Case-I) One-Way Slab Consider Floor ABDFEC. The floor is composed of two one-way

slabs: ABDC and CDFE.


5 m

5 m

10 m

100 N/m2

2.5 m

2.5 m

2.5 m

2.5 m

500 N/m

2500 N/m 2500 N/m

10 m

1250 N 1250 N 2500 N

5 m 5 m

Case-II-a) Two-Way Square Slab Consider Floor ABDC. The floor is composed of one two-way

square slab: ABDC.


100 N/m2

10 m

10 m

10 m

5 m

10 m

500 N/m

5 m 5 m

Case-II-b) Two-Way Rectangular Slab Consider Floor ABDC. The floor is composed of one two-way

rectangular slab: ABDC.


100 N/m2

15 m

10 m

500 N/m

5 m

15 m

5 m

5 m 5 m 5 m

10 m

5 m 5 m

500 N/m

5 m 5 m

Adjacent Slabs Slabs are typically built adjacent to each others. Thus, loads must

be distributed accordingly.


Problem SA2.1 The frame is used to support a wood dick (not shown) that is to be subjected to a uniform load of 6.5 kN/m2. Sketch the loading that acts along members BG and ABCD by taking:

a) b = 3 , a = 1.5 m b) b = 2.5 m, a = 2.5 m c) b = 4.5, a = 3 m


SA2.2) Principle of Superposition The principle of superposition states that the total

displacements or stresses at a point in a structure subjected to several external loadings equals the combination of displacements and stresses caused by each individual external loading, as long as: → The material behaves in a linear-elastic manner (Hooke’s

law is valid). → The geometry of the structure does not undergo significant

chage when the loads are applied (small displacements theory applies).


SA2.3-2.4) Equilibrium, Determinacy, and Stability A coplanar element is controlled by 3 equations of equilibrium.

→ ∑𝐹𝑥 = 0. → ∑𝐹𝑦 = 0. → ∑𝑀𝑂 = 0.

Thus, when analyzing a coplanar structure composed of n elements, we can determine up to 3n force and moment reactions r using statics. i.e. when r = 3n, the structure is statically determinate. If r > 3n, the structure becomes statically indeterminate and the level of indeterminacy is r - 3n.

A structure is unstable when r < 3n or r >= 3n and the member reactions are concurrent or parallel or some of the components form a collapsible mechanism.


Causes of Instability To ensure the stability of the structure, it must be properly held or

constrained by its supports. Two conditions of instability might arise: → Partial Constraints: The structure or one of its members have

fewer reactive forces thatn equations of equilibrium that must be satisfied.

→ Improper Constraints: All the support reactions are concurrent at a point or all parallel.


Examples


(a) (b)

(c) (d)

(e) (f) (g)

(h) (i)

(k)

(l)

(m)

(n)

(j)

Load-Shear-Moment Relationship Note the following:

Also note, the V will jump upward when the applied force is positive and M will jump upward when the applied moment is clockwise when drawing the diagrams from the left. Why?


SA4.5) Moment Diagrams by Superposition Method Most loadings on beams can be a combination of the following

loadings, so superposition can be used to construct the moment diagram for a beam.


(m)

(m)

(m)

(m)

M (kN.m)

M (kN.m)

M (kN.m)

M (kN.m)

4 kN/m

300 kN.m

10 m

15 kN 25 kN 10 m

4 kN/m

300 kN.m

25 kN 10 m 10 m

15 kN

4 kN/m 40 kN

200 kN.m

300 kN.m

500 kN.m

25kN

300 kN.m

10 m

10 m

20 m

25kN

SA4.4) Shear and Moment Diagrams for a Frame Drawing V and M diagrams for frames follow the same

procedure for drawing V and M diagrams for beams. However, when drawing the diagrams for a frame, each member of the frame is dissembled first and end forces for that member are determined. Then, the diagrams can be drawn on that member with the positive side on the compression fiber.


Problem SA4.63 Draw the moment diagram for the each member of the frame. Assume A is fixed, the joint at B is a pin, and support C is a roller.


100 kN

5 kN/m

4 m

3 m 3 m

Problem SA4.65 Draw the moment diagram for each member of the frame. The joints A, B, and C are pin connected.


Problem SA4.66 Draw the moment diagram for each member of the frame. The joints A, B, and C are pin connected.


2.4 N/m

5 N/m

1.5 N/m 1.5 N/m

2 m

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